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Physique des particules avanc´ ee PHY 5802 An introduction to the Standard Model of Particle Physics Master HEP Ecole Polytechnique Pascal Paganini Laboratoire Leprince Ringuet, Ecole Polytechnique, Palaiseau France Year 2014-2015

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Page 1: Physique des particules avanc ee PHY 5802 An introduction ...polypaganini/SM.pdf · Physique des particules avanc ee PHY 5802 An introduction to the Standard Model of Particle Physics

Physique des particules avanceePHY 5802

An introduction to theStandard Model of Particle Physics

Master HEP Ecole Polytechnique

Pascal PaganiniLaboratoire Leprince Ringuet, Ecole Polytechnique, Palaiseau France

Year 2014-2015

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Page 3: Physique des particules avanc ee PHY 5802 An introduction ...polypaganini/SM.pdf · Physique des particules avanc ee PHY 5802 An introduction to the Standard Model of Particle Physics

1

The question of the structure of the matter and the understanding of the different forcesacting on this matter have always been at the heart of the scientific curiosity. What is the worldmade of? More than 2500 years ago, it was believed that air (Anaximenes) or water (Thales)was the primary substance of which all other things are made. Simple but wrong. Mendeleev,150 years ago, gave a more appropriate answer, but at the price of an increase of complexity.Today, this question has even been extended to the origin of the universe itself. Particle physicsconstitutes a possible path to (partially) satisfy this curiosity. It has strongly evolved these last 50years with many experimental discoveries and now a very well established theoretical framework,the standard model of particle physics, which from a certain point of view recovers a form ofsimplicity.

The purpose of this course is to give an introduction to the standard model of particle physics.The theoretical framework is mainly based on gauge theories with its two pillars, the quantumfield theory and the group theory. I do not assume the reader to be an expert in quantum fieldtheory nor in group theory, but a basic knowledge is preferable. Nevertheless, some reminderswill be given. However, I do assume that Special Relativity and non relativistic quantum me-chanics are well known.

This course is designed for students in physics in the third year (3A) of Ecole Polytechniquebut it is also appropriate for M2 students. Most of them should have followed an introduction toparticle physics (for example PHY554 at Ecole Polytechnique) mainly based on a phenomeno-logical description of this discipline. Here, I try to make the connections between the theory andthe experimental facts. From this point of view, both this document and the slides shown duringthe lectures (which insist more on the experiments) are complementary.

There are probably (too many) mistakes and typos in this document. I would really appreciateif the reader sends his/her remarks and corrections to my email address:[email protected].

Finally, I am grateful to the students in general, for their excellent questions that contributedto improve my knowledges and thus contribute in improving this document. In particular, I wouldlike to thank Thomas Strebler and Matthieu Licciardi.

P.Paganini Ecole Polytechnique Physique des particules avancee

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2

P.Paganini Ecole Polytechnique Physique des particules

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Contents

1 Preliminary notions about particles 9

1.1 Covariance, contravariance and 4-vector notation . . . . . . . . . . . . . . . . . . 9

1.2 Lorentz transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2.1 Special Lorentz transformation . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2.2 Proper time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2.3 Useful 4-vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2.3.1 Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.2.3.2 4-velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.2.3.3 4-momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.2.3.4 4-acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.2.3.5 4-current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.2.3.6 Electromagnetic 4-potential . . . . . . . . . . . . . . . . . . . . . 13

1.2.4 Rapidity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.3 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.4 Angular momentum of particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.4.1 Orbital angular momentum in quantum mechanics . . . . . . . . . . . . . 16

1.4.2 Spin angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.4.3 Total angular momentum and addition of angular momenta . . . . . . . . 19

1.4.4 Effect of rotations on angular momentum states . . . . . . . . . . . . . . 21

1.5 Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.5.1 4-momentum conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.5.2 Two-particles scattering and Mandelstam variables . . . . . . . . . . . . . 23

1.5.2.1 Determination of center-of-mass variables . . . . . . . . . . . . . 24

1.5.2.2 Case of fixed target . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.5.3 Crossed reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.6 Elementary particles dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.6.1 Phase space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.6.2 Transition rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.6.3 Interlude: calculating with Dirac distribution . . . . . . . . . . . . . . . . 27

1.6.4 Decay width and lifetime . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

1.6.5 Cross-sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

1.6.6 Decomposition of the phase space . . . . . . . . . . . . . . . . . . . . . . 31

1.6.7 Few applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

1.6.7.1 Decay to 2 particles . . . . . . . . . . . . . . . . . . . . . . . . . 32

1.6.7.2 Scattering cross section 2→ 2 . . . . . . . . . . . . . . . . . . . 33

1.6.7.3 The three particles final state: Dalitz plots . . . . . . . . . . . . 34

3

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4 CONTENTS

1.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2 From wave functions to quantum fields 37

2.1 Schrodinger equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.1.1 Correspondence principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.1.2 4-current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.2 Spin 0 particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.2.1 Klein-Gordon equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.2.2 Re-interpretation of the 4-current . . . . . . . . . . . . . . . . . . . . . . . 39

2.2.3 Few words about the quantized field . . . . . . . . . . . . . . . . . . . . . 40

2.3 Spin 1/2 particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

2.3.1 Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

2.3.2 4-current and the adjoint equation . . . . . . . . . . . . . . . . . . . . . . 45

2.3.3 Free-particles solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

2.3.3.1 Solutions for particle at rest . . . . . . . . . . . . . . . . . . . . 47

2.3.3.2 General solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.3.3.3 Interpretation of negative energies . . . . . . . . . . . . . . . . . 48

2.3.3.4 Solutions with normalization . . . . . . . . . . . . . . . . . . . . 49

2.3.3.5 Interpretation in terms of spin and helicity . . . . . . . . . . . . 50

2.3.4 Operations on spinors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

2.3.4.1 Charge conjugation . . . . . . . . . . . . . . . . . . . . . . . . . 52

2.3.4.2 Parity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

2.3.4.3 Chirality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.3.4.4 Useful formulas and completeness relations . . . . . . . . . . . . 55

2.3.5 Few words about the quantized field . . . . . . . . . . . . . . . . . . . . . 56

2.4 Spin 1 particle: the photon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

2.4.1 Solutions of Maxwell equations . . . . . . . . . . . . . . . . . . . . . . . . 58

2.4.2 Few words about the quantized field . . . . . . . . . . . . . . . . . . . . . 59

2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3 A brief view of Quantum Electrodynamic 63

3.1 The action and Lagrangians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.1.1 The least action principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.1.2 Free electron Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.1.3 Free photon Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

3.1.4 Gauge invariance consequences . . . . . . . . . . . . . . . . . . . . . . . . 66

3.1.5 QED Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

3.2 Perturbation with interacting fields: the S-matrix . . . . . . . . . . . . . . . . . . 69

3.2.1 Schrodinger, Heisenberg and Interaction representations . . . . . . . . . . 69

3.2.2 Dyson expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

3.2.3 Connection with the general formula of transition rate . . . . . . . . . . . 72

3.3 Feynman rules and diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

3.3.1 Electron-photon vertex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

3.3.2 Photon and electron propagators . . . . . . . . . . . . . . . . . . . . . . . 78

3.3.3 Summary of QED Feynman rules . . . . . . . . . . . . . . . . . . . . . . . 79

3.4 QED and helicity/chirality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

3.5 Simple examples of graph calculation . . . . . . . . . . . . . . . . . . . . . . . . . 82

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CONTENTS 5

3.5.1 Spin and polarisation summations: traces theorems . . . . . . . . . . . . . 82

3.5.2 Electron-muon scattering: e−µ− → e−µ− . . . . . . . . . . . . . . . . . . 85

3.5.3 Electron-positron annihilation: e−e+ → µ+µ− . . . . . . . . . . . . . . . . 86

3.5.4 Compton scattering: e−γ → e−γ . . . . . . . . . . . . . . . . . . . . . . . 87

3.6 Few words about the renormalization . . . . . . . . . . . . . . . . . . . . . . . . . 92

3.7 A major test of QED: g-2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

3.7.1 Prediction with Dirac’s equation . . . . . . . . . . . . . . . . . . . . . . . 95

3.7.2 Higher order corrections . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

3.7.3 Measurement of the g-factor . . . . . . . . . . . . . . . . . . . . . . . . . . 100

3.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

4 From hadrons to partons 105

4.1 Electron-proton scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

4.1.1 Elastic scattering with a point-like proton . . . . . . . . . . . . . . . . . . 106

4.1.2 Elastic scattering with a non-point like proton: nucleon form-factor . . . 108

4.1.3 Inelastic electron-proton scattering . . . . . . . . . . . . . . . . . . . . . . 110

4.1.4 Recapitulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

4.2 The parton model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

4.2.1 Bjorken scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

4.2.2 The partons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

4.3 Parton distribution functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

4.4 What is a proton? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

4.4.1 PDFs of the proton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

4.4.2 Proton spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

4.5 Scaling violation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

5 Quantum Chromodynamic 123

5.1 Quark model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

5.1.1 Baryon number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

5.1.2 Isospin and SU(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

5.1.3 Strangeness and SU(3) flavours . . . . . . . . . . . . . . . . . . . . . . . . 125

5.1.4 Need for colors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

5.1.5 Summary of the quark model . . . . . . . . . . . . . . . . . . . . . . . . . 130

5.2 The colour as a gauge theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

5.2.1 Experimental evidences of the colour . . . . . . . . . . . . . . . . . . . . . 131

5.2.2 The symmetry group SU(3)c . . . . . . . . . . . . . . . . . . . . . . . . . 133

5.2.3 Gauge invariance consequences . . . . . . . . . . . . . . . . . . . . . . . . 135

5.2.4 QCD Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

5.3 QCD Feynman rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

5.4 QCD colour factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

5.4.1 Quark-antiquark interaction . . . . . . . . . . . . . . . . . . . . . . . . . . 142

5.4.2 Quark-quark interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

5.5 QCD properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

5.5.1 Asymptotic freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

5.5.2 Non perturbative QCD . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

5.6 Experimental evidences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

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6 CONTENTS

5.6.1 Discovery of the gluon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

5.6.2 Measurement of αs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

5.6.3 Measurement of colour factors . . . . . . . . . . . . . . . . . . . . . . . . 149

5.7 Back to the parton model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

5.7.1 QCD corrections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

5.7.2 Application to hadrons collider . . . . . . . . . . . . . . . . . . . . . . . . 154

5.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

6 Weak interaction 159

6.1 The Fermi’s theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

6.1.1 Birth of a new interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

6.1.2 The θ+ and τ+ mystery . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

6.1.3 The Wu’s experiment: parity is violated . . . . . . . . . . . . . . . . . . . 161

6.2 Modification of Fermi’s theory: the V −A current . . . . . . . . . . . . . . . . . 161

6.2.1 Parity conservation in QED and QCD . . . . . . . . . . . . . . . . . . . . 161

6.2.2 The V −A current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

6.2.3 Evidence for V −A current . . . . . . . . . . . . . . . . . . . . . . . . . . 164

6.2.4 The four-fermion contact interaction failure . . . . . . . . . . . . . . . . . 164

6.3 A more modern weak interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

6.3.1 The W boson . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

6.3.2 The charged current of leptons . . . . . . . . . . . . . . . . . . . . . . . . 167

6.3.3 The charged current of quarks . . . . . . . . . . . . . . . . . . . . . . . . 170

6.3.3.1 The Cabibbo angle . . . . . . . . . . . . . . . . . . . . . . . . . 170

6.3.3.2 The GIM mechanism . . . . . . . . . . . . . . . . . . . . . . . . 170

6.3.3.3 The CKM matrix . . . . . . . . . . . . . . . . . . . . . . . . . . 171

6.3.4 The neutral current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

6.3.5 Strength of the weak interaction . . . . . . . . . . . . . . . . . . . . . . . 174

6.3.6 Failure of the model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

6.4 The neutrinos case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

6.4.1 The PMNS matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

6.4.2 Neutrino oscillations formalism . . . . . . . . . . . . . . . . . . . . . . . . 177

6.4.3 Considerations on the measurement of neutrinos masses and the oscilla-tions phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

6.4.4 Lepton number: conserved or not conserved? . . . . . . . . . . . . . . . . 181

6.5 What is violated by weak interaction? . . . . . . . . . . . . . . . . . . . . . . . . 181

6.5.1 Charge conjugation C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

6.5.2 CP transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

6.5.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

6.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

7 The Standard Model 189

7.1 The Electroweak or Glashow-Salam-Weinberg theory . . . . . . . . . . . . . . . . 189

7.1.1 Weak isospin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

7.1.2 The weak hypercharge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

7.1.3 The electroweak unification . . . . . . . . . . . . . . . . . . . . . . . . . . 194

7.1.4 The Feynman rules for fermions-gauge bosons interactions . . . . . . . . . 197

7.1.4.1 Interactions with W± . . . . . . . . . . . . . . . . . . . . . . . . 197

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CONTENTS 7

7.1.4.2 Interactions with γ . . . . . . . . . . . . . . . . . . . . . . . . . 1987.1.4.3 Interactions with Z0 . . . . . . . . . . . . . . . . . . . . . . . . . 198

7.1.5 The gauge field transformations . . . . . . . . . . . . . . . . . . . . . . . . 2007.1.6 The Feynman rules for the interactions among gauge bosons . . . . . . . . 202

7.1.6.1 Gauge fields trilinear couplings . . . . . . . . . . . . . . . . . . . 2037.1.6.2 Gauge fields quadrilinear couplings . . . . . . . . . . . . . . . . . 204

7.1.7 Recapitulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2067.2 The Electroweak symmetry breaking . . . . . . . . . . . . . . . . . . . . . . . . . 206

7.2.1 A simple example: spontaneous U(1) symmetry breaking . . . . . . . . . 2067.2.2 Spontaneous breaking of the global SU(2)L × U(1)Y symmetry . . . . . . 2107.2.3 Spontaneous breaking of the local SU(2)L × U(1)Y symmetry . . . . . . . 2127.2.4 Generation of the bosons masses . . . . . . . . . . . . . . . . . . . . . . . 2137.2.5 Consideration about the choice of the gauge . . . . . . . . . . . . . . . . . 2157.2.6 Higgs couplings to bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . 2167.2.7 Generation of the fermions masses . . . . . . . . . . . . . . . . . . . . . . 217

7.2.7.1 Masses without fermions mixing . . . . . . . . . . . . . . . . . . 2177.2.7.2 Masses with fermions mixing . . . . . . . . . . . . . . . . . . . . 221

7.3 The Standard Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2237.3.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2237.3.2 Strength and weakness of the theory . . . . . . . . . . . . . . . . . . . . . 2257.3.3 Tests of the theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

7.4 Changes of concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2307.4.1 Mass in modern physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2307.4.2 The vacuum and the Higgs field . . . . . . . . . . . . . . . . . . . . . . . 232

7.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234

8 Experimental aspects of the famous Higgs boson 237

Bibliography 238

A Fermi’s golden rule and time dependent perturbation theory 241A.1 Fermi’s golden rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

A.1.1 Quantum mechanics approach . . . . . . . . . . . . . . . . . . . . . . . . . 241A.1.2 Quantum field approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

A.2 Transition rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

Index 247

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Chapter 1

Preliminary notions about particles

Few references:Andre Rouge, “Introduction a la physique subatomique”, Chapter 3, 6 and 7, Les editions del’Ecole polytechniqueD. Griffiths, “Introduction to elementary particles”, Chapter 3 and 4, Harper & Row, 1987

In this first chapter, I give some reminders about Special Relativity and Quan-tum Mechanics that are relevant for particle physics. No real justifications willbe given. At the end of the chapter, we derive very useful formulas to computedecay rates and branching ratio of particles.

1.1 Covariance, contravariance and 4-vector notation

An event is defined in a given frame of the spacetime thanks to 4 numbers (ct, x, y, z). Havinga standard basis ~eµ, the vector “event” is then ~X =

∑xµ~eµ = xµ~eµ where in the last equation

the summation on µ is implict (Einstein notation). The xµ are the contravariant coordinates ofthe 4-vector position:

xµ = (x0 = ct, x1 = x, x2 = y, x3 = z) = (ct, ~x) (1.1)

Let us now define a scalar product of 2 vectors ~X = xµ~eµ and ~Y = yµ~eµ:

~X.~Y = xµyν~eµ. ~eν

We wish this scalar product to be consistent with the definition of the spacetime interval of theSpecial Relativity:

∆s2 = ( ~X1 − ~X2).( ~X1 − ~X2) = c2(t1 − t2)2 − (x1 − x2)2 − (y1 − y2)2 − (z1 − z2)2

Recall that the spacetime interval is basically the relativistic equivalent of the (squared) Eu-clidian distance. ∆s2 can be positive, negative or null. Since the celerity of light is assumed tobe constant (at c) in any frame, ∆s2 = 0 for light-like interval, positive for time-like interval(meaning that the 2 events can be causally related), and negative for space-like interval (the 2events cannot be causally related). Hence, by writing gµν = ~eµ. ~eν , the scalar product of two4-vectors consistent with the spacetime interval is then:

~X.~Y = gµνxµyν (1.2)

9

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10 Preliminary notions about particles

where gµν , the so-called metric tensor of the Minkowski space is:

gµν =

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

(1.3)

We can define a new set of numbers:xµ = gµνx

ν (1.4)

that are called covariant coordinates with:

xµ = (ct,−~x) (1.5)

xµ can be expressed from xµ by writing:

xµ = gµνxν (1.6)

wheregµν = gµν (1.7)

Thus, a scalar product is defined by:

~X.~Y ≡ x.y = gµνxµyν = gµνxµyν = xµyµ = xµy

µ (1.8)

More rigorously, xµ are the components of the dual vector ~X? of ~X in the dual basis and are

called covariant coordinates of ~X while xµ components are contravariant coordinates1. Lowerand upper indices indicate always covariant and contravariant coordinates, respectively.

From now on, we will write a 4-vector without the arrow and using as a symbol indifferentlythe name of the 4-vector or its components (it’s a common abuse of language). Ex: the 4-vectorposition can be written simply x or xµ.

1.2 Lorentz transformation

A Lorentz transformation conserves the scalar product, and hence the spacetime interval between2 events.

x′µ = Λµνxν (1.9)

where Λ is a matrix of Lorentz transformation. Using matrix notation with g the matrix repre-sentation of the Minkowski tensor:

x′2 = x2 ⇒ (Λx)tg(Λx) = xtgx⇒ xtΛtgΛx = xtgx

Since the equality is true for any x, one has ΛtgΛ = g. Taking the determinant, it comesdet(Λt)det(Λ) = 1 ⇒ det2(Λ) = 1, and so Λ has a determinant of ±1. When det(Λ) = 1,the Lorentz transformation conserves the space-time orientation and is called proper Lorentztransformation. Furthermore, if Λ0

0 ≥ 1, it is orthochronous, meaning that time direction isconserved. A proper-orthochronous transformation consists purely of boosts and rotations in3D space, and forms the special (or restricted) Lorentz group.

1In mathematics, any vector space, V, has a corresponding dual space consisting of all linear functionals on V.

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Special Lorentz transformation 11

1.2.1 Special Lorentz transformation

In the framework of special relativity, the Lorentz transformations correspond to the law ofchange of inertial (or Galilean) reference frame, for which the physics equations must be pre-served and the speed of light must be the same for any Galilean frame while preserving theorientations of space and time. With these assumptions, one can derive the following Lorentzmatrix (see any Special relativity course):

x′0

x′1

x′2

x′3

=

γ −βγ 0 0−βγ γ 0 0

0 0 1 00 0 0 1

x0 = ctx1 = xx2 = yx3 = z

(1.10)

where x′µ are the coordinates in the <′ frame moving at v along the x axis with respect to < (xand x′ being aligned) and

β =v

c, γ =

1√1− β2

(1.11)

γ being called the boost of the Lorentz transformation. The coordinates in < are deduced fromthe ones in <′ by replacing β → −β. In case of an arbitrary direction of the <′ frame ~β = ~v/c,the general special Lorentz transformation becomes:

x′0 = γ(x0 − ~β.~x)

~x′ = ~x+ (γ − 1)(~β.~x)~ββ2 − ~βγx0

(1.12)

where ~x = (x1, x2, x3). In terms of matrix representation, it corresponds to:

x′0

x′1

x′2

x′3

=

γ −βxγ −βyγ −βzγ−βxγ 1 + (γ − 1)β

2xβ2 (γ − 1)

βxβyβ2 (γ − 1)βxβz

β2

−βyγ (γ − 1)βyβxβ2 1 + (γ − 1)

β2y

β2 (γ − 1)βyβzβ2

−βzγ (γ − 1)βzβxβ2 (γ − 1)

βzβyβ2 1 + (γ − 1)β

2zβ2

x0 = ctx1 = xx2 = yx3 = z

(1.13)

1.2.2 Proper time

In the center-of-mass frame of a particle, the space-time distance ds2 is simply reduced to itsproper time contribution c2dτ2 when the particle decays, while in the lab frame it is c2dt2 −dx2 − dy2 − dz2 = c2dt2(1− v2/c2), and hence dτ =

√(1− v2/c2)dt i.e:

dτ =dt

γ(1.14)

The proper time of a particle (namely the lifetime in the center-of-mass frame) is always shorterthan its lifetime measured in another frame.

1.2.3 Useful 4-vectors

So far, only the 4-vector position was considered. But any 4 numbers that follow the sametransformation as the 4-vector position in a change of frame is by definition a 4-vector.

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12 Preliminary notions about particles

1.2.3.1 Derivative

For instance, if F is a scalar function, the quantity:

dF =∂F

∂xµdxµ

is a scalar as well, so that the 4 components of :

∂µ =∂

∂xµ= (

1

c

∂t,∂

∂x,∂

∂y,∂

∂z) (1.15)

are covariant components. And equivalently:

∂µ =∂

∂xµ= (

1

c

∂t,− ∂

∂x,− ∂

∂y,− ∂

∂z) (1.16)

so that:

∂µ∂µ = ∂µ∂µ = =

1

c2

∂2

∂t2− ∂2

∂x2− ∂2

∂y2− ∂2

∂z2(1.17)

is the d’Alembertian operator.

1.2.3.2 4-velocity

In order to define the relativistic velocity 4-vector uµ , one has to divide the increase of the4-vector position by a time interval. The obvious choice is to use the proper time since it isindependent of the choice of the reference frame. Hence,

uµ =dxµ

dτ= γ

dxµ

dt= γ(c,~v) (1.18)

As expected, one can easily check that u2 = c2.

1.2.3.3 4-momentum

The mass m being a 4-scalar2, the 4-momentum:

pµ = muµ = (γmc, γm~v) = (E/c, ~p) (1.19)

is still a 4-vector with a norm p2 = m2u2 = m2c2. Since p2 is also E2/c2 − |~p|2, one has:

E2 = |~p|2c2 +m2c4 (1.20)

A particle satisfying p2 = m2 is said to be on mass-shell. Equation 1.20 is usually called themass-shell condition. Incidentally, we notice that equality 1.19 implies:

p =β

cE (1.21)

which is valid for any mass. When β = 1, p = E/c which injected in 1.20 implies m = 0.

2There is sometimes a source of confusion in some old textbooks (an historical artifact of special relativity),where it is sometimes said that the mass increases as the velocity of the particle. This approach is now completelyobsolete and better to avoid: the mass is invariant, but not the energy and the momentum which do increase withthe velocity. For more details about the mass, see section 7.4.1.

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Useful 4-vectors 13

1.2.3.4 4-acceleration

Following the same argument as for the 4-velocity, one has:

Γµ =duµ

dτ= γ(

dtc,dγ

dt~v + γ~a) (1.22)

1.2.3.5 4-current

We wish to combine the charge density ρ with the current density ~j in a 4-vector. Let us considera point-like charge having a constant density ρ in the small element of volume dV and movingat velocity ~v. The current density is thus ~j = ρ~v while the charge q = ρdV . The charge beingan intrinsic property of the particle must not depend on the reference frame (so be a 4-scalar)and we must have ρdV = ρ′dV ′. In a frame <, during dt the particle moves (in the spacetime)by dxµ = (cdt, d~r) which is obviously a 4-vector. So the quantity ρdV dxµ = ρdtdV (dxµ/dt) =ρ(dΩ/c)(dxµ/dt) is a 4-vector as well, where dΩ = cdtdV is a small “volume” (in 4 D) of thespacetime, considered between 2 infinitely close moments. dΩ being a 4-scalar (see exercise 1.1),ρ(dxµ/dt) must be a 4-vector. The 4-current is then defined as:

jµ = ρdxµ

dt= (cρ,~j) (1.23)

We will see in chapter 3 the importance of the 4-current for the description of the interactionbetween charged particles and photons.

1.2.3.6 Electromagnetic 4-potential

In classical electrodynamic, the electric field ~E and magnetic field ~B are solution of the Maxwell’sequations: (

(1) ~∇. ~E = ρ/ε0 (3) ~∇. ~B = 0

(2) ∂ ~B∂t + ~∇× ~E = 0 (4) ~∇× ~B − 1

c2∂ ~E∂t = µ0

~j

)(1.24)

where ρ and ~j are respectively the charge density and current density. Let us define the 4-potential:

Aµ = (V/c, ~A) (1.25)

with V the scalar potential and ~A the vector potential. Now defining the electromagnetic tensorof rank 2 by:

Fµν = ∂µAν − ∂νAµ (1.26)

we will see that we can easily recover the Maxwell’s equations. The main interest of usingthe expression of Maxwell’s equations with Fµν is that the formalism is explicitly covariant.We notice first that Fµν is obviously antisymmetric and hence depends on only 6 independentparameters. Developing:

F 01 = ∂0A1 − ∂1A0 = 1/c(∂Ax/∂t+ ∂V/∂x)F 02 = ∂0A2 − ∂2A0 = 1/c(∂Ay/∂t+ ∂V/∂y)F 03 = ∂0A3 − ∂3A0 = 1/c(∂Az/∂t+ ∂V/∂z)

we can set F 01 = 1/c(−Ex), F 02 = 1/c(−Ey), F 03 = 1/c(−Ez) with ~E verifying:

~E = −∂~A

∂t− ~∇V (1.27)

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14 Preliminary notions about particles

where we call ~E the electric field. We can set the 3 other independent parameters:

F 12 = −(∂Ax/∂y − ∂Ay/∂x) ≡ −BzF 13 = −(∂Ax/∂z − ∂Az/∂x) ≡ ByF 23 = −(∂Ay/∂z − ∂Az/∂y) ≡ −Bx

the components of the magnetic field ~B satisfying

~B = ~∇× ~A (1.28)

Fµν then reads:

Fµν =

0 −Ex/c −Ey/c −Ez/cEx/c 0 −Bz ByEy/c Bz 0 −BxEz/c −By Bx 0

(1.29)

Calculating ∂ ~B∂t + ~∇ × ~E from equ. 1.27 and 1.28 we do check that it is 0 ie the Maxwell’s

equation (2). And the same for ~∇. ~B = 0 (equation (3)). In order to find (1) and (4) we justhave to write:

∂µFµν = µ0j

ν (1.30)

jν being the 4-current. Let us check:

∂µFµ0 = µ0j

0 ⇒ 0 +1

c

∂Ex∂x

+1

c

∂Ey∂y

+1

c

∂Ez∂z

= µ0cρ⇔ ~∇. ~E = ρ/ε0

(using ε0µ0 = 1/c2). For the other component:

∂µFµ1 = µ0j

1 ⇒ − 1

c2

∂Ex∂t

+∂Bz∂y− ∂By

∂z= µ0jx

and similar result for ∂µFµ2 and ∂µF

µ3 (with permuted x, y and z) so that one actually findsequation (4).

It is easy to realize that the electric field and magnetic field are entangled. Suppose forexample A0 = 0 and A1,2,3 = (0, 0, sinωt). According to 1.28 and 1.27: ~B = 0 and ~E =(0, 0, ω cosωt). Since Aµ is a 4-vector with no component along the x-axis, in a frame movingat β with respect to the x-axis, we have A′µ = Aµ. However, in this frame the magnetic fieldis now3 ~B′ = ~∇′ × ~A′ = (0,−βγω cosωt, 0) 6= 0. A pure electric field can be seen as a magneticfield (or a mix of electric and magnetic fields) in another frame and vice-versa. Both are clearlyentangled and inseparable. A well-known result.

1.2.4 Rapidity

The rapidity ζ is given by:

ζ = tanh−1 β ⇒ γ = cosh ζ ⇔ βγ = sinh ζ (1.31)

The rapidities of 2 Lorentz transformations just add up: ζ<′′/< = ζ<′′/<′ + ζ<′/< (assuming thatthe two frames <′ and <” are moving in the same direction with respect to <). That means

3Alternatively one may transform the electromagnetic tensor F ′µν = ΛµρΛνσF

ρσ = Λµρ(Λt)σνF

ρσ =ΛµρF

ρσ(Λt)σν ⇒ F ′ = ΛFΛt to deduce the new components of the electric and magnetic fields.

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Rapidity 15

that if the rapidity is known in a given frame, the value in another frame moving at β will justbe translated: ζ ′ = ζ + tanh−1 (β). In other words, dζ ′ = dζ is Lorentz invariant and so therapidity distribution dN/dζ as well. The addition of rapidities gives an easy way to rememberthe addition of relativistic velocities: β<′′/< = tanh(tanh−1 β<′′/<′ + tanh−1 β<′/<) leading to:

β<′′/< =β<′′/<′ + β<′/<1 + β<′′/<′β<′/<

(1.32)

Knowing that β = tanh ζ = (eζ − e−ζ)/(eζ + e−ζ), equation 1.31 can be rewritten:

ζ =1

2ln

(1 + β

1− β

)

Hence, at low velocity,

ζ =1

2[ln(1 + β)− ln(1− β)] ≈ β (β 1)

rapidity converges to velocity. We recover the usual Euclidian addition of velocities at lowvelocity.

Now, let us consider a non zero mass particle moving at β = v/c. Combining E = γmc2 andp = γmv, the rapidity becomes:

ζ =1

2ln

(E + pc

E − pc

)

However, experimental particle physicists often use a modified definition of rapidity relative toa beam axis where pL is the longitudinal component of momentum along the beam axis:

ζ =1

2ln

(E + pLc

E − pLc

)(1.33)

Any Lorentz transformation along the beam axis will just add a term tanh−1 β. This is typicallythe case of asymmetric collisions such as with fixed-target experiments or in hadron colliderswhere the elements colliding (so called partons) take a varying fraction of the initial hadrons’energy. Hence, with this definition, the rapidity distribution dN/dζ is the same in the center ofmass frame of the collision or in the lab frame.

In formula 1.33, the mass of the particle has to be known (or both E and pL have to bemeasured independently). However, as soon as E mc2, E ' pc, pL = p cos θ ⇒ pLc ' E cos θand the previous definition becomes

ζ ' η =1

2ln

(1 + cos θ

1− cos θ

)=

1

2ln

(cos2 θ

2

sin2 θ2

)⇒ η = − ln

(tan

θ

2

)

where η is called the pseudorapidity. The advantage of the pseudorapidity is that it doesn’tdepend on the mass, and thus on the nature of the particle which is usually unknown in exper-iments. When θ is close to zero (on the beam axis), η diverges to the infinity. η and ζ are veryclose as soon as E mc2 and the angle of the particle w.r.t beam axis is large enough. A noteof caution with the pseudorapidity: while the behavior of the rapidity is clear under a Lorentzboost (ζ ′2 − ζ ′1 = ζ2 − ζ1), it is not the case for the pseudorapidity where η′2 − η′1 6= η2 − η1.

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16 Preliminary notions about particles

1.3 Units

The two fundamental constants in High Energy Physics are the speed of light c = 2.998 ×108 ms−1 and the Plank constant ~ = h

2π = 1.0546 × 10−34 Js (reflecting that we usually dealwith relativistic quantum objets). In the rest of the document, we are going to use the so-callednatural units with c = ~ = 1, so that length units are equivalent to time units and energy unitsequivalent to mass or inverse of time. Hence:

~ = 1⇒1 GeV−1 ≡ 6.58× 10−25 s (1.34)

~c = 1⇒1 fm ≡ 5 GeV−1 ≈ (1/5 GeV)−1 = (197 MeV)−1 (1.35)

This will strongly simplify the formulas.

1.4 Angular momentum of particles

The purpose of this section is to remind the reader of important properties of the angularmomentum in quantum mechanics that are often used in particle physics. They will not bejustified, the reader can refer to any quantum mechanics book (for example [2] or [3]).

1.4.1 Orbital angular momentum in quantum mechanics

In classical mechanics, the angular momentum is defined as ~L = ~r∧ ~p. It depends on 6 numbersrx, ry, rz, px, py and pz. In quantum mechanics, due to the Heisenberg uncertainty principle, it’snot possible to measure simultaneously these 6 numbers with an arbitrary precision. It turns outthat the best that one can do is to simultaneously measure both the magnitude of the angularmomentum vector and its component along one axis. The angular momentum is quantized andis defined as an operator (symbolˆbelow) acting on the wave function. Its expression is:

~L = ~r ∧ ~p = −i~ ~r ∧ ~∇ (1.36)

where we used the correspondence principle stating that ~p = −i~~∇ (see next chapter). Itsatisfies the following canonical commutation relations expressing the impossibility to measuresimultaneously all the angular momentum components:

[Lx, Ly] = i~Lz, [Ly, Lz] = i~Lx, [Lz, Ly] = i~Ly (1.37)

which can be simply written [Li, Lj ] = i~εijkLk, εijk being the antisymmetric Levi-Civita symbol.

Then, one can easily show that L2 = L2x + L2

y + L2z satisfies:

[L2, Lx,y,z] = 0 (1.38)

Thus, one can find a basis of eigenvectors of L2 and Lz4. One can show that the eigenvectors

|l,m〉 satisfy:

L2 |l,m〉 = l(l + 1)~2 |l,m〉Lz |l,m〉 = m~ |l,m〉 , m ∈ [−l, l] (1.39)

where l is the quantum number of the orbital momentum and m is the quantum number of theprojection of the orbital momentum on the Z-axis. In case of orbital angular momentum, l,mare integers.

4Z axis is chosen by convention.

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Spin angular momentum 17

Using the cylindrical coordinates (x = r sin θ cosφ, y = r sin θ sinφ, z = r cos θ with θ and φthe polar and azimuthal angles), we have from 1.36:

L2 = −~2

(∂2

∂θ2+

1

tan θ

∂θ+

1

sin2 θ

∂2

∂φ2

)

and

Lz = −i~ ∂

∂φ

The spherical harmonics Y ml (θ, φ) = 〈θ, φ|l,m〉 are eigenfunctions of the orbital angular momen-

tum and thus satisfy the equation 1.39 i.e.:

L2 Y ml (θ, φ) = l(l + 1)~2 Y m

l (θ, φ)

Lz Yml (θ, φ) = m~ Y m

l (θ, φ), m ∈ [−l, l]

The expression of the most useful spherical harmonics is given in the table below.

l/m 0 1 2

0√

14π

√3

4π cos θ√

54π

(32 cos2 θ − 1

2

)

1√

38π sin θeiφ −

√158π sin θ cos θeiφ

2 14

√152π sin2 θe2iφ

Table 1.1: Spherical harmonics Y ml .

The two following properties of the Y ml are useful:

Y −ml (θ, φ) = (−1)m [Y ml (θ, φ)]∗ m > 0

Y ml (π − θ, φ+ π) = (−1)l Y m

l (θ, φ)

The last property explains why under parity transformation (which inverts all spatial coordinatesand thus θ → π − θ and φ→ φ+ π), the orbital angular momentum gets an extra factor (−1)l.

1.4.2 Spin angular momentum

As the name suggests, spin was originally conceived as the rotation of a particle about its axis.This name is very unfortunate: elementary particles are described as point-like objects for which“rotation about its axis” is meaningless. One has to resign with the classical picture and justadmit that spin is a kind of angular momentum in the sense that it obeys the same mathematicallaws as quantized orbital angular momenta do i.e.:

S2 |s,m〉 = s(s+ 1)~2 |s,m〉Sz |s,m〉 = m~ |s,m〉 , m ∈ [−s, s] (1.40)

However, spins have some peculiar properties that distinguish them from orbital angular mo-menta. Spin quantum number s may take an half-odd-integer positive values (for fermions).

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18 Preliminary notions about particles

Spin is usually presented as the intrinsic angular momentum suggesting that it is the residualangular momentum that still remains even if the particle is at rest (and hence p = 0) or if therestframe of the particle is used. This image is obviously not appropriate for massless particlealways propagating at the velocity of light.

A case of most importance is the spin 1/2. A particle having a spin 1/2 has two possibleprojections |12 ,+1

2〉 and |12 ,−12〉 that can be gathered into a 2-component notation named Pauli

spinor:

|12,+

1

2〉 =

(10

)= |↑〉 , |1

2,−1

2〉 =

(01

)= |↓〉

with:

Sz |↑〉 =1

2|↑〉 , Sz |↓〉 = −1

2|↓〉

while the projection on the two other axis gives (see for instance [2, p. 252]):

Sx |↑〉 =1

2|↓〉 , Sx |↓〉 =

1

2|↑〉 , Sy |↑〉 = i

1

2|↓〉 , Sy |↓〉 = −i1

2|↑〉

where we have dropped the ~. We see that the mean value of the spin on the x and y-axes fora system prepared in |↑〉 or |↓〉 gives 0 as expected (〈↑ |Sx| ↑〉 = 〈↓ |Sx| ↓〉 = 0). In general, thestate of a spin 1/2 is a linear combination of the 2 spin-up |↑〉 and spin-down |↓〉 eigenstates:

α |↑〉+ β |↓〉 =

(αβ

)

so that the probability to find after the measurement of the z-axis the spin-up state is |α|2 and|β|2 for the spin-down one (with |α|2 + |β|2 = 1). We see that using the spinor notation, theprojection on the different axes is obtained with 2× 2 matrices, called the Pauli matrices:

~S =1

2~σ (1.41)

meaning:

Sx =1

2σ1 , Sy =

1

2σ2 , Sz =

1

2σ3

where σi=1,2,3 are given by:

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

)(1.42)

~S verifies the commutation relations:

[Sx, Sy] = iSz

Moreover, we do have, as eigenstates of S2 = S2x + S2

y + S2z , the spin-up and spin-down states

with the expected value s(s+ 1) with s = 12 . Conclusion: the ~S operators made with the Pauli

matrices verify all the conditions of the quantum mechanical angular momentum operators, withS = 1

2 , ie the spin. They represent the operators for spin 12 .

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Total angular momentum and addition of angular momenta 19

1.4.3 Total angular momentum and addition of angular momenta

As for classical mechanics, the total angular momentum is the sum of the orbital angular orbital

momentum and spin (which in classical mechanics is the rotation of the object on itself): ~J =

~L+ ~S, J satisfying:J2 |j,m〉 = j(j + 1)~2 |j,m〉Jz |j,m〉 = m~ |j,m〉 , m ∈ [−j, j] (1.43)

j and m being integers or half-odd-integers.The spin of composite particles is usually understood to mean the total angular momentum.

This is the sum of the spins and orbital angular momenta of the constituent particles5 and isunderstood to refer to the spin of the lowest-energy internal state of the composite particle (i.e.,a given spin and orbital configuration of the constituents).

The dimension of the Hilbert space H(j) generated by the states of the angular momentumJ is 2j + 1. When two angular momenta ~J1 and ~J2 are added, the new space is the tensorialproduct H1(j1) ⊗ H2(j2) of dimension (2j1 + 1) × (2j2 + 1) for which a complete basis can bedenoted by the following eigenvectors:

|j1, j2,m1,m2〉 ≡ |j1,m1〉 ⊗ |j2,m2〉The eigenvectors obviously satisfy:

J21 |j1, j2,m1,m2〉 = j1(j1 + 1)~2 |j1, j2,m1,m2〉 , J1z |j1, j2,m1,m2〉 = m1~ |j1, j2,m1,m2〉J2

2 |j1, j2,m1,m2〉 = j2(j2 + 1)~2 |j1, j2,m1,m2〉 , J2z |j1, j2,m1,m2〉 = m2~ |j1, j2,m1,m2〉One can show that there exists another complete basis of this space which corresponds to thetotal angular momentum ~J = ~J1 + ~J2. This basis is generated by the eigenvectors of J2

1 , J22 , J

2,and Jz:

|j1, j2, j,m〉 with j ∈ j1 + j2, j1 + j2 − 1, · · · , |j1 − j2| and m = m1 +m2 ∈ [−j, j]and thus satisfy:

J21 |j1, j2, j,m〉 = j1(j1 + 1)~2 |j1, j2, j,m〉 , J2

2 |j1, j2, j,m〉 = j2(j2 + 1)~2 |j1, j2, j,m〉J2 |j1, j2, j,m〉 = j(j + 1)~2 |j1, j2, j,m〉 , Jz |j1, j2, j,m〉 = m~ |j1, j2, j,m〉

In other words, the resulting space is decomposed as:

H1(j1)⊗H2(j2) =

j1+j2⊕

j=|j1−j2|H(j)

The relationship between the basis |j1, j2j,m〉 and |j1, j2,m1,m2〉 is given by the Clebsch-Gordancoefficients (chosen as real numbers and denoted below as 〈j1, j2,m1,m2|j1, j2, j,m〉):

|j1, j2, j,m〉 =

j1∑

m1=−j1

j2∑

m2=−j2〈j1, j2,m1,m2|j1, j2, j,m〉 |j1, j2,m1,m2〉

where only coefficients for which m = m1 + m2 are non-zero. The most useful coefficients forparticle physics are given in a concise way in figure 1.1. Looking for example at the table labelled

5The notion of constituent is here very large: for instance, it includes valence quarks of hadrons as well asquarks or gluon from quantum fluctuations. It is known now that the spin of the proton cannot be explained bythe contribution of the valence quarks only.

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20 Preliminary notions about particles

40. Clebsch-Gordan coe!cients 1

40. CLEBSCH-GORDAN COEFFICIENTS, SPHERICAL HARMONICS,

AND d FUNCTIONS

Note: A square-root sign is to be understood over every coe!cient, e.g., for !8/15 read !!

8/15.

Y 01 =

"3

4!cos "

Y 11 = !

"3

8!sin " ei!

Y 02 =

"5

4!

#3

2cos2 " ! 1

2

$

Y 12 = !

"15

8!sin " cos " ei!

Y 22 =

1

4

"15

2!sin2 " e2i!

Y !m" = (!1)mY m"

" "j1j2m1m2|j1j2JM#= (!1)J!j1!j2"j2j1m2m1|j2j1JM#d "

m,0 =

"4!

2# + 1Y m

" e!im!

djm!,m = (!1)m!m!

djm,m! = d

j!m,!m! d 1

0,0 = cos " d1/21/2,1/2

= cos"

2

d1/21/2,!1/2

= ! sin"

2

d 11,1 =

1 + cos "

2

d 11,0 = ! sin "$

2

d 11,!1 =

1 ! cos "

2

d3/23/2,3/2

=1 + cos "

2cos

"

2

d3/23/2,1/2

= !$

31 + cos "

2sin

"

2

d3/23/2,!1/2

=$

31 ! cos "

2cos

"

2

d3/23/2,!3/2

= !1 ! cos "

2sin

"

2

d3/21/2,1/2

=3 cos " ! 1

2cos

"

2

d3/21/2,!1/2

= !3 cos " + 1

2sin

"

2

d 22,2 =

#1 + cos "

2

$2

d 22,1 = !1 + cos "

2sin "

d 22,0 =

$6

4sin2 "

d 22,!1 = !1 ! cos "

2sin "

d 22,!2 =

#1 ! cos "

2

$2

d 21,1 =

1 + cos "

2(2 cos " ! 1)

d 21,0 = !

"3

2sin " cos "

d 21,!1 =

1 ! cos "

2(2 cos " + 1) d 2

0,0 =#3

2cos2 " ! 1

2

$

Figure 40.1: The sign convention is that of Wigner (Group Theory, Academic Press, New York, 1959), also used by Condon and Shortley (TheTheory of Atomic Spectra, Cambridge Univ. Press, New York, 1953), Rose (Elementary Theory of Angular Momentum, Wiley, New York, 1957),and Cohen (Tables of the Clebsch-Gordan Coe!cients, North American Rockwell Science Center, Thousand Oaks, Calif., 1974).

Figure 1.1: The Clebsch-Gordan coefficients, spherical harmonics and d-functions. From [19].

1/2× 1/2, it is easy to check that the addition of 2 spins 12 (j1 = j2 = 1

2) gives the well-knownresult:

|12 , 12 , 1, 1〉 = |12 , 1

2 ,+12 ,+

12〉 ≡ |↑↑〉

|12 , 12 , 1, 0〉 = 1√

2|12 , 1

2 ,+12 ,−1

2〉+ 1√2|12 , 1

2 ,−12 ,+

12〉 ≡ 1√

2|↑↓〉+ 1√

2|↓↑〉

|12 , 12 , 0, 0〉 = 1√

2|12 , 1

2 ,+12 ,−1

2〉 − 1√2|12 , 1

2 ,−12 ,+

12〉 ≡ 1√

2|↑↓〉 − 1√

2|↓↑〉

|12 , 12 , 1,−1〉 = |12 , 1

2 ,−12 ,−1

2〉 ≡ |↓↓〉

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Effect of rotations on angular momentum states 21

1.4.4 Effect of rotations on angular momentum states

Consider a rotation of a quantum mechanic system. A rotation in general is defined by 3numbers: they could be the two polar coordinates φ, θ of the rotation axis ~n and the rotationangle Θ around the axis or they could be the three Euler angles φ, θ, ψ respectively around thez, y and z axes as shown in figure 1.2 [33]. In the former definition, the rotation is defined by

Physique des particules M1 HEP X Pascal Paganini LLR-IN2P3-CNRS 34

4) Angular momentum of particles

Figure 1.2: The 3 Euler angles for an extrinsic rotation (i.e. with respect to fixed axes).

the operator6:

R(n,Θ) = e−iΘ~ ~n.

~J (1.44)

while in the later definition:

R(φ, θ, ψ) = e−iφ~ Jze−i

θ~ Jye−i

ψ~ Jz (1.45)

When applied to the angular momentum states, the norm j remains unchanged but the projec-tion on an axis is obviously affected -since the frame (passive transformation) or the angularmomentum (active transformation) is rotated-, and hence:

R |j,m〉 =∑j

m′=−j |j,m′〉 〈j,m′|R|j,m〉=∑j

m′=−j |j,m′〉Djm′,m

(1.46)

where Djm′,m (which depend on the angles) are the so-called Wigner rotation D-functions. The

interpretation of D-functions in case of a passive transformation is easier: in equation 1.46, theleft hand side ket is an eigenvector of Jz (with eigenvalue m~) where the z-axis is the rotatedaxis. The right hand-side ket is an eigenvector of Jz (with eigenvalue m′~) where now the z-axisis the original axis. Thus, Dj

m′,m is simply the amplitude to measure the value m′~ along thez-axis when the angular momentum points in the direction of the rotated axis with componentm~. Using the Euler angle rotation 1.45, the expression of Dj

m′,m can be simplified since |j,m〉and |j,m′〉 are eigenvector of Jz:

Djm′,m(φ, θ, ψ) = 〈j,m′|e−iφ~ Jze−i θ~ Jye−iψ~ Jz |j,m〉

= e−iφm′djm′,m(θ) e−iψm

(1.47)

where the d-functions:djm′,m(θ) = 〈j,m′|e−i θ~ Jy |j,m〉 (1.48)

are reduced rotation matrices and the most useful expressions are given in figure 1.1. Since weare interested in the angular momentum, only rotations that affect the z-axis matter. Therefore,

6The angular momentum operators are the (infinitesimal) generators of the rotations.

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22 Preliminary notions about particles

we can simply ignore the rotation with the ψ angle i.e. choose ψ = 0 (it is the usual convention):such rotation R(φ, θ, 0) would bring the z-axis in the same position as R(φ, θ, ψ) (only x andy-axes would differ). Consequently, equation 1.47 finally reduces to:

Djm′,m(φ, θ, 0) = e−iφm

′djm′,m(θ) (1.49)

As an example, it is interesting to consider a spin 1/2. Using formulae in figure 1.1, we find:

D1212, 12

= e−iφ2 cos θ2 , D

12

− 12, 12

= e+iφ2 sin θ

2 , D1212,− 1

2

= −e−iφ2 sin θ2 , D

12

− 12,− 1

2

= e+iφ2 cos θ2 ,

and thus:

R |12 , 12〉 = |12 ,−1

2〉D12

− 12, 12

+ |12 , 12〉D

1212, 12

= |12 ,−12〉 e+iφ

2 sin θ2 + |12 , 1

2〉 e−iφ2 cos θ2

R |12 ,−12〉 = |12 ,−1

2〉D12

− 12,− 1

2

+ |12 , 12〉D

1212,− 1

2

= |12 ,−12〉 e+iφ

2 cos θ2 − |12 , 12〉 e−i

φ2 sin θ

2

Using the spinor notation introduced in section 1.4.2, the previous equation reads:

R

(10

)= e−i

φ2 cos θ2

(10

)+ e+iφ

2 sin θ2

(01

)=

(e−i

φ2 cos θ2

e+iφ2 sin θ

2

)

R

(01

)= −e−iφ2 sin θ

2

(10

)+ e+iφ

2 cos θ2

(01

)=

(−e−iφ2 sin θ

2

e+iφ2 cos θ2

)

meaning that R corresponds to the matrix representation7:

R(θ, φ) =

(e−i

φ2 cos θ2 −e−iφ2 sin θ

2

e+iφ2 sin θ

2 e+iφ2 cos θ2

)

Let us see the effect of the rotation R of angles θ = 2π and φ = 0 which brings back the systemin its original position:

R(2π, 0) = − l1

where l1 denotes the identity matrix. We find the famous result R(2π, 0) |ψ〉 = − |ψ〉: forspin 1/2 particles, a rotation of 2π is not enough to recover the initial state (that would haveincidence on interferences properties), but 2 turns (4π) are actually needed. There is a directconnection with the spin-statistic theorem: interchanging two particles is topologically the sameas rotating either one of them by 360 degrees as shown in figure 1.3. Thus, we see that theinterchange of two spin 1/2 particles will generate a minus sign. For example, if ψ(1)χ(2)describes particle 1 in state ψ and particle 2 in state χ, then the particle interchange must bedescribed by −ψ(2)χ(1) and since both particles are identical, the correct wave-function is thelinear superposition ψ(1)χ(2)−ψ(2)χ(1) which is antisymmetric as expected for fermions. Thereader can check that with spin 1 particles, only one turn is enough to recover the initial state.

7The reader could have found directly this result by expanding the formula e−iφ~ Sze−i

θ~ Sy with Sz and Sy

given by the Pauli matrix (equation 1.41).

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Collisions 23

Physique des particules M1 HEP X Pascal Paganini LLR-IN2P3-CNRS 36

21 2 1

Figure 1.3: The rotation with an angle 2π of one of the 2 particles is equivalent to the interchange ofthe 2 particles.

1.5 Collisions

1.5.1 4-momentum conservation

In physics, the invariance of spacetime by translation implies that an isolated system has a 4-momentum which is conserved. For a system consisted of particles without interaction, the globalmomentum of the system is just the addition of the individual particles momenta. However, forparticles in interaction with fields, the situation is more complicated and one should take intoaccount the momenta of the fields themselves which can be complicated. Fortunately, in case ofcollision of particles, one makes the assumption that the initial state is described by particleswithout interaction, which then interact and produce a final state in which the particles areconsidered again without interaction. Hence, even if we don’t try to describe what happensduring the collision (the dynamic of the process), we can still write that the sum of particles’momenta of the initial state is equal to the sum of particles’ momenta of the final state.

1.5.2 Two-particles scattering and Mandelstam variables

Let us consider the collision sketched on figure 1.4: (1) + (2) → (3) + (4) with 4-momentap1, p2, p3, p4. The amplitude (-probability) of the process must not depend on the frame,

(2) p2

(1) p1

(4) p4

(3) p3

Figure 1.4: A scattering with two particles.

so only 4-scalar must be involved i.e. scalar product of 4-momenta. There are 10 possiblecombinations but, because of momentum conservation and the on mass-shell constraint, only 2independent combination remain. It is often convenient to express the amplitude as function of

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24 Preliminary notions about particles

other Lorentz-invariant variables called Mandelstam variables and defined as:

s = (p1 + p2)2 = (p3 + p4)2

t = (p1 − p3)2 = (p4 − p2)2

u = (p1 − p4)2 = (p3 − p2)2(1.50)

where s is the square of the center-of-mass energy (in the center-of-mass frame8 ~p∗1 + ~p∗2 = 0,

thus (p∗1 + p∗2)2 = (E∗1 + E∗2)2 − (~p∗1 + ~p∗2)2 = (E∗1 + E∗2)2) and t the square of the 4-momentumtransfer (trivial in case of elastic scattering). By developing the expressions of equation 1.50,and using the relation p1 + p2 = p3 + p4, one finds

s+ t+ u =4∑

1

m2i (1.51)

reflecting that only 2 variables are indeed really independent.

1.5.2.1 Determination of center-of-mass variables

Let us find now the momentum and energy in the center-of-mass frame (labelled CM in whatfollows) where |~p∗1| = |~p∗2| = |~p∗|:

E∗21 = m21 + ~p∗2 ⇒ ~p∗2 = E∗21 −m2

1

E∗22 = m22 + ~p∗2 ⇒ E∗22 = m2

2 + E∗21 −m21√

s = E∗1 + E∗2 ⇒ √s = E∗1 +√m2

2 −m21 + E∗21

⇒ (√s− E∗1)2 = m2

2 −m21 + E∗21

Finally:

E∗1 =s+m2

1 −m22

2√s

(1.52)

And thus:

E∗2 =√s− E∗1 ⇒ E∗2 =

s+m22 −m2

1

2√s

(1.53)

And:

~p∗2 = E∗21 −m21 =

(s+m21 −m2

2)2

4s−m2

1 =s2 +m4

1 +m42 − 2m2

1m22 − 2sm2

1 − 2sm22

4s

~p∗2 =λ(s,m2

1,m22)

4s=

[s− (m1 +m2)2][s− (m1 −m2)2]

4s(1.54)

with λ(x, y, z) the triangle function (also known as the Kallen function):

λ(x, y, z) = x2 + y2 + z2 − 2xy − 2xz − 2yz = [x− (√y +√z)2][x− (

√y −√z)2] (1.55)

For the energies and momenta of the particles in the final state, one has just to replace m1 → m3

and m2 → m4 in equ. 1.52, 1.53 and 1.54. In case of elastic scattering where m1 = m3 andm2 = m4 the final and initial quantities in the CM are obviously the same.

One should notice that formulas 1.52, 1.53 or 1.54 do not depend on the final state. Hence,they are valid whatever the number of particles in the final state. Inversely, the 3 correspondingformulas for the final state are independent of the initial state and are valid whatever the numberof particles in the initial state.

8In center-of-mass frame, the quantities are labelled with a *.

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Crossed reactions 25

1.5.2.2 Case of fixed target

We can apply these formulas in case of a collision of a particle (1) to a nucleus of fixed target (2).We have ~p2 = 0 and hence E2 = m2. Therefore s = (p1 +p2)2 = m2

1 +m22 +2E1m2 and including

this result in the numerator of 1.54, we get ~p∗2 = [2E1m2 − 2m1m2][2E1m2 + 2m1m2]/4s =m2

2(E21 −m2

1)/s. Since E21 −m2

1 = ~p21, it remains9:

|~p∗| = m2|~p1|√s, s = m2

1 +m22 + 2E1m2 (1.56)

Usually, E1 m1,m2, so that√s ' √2E1m2. For example, when the Tevatron collider (near

Chicago, USA) ran with fixed target mode with E1 = 980 GeV and m2 ' 1 GeV, we have√s ' 44 GeV to be compared to the collider mode where

√s = 2× 980 = 1960 GeV.

1.5.3 Crossed reactions

So far, the 3 Mandelstam variables were simply presented as a combination of 4-momenta withs being the squared energy in the CM, t the squared transfer momenta and u without a clearphysical meaning. However, we can go further as soon as we re-interprete the negative signs inequations 1.50. We will see in the chapter devoted to Quantum ElectroDynamic that a particlewith 4-momentum −pµ is interpreted as an antiparticle:

pµ ≡ −pµ (1.57)

so that the amplitude M (a complex function which completely describes the scattering pro-cess) involving an antiparticle can be described by the amplitude of the reaction involving theparticle but using instead −pµ. Actually, this is a general property of Quantum Field Theorythat the amplitude involving a particle in the initial state with 4-momentum pµ is identical tothe amplitude of the otherwise same process but with an antiparticle in the final state withmomentum −pµ. Hence, the amplitudes of the following reactions can be derived from the sameamplitude:

(1) + (2)→ (3) + (4) (1.58)

(1) + (3)→ (2) + (4) (1.59)

(1) + (4)→ (3) + (2) (1.60)

and even for (1) → (2) + (3) + (4). However, even if the amplitude would be the same, it doesnot mean that all these reactions are allowed: conservation of 4-momentum must be respected,or in other words the kinematics must be adequate.

M depends on the four 4-momenta which are not independent because of the energy-momentum conservation. Equivalently, M depends on s, t, u. Let us denote by M(s = (p1 +p2)2, t = (p1 − p3)2, u = (p1 − p4)2) the amplitude of 1.58 and M′(s′ = (p1 + p3)2, t′ = (p1 −p2)2, u′ = (p1− p4)2) the amplitude of 1.59. Applying 1.57 rule, we see on figure 1.5 that t = s′,s = t′ and u = u′. Hence, the amplitude of reaction 1.59 is just M′(s′, t′, u′) = M(t′, s′, u′).Similarly for reaction 1.60, we haveM(u′, t′, s′). The 3 processes in which s, t, u are the squaredCM energy are called respectively s, t, and u-channel and are displayed in figure 1.6. Reaction1.59 and 1.60 are then respectively the t and u-channel of reaction 1.58. A summary of thesituation is given in the following table:

9you can find this result by using a Lorentz transformation.

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26 Preliminary notions about particles

(2) p2

(1) p1

(4) p4

(3) p3

(3) −p3

(1) p1

(4) p4

(2) −p2

(4) −p4

(1) p1

(2) −p2

(3) p3

Figure 1.5: Crossed-reactions: the middle and right reactions are respectively the t-channel and u-channel of the reaction on the left.

channel reaction s-variable t-variable u-variable

s-channel (1) + (2)→ (3) + (4) s t u

t-channel (1) + (3)→ (2) + (4) t s u

u-channel (1) + (4)→ (3) + (2) u t s

Table 1.2: The Mandelstam variables as function of the channels of the reaction (1) + (2)→ (3) + (4).

1.6 Elementary particles dynamics

1.6.1 Phase space

Let us consider the following reaction with ni particles in the initial state and nf in the finalstate:

p1 + p2 + ...+ pni → p′1 + p′2 + ...+ p′nf (1.61)

The particles are considered on mass shell: E2 = p2 +m2 so that the final state is represented by4nf−nf = 3nf parameters. Because of the conservation of the 4-momemtum between the initialand final state, there are 3nf − 4 independent parameters. A given reaction will correspond toa single point on an hyper-surface of dimension 3nf − 4 of the space of dimension 3nf . Thishyper-surface is the phase space: it corresponds to all possible values of 4-momenta componentsof the particles in the final state involved in a reaction.

1.6.2 Transition rate

In chapter 3, we will show that the general formula of the elementary transition rate (probabilityper unit time) from an initial state |i〉made of ni particles to a final state |f〉made of nf particlesis given by10:

dΓi→f = V1−ni(2π)4δ(4)(p′1 + ...+ p′nf − p1 − ...− pni)|Mfi|2ni∏

k=1

1

2Ek

nf∏

k=1

d3~p′k(2π)32E′k

(1.62)

where the prime symbols denote final states quantities and Mfi is the invariant scatteringamplitude (or probability amplitude or often called matrix element) that can be computed with

10We also provide in appendix A a “classical” justification of this formula based on quantum mechanics andnot on quantum fields theory.

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Interlude: calculating with Dirac distribution 27

Figure 1.6: The Mandelstam variables.

Feynman diagrams as we will learn in the third chapter. In general, we square |Mfi| and sum oraverage over various degrees of freedom that are not observed, such as spin or color. The |Mfi|appearing here implicitly includes these factors. The symbol V is the volume of the 3-space.This arbitrary parameter appears here to avoid diverging quantities but as we will see in thenext sections, all physical quantities such as decay rate (and thus lifetime of a particle) andcross-sections do not depend on it. Therefore, the limit to infinite space volume can be safelytaken.

The quantity:

dΦ = (2π)4δ(4)(p′1 + ...+ p′nf − p1 − ...− pni)nf∏

k=1

d3~p′k(2π)32E′k

(1.63)

is called the Lorentz Invariant Phase Space (often denoted dLIPS).

1.6.3 Interlude: calculating with Dirac distribution

Let us remind few useful properties of Dirac distribution:

∫ +∞

−∞f(x)δ(x− x0) dx = f(x0) (1.64)

for all smooth functions f (that have derivatives to all orders) on R with compact support. Onecan define the composition δ(g(x)) for continuously differentiable functions g by:

δ(g(x)) =∑

i

δ(x− xi)∣∣∣ ∂g∂x(xi)∣∣∣⇒∫ +∞

−∞δ(g(x)) dx =

i

1∣∣∣ ∂g∂x(xi)∣∣∣

(1.65)

where xi are the real roots of the g function (g(xi) = 0). If g has no roots, then δ(g(x)) = 0.From equ. 1.65, one finds the well known properties:

δ(αx) = δ(x)|α| , δ(−x) = δ(x) (1.66)

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28 Preliminary notions about particles

Applying equ. 1.65 to p2−m2 with g(p0) = (p0)2−|~p|2−m2 = (p0)2−E2 = (p0−E)(p0 +E),one gets: ∫ +∞

−∞δ(p2 −m2) dp0 =

1

E

But since δ(p2 −m2) is an even function versus p0, we have:

∫ +∞

−∞δ(p2 −m2) dp0 = 2

∫ +∞

0δ(p2 −m2) dp0 = 2

∫ +∞

−∞δ(p2 −m2)θ(p0) dp0

θ being the Heaviside function. Hence:

∫ +∞

−∞δ(p2 −m2)θ(p0) dp0 =

1

2E(1.67)

Thus, one has the useful equivalence:

∫ +∞

−∞d4p δ(p2 −m2)θ(p0) =

∫ +∞

−∞

d3~p

2E(1.68)

Applying the Fourier transformations with the following convention ˆf(ξ) =∫ +∞−∞ f(x)e−iξxdx⇔

f(x) = 12π

∫ +∞−∞

ˆf(ξ)eiξxdξ to the Dirac distribution

δ = 1⇔ δ(x) =1

∫ +∞

−∞eiξxdξ ⇒ δ(n)(x) =

1

(2π)n

∫eiξ.xdnξ (1.69)

1.6.4 Decay width and lifetime

We apply equ. 1.62 with ni = 1 and nf = n particles:

dΓ1→n = (2π)4δ(4)(p′1 + ...+ p′n − p1)|M|2 1

2E1

n∏

k=1

d3~p′k(2π)32E′k

And hence the transition rate to a set of particles is obtained by integrating over the phasespace:

Γ1→n = (2π)4 1

2E1

∫δ(4)(p′1 + ...+ p′n − p1)|M|2

n∏

k=1

d3~p′k(2π)32E′k

(1.70)

The rate is in s−1, and so equivalent to GeV in natural unit. We call it the partial decaywidth. The total decay width (or total transition rate) is obtained by summing up on allpossible decays:

Γtot =∑

Γ1→fi (1.71)

and the branching ratio which represents the fraction of decay in a given channel is simply:

BR1→fi =Γ1→fiΓtot

(1.72)

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Cross-sections 29

A particle that decays has a given lifetime. When a particle is stable, we can write its wavefunction as ψ(t) = ψ(0)e−iEt, so that |ψ(t)|2 = |ψ(0)|2. When the particle is unstable, we expect|ψ(t)|2 = |ψ(0)|2e−t/τ , τ being the lifetime. Let us write:

E = E0 − iΓ/2⇒ |ψ(t)|2 = |ψ(0)|2e−Γt

Clearly, the particle lifetime has to be identified as the inverse of the total decay width:

τ =1

Γtot(1.73)

But there is another consequence: in order to find the probability of finding the particle statewith energy E, let us take the Fourier transform:

ψ(E) =1√2π

∫ +∞

−∞dteiEtψ(t) =

ψ(0)√2π

∫ +∞

0dtei(E−E0)t−Γ

2t =

iψ(0)√2π

1

(E − E0) + iΓ2

where ψ(t < 0) = 0 (the particle doesn’t exist yet). The probability of finding the energy E isthen:

|ψ(E)|2 =|ψ(0)|2

1

(E − E0)2 + Γ2

4

The distribution of the energy (or mass) follows a Breit-wigner law with a spread of energiesgiven by the decay width. The relationship between lifetime and decay-width can be found againvia the Heisenberg’s uncertainty principle:

Γtot ≥ ~/τ

The greater the mass (or energy) is shifted from its nominal value, the shorter the particle lives.Note that according to 1.70, the lifetime depends on E1 and so on the reference frame. By

convention, the lifetime is measured in the particle’s rest frame.

1.6.5 Cross-sections

In particle physics, cross-sections are measured in barns = 10−24 cm2, a unit of surface area, byreference to the image of a particle (1) having a radius r1 impinging on the surface S of a targetmade of particle (2) of radius r2 and containing n2 particles per unit of volume. Two particlescan hit only if the distance between their centre is below r1 + r2. If around each target particle,we draw a disc of area σ = π(r1 + r2)2, the probability of interaction is given by the numberof particles (2) seen by (1) times the ratio of σ over the surface of the target. If the particle(1) travels at speed |~v1|, during dt, (1) is going to go through a region of the target containingn2S|~v1|dt particles of the target so that the probability of interaction is:

dP =σ

Sn2 S|~v1|dt = σn2|~v1|dt

which leads to the rate:

Γ =dP

dt= σn2|~v1|

Now, instead of a single particle, we have a beam, larger than the surface S of the target witha density of n1 particle per unit of volume. Considering a volume V of the target, the rate willbe multiplied by the number of particle of the beam that can interact in that volume:

Γ = σn2|~v1| n1V = σn2V n1|~v1| = σN2φ (1.74)

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30 Preliminary notions about particles

where φ = n1|~v1| is the flux of incident particles and N2 the number of particles considered inthe target. If particles (2) are also in a beam moving at ~v2, the flux becomes: φ = n1|~v1 − ~v2|.According to 1.74, the cross section is defined as:

σ =Γ

N2φ

Let us apply this definition to the experiments in particle physics where we are interested in theinteraction between two single particles. In that case, the cross section, σ has to be understoodas an effective area. The flux of the incoming particle is φ = 1

V |~v1−~v2| (using n1 = 1/V), leadingto the cross-section formula:

σ =number of interactions per unit time

flux incident particles=

Γ

φ

Applying equ. 1.62 with ni = 2 and nf = n particles to get the number of interaction per unittime:

dΓ2→n =1

V (2π)4δ(4)(p′1 + ...+ p′n − p1 − p2)|M|2 1

2E1

1

2E2

n∏

k=1

d3~p′k(2π)32E′k

so that the cross section is:

dσ =1

4|~v1 − ~v2|E1E2(2π)4δ(4)(p′1 + ...+ p′n − p1 − p2)|M|2

n∏

k=1

d3~p′k(2π)32E′k

Let us denote by F , the flux factor: F = |~v1 − ~v2|E1E2. In general , it is not Lorentz invariantwhereas the rest of the previous equation is. However, if we consider only cases of 2 collinearbeams, then F is Lorentz invariant. Rearranging F , it’s going to be clear:

F = |~v1 − ~v2|E1E2 = | ~p1

E1− ~p2

E2|E1E2 = ( |~p1|

E1+ |~p2|

E2)E1E2 = (|~p1|E2 + |~p2|E1)

⇒ F 2 = |~p1|2E22 + |~p2|2E2

1 + 2|~p1|E2|~p2|E1

But p1p2 = E1E2 + |~p1||~p2| ⇒ (p1p2)2 = E21E

22 + |~p1|2|~p2|2 + 2E1E2|~p1||~p2| and hence:

F 2 − (p1p2)2 = |~p1|2(E22 − |~p2|2) + E2

1(|~p2|2 − E22) = |~p1|2m2

2 − E21m

22 = −m2

1m22

Therefore:

F = E1E2

( |~p1|E1

+|~p2|E2

)=√

(p1.p2)2 − (m1m2)2 (1.75)

which is obviously Lorentz invariant. The elementary cross-section is finally:

dσ2→n = 1

4√

(p1.p2)2−(m1m2)2(2π)4δ(4)(p′1 + ...+ p′n − p1 − p2)|M|2∏n

k=1d3~p′k

(2π)32E′k

= |M|24F dΦ

(1.76)

If the 2 beams are not collinear, the cross section is by definition the one that would be measuredin the rest-frame of the 2 particles.

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Decomposition of the phase space 31

1.6.6 Decomposition of the phase space

Let us denote by P the 4-momentum of the initial state (potentially made of several particles)and p1, p2, · · · , pn the ones of the n particles in the final state. For simplicity, we will startwith n = 3. The Lorentz invariant phase space (equ. 1.63) corresponds to:

dΦ(P → p1p2p3) = (2π)4δ(4)(p1 + p2 + p3 − P )d3~p1

(2π)32E1

d3~p2

(2π)32E2

d3~p3

(2π)32E3

The particles in the final state are often produced via intermediate resonances and sometimes,it simplifies the calculation to make it appear explicitly. For example let us suppose that thereaction is actually P → q12p3 → p1p2p3 where q12 denotes the 4-momentum of the intermediatestate which further decays in p1p2. One can always introduce the 2 identity integrals:

1 =

∫d4q12 δ

(4)(q12 − p1 − p2)θ(q012) , 1 =

∫dm2

12 δ(m212 − q2

12)

m212 being the mass-squared of the particle described by the 4-momentum q12. Putting together

these 2 identity integrals and using the equivalence of equ. 1.68 gives:

1 =

∫∫d4q12 δ

(4)(q12 − p1 − p2)θ(q012) dm2

12 δ(m212 − q2

12) =

∫∫d3~q12

2E12δ(4)(q12 − p1 − p2) dm2

12

And hence:

dΦ(P → p1p2p3) = (2π)4δ(4)(q12 + p3 − P ) d3~p1

(2π)32E1

d3~p2

(2π)32E2

d3~p3

(2π)32E3

d3~q12

2E12δ(4)(q12 − p1 − p2) dm2

12

=dm2

122π (2π)4δ(4)(q12 + p3 − P ) d3~p3

(2π)32E3

d3~q12

(2π)32E12

×(2π)4δ(4)(q12 − p1 − p2) d3~p1

(2π)32E1

d3~p2

(2π)32E2

where the integral sign is implicit. Thus, we have decomposed the 3 particles phase space intosmaller (2-body) phase spaces:

dΦ(P → p1p2p3) =dm2

12

2πdΦ(P → q12p3) dΦ(q12 → p1p2) (1.77)

Note that the number of degrees of freedom stays the same: initially, we had 3 × 3 (the 33-momenta) reduced by the delta function to 3 × 3 − 4 = 5. Each 2-body phase space has2× 3− 4 = 2 degrees of freedom, and thus 2 + 2 + 1 = 5 (1 being due to dm2

12).

The decomposition of the 3-particles final state can be easily extended to larger states con-taining n particles:

dΦ(P → p1 · · · pn) =dm2

[1···j]2π

dΦ(P → q[1···j]pj+1 · · · pn) dΦ(q[1···j] → p1 · · · pj) (1.78)

with m2[1···j] = (p1 + · · · pj)2 = q2

[1···j]. Since each dΦ is Lorentz invariant, they can be evaluatedin different frames. The fact that the particle described by q is a real resonance or not does notmatter in this decomposition. q can correspond to any sum of 4-momenta final state particles.However, in case of a resonance, q2 distribution follows a Breit-Wigner as seen previously.

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32 Preliminary notions about particles

1.6.7 Few applications

1.6.7.1 Decay to 2 particles

(i)→ (1) + (2)

Let us denote by p = (E, ~p) the initial particle 4-momenta and pf=1,2 = (Ef , ~pf ) the two4-momenta of the decay products. Applying equ. 1.70:

Γ1→2 = (2π)4 1

2E

∫δ(4)(p1 + p2 − p)|M|2

d3~p1

(2π)32E1

d3~p2

(2π)32E2

In the initial particle’s rest frame, we have:

~p1 + ~p2 = 0, p = (m, 0), p1 = (√|~p∗|2 +m2

1, ~p∗), p2 = (

√|~p∗|2 +m2

2,−~p∗)

Integrating over ~p2, we get:

Γ1→2 = (2π)−2 1

2m

∫δ(√|~p∗|2 +m2

1 +√|~p∗|2 +m2

2 −m)|M|2 d3~p∗

2√|~p∗|2 +m2

1

1

2√|~p∗|2 +m2

2

where now |M| depends on ~p∗. Using spherical coordinates d3~p∗ = p∗2 sin θ dp∗ dθ dφ =p∗2 dp∗ dΩ:

Γ1→2 =1

32π2m

∫dΩ

∫ +∞

0dp∗

δ(√p∗2 +m2

1 +√p∗2 +m2

2 −m)√p∗2 +m2

1

√p∗2 +m2

2

p∗2|M|2

We could use the formula 1.65 to perform the integration over p∗ but it is simpler to make thechange of variables:

E =√p∗2 +m2

1 +√p∗2 +m2

2 ⇒ dE =Ep∗√

p∗2 +m21

√p∗2 +m2

2

dp∗

E representing the total energy√s. Then it comes:

Γ1→2 =1

32π2m

∫dΩ

∫ +∞

m1+m2

dE

Eδ(E −m)p∗|M|2

Γ1→2 =|~p∗|

32π2m2

∫dΩ|M|2 if m > m1 +m2, 0 otherwise (1.79)

p∗ satisfying m =√p∗2 +m2

1 +√p∗2 +m2

2 and corresponds to the momentum in the rest-frameand hence, its expression is the one we calculated previously in equ. 1.54. M must be evaluatedat ~p1 = ~p∗ and ~p2 = −~p∗. This result is general, no assumptions were made on the detailedshape of M. If, we assume in addition that the initial particle is spinless, by symmetry, theamplitude M cannot depend on the solid angle. Then

∫sin θdθ dφ = 4π and finally we get:

Γ1→2 =|~p∗|

8πm2|M|2 if m > m1 +m2, 0 otherwise

Following the same kind of calculations the phase space for the 2-body is easy to obtain.Looking at formula 1.79, it is simply:

dΦ(p→ p1p2) = dΓ1→22E

|M|2 =⇒ dΦ(p→ p1p2) =|~p∗1|

16π2m12dΩ∗1 (1.80)

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Few applications 33

where m2 has been changed to m212 to be more general, E = m12 and all variables have to be

understood in the rest frame of particles 1 and 2 and denoted with a star index. Finally, usingformula 1.54 for |~p∗1|, it gives:

dΦ(p→ p1p2) =

√1− 2

(m21 +m2

2)

m212

+(m2

1 −m22)2

m412

dΩ∗132π2

(1.81)

1.6.7.2 Scattering cross section 2→ 2

(1) + (2)→ (1′) + (2′)

We apply the formula 1.76 with nf = 2:

dσ2→2 =1

4√

(p1.p2)2 − (m1m2)2(2π)4δ(4)(p′1 + p′2 − p1 − p2)|M|2 d3~p′1

(2π)32E′1

d3~p′2(2π)32E′2

(1.82)

In the Center-of-mass frame, p1 = (E∗1 =√|~p∗|+m2

1, ~p∗) and p2 = (E2 =

√|~p∗|+m2

2,−~p∗) withE1 + E2 =

√s. Hence F simplifies to F = (E1 + E2)|~p∗| (using 1.75) and thus:

dσ2→2 =1

4√s|~p∗|(2π)2

δ(E′1 + E′2 −√s)δ(3)(~p′1 + ~p′2)|M|2 d3~p′1

2E′1

d3~p′22E′2

(1.83)

which gives the cross-section for a process in which ~p′1,2 lies in the range d3~p′1,2. In general,we are interested in the probability of observing one of the final products in certain directionwithin an element of the solid angle dΩ∗ = sin θ∗dθ∗dφ∗. Let us integrate over the other decayproduct, let us say ~p′2:

dσ2→2 =1

4√s|~p∗|(2π)2

δ(E′∗1 + E′∗2 −√s)|M|2 |

~p′∗|2d| ~p′∗|dΩ∗

2E′∗1

1

2E′∗2(1.84)

Replacing E′∗i by√| ~p′∗|2 +m′i

2:

dσ2→2dΩ∗ = 1

16√s|~p∗|(2π)2

∫ +∞0 δ

(√|~p∗|2 +m′1

2 +√|~p∗|2 +m′2

2 −√s)|M|2

× | ~p′∗|2d| ~p′∗|√|~p∗|2+m′1

2√|~p∗|2+m′2

2

(1.85)

Using the variable E∗ =√|~p∗|2 +m′1

2 +√|~p∗|2 +m′2

2, we have

dE∗ = d| ~p′∗|| ~p′∗|

√|~p∗|2 +m′1

2 +√|~p∗|2 +m′2

2

√|~p∗|2 +m′1

2√|~p∗|2 +m′2

2

so that:dσ2→2

dΩ∗=

1

16√s|~p∗|(2π)2

∫ +∞

m′1+m′2

δ(E∗ −√s

)|M|2dE

E∗|~p′∗|

and thus the differential cross-section is:

dσ2→2

dΩ∗=

1

64π2s|M|2 |

~p′∗||~p∗| if

√s > m′1 +m′2 , 0 otherwise (1.86)

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34 Preliminary notions about particles

where |~p′∗| and |~p∗| have been calculated with equ. 1.54. To calculate the total cross sec-tion (which is Lorentz invariant with collinear beams), it is sufficient to integrate the pre-vious expression. However, it is not so useful for the differential cross-section, since thisformula is only available in the CM frame. We can go a bit further by expressing dΩ∗ asfunction of the Mandelstam variable t = (p1 − p′1)2 = m2

1 + m′21 − 2E∗1E′∗1 + 2~p∗1.~p′

∗1. With

~p∗ = (0, 0, |~p∗|) and ~p′∗

= (|~p′∗| sin θ∗, 0, |~p′∗| cos θ∗), t = m21 + m′21 − 2E∗1E

′∗1 + 2|~p∗||~p′∗| cos θ∗,

giving dt = 2|~p∗||~p′∗|d(cos θ∗) and since dΩ∗ = d(cos θ∗)dφ (the sign doesn’t matter), one has:

dσ2→2

dt=dσ2→2

dΩ∗dΩ∗

dt=

1

64π2s|M|2 |

~p′∗||~p∗|

2|~p∗||~p′∗|=

1

128π2s|M|2 dφ

|~p∗|2 (1.87)

Let us assume M doesn’t depend on φ (it’s not systematically true because the symmetry isbroken by the beam axis). After integration over the angle:

dσ2→2

dt=

1

64πs

|M|2|~p∗|2 (1.88)

which gives a formula valid in all frames with |~p∗| from 1.54.

1.6.7.3 The three particles final state: Dalitz plots

Let us consider the decay of a particle with a mass m and 4-momentum p into 3 particleswith mass mi=[1,3] and 4-momenta pi=[1,3]. Using the rest-frame of the decaying particle, thedifferential decay rate is given by (equ. 1.70):

dΓ1→3 = (2π)4 1

2mδ(4)(p1 + p2 + p3 − p)|M|2

d3~p1

(2π)32E1

d3~p2

(2π)32E2

d3~p3

(2π)32E3

Integrating over ~p3, it gives:

dΓ1→3 =1

16m(2π)5δ(E1 + E2 + E3 −m)|M|2 |~p1|2d|~p1|dΩ1

E1

|~p2|2d|~p2|dΩ2

E2

1

E3

Let us assume that we are not interested by the angular distributions and that the matrix elementdoes not depend on them (in other words, we suppose that there is no spin polarization). Thenthe integration over all angles except the one between particle 1 and 2 will give:

∫dΩ1 = 4π

and dΩ2 = 2πd(cos θ12) leading to:

dΓ1→3 =1

8m(2π)3δ(E1 + E2 + E3 −m)|M|2 |~p1|2|~p2|2d|~p1|d|~p2|d(cos θ12)

E1E2E3

Since E2i = m2

i + |~pi|2, we have EidEi = |~pi|d|~pi|. In addition, |~p3|2 = |~p1 + ~p2|2 = |~p1|2 + |~p2|2 +2|~p1||~p2| cos θ12 and thus E3dE3 = |~p3|d|~p3| = |~p1||~p2|d(cos θ12). The differential decay width isthen reduced to:

dΓ1→3 =1

8m(2π)3δ(E1 + E2 + E3 −m)|M|2dE1dE2dE3

which simply gives after integration over E3:

dΓ1→3 =1

8m(2π)3|M|2dE1dE2

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Exercises 35

The 3-body decay is then described by 2 variables, here E1 and E2. The Dalitz plot is a planeshowing the density (i.e. relative frequency) of these two variables, with one variable on thex-axis and the other on the y-axis. Traditionally, instead of the energies, the squares of themasses of two pairs of the decay products are used. It has the advantage not to depend on the

frame. Since m213 = (p1 + p3)2 = (p − p2)2 = m2 + m2

2 − 2mE2, we have dE2 = −dm213

2m and

similarly dE1 = −dm223

2m leading to:

dΓ1→3 =1

32m3(2π)3|M|2dm2

13dm223

If the matrix element |M|2 is a constant, then the plan (m213,m

223) should be uniformly pop-

ulated. In this case, the differential decay width only depends on the phase space and thecontour of the Dalitz plot is shaped by the energy-momentum conservation and the masses ofthe particles involved in the reaction. However, as soon as there are departures from a uniformdensity, it is a sign of a underlying dynamic in the process: |M|2 depends on m2

13 or m223. This

is the case for example when the 3-body reaction occurs via an intermediate resonance (|M|2depending on a Breit-Wigner).

An example of Dalitz plot is shown on figure 1.7 in which the reaction K−+p→ Λ+π+ +π−

was studied [34]. Clearly, the density is not uniform in the distorded ellipse. An accumulationof data is seen in horizontal and vertical bands for masses mΛπ+ ≈ mΛπ− ≈ 1.38 GeV. Thisaccumulation turns out to be due to the presence of the Σ±(1385) resonance having a mass of1385 MeV. The reaction was actually: K−+p→ Σ+ +π− → Λ+π+ +π− (and the C-conjugatereaction)

1.7 Exercises

Exercise 1.1 Using Lorentz transformation, show that dΩ = cdtdV is a 4-scalar. Hint: do notforget the jacobian of the transformation!

Exercise 1.2 What is the appropriate factor to convert cross-sections expressed in natural units( GeV−2) to µb?

Exercise 1.3 π+ decay in µ+νµConsider N0 = 600 π+ (spin 0, m ≈ 140 MeV/c2) with a kinetic energy of 140 MeV. The π+’sdecay in µ+νµ (mµ ≈ 106 MeV/c2, mν ≈ 0).

1. Determine the angular distribution dNd cos θ∗ of the µ in the π rest frame? (θ∗ is the angle

between the π direction and µ in the π rest frame)

2. What are the energy and momentum of the µ in the π rest frame?

3. What is the minimum and maximum energy of the µ in the lab frame?

4. Draw the shape of the energy distribution dNdEµ

of the µ in the lab frame assuming a binningof 1 MeV.

Exercise 1.4 ∆0 production and decayThe ∆0 resonance has a spin 3/2 and an intrinsic parity +1. The reaction π−p → ∆0 → π0n

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36 Preliminary notions about particles

VOLUME 10, NUMBER 5 PHYSICAL REVIEW LETTERS 1 MARCH 1963

reactions are produced according to phase space andthat the effect of Zo contamination in the Y&* peak isnegligible.SR. H. Dalitz, Brookhaven National Laboratory Report

BNL 735 (T-264) (unpublished).

~OJ. W. Cronin and D. E. Overseth, International Con-ference on High-Energy Nuclear Physics, Geneva, 1962(CERN Scientific Information Service, Geneva, Switzer-land, 1962), p. 453.~D. Colley et al. , Phys. Rev. 128, 1930 (1962).

SPIN AND PARITY OF THE 1385-MeV Y,* RESONANCE~Janice B. Shafer, Joseph J. Murray, and Darrell O. Huwe

Lawrence Radiation Laboratory, University of California, Berkeley, California(Received 24 January 1963)

Study of the reaction

at a momentum of 1.22 BeVy'c has shown conclu-sively that the spin of the 1385-MeV Y,* is «&3,

as reported earlier by Ely et al. ', it has also in-dicated that the Y,* state is P» (even Y~ —Aparity) rather than D» (odd parity). These con-clusions result from the angular distribution ofthe lambda and the angular dependence of thelambda polar ization.The events were obtained through the use of

the 72-inch hydrogen bubble chamber placed ina beam of high-energy K mesons extracted fromthe Bevatron. The momentum spread of the beamwas +2. 5 $, and the pion background was about10$Approximately 1650 events were analyzed which

satisfied the E +p —A+ n++ n hypothesis. Someresults from a partial sample were reportedearlier. The events were identified by kine-matic fitting, first at the decay vertex and thenat the production vertex, with the IBM-7090 pro-gram "PACKAGE, " After this two-step fitting,the number of ambiguous events that might havebeen F' production rather than lambda productionwas approximately 1 $ of the total. Almost com-plete separation of the Z n+v from the Av+nevents was accomplished by examination of theratio y2(Am~) j2y'(Zmw) for each event, the Z'vvy' being weighted by a factor of two because ofthe difference in average g' values. The finalAm+a- sample included about 93 $ of the true Annevents and about 5$ of the true Zo~m events. TheA3v events were excluded from the Ann sampleby eliminating all events with y2(A3w) less than 10.The cross section for production of the A~+~

final state by 1.22-BeV/c Z mesons was de-termined by comparison with the observed num-ber of tau decays in the same film sample; its

value is 2. 2~0. 2 mb. The angular distributionsof Y*+ and Y~ production were found similar tothat of Y*+ production in the work of Ely et al. '(see reference 3).The Dalitz plot for the Av v events is shown

in Fig. 1. Projection onto the Av+ mass axis isdisplayed; the Am projection (not shown) is sim-ilar. These mass spectra of the A~+ and the Avsystems are well fitted by Y*+ and Y~ reso-nance curves alone, without background; valuesof M(Y*) = 1385 MeV and f' = 50 MeV are required.The production ratio of Y*+ to Y* is 0.80. Back-ground of 5 to 10$ cannot be ruled out.On the basis of these mass spectra, the limits

1340 MeV ~M(Y*) ~ 1430 MeV were utilized inthe analysis discussed below. Only events withlarge production angles in the center of mass

Mass of Aw- pair fMeV)I 300 l 400 l 500 l600 1700

—0Q. ~

C4

V0

0l.50 2.00 2.50

Mass squared of A~- pair (BeV2j3.00

0OgCO

0 a0 ~0~0 o

EA

OX00Number per l0 MeV

FIG. 1. Dalitz plot of A~ m events from E P inter-actions at 1.22 BeV/c. The square of A&+ effectivemass is plotted against the square of Ax effective mass.Scales giving the masses in MeV are also shown. Pro-jection of the events onto the A~+ mass axis is displayedto the right of the figure; the curve represents the fittingof Breit-Wigner resonance expressions to the An+ andAm systems.

Figure 1.7: The Dalitz plot of the reaction K− + p→ Λ + π+ + π−. From [34].

is considered. The decay occurs via the strong interaction which conserves the parity. We recallthat both the proton and the neutron have spin 1/2 and parity +1 while the pion has a spin 0and parity -1.

1. Considering the Z axis as the direction of the π−, show that the angular momentum con-servation imposes that the projection of the ∆0 spin is Sz(∆

0) = ±12 .

2. What is the value of the orbital angular momentum l of the final system π0n?

3. Assuming Sz(∆0) = +1

2 , show that the neutron angular distribution is (1+3 cos2 θ)8π . Hint:

consider successively the neutron in the spin-up and spin-down state and determine thecorresponding angular distribution. Use the figure 1.1 for the Clebsch-Gordan coefficientsand the expression of the spherical harmonics.

4. Same question if Sz(∆0) = −1

2 . Can you guess the answer without any calculation?

Exercise 1.5 Show that the Lorentz invariant phase space:

dΦ = (2π)4δ(4)(p′1 + ...+ p′nf − p1 − ...− pni)nf∏

k=1

d3~p′k(2π)32E′k

is indeed Lorentz invariant. Hint: for the δ-Dirac function, use Fourier transformation.

Exercise 1.6 Decompose the phase space of a 4-particles final state into product of 2-particlesfinal states.

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Chapter 2

From wave functions to quantumfields

Few references:F. Halzen and D. Martin, “Quarks & Leptons”, Chapter 3 and 5, Wiley.P. Fayet, “Champs relativistes”, Chapter 8, 10, 11, les editions de l’ecole polytechnique.

The purpose of this chapter is to clearly define the mathematical objects thatdescribe particles of various nature: bosons (spin 0 and spin 1) or spin 1/2fermions. This is an important step toward the calculation of Feynman dia-gram.

2.1 Schrodinger equation

2.1.1 Correspondence principle

In quantum mechanics, the correspondence principle states that a quantum of momentum ~p is awave with a wavevector ~k through the relation postulated by De Broglie ~p = ~~k and its energyE to the angular frequency ω: E = ~ω (initially postulated by Einstein for photons). Then,Schrodinger expressed the phase of a plane wave as a complex phase factor using:

ψ(~x, t) = Ne−i(ωt−~k.~x) = Ne−

i~ (Et−~p.~x) = Ne−

i~pµx

µ(2.1)

Realizing that such wave is an eigenstate of the operators:

H = i~∂

∂t, ~P = −i~~∇ (2.2)

with the corresponding eigenvalue E and ~p, leads immediately to the identification of H as the

energy operator and ~P as the momentum operator, or in a covariant way (see definition 1.16and 1.15):

Pµ = i∂µ = i∂

∂xµ=

(H

− ~P

)(2.3)

37

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38 From wave functions to quantum fields

Pµ being the operator 4-momentum. Moreover, the conservation of the energy (non relativistic)

E = p2

2m + V , V being a potential, implies the Schrodinger equation:

H =1

2m~P 2 + V ⇒ i~

∂tψ(~x, t) = − ~2

2m~∇2ψ(~x, t) + V ψ(~x, t) (2.4)

The plane wave is solution of the equation for free particles (namely with V=0).

2.1.2 4-current

In what follows we are going to use probability density ρ and probability current (i.e. densityflux) ~j, since we are generally interested in moving particles. Probabilities are conserved quanti-ties in the sense that ∂

∂t

∫universe ρd

3~x = 0 which leads to the local version through the continuityequation:

∂ρ

∂t+ ~∇.~j = 0 (2.5)

Indeed, this equation simply states that the rate of decrease of the number of particles in avolume V: − ∂

∂t

∫V ρ d

3~x must be equal to the total flux of particles escaping from that volume∮S~j.~n dS, ~n, being a unit vector normal to the element dS of the surface S enclosing the volume

V. Because of Gauss theorem, the closed path integral is equal to∫V~∇.~j d3~x so that:

− ∂

∂t

Vρ d3~x =

V~∇.~j d3~x

Since this equation must be true for any volume V, we find the continuity equation 2.5. In orderto clearly identify ρ and ~j, we first conjugate the Schrodinger equation and multiply by ψ:

−i~ψ(~x, t)∂

∂tψ∗(~x, t) = − ~2

2mψ(~x, t)~∇2ψ∗(~x, t) + V ψ∗(~x, t)ψ(~x, t)

while multiplying by ψ∗(~x, t) the Schrodinger equation we obtain:

i~ψ∗(~x, t)∂

∂tψ(~x, t) = − ~2

2mψ∗(~x, t)~∇2ψ(~x, t) + V ψ∗(~x, t)ψ(~x, t)

Subtracting the 2 previous results leads to:

i~(ψ∗∂

∂tψ + ψ

∂tψ∗) = − ~2

2m(ψ∗~∇2ψ − ψ~∇2ψ∗)⇒ i~

∂t(ψψ∗) +

~2

2m~∇.(ψ∗~∇ψ − ψ~∇ψ∗) = 0

A simple comparison with the continuity equation 2.5 allows the identification:

ρ = |ψ|2 , ~j = i~

2m(ψ~∇ψ∗ − ψ∗~∇ψ) (2.6)

Therefore, for a plane wave (equ. 2.1), the number of particles per unit volume is ρ = |N |2 and~j = |N |2 ~p

m = |N |2~v.

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Spin 0 particles 39

2.2 Spin 0 particles

2.2.1 Klein-Gordon equation

We can apply the correspondence principle equations 2.2 to the relativistic equation for energy:E2 = |~p|2 +m2 (going back to natural units). Thus:

− ∂2

∂t2φ = −∇2φ+m2φ⇒ (+m2)φ = 0 (2.7)

which is the formula of the Klein-Gordon equation for a free particle. What are the density andthe current with such equation? Following the same procedure as for the Schrodinger equation(namely taking the conjugate, multipling by φ and subtracting the result to the Klein-Gordonequation times φ∗, we obtain:

∂2

∂t2φ−∇2φ+m2φ = 0

∗−→ ∂2

∂t2φ∗ −∇2φ∗ +m2φ∗ = 0

×φ−−→ φ ∂2

∂t2φ∗ − φ∇2φ∗ +m2φ∗φ = 0

∂2

∂t2φ−∇2φ+m2φ = 0

×φ∗−−→ φ∗ ∂2

∂t2φ− φ∗∇2φ+m2φ∗φ = 0

⇒ φ∗ ∂2

∂t2φ− φ ∂2

∂t2φ∗ − φ∗∇2φ+ φ∇2φ∗ = 0

⇒ ∂∂t(φ

∗ ∂∂tφ− φ ∂

∂tφ∗) + ~∇.(φ~∇φ∗ − φ∗~∇φ) = 0

The last equation can be multiplied by any constant. In order to find a formula similar to theSchrodinger case, we multiply it by i, so that finally:

ρ = i(φ∗∂

∂tφ− φ ∂

∂tφ∗) , ~j = i(φ~∇φ∗ − φ∗~∇φ) (2.8)

Let us examine the consequences with plane wave 2.1. Applying the Klein-Gordon equation onplane wave:

−E2φ+ |~p|2φ+m2φ = 0⇒ E = ±√|~p|2 +m2

The 2 values for the energy are possible: +E and −E where E =√|~p|2 +m2. Now, injecting

the plane wave into the density and current equation:

ρ = 2E|N |2 , ~j = 2~p|N |2

or in a more compact way:

jµ =

(ρ~j

)= 2|N |2

(E~p

)= 2|N |2pµ (2.9)

Using the negative solution for the energy and injecting it into the density, we see that thedensity itself can be negative! It excludes the interpretation of the density as a probabilitydensity!

2.2.2 Re-interpretation of the 4-current

The 2 solutions for the energy means that both wave functions φ1 = Ne−i(Et−~p.~x) and φ2 =Ne−i(−Et−~p.~x) are solution of the Klein-Gordon equation leading to the 4-current:

jµ(φ1) = 2|N |2(E~p

), jµ(φ2) = 2|N |2

(−E~p

)

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40 From wave functions to quantum fields

The probability interpretation is impossible. However, Pauli and Weisskopf suggested (6 yearsafter the development of Dirac’s equation) to interpret the 4-current as a 4-current chargedensity. In order to do so, the direction of the current has to change as the charge changes. Itis possible if the direction of the momentum is changed. So let us define the 2 wave functions:

φ(+)~p = Ne−i(Et−~p.~x) = Ne−ip.x , φ

(−)~p = Ne−i((−E)t−(−~p).~x) = Ne+ip.x

where p.x is a shorthand notation for pµxµ. The 2 functions φ

(+)~p and φ

(−)~p are still solution of

the Klein-Gordon equation and now lead to the 4-current:

jµem = i(φ∗∂µφ− φ∂µφ∗)⇒ jµem(φ(±)~p ) = ±2|N |2

(E~p

)(2.10)

Hence, the 2 wave functions φ(±)~p can now be interpreted as states of opposite electric charges.

At the time of Pauli and Weisskopf interpretation, Dirac had already proposed his equation andCarl D. Anderson had discovered the anti-electron, namely the positron in 1932. It was then

logical to consider φ(−)~p as a state of an antiparticle with a positive energy. Antiparticle means

a charge conjugate state to the positive-energy state (φ(−)~p = φ

(+)~p

∗) with the same mass, with

opposite charge and reversed momentum direction. There is no place here for negative energiesor propagation backward in time. We can just say that an antiparticle can be represented as aparticle as soon as pµ is reversed. It is going to be even more clear with a quantized field.

2.2.3 Few words about the quantized field

In this section, we just want to give some basic ideas without demonstration. We refer to theQuantum Field Theory courses for the details. See for example [4].

So far, we have implicitly considered that the Klein-Gordon equation describes a single rela-tivistic particle of spin 0. Indeed, if it were describing particles with non-zero spin, the solutionsof the Klein-Gordon equation should be able to distinguish particles with different spin projec-tion. However, there is no place in the solution for such extra-degree of freedom. Thus, theKlein-Gordon equation seems consistent with the description of a single relativistic particle ofspin 0. However, as soon as we are in the relativistic regime, this interpretation cannot hold.Indeed, imagine a single particle of mass m in a box of size L. Because of Heisenberg uncer-tainty, ∆p ≥ ~/L. Since the particle is relativistic, E ≈ pc and hence ∆E ≥ ~c/L. But thisuncertainty can potentially exceed the threshold of production from the vacuum of a pair ofparticle anti-particule ∆E = 2mc2. We can then conclude, that as soon as a particle is localizedwithin a distance called the Compton wavelength of order λ = ~/mc (forgetting the factor 2),the probability to detect a pair of particle anti-particle becomes large. Thus, the notion of asingle particle within a volume becomes a non-sense. The formalism must be able to describestates of any number of particles. This is precisely what does the quantum field theory bychanging the wave functions with operators.

Let us use N = 1 as the normalization of the wave functions

φ(±)~p (x) = e−i(±p.x) (2.11)

The wave functions φ(±)~p (x) constitute a basis, and hence any wave function can be expressed

as:

φ(x) =

∫d3~p

(2π)32Ep

[apφ

(+)~p (x) + b∗pφ

(−)~p (x)

], Ep =

√|~p|2 +m2 (2.12)

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Few words about the quantized field 41

where the volume of integration is large enough (infinity) to consider the quantization of mo-mentum to be continuous. ap and b∗p are the complex amplitudes of the eigenmodes of φ andcorrespond to the Fourier transform coefficients of the wave functions basis.

The quantisation of the wave function (classical field) transforms φ in an operator acting on theHilbert space of state vectors. φ still obeys the same dynamic equation (Klein Gordon) but apand b∗p are now operators. To better emphasize this, b∗p is now written b†p. The quantum fieldwith its hermitian adjoint are then:

φ(x) =∫ d3~p

(2π)32Ep

[apφ

(+)~p (x) + b†pφ

(−)~p (x)

]

φ†(x) =∫ d3~p

(2π)32Ep

[a†pφ

(+)∗~p (x) + bpφ

(−)∗~p (x)

] (2.13)

One can show that ap and a†p obey the same commutation rules as the one of the harmonic

oscillator and thus, are respectively annihilation and creation operator. a†p creates a quantum

of excitation associated to the plane wave φ(+)~p which plays the role of a propagation mode

with vector ~p (or ~k). This quantum is interpreted as a particle of mass m propagating with

momentum ~p. bp and b†p form another set of annihilation and creation operators associated

to the plane wave φ(−)~p and similarly, b†p creates a particle with the same mass m as a†p. As

annihilation and creation operators they satisfy the commutation relations:

[ap′ , a

†p

]=[bp′ , b

†p

]= (2π)32Ep δ

(3)(~p ′ − ~p)[ap′ , ap

]=[bp′ , bp

]=[a†p′ , a

†p

]=[b†p′ , b

†p

]= 0[

a†p′ , b†p

]=[ap′ , b

†p

]=[a†p′ , bp

]= 0

(2.14)

The vacuum state |0〉 is the state with 0 particle properly normaized 〈0|0〉 = 1. With ournormalization1, we have:

ap |0〉 = 0

a†p |0〉 = |1a, ~p〉ap |1a, ~p〉 = |0〉

where |1a, ~p〉 correspond to a state with 1 particle of type a having a momentum ~p. Similarrelations are obtained applying |0〉 on b operators. Since the operators a and b commute,bp |1a, ~p〉 = 0. one can then show that the hamiltonian, momentum and charge operators are:

H =∫ d3~p

(2π)32EpEp (a†pap + b†pbp)

P =∫ d3~p

(2π)32Ep~p (a†pap + b†pbp)

Q =∫ d3~p

(2π)32Ep(a†pap − b†pbp)

1In some textbooks, the operators used are of the kind a†~p i.e for the 3-momentum. The connection to the

notation used here is a†p =√

2Epa†~p .

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42 From wave functions to quantum fields

Let us see the value of the energy, momentum and charge of the state |1a, ~p〉.

H |1a, ~p〉 =∫ d3~p ′

(2π)32E′pE′p (a†p′ap′ + b†p′bp′) |1a, ~p〉

=∫ d3~p ′

(2π)32E′pE′p a

†p′ap′a

†p |0〉

=∫ d3~p ′

(2π)32E′pE′p a

†p′([ap′a

†p] + a†pap′) |0〉

=∫ d3~p ′

(2π)32E′pE′p a

†p′(2π)32Ep δ

(3)(~p ′ − ~p) |0〉= Ep a

†p |0〉

= Ep |1a, ~p〉

The same procedure can be followed for P and Q with the states |1a, ~p〉 and |1b, ~p〉 and we find:

H |1a, ~p〉 = Ep |1a, ~p〉 , H |1b, ~p〉 = Ep |1b, ~p〉P |1a, ~p〉 = ~p |1a, ~p〉 , P |1b, ~p〉 = ~p |1b, ~p〉Q |1a, ~p〉 = 1 |1a, ~p〉 , Q |1b, ~p〉 = −1 |1b, ~p〉

Hence, the 2 states |1a, ~p〉 and |1b, ~p〉 have the same energy, same momentum (and thus samemass since m =

√E2 − |~p|2) but opposite charge: |1b, ~p〉 is just the antiparticle of |1a, ~p〉 (and

vise-versa). Moreover, since we can apply the creation operator several times, we can havestates like |na, ~p〉 with na > 1. Several particles in the same quantum state of propagation isonly possible for bosons (Pauli exclusion principle). Here is another way of realizing the bosonnature: consider a Fock space with only 2 particles. The wave function is related to:

ψ~p1,~p2( ~x1, ~x2) = 〈 ~x1, ~x2|1, ~p1; 1, ~p2〉

= (〈 ~x1| ⊗ 〈 ~x2|)(|1, ~p1〉 ⊗ |1, ~p2〉)= (〈 ~x1| ⊗ 〈 ~x2|)a†p1a

†p2 |0〉

= (〈 ~x1| ⊗ 〈 ~x2|)a†p2a†p1 |0〉

= 〈 ~x1, ~x2|1, ~p2; 1, ~p1〉= ψ~p1,~p2

( ~x2, ~x1)

As expected, the wave function of two identical bosons is symmetric under the interchange ofthe two particles.

We notice that a spatial rotation or a change of reference frame doesn’t affect the internalstructure of the field since it is a scalar field. Hence, the theory describes particles with spin 0that can have an electric charge. A remark about neutral particles: according to the expressionof Q, the charge of a state of a single particle can be zero if a = b. Hence φ(x) = φ†(x): thescalar field is hermitian. The particle is then its own antiparticle: a well known example is theπ0.

Now that we have a better understanding of the operators, looking at the field formula 2.13,we see that φ destroys a positive charge or creates a negative charge corresponding to the an-tiparticle. The global effect of φ is to decrease the total charge of the system by 1 unit. On thecontrary, the effect of φ† is to increase the total charge of the system by 1 unit. A term involvingφ†(x)φ(x) (appearing in the Lagrangian) doesn’t change the global charge of the system. Suchquadratic term can describe at the spacetime x any of the following charge-conserving processes:1) creation and annihilation of a particle, 2) creation and annihilation of an antiparticle, 3) cre-ation of a pair particle-antiparticle, 4) annihilation of a pair particle-antiparticle.

Finally, one may wonder what is the connection between the usual wave functions φ±(x) =e−i(±p.x) and the quantum field φ(x) of formula 2.13? Since usual plane wave functions deal

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Spin 1/2 particles 43

with a single particle, one has to project the field into a single state particle. More precisely, itis easy to check that:

〈0|φ(x)|1a, ~p〉 = e−ip.x

〈1b, ~p |φ(x)|0〉 = e+ip.x

For instance,

〈0|φ(x)|1a, ~p〉 =∫ d3~p ′

(2π)32E′p

[〈0|ap′ |1a, ~p〉φ(+)

~p′(x) + 〈0|b†p′ |1a, ~p〉φ

(−)~p′

(x)]

=∫ d3~p ′

(2π)32E′p

[〈0|ap′a†p|0〉φ(+)

~p′(x) + 〈0|b†p′a

†p|0〉φ(−)

~p′(x)]

=∫ d3~p ′

(2π)32E′p

[〈0|0〉 (2π)32Ep′δ

(3)(~p′ − ~p)φ(+)~p′

(x)]

= φ(+)~p (x) = e−ip.x

and similarily for 〈1b, ~p |φ(x)|0〉.

2.3 Spin 1/2 particles

2.3.1 Dirac equation

In 1928, P.A.M Dirac proposed his famous equation in order to avoid the solution with a negativeenergy and a negative density probability of the Klein-Gordon equation. It was 6 years before thecorrect interpretation of the Klein-Gordon equation. Even if it was not his initial goal, Diracfound that his equation was able to describe particles and antiparticles with half spin unit.Starting from the Klein-Gordon equation, he realized that the solution with negative energywas due to the second derivative ∂2/∂t2 leading to a probability density (old interpretation)involving a single derivative ∂/∂t, and thus allowing a negative probability. Hence he looked foran equation having a ∂/∂t dependency as in the Schrodinger equation. The equation should becovariant under Lorentz transformations, and hence, the dependency must be also linear with∇. Moreover,we want the wave function to still satisfy the Klein-Gordon equation:

(∂µ∂µ +m2)ψ = 0

So basically, the idea is to factorize the previous equation:

(γκ∂κ − im)(βλ∂λ + im)ψ = 0

Where γµ and βλ are a priori 2 sets of 4 numbers. If ψ satisfies

(γκ∂κ − im)ψ = 0 (2.15)

or

(βλ∂λ + im)ψ = 0 (2.16)

then the Klein-Gordon equation will be satisfied as well. Equations 2.15 or 2.16 fulfill thelinearity condition with the derivatives. Let us develop and identify the different terms:

(∂µ∂µ +m2)ψ = (γκ∂κ − im)(βλ∂λ + im)ψ= (γκβλ∂κ∂λ + im(γκ∂κ − βλ∂λ) +m2)ψ

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44 From wave functions to quantum fields

In order to cancel the linear term with m, we see that γκ = βκ. The identification with theD’Alembertian term (∂µ∂µ) then imposes:

∂µ∂µ = γκγλ∂κ∂λ

namely:

∂20 − ∂2

1 − ∂22 − ∂2

3 = (γ0)2∂20 + (γ1)2∂2

1 + (γ2)2∂22 + (γ3)2∂2

3

+(γ0γ1 + γ1γ0)∂0∂1 + (γ0γ2 + γ2γ0)∂0∂2 + (γ0γ3 + γ3γ0)∂0∂3

+(γ1γ2 + γ1γ2)∂1∂2 + (γ1γ3 + γ1γ3)∂1∂3

+(γ2γ3 + γ3γ2)∂2∂3

If the γ’s were complex numbers, the first line of the equality would impose γ0 = ±1 andγk=1,2,3 = ±i. However, It would be impossible to cancel the last 3 lines. Dirac then proposedto interpret the γ’s as matrices satisfying:

(γ0)2 = l1 , (γk=1,2,3)2 = − l1γµγν + γνγµ = 0 for µ 6= ν

which can be compacted with:

γµ, γν = 2gµν l1 (2.17)

where is the anticommutator a, b = ab + ba. The relation 2.17 is known as the Cliffordalgebra. What is the minimal rank of the γ’s? Knowing that Tr(AB) = Tr(BA) and sinceγ0γ0 = 1 we have

Tr(γk) = Tr(γkγ0γ0) = Tr(γ0γkγ0) = −Tr(γ0γ0γk) = −Tr(γk)

where the negative sign is due to the γ matrices properties. Thus Tr(γk) = 0. Similarly,

Tr(γ0) = −Tr(γ0γkγk) = −Tr(γkγ0γk) = Tr(γkγkγ0) = −Tr(γ0)

where we applied successively, γkγk = − l1, Tr(AB) = Tr(BA), γµγν + γνγµ = 0 and γkγk = − l1.So, Tr(γµ) = 0. But the trace of a matrix is equal to the sum of its eigenvalues. Since γ0γ0 = l1,the eigenvalues of γ0 are ±1. Similarly, the eigenvalues of γk are ±i. The trace being 0, theremust be as many positive eigenvalues as negative ones. Hence the rank n is necessarily an evennumber. Rank n = 2 is not possible, because there are only 3 independent traceless matricesin the vector space generated by the 2 × 2 matrices. But we need 4 independent matrices!(because of the dimension of the Minkowski space). Hence, the γµ’s with minimal rank are 4×4hermitian matrices. Several choices are possible and the most common one, called the Dirac’srepresentation is:

γ0 =

(l1 0000 − l1

), γi =

(00 σi

−σi 00

)(2.18)

where the symbols l1 and 00 denote respectively the 2× 2 unit matrix and zero matrix, and σi arethe Pauli spin matrices seen in the first chapter:

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

)

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4-current and the adjoint equation 45

Actually, any matrix UγU−1 where U is a 4× 4 unitary matrix is a valid choice. We can easilycheck that the definition of the γ’s matrices implies for the hermitian conjugates2:

γ0† = γ0 , γ1† = −γ1 , γ2† = −γ2 , γ3† = −γ3 (2.19)

or in more concise:㵆 = γ0γµγ0 (2.20)

We can now come back to the 2 “square root” equations 2.15 and 2.16 of the Klein-Gordonequation. By convention, the Dirac equation is 2.16 (with β = γ) multiply by i (remember thati∂µ = Pµ):

(i/∂ −m)ψ = 0 with /∂ = γµ∂µ (2.21)

Since the γ’s are 4× 4 matrices, the ψ wave function is actually a wave function with 4 compo-nents:

ψ =

ψ1

ψ2

ψ3

ψ4

It is called a Dirac-spinor or 4-spinor. At this point, we can already begin to see that the extramultiplicity is likely to have something to do with an angular momentum degree of freedom assuggested by the presence of the Pauli matrices in the γ’s definition.

2.3.2 4-current and the adjoint equation

In order to find the 4-current, let us proceed as for the Klein-Gordon or Schrodinger equations.This time however, the complex conjugate must be replaced by the hermitian conjugate becauseof the presence of matrices. The hermitian conjugate of 2.21 is

(−i∂µψ†γµ† −mψ†) = 0 (2.22)

where the hermitian adjoint ψ† is:

ψ† = (ψ∗1, ψ∗2, ψ

∗3, ψ

∗4) (2.23)

Now, using equality 2.19, equation 2.22 becomes:

(−i∂0ψ†γ0 + i∂kψ

†γk −mψ†) = 0

and multiplying by γ0 from the right side:

(−i∂0ψ†γ0γ0+i∂kψ

†γkγ0−mψ†γ0) = −(i∂0ψ†γ0γ0+i∂kψ

†γ0γk+mψ†γ0) = −(i∂µψ†γ0γµ+mψ†γ0) = 0

where the anticommutation relations have been used. Now let us define the row Dirac adjoint:

ψ = ψ†γ0 = (ψ∗1, ψ∗2, − ψ∗3, − ψ∗4) (2.24)

so that ψ satisfies the adjoint equation:

i∂µψγµ +mψ = 0 (2.25)

2A† = (A∗)t where A∗ means complex conjugate and t the transpose matrix.

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46 From wave functions to quantum fields

We can now derive the continuity equation by multiplying the previous equation from the rightby ψ, equation 2.21 from the left by ψ and add the result:

i(∂µψ)γµψ + iψγµ∂µψ = i∂µ(ψγµψ) = 0

so that the conserved 4-current satisfying the previous continuity equation is:

jµ = ψγµψ (2.26)

leading to the density:

ρ = ψγ0ψ = ψ†γ0γ0ψ = ψ†ψ ⇒ ρ =4∑

i=1

|ψi|2 (2.27)

This time the density is always positive and can be interpreted as a probability density.

As for boson, we can define a charge current by simply put by hands the charge of the particleq = ±1:

jµq = qjµ = qψγµψ (2.28)

Contrary to the boson case, changing the sign of p won’t change jµ. Here, we have to explicitlyput the correct charge.

2.3.3 Free-particles solutions

The way the Dirac’s equation was built shows that a solution of the equation is also a solutionof the Klein-Gordon equation, or more precisely, each component of the 4-spinor is a solutionof the Klein-Gordon equation. Then, it is natural to look for solutions in which the space-timebehaviour is the one of the plane-wave:

ψ(x) = ue−ip.x (2.29)

where, this time, u is a 4-spinor that doesn’t depend on x and p.x the 4-momentum product(we omit the index µ for simplicity). A general solution of the Dirac’s equation can alwaysbe expressed as a linear combination of plane-wave solutions. Injecting 2.29 into the Dirac’sequation 2.21, we get the so-called momentum space Dirac equation:

(iγµ ×−ipµ −m)u = 0⇒ (/p−m)u = 0 (2.30)

Let us write this 4-spinor as a 2-component spinor:

u =

(uaub

)(2.31)

ua and ub are bi-spinor.

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Free-particles solutions 47

2.3.3.1 Solutions for particle at rest

It is very instructive to first examine the solutions for a particle at rest where only p0 = E isnon zero. Equation 2.30 simplifies to:

Eγ0u−mu = 0⇒ E

(ua−ub

)= m

(uaub

)⇒Eua = muaEub = −mub

The equation satisfied by ua corresponds to 2 independent orthogonal solutions

(10

)and

(01

)

both with E = m while the 2 independent orthogonal solutions for ub have E = −m. Comingback to 2.29, the 4 solutions are then:

ψ1 =

1000

e−imt , ψ2 =

0100

e−imt , ψ3 =

0010

e+imt , ψ4 =

0001

e+imt (2.32)

ψ1 and ψ2 having E = m > 0 and ψ3 and ψ4 having E = −m < 0. Finally, even in Dirac’sequation we have negative energy solutions! However, this time, they don’t lead to a negativeprobability density. These negative energy solutions will be interpreted in the next section assolutions for antiparticles.

2.3.3.2 General solutions

Let us come back to the general case with ~p 6= 0. Injecting 2.31 into 2.30, we have:

(γµpµ −m)

(uaub

)= (γ0E − ~γ.~p−m)

(uaub

)

=

((E.l1 0000 −E.l1

)−(

00 ~σ.~p−~σ.~p 00

)−(m.l1 0000 m.l1

))(uaub

)

=

((E −m).l1 −~σ.~p

~σ.~p −(E +m).l1

)(uaub

)

(2.33)so ua and ub must satisfy the 2 coupled equations:

ua =~σ.~p

E −m ub (2.34)

ub =~σ.~p

E +mua (2.35)

These equations impose a constraint on E and ~p. Indeed, let us inject 2.35 into 2.34:

ua =(~σ.~p)2

E2 −m2ua

but:

~σ.~p =

(0 11 0

)px +

(0 −ii 0

)py +

(1 00 −1

)pz =

(pz px − ipy

px + ipy −pz

)(2.36)

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48 From wave functions to quantum fields

And thus:(~σ.~p)2 = |~p|2 l1

so that, the non trivial solution (ua 6= 0) requires:

|~p|2E2 −m2

= 1⇒ E = ±√|~p|2 +m2

ua is a spinor with 2 components: thus, we can choose for ua any 2 orthogonal spinors φ1 and φ2,ub will have then to satisfy 2.35. Similarly, we can decide to chose first ub from any 2 orthogonalspinors χ1 and χ2, and then ua will have to satisfy 2.34. In both cases, the relativistic energyconstraint must be fulfilled. Hence, 4 independent solutions of the Dirac’s equation are givenby these 4-spinors:

u1 = N1

(φ1

~σ.~pE+m φ1

), u2 = N2

(φ2

~σ.~pE+m φ2

), u3 = N3

( ~σ.~pE−m χ1

χ1

), u4 = N4

( ~σ.~pE−m χ2

χ2

)

(2.37)whereNi are normalization factors of the spinors. We have to identify which 4-spinors correspondto positive or negative energy. In the limit where ~p → 0, we should recover the solutions for aparticle at rest, so that we can conclude that u1 and u2 are the solutions with E = +

√|~p|2 +m2

while u3 and u4 are the solutions with E = −√|~p|2 +m2.

2.3.3.3 Interpretation of negative energies

Historically, Dirac considered that the vacuum was full (!) of negative energy states satisfyingthe Pauli exclusion principle. This theory is usually referred as the “Dirac sea”. A hole in thenegative energy states was then interpreted as an antiparticle with positive energy and oppositecharge. This interpretation was successful in predicting the positron (discovered by Anderson in1932) but had many drawbacks. For instance, the vacuum being full of fermions with negativeenergy states didn’t give any explanations why bosons couldn’t populate the vacuum as well(which would be dramatic since they are not concerned by the Pauli exclusion principle). Infact, Stuckelberg in 1941 and Feynman in 1948 proposed the correct interpretation still valid withquantum field theory and briefly mentioned in the previous case of the Klein-Gordon equation.Consider the 2 solutions u3 and u4 with negative energy in 2.37. They can be rewritten:

u3 = N3

(− ~σ.~p|E|+m χ1

χ1

), u4 = N4

(− ~σ.~p|E|+m χ2

χ2

)

and both are associated to a propagation term e−i(−|E|t−~p.~x). But this term can be simplywritten e−i(|E|(−t)−~p.~x), so that it looks as a particle travelling backward in time form (t2, x1) to(t1, x2) where t2 > t1 as shown on schema b of figure 2.1. The picture is clearly not equivalentto an electron travelling “normally” from (t1, x2) to (t2, x2) as depicted in schema a. Moreover,the propagation term is no more Lorentz invariant. In order to restore the invariance, we alsohave to change direction of the momentum (to have a Lorentz scalar p.x) so that we obtain theschema c except that the final point in the spacetime is exchanged with the initial point.Feynman’s approach consists in interpreting schema c as the antiparticle of schema a withpositive energy and travelling “normally” from (t1, x2) to (t2, x2). We can now define the twonew spinors v1 and v2 for the antiparticle, interpreting E as a positive quantity and invertingthe momentum:

u3,4(−E,−~p)e−i((−E)t−(−~p).~x) = v1,2(E, ~p)e+i(Et−~p.~x)

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Free-particles solutions 49

Physique des particules M1 HEP X Pascal Paganini LLR-IN2P3-CNRS 21

3) Spin 1/2 particles

x1 x2

t1

t2

x1 x2

t1

t2

x1 x2

t1

t2 a) b) c)

Figure 2.1: Feynman approach of the antiparticles: left, a particle travelling from x1 to x2. Centre:time is reversed. Right: antiparticle as a particle with −t and −~p.

with:

v1 = N ′1

( ~σ.~pE+m χ1

χ1

), v2 = N ′2

( ~σ.~pE+m χ2

χ2

)(2.38)

Injecting ve+ip.x into the Dirac equation 2.21, we see that v satisfies the momentum space Diracequation:

(/p+m)v = 0 (2.39)

2.3.3.4 Solutions with normalization

So far, we left over the normalization of the spinors. As for the boson case, we are going tonormalize to 2E particles per unit volume 3. Using the density 2.27, we require:

V=1ψ†ψ = u†u = 2E

where u can be u1, u2, v1 or v2. Doing the job for vi=1,2 as an example:

u†u = |N ′i |2(( ~σ.~pE+m χi)

†, χ†i )( ~σ.~p

E+m χiχi

)

= |N ′i |2(χ†i (~σ.~pE+m)† ~σ.~p

E+mχi + χ†iχi)= |N ′i |2(χ†i (

~σ.~pE+m)2χi + 1)

= |N ′i |2( |~p|2(E+m)2 + 1)

= |N ′i |2 2EE+m

where moving from the second line to the third, we used (~σ.~p)† = ~σ.~p and χ†iχi = 1. Thus, weconclude N ′i =

√E +m. And similarly for Ni. Hence the 4 solutions of the Dirac equation are:

E = +√|~p|2 +m2

ψ1 =√E +m

(φ1

~σ.~pE+m φ1

)e−ip.x, ψ2 =

√E +m

(φ2

~σ.~pE+m φ2

)e−ip.x

ψ1 =√E +m

( ~σ.~pE+m χ1

χ1

)e+ip.x, ψ2 =

√E +m

( ~σ.~pE+m χ2

χ2

)e+ip.x

(2.40)

3When we chose N = 1 in 2.11, it implies for the 4-current 2.10, jµ = ±2

(E~p

)and hence 2E particles per

unit volume

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50 From wave functions to quantum fields

where ψ1 and ψ2 correspond to a particle and ψ1 and ψ2 to an antiparticle.

Example of explicit formula: let us consider for example, φ1 = χ2 as a spin-up state andφ2 = χ1 as a spin-down state4 (and thus eigenstates of σ3):

φ1 = χ2 =

(10

), φ2 = χ1 =

(01

)

The 4 solutions become:

ψ1 =√E +m

10pz

E+mpx+ipyE+m

e−ip.x, ψ2 =

√E +m

01

px−ipyE+m

− pzE+m

e−ip.x

ψ1 =√E +m

px−ipyE+m

− pzE+m

01

e+ip.x, ψ2 =

√E +m

pzE+mpx+ipyE+m

10

e+ip.x

(2.41)

where we used equality 2.36.

2.3.3.5 Interpretation in terms of spin and helicity

For a particle or an antiparticle, we always have two solutions degenerated in energy. Theremust be an operator that commutes with the energy operator HD of the Dirac’s equation whoseeigenvalues would distinguish the solutions. First, let us determine HD. Expanding the Dirac’sequation 2.21, we have:

iγ0 ∂

∂tψ + (i~γ.~∇−m)ψ = 0

Multiplying from the left by γ0, and using ~P = −i~∇ we get:

i∂

∂tψ = HD ψ

with the Dirac’s Hamiltonian:

HD = γ0~γ. ~P + γ0m =

(00 ~σ~σ 00

). ~P + γ0m (2.42)

Spin states: Now, we have to find an operator that commutes with HD. Let us try to see ifthe spin operators are satisfactory. A simple generalization of the operators 1.41 based on Paulimatrices for the 4-spinors case is clearly:

~S =1

2~Σ =

1

2

(~σ 0000 ~σ

)(2.43)

4with this choice solution for particle and antiparticle with the same index, will have a similar interpretationin term of spin.

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Free-particles solutions 51

However, HD and ~S don’t commute because of the presence of ~P . Indeed, for instance with[HD, Sz], we would have a term5:

[~σ. ~P, σz

]=[σxPx + σyPy + σzPz, σz

]= [σx, σz]Px + [σy, σz]Py 6= 0

In case of zero momentum, HD reduced to γ0m that commutes with ~S and the explicit solutions2.41 (with px = py = pz = 0) are eigenstates of Sz:

~p = 0⇒ Szψ1 = +1

2ψ1 , Szψ2 = −1

2ψ2 , ˆSzψ1 = +

1

2ψ1 , ˆSzψ2 = −1

2ψ2

Note that the spin operator for the antiparticles is defined as:

ˆS = −S

It is justified because with antiparticles, the operator returning the physical momentum (positive

energy, appropriate direction for ~p) is changed from ~P → − ~P (because antiparticles have a

propagation mode e+ip.x), so that ~L = ~r ∧ ~P → −~L. Thus the conservation of the total angular

momentum ~J = ~L + ~S, requires ~S → − ~S. Finally, we see that ψ1 and ψ1 would have spin upwhile ψ2 and ψ2 would have spin down when px = py = pz = 0. This result also holds if the(anti)particle travels in the z-direction (px = py = 0, pz = ±|~p|).

Helicity states: For any ~p, the spin alone cannot distinguish the 2 degenerated solutions(since it’s not a conserved number because the spin operator in general does not commute with

the Hamiltonian). One can show that the total angular momentum ~J = ~L+ ~S with ~L = ~r ∧ ~Pdoes commute. It is fortunate because otherwise, the total angular momentum would not beconserved! Another possibility is the so-called helicity defined as the projection of the spin onthe direction of the momentum. Hence, the operator is simply defined for fermion spinor as:

h = ~S.~P

|~p| =1

2~Σ.

~P

|~p| =1

2

~σ.

~P|~p| 00

00 ~σ.~P|~p|

(2.44)

Note that for antifermions, the same operator can be used since ~S → − ~S and ~P → − ~P . Thehelicity operator clearly commutes with HD, so there exists a basis of 4-spinors which is botheigenstates of HD and of the helicity. Since the spin on any axis (and thus on the direction of themomentum) can only be ±1/2, the helicity quantum number6 λ can only be ±1/2. Dependingon λ, the state is called:

λ = −12 ⇒ Left handed helicity

λ = +12 ⇒ Right handed helicity

5I use indifferently σx for σ1, σy for σ2 and σz for σ3.6Note that some authors, define the helicity as twice that number in order to have an integer for both fermions

and bosons.

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52 From wave functions to quantum fields

After some math (see exercise 2.3), one can determine the general solution for helicity states:

ψ+ 12

=√E +m

cos θ2eiφ sin θ

2|~p|

E+m cos θ2|~p|

E+meiφ sin θ

2

e−ip.x, ψ− 1

2=√E +m

− sin θ2

eiφ cos θ2|~p|

E+m sin θ2

− |~p|E+me

iφ cos θ2

e−ip.x

ψ+ 1

2

=√E +m

|~p|E+m sin θ

2

− |~p|E+me

iφ cos θ2− sin θ

2

eiφ cos θ2

e+ip.x, ψ− 1

2

=√E +m

|~p|E+m cos θ2|~p|

E+meiφ sin θ

2

cos θ2eiφ sin θ

2

e+ip.x

(2.45)where θ and φ are the polar angles.

A remark: beware that the helicity is not Lorentz invariant in general. Actually, fora massive particle, one can always find a reference frame where the particle appears to reverseits relative direction of motion.

2.3.4 Operations on spinors

2.3.4.1 Charge conjugation

In classical electrodynamic, the motion of a charged particle with a charge q in an electromagneticfield Aµ = (V, ~A) is obtained by making the substitution:

~p→ ~p− q ~A , E → E − qV

The quantum version is then simply from the correspondence principle:

pµ → pµ − qAµ ⇒ i∂µ → i∂µ − qAµ ⇒ ∂µ → ∂µ + iqAµ (2.46)

so that the Dirac’s equation becomes:

[γµ(i∂µ − qAµ)−m]ψ = 0 (2.47)

Now the charge conjugate state:

ψ′ = Cψ (2.48)

is supposed to satisfy:

[γµ(i∂µ + qAµ)−m]ψ′ = 0 (2.49)

In order to find C, let us start by conjugating 2.47:

[γµ∗(−i∂µ − qAµ)−m]ψ∗ = 0[γ0(−i∂0 − qA0) + γ1(−i∂1 − qA1)− γ2(−i∂2 − qA2) + γ3(−i∂3 − qA3)−m

]ψ∗ = 0

In order to have the correct sign in front of all γ’s we can multiply by γ2 from the left and usethe anticommutation relation:[−γ0(−i∂0 − qA0)− γ1(−i∂1 − qA1)− γ2(−i∂2 − qA2)− γ3(−i∂3 − qA3)−m

]γ2ψ∗ = 0

[γµ(i∂µ + qAµ)−m] γ2ψ∗ = 0

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Operations on spinors 53

Thus, γ2ψ∗ seems a good candidate for ψ′. If ψ represents the wave function of an electron, weexpect ψ′ to be the one of a positron. Let us check for instance with the electron described byψ1 in 2.41:

γ2ψ∗ =√E +m

0 0 0 −i0 0 i 00 i 0 0−i 0 0 0

10pz

E+mpx+ipyE+m

e−ip.x

= −i√E +m

px−ipyE+m

− pzE+m

01

e+ip.x

It would be the positron solution described by ψ1 except that there is this extra −i. Hence, byusing:

Cψ = iγ2ψ∗ (2.50)

we get the correct description.

2.3.4.2 Parity

The parity transformation changes the space coordinates into their opposite:

x = (t, ~x)→ x′ = (t,−~x) (2.51)

How the spinors satisfying the Dirac’s equation 2.21 must be transformed to still satisfy:

ψ(x)→ ψ′(x′) = Pψ(x) with (iγµ∂′µ −m)ψ′(x′) = 0

Let us express the Dirac’s equation with the new coordinates:

(iγµ∂µ −m)ψ = 0

(iγ0 ∂∂t′

∂t′∂t + iγ1 ∂

∂x′∂x′∂x + iγ2 ∂

∂y′∂y′∂y + iγ3 ∂

∂z′∂z′∂z −m)ψ = 0

(iγ0 ∂∂t′ − iγ1 ∂

∂x′ − iγ2 ∂∂y′ − iγ3 ∂

∂z′ −m)ψ = 0

We need to revert the sign in front of γi=1,2,3. Thus, let us multiply by γ0 from the left and usethe anticommutation algebra:

(iγ0 ∂∂t′ + iγ1 ∂

∂x′ + iγ2 ∂∂y′ + iγ3 ∂

∂z′ −m)γ0ψ = 0

(iγµ∂′µ −m)ψ′(x′) = 0

where we identify:

ψ′ = γ0ψ ⇒ P = γ0 (2.52)

Now, according to spinors 2.40 solution of the Dirac’s equation, we see that:

u(−~p) = P u(~p) , v(−~p) = −P v(~p)

meaning that the intrinsic parity is +1 for fermion and −1 for antifermion:

ηPf = +1 , ηPf = −1 (2.53)

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54 From wave functions to quantum fields

2.3.4.3 Chirality

consider the so-called chirality matrix:

γ5 = iγ0γ1γ2γ3 (2.54)

which in the Dirac representation reads:

γ5 =

(00 l1l1 00

)(2.55)

It’s easy to show that γ5 satisfies:

(γ5)2 = 1 , γ5† = γ5 (2.56)

Since (γ5)2 = 1, the eigenvalues of γ5 are ±1. The eigenstates of γ5 are called chirality statesand are denoted with the Left L and Right R label for a reason that is going to be clarifiedin few lines. Eigen-spinors with positive (negative) chirality have right (left) chirality and theopposite for eigen-antispinors:

γ5uR = +uR , γ5uL = −uL , γ5vR = −vR , γ5vL = +vL (2.57)

The reason why chiral antispinors have opposite eigenvalues with respect to chiral spinor isbecause we want that chirality states match the helicity states in the massless approximation(when m = 0, for both particles and anti-particles, a left-handed chiral state is equal to a left-handed helicity state as shown few lines below). Note that due to the anticommutation algebraof the γ’s matrices, we have:

γ5, γµ

= 0 (2.58)

And hence, the commutator with the Dirac’s Hamiltonian 2.42 reduces to:[HD, γ

5]

= m[γ0, γ5

]

which is different than zero, except when m = 0. Therefore, for massive particles, the chiralityeigenvalues are not conserved and the physical states having a defined energy (so eigenstates ofthe hamiltonian) will be a mixture of chirality states.

Now, Consider the helicity states 2.45 in the ultrarelativistic limit E m or equivalently whenm = 0:

u+ 12≈√E

cos θ2eiφ sin θ

2

cos θ2eiφ sin θ

2

, u− 1

2≈√E

− sin θ2

eiφ cos θ2sin θ

2

−eiφ cos θ2

v+ 12≈√E

sin θ2

−eiφ cos θ2− sin θ

2

eiφ cos θ2

, v− 1

2≈√E

cos θ2eiφ sin θ

2

cos θ2eiφ sin θ

2

(2.59)

It is easy to check that these helicity states are now eigenstates of γ5 using the representation2.55. Hence, in the limit E m, the helicity states coincide with the chirality states:

u+ 12

= uR , u− 12

= uL , v+ 12

= vR , v− 12

= vL (2.60)

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Operations on spinors 55

and we now understand why the chiral states are called left or right handed. Looking at 2.59, wesee that, in the limit E m, left handed fermions are equivalent to right handed antifermions(the difference of sign has no physical effects). We will see later that the weak interaction viacharged current (mediated by W bosons) only acts on the chirality left-handed states of particlesand chirality right-handed states of antiparticles. Thus, it is useful to define the projectors onthese states:

PL =1

2(1− γ5) , PR =

1

2(1 + γ5) (2.61)

It’s trivial to check that PL and PR have the projectors properties: P 2L = PL, P

2R = PR,

PLPR = 0, PL + PR = l1. And using 2.59 and 2.60, we see that:

PR uR = uR PR uL = 0 PL uR = 0 PL uL = uLPR vR = 0 PR vL = vL PL vR = vR PL vL = 0

meaning for example that for spinors, PR projects out right handed states while for antispinors,it projects out left handed states. Therefore, for any (anti)spinors, we can always write:

u = uL + uR uL = PL u uR = PR uv = vL + vR vL = PR v vR = PL v

(2.62)

Now consider the parity transformation of the projector on left-handed state7

PPLP† = γ0 1

2(1− γ5)γ0† =

1

2(1 + γ5)γ0γ0 = PR

meaning that the chirality changes under a parity transformation as we would expect from itsdenomination (chirality comes from the greek, meaning “hand”: a left hand in a mirror becomesa right hand).

Spin states, helicity states and chirality states are often confused. We invite the reader toconsult the reference [5] for detailed explanations of the differences.

2.3.4.4 Useful formulas and completeness relations

Momentum space equations: We already established the momentum space equations (2.30and 2.39):

(/p−m)u = 0 , (/p+m)v = 0 (2.63)

We wish to see which equations satisfy u and v. For instance, taking the hermitian conjugateof 2.30:

u†(㵆pµ −m) = 0u†(γ0γµγ0pµ −m) = 0

where we used 2.20. Multiplying from the right by γ0 and using (γ0)2 = 1:

u†(γ0γµpµ −mγ0) = 0

7The transformation of a wave function ψ under P is Pψ while the transformation of an operator A is PAP †.Hence, if a state |a〉 is transformed in |a′〉 = U |a〉 under a unitary transformation U (i.e. U† = U−1 because〈a|b〉 = 〈a′|b′〉 ⇒ U†U = 1), the variable 〈a|A|b〉 must be equal to 〈a′|A′|b′〉 and thus: 〈a|A|b〉 = 〈a|U†A′U |b〉 andso for any state |a〉 or |b〉. We thus have: UAU† = A′. Requiring 〈a′|A′|b′〉 = 〈a|A|b〉, is equivalent for a classicalscalar field (temperature for instance), as saying φ′(x′) = φ(x): the temperature at a rotated point seen by therotated field is the same as the temperature at the initial point.

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56 From wave functions to quantum fields

Applying the same procedure for v we finally get:

u(/p−m) = 0 , v(/p+m) = 0 (2.64)

Normalisations: Using the normalized spinors appearing 2.40, and following a similar ap-proach as for the spinors normalization, we have:

uiuj = u†iγ0uj = (E +m)

(φ†i ,

(~σ.~p)†

E +mφ†i

)(l1 0000 −l1

)(φj

~σ.~pE+m φj

)

uiuj = (E +m)(φ†i ,

~σ.~pE+mφ

†i

)( φj− ~σ.~pE+m φj

)

= (E +m)(δij − |~p|2

(E+m)2 δij

)

And finally:

uiuj = 2m δij , vivj = −2m δij (2.65)

with a similar calculus for v.

Completeness relations: Following same kind of calculus, it’s easy to show:

i=1,2

uiui = /p+m ,∑

i=1,2

vivi = /p−m (2.66)

These matrices are extensively used with Feynman diagram.

2.3.5 Few words about the quantized field

As for the scalar case, we give here basic ideas without rigorous demonstration. Consult [4]for the details. We insist again, that we have to give up with the notion of a relativistic singleparticle described by the Dirac equation (see explanations in Klein-Gordon section).

The wave functions:

ur(p)e−ipx , vr(p)e

+ipx

with r = 1, 2 constitute a basis of solutions of the Dirac’s equation. Hence, a general functioncan be written:

ψ(x) =

∫d3~p

(2π)32Ep

r=1,2

[c~p,rur(p)e

−ipx + d∗~p,rvr(p)e+ipx

]

where c~p,r and d∗~p,r are complex Fourier coefficients. Quantization of the field transforms thesenumbers in operators:

ψ(x) =

∫d3~p

(2π)32Ep

r=1,2

[c~p,rur(p)e

−ipx + d†~p,rvr(p)e+ipx

](2.67)

ψ(x) =

∫d3~p

(2π)32Ep

r=1,2

[c†~p,rur(p)e

+ipx + d~p,rvr(p)e−ipx

](2.68)

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Few words about the quantized field 57

We obviously wish to interpret c†~p,r as an operator creating a fermion with a given momentum~p and polarization r. Now consider the wave function of a state with 2 such particles:

ψ~p1,~p2( ~x1, ~x2) = 〈 ~x1, ~x2|1, ~p1, r1; 1, ~p2, r2〉

= (〈 ~x1| ⊗ 〈 ~x2|)c†p1,r1c†p2,r2 |0〉

Clearly if the c† operators commute, we will end-up with a symmetric function as for the bo-son case. The only way to get an asymmetric function is to postulate that fermion operatorsanticommute. In that case:

ψ~p1,~p2( ~x1, ~x2) = (〈 ~x1| ⊗ 〈 ~x2|)c†p1,r1c

†p2,r2 |0〉

= −(〈 ~x1| ⊗ 〈 ~x2|)c†p2,r2c†p1,r1 |0〉

= −ψ~p1,~p2( ~x2, ~x1)

With the normalization we use, the complete set of anticommutation rules then reads:

cp′,r′ , c†p,r = dp′,r′ , d†p,r = (2π)32Ep δ(3)(~p ′ − ~p)δr′r

cp′,r′ , cp,r = dp′,r′ , dp,r = c†p′,r′ , c†p,r = d†p′,r′ , d

†p,r = 0

c†p′,r′ , d†p,r = cp′,r′ , d†p,r = c†p′,r′ , dp,r = 0

(2.69)

The interpretation of the operators is the following: c†~p,r creates a fermion with momentum ~p

and polarization r, while c~p,r destroys it. And d†~p,r creates a antifermion with momentum ~p andpolarization r, while d~p,r destroys it.We can easily check that we can’t create 2 particles or antiparticles in the same states: sincec†~p,r, c

†~p,r = 2c†~p,rc

†~p,r = 0. Thus c†~p,rc

†~p,r |0〉 = |0〉.

The hamiltonian, momentum and charge operators expressed in terms of creation annihilationoperators are then:

H =∫ d3~p

(2π)32Ep

∑r=1,2 Ep (c†p,rcp,r + d†p,rdp,r)

P =∫ d3~p

(2π)32Ep

∑r=1,2 ~p (c†p,rcp,r + d†p,rdp,r)

Q =∫ d3~p

(2π)32Ep

∑r=1,2 q (c†p,rcp,r − d†p,rdp,r)

As for the scalar field, the energy, momentum and charge of a state with 1 fermion:

|1~p, r〉 = c†p,r |0〉 (2.70)

is Ep, ~p and q while for the antifermion

|1~p, r〉 = d†p,r |0〉 (2.71)

we have Ep, ~p and −q.

Finally, as for scalar fields, one may wonder what is the connection between the usual wavefunctions ur(p)e

−ip.x, vr(p)e+ip.x and the quantum field ψ(x) of formula 2.67? The answer is

very similar to the scalar case and can be easily checked using the commutation rules of theoperators:

〈0|ψ(x)|1~p, r〉 = ur(p)e−ip.x

〈1~p, r |ψ(x)|0〉 = vr(p)e+ip.x

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58 From wave functions to quantum fields

A last word concerning the handedness of the Dirac field. By definition, a field ψL is left-handed (for chirality) and ψR right-handed if:

PLψL = ψL , PRψL = 0 , PRψR = ψR , PLψR = 0

Let us apply the left-handed chiral projector PL on the field ψ (2.67):

ψL = PLψ =∫ d3~p

(2π)32Ep

∑r=1,2

[c~p,rPLur(p)e

−ipx + d†~p,rPLvr(p)e+ipx

]

=∫ d3~p

(2π)32Ep

∑r=1,2

[c~p,ruL,r(p)e

−ipx + d†~p,rvR,r(p)e+ipx

]

We see that ψL annihilates a particle associated with uL, a (chiral) left-handed particle butcreates an anti-particle associated with vR, a (chiral) right handed antiparticle8. Naturally,ψL creates a left-handed particle and annihilates a right-handed antiparticle. Similarly, ψRannihilates a right-handed particle and creates a left-handed antiparticle while ψR creates aright-handed particle and annihilates a left-handed helicity antiparticle. For massless particles,the chirality of the particle or antiparticle created/annihilated by the field is equal to its helicity.

2.4 Spin 1 particle: the photon

2.4.1 Solutions of Maxwell equations

In the first chapter, we have already seen that the Maxwell equations can be derived from theelectromagnetic tensor Fµν (see section 1.2.3.6) where:

Fµν = ∂µAν − ∂νAµ

depends on the 4-potential Aµ = (V/c, ~A) and satisfies:

∂µFµν = µ0j

ν

jν being the 4-current. Hence, Aµ, satisfies the wave equation:

∂µ(∂µAν − ∂νAµ) = µ0jν ⇒ Aν − ∂ν(∂µA

µ) = µ0jν (2.72)

However, Aµ is not uniquely defined. Consider an arbitrary scalar function χ. The new potential:

Aµ → A′µ = Aµ + ∂µχ (2.73)

still leads to the same electromagnetic tensor. Transformation 2.73 is called a gauge transfor-mation. And because the electromagnetic tensor is invariant under this gauge transformation,the Maxwell equations will be as well. Now, χ can be chosen so that:

∂µA′µ = 0 (2.74)

This gauge choice is called Lorenz gauge (∂µAµ + χ = 0). Now the wave equation simplifies

to:Aµ = µ0j

µ (2.75)

8What is done here is not a proof. One should establish how the annihilation/creator operators behave underchirality operation. In addition, some authors called vL = PLv. It has the advantage of using consistent notationsin the expression of ψL field but has the drawback of being called “left” while the created particles is actuallyright-handed!

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Few words about the quantized field 59

where the new electromagnetic field is written again without the ’ symbol. The solution for afree photon (jµ = 0) has a plane-wave form:

Aµ = Nεµe−ip.x

εµ being a polarization 4-vector and N a normalization factor. A possible basis of 4-vectorspolarization requires 4 such vectors ελ=0,1,2,3 with:

ελ=0 =

1000

ελ=1 =

0100

ελ=2 =

0010

ελ=3 =

0001

However, inserting the previous plane-wave into the wave equation 2.75 (with jµ = 0), we getp2 = 0, as expected since photons are massless. Therefore, the Lorenz gauge condition 2.74translates to:

ε.p = 0

Thus the polarization has apparently 3 degrees of freedom (4 components of 4-vector - 1 due tothe condition above). For instance, choosing the z-axis as the direction of propagation of thephoton, only a basis with ε1, ε2 and ε0 + ε3 would be needed. Actually, it is not true. Nothingprevents us to redo the same kind of gauge transformation:

Aµ → A′µ = Aµ + ∂µχ

but this time χ must satisfies χ = 0 since the field Aµ now satisfies the Lorenz gauge. Let uschoose:

χ = iNae−ip.x

with a a constant that doesn’t depend on x. The new field definition then becomes:

A′µ = N(εµ − apµ)e−ip.x

meaning that A′ has the new polarization vector ε′µ = εµ−apµ. We can choose a so that ε′0 = 0,and forgetting again the ’, the field polarization vector satisfies now:

~ε.~p = 0

and therefore, there are now only 2 degrees of freedom. With a photon traveling along z,a basis with only ε1, ε2 would be enough. As expected from classical electromagnetism, thepolarization of a (free-)photon is orthogonal to the direction of motion. By convention, even ifthe photon travels in another direction, ε1, ε2 denote the two 4-vectors polarizations transverseto the direction of motion. The choice imposing ε0 = 0 (which is clearly not covariant) is calledthe Coulomb gauge.

2.4.2 Few words about the quantized field

The photon polarization is orthogonal to the direction of motion. Any polarisation can then beexpressed as the linear combination of the 2 basis polarization-vectors ~ε(~p, λ = 1) and ~ε(~p, λ = 2)namely ~ε(~p, 1)α~p,1 + ~ε(~p, 2)α~p,2 . The general solution of the photon (classical) field is then:

Aµ(x) =

∫d3~p

(2π)32Ep

λ=1,2

εµ(p, λ)α~p,λe−ip.x + ε∗µ(p, λ)α∗~p,λe

+ip.x

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60 From wave functions to quantum fields

where Ep = |~p| and ε0(p, 1) = ε0(p, 2) = 0 because of the Coulomb gauge. The norm was chosenas before with the scalar and spin 1/2 cases. Notice that Aµ(x), the photon field is a real field,reflecting the fact that the photon is its own anti-particle.

The quantization of the field promotes the coefficients α~p,λ to operators but there are manysubtleties in the quantization procedure of the photon field mainly due to translation of gaugecondition in the quantum field language. The interesting reader is invited to consult the reference[6, p. 189-195] to know more. The field after quantization then reads:

Aµ(x) =

∫d3~p

(2π)32Ep

3∑

λ=0

εµ(p, λ)α~p,λe−ip.x + ε∗µ(p, λ)α†~p,λe

+ip.x (2.76)

where the operator α†~p,λ creates a photon of momentum ~p and polarization λ and α~p,λ is thecorresponding destructor. One of the subtleties we mentioned above leads to the (strange)commutation rules: [

α~p′,λ′ , α†~p,λ

]= −gλλ′(2π)32Epδ

(3)(~p′ − ~p)[α~p′,λ′ , α~p,λ

]=[α†~p′,λ′ , α

†~p,λ

]= 0

(2.77)

It looks pretty similar to the scalar boson case (commutator) except that there is the extra−gλλ′ . Note also, that we sum-up on 4 polarization states: λ = 0 is a “time-like” polarisation,λ = 1, 2 are the 2 transverse polarizations while λ = 3 is the longitudinal polarization. You seethat for the polarization λ = 1, 2, 3, the corresponding operators do correspond to the ones ofthe harmonic oscillators as in the scalar case (−gλλ = 1). However, it’s note the case for λ = 0(−g00 = −1)! In fact, after properly taking into account the gauge constraint, one can showthat for physical states (on which we can do measurements), only the transverse polarizationsmatter.

2.5 Exercises

Exercise 2.1 Using the spinors v1 and v2 given by 2.38, check that v1e+ip.x and v2e

+ip.x satisfy(/p+m)v = 0.

Exercise 2.2 Covariance of the Dirac’s equation. We recall that establishing the covariance ofthe Dirac’s equation means that under a Lorentz transformation, ψ(x) becomes ψ′(x′) = Sψ(x)and satisfies the same mathematical form of the Dirac’s equation: (/∂

′ − m)ψ′(x′) = 0. Thequestion is to establish the existence of the matrix S which should only depend on the Lorentztransformation parameters (i.e. γ and β).

1. Show that S must satisfy Sγµ = γνS∆µν where ∆µ

ν is the Lorentz transformation matrix.

2. Restricting to a transformation along the x-axis, show that S =√

γ+12 l1 +

√γ−1

2 γ0γ1

satisfies the previous relation.

Exercise 2.3 Determination of the helicity states. In the following, uλ denotes a spinor for aparticle eigenstate of the helicity operator (equation 2.44) with an helicity λ = ±1/2. We recallthe general formula of a u-spinor:

uλ =√E +m

(φλ

~σ.~pE+m φλ

)

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Exercises 61

1. Show that ~σ.~p φλ = 2λ|~p|φλ.

2. Using polar coordinates, show that φλ can take the form for λ = +1/2:

φ 12

=

(cos θ2eiφ sin θ

2

)

3. Finally, show that the helicity state is given by:

ψ+ 12

=√E +m

cos θ2eiφ sin θ

2|~p|

E+m cos θ2|~p|

E+meiφ sin θ

2

e−ip.x

4. Using the same procedure, show that the 3 other helicity states are given by formula 2.45.

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62 From wave functions to quantum fields

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Chapter 3

A brief view of QuantumElectrodynamic

Few references:I.J.R Aitchison & A.J.G. Hey,“Gauge theories in particle physics”, vol1, Chapters 6,7,8, IoP2003.M.E. Peskin & D.V. Schroeder, “An introduction to Quantum Field Theory”, Chapter 4,5,Westview Press Inc 1995

With the first two chapters, we saw that the quantization of the solutions ofthe wave equations leads to quantum fields, a well adapted framework to treatstates made of many particles that can be created or annihilated because ofinteractions. This chapter will briefly try to explain how this can be adaptedto the interaction of electrons with photons. I want to warn the reader: thiscourse is not a quantum field course. The goal here, is to give the conceptswithout complete demonstrations and prepare the reader to be able to do simplecalculation of scattering processes at the lowest order.

3.1 The action and Lagrangians

3.1.1 The least action principle

Classical case: In the hamiltonian formulation of classical mechanics, the equations of motionare deduced by minimizing a quantity called the action S. Let us recall how it works using asimple example. Consider a non-relativistic particle moving in 1 dimension x with a kineticenergy T = 1

2mx2 and with a potential energy V (that can be mgx for instance). The quantity:

L = T − V

is called a Lagrangian with the dimension of the energy and depends obviously on x and x. Themomentum and energy are given by:

p = ∂L∂x

E = px− L

63

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64 A brief view of Quantum Electrodynamic

The equation of motion between the time t1 and t2 is deduced by minimizing the action Sdefined by:

S =

∫ t2

t1

L(x, x)dt

The meaning of minimization is that among all trajectories x(t) that can be imagined startingfrom x1 at t1 and finishing at x2 at t2, the one which will have the lowest value of the actionwill be the one adopted by the particle. Let us call x0(t) the trajectory we want to find. Anytrajectory can be written x(t) = x0(t) + δx(t). The action becomes:

S =

∫ t2

t1

L(x0 + δx, x0 +d

dtδx)dt =

∫ t2

t1

(L(x0, x0) +

∂L

∂xδx+

∂L

∂x

d

dtδx

)dt

Integrating by part:

∫ t2

t1

∂L

∂x

d

dtδx dt =

[∂L

∂xδx

]t2

t1

−∫ t2

t1

d

dt

(∂L

∂x

)δx dt = −

∫ t2

t1

d

dt

(∂L

∂x

)δx dt

where δx(t1) = δx(t2) = 0 has been used (all trajectories start from the same point and finishat the same point). Hence the variation of the action is just:

δS =

∫ t2

t1

(∂L

∂x− d

dt

∂L

∂x

)δx dt

and thus:

δS = 0⇒ ∂L

∂x− d

dt

∂L

∂x= 0

which is the Euler-Lagrange equation. Applying it to our simple example:

∂L∂x = −∂V

∂xddt∂L∂x = mx

mx = −∂V

∂x= F

So, we find the usual Newton’s law.

Simple quantum case: Let us call qr (r = 1, n) the generalized coordinates (was x before,with n = 1). In quantum mechanics, we have the additional constraint:

[qr, pr′ ] = iδrr′

(with ~ = 1), qr and pr′ being operators. The Lagrangian L(q1, . . . , qn, q1, . . . , qn) is now anoperator and we still have:

pr = ∂L∂qr

H = (∑n

r=1 pr qr)− Lwhere H the hamiltonian, gives the total energy of the system. Since the hamiltonian is her-metian (the energy is real), the Lagrangian must be hermetian too. Now, the Euler-Lagrangeequation is simply:

∂L

∂qr− d

dt

∂L

∂qr= 0

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Free electron Lagrangian 65

Quantum field case: in the continuous case ie n → ∞, the n qr operators become a fieldφ(~x, t) depending on spacetime coordinates, and the summation must become an integral. There-fore, instead of considering the Lagrangian L itself, we consider the Lagrangian density L:

L =

∫L d3~x (3.1)

and hence, the action becomes:

S =

∫L d4x (3.2)

As before L depends on φ and φ but since φ is a continious function of spacetime, L can alsodepend on ∂φ/∂x, ∂φ/∂y, ∂φ/∂z so that L is now L(φ, ∂µφ).The momentum p becomes a “momentum field” π(xµ):

π(xµ) =∂L

∂φ(xµ)(3.3)

and the hamiltonian density field is:

H = π(xµ)φ(xµ)− L (3.4)

The quantization generalizes what we had in the previous case1:

[φ(xµ), π(xµ)] = iδ(3)(~x− ~y) (3.5)

This is this procedure which yields the quantized field of the previous chapter, the commutatorbetween the creation and annihilation operators being a simple consequence of the previousrelation. Finally, the Euler-Lagrange equation becomes:

∂L∂φ− ∂µ

∂L∂(∂µφ)

= 0 (3.6)

3.1.2 Free electron Lagrangian

When the wave equation is known (in our case the Dirac equation), it is not too difficult to findthe Lagrangian. The free electron Lagrangian, namely the Dirac Lagrangian density, is:

LD = ψ(i/∂ −m)ψ (3.7)

Let us check that it gives the Dirac’s equation. The variables that have to be considered arethe 4 components of the Dirac’s field ψα, its adjoint field ψα (since ψ is complex, we can vary ψand ψ independently) and the derivatives ∂µψα and ∂µψα with α = 1, 4. Making explicit theLagrangian in terms of components:

L =∑

α,β

ψαiγµαβ∂µψβ −mδαβψαψβ

where δαβ is the Kronecker symbol. Applying the Euler-Lagrange equation for φ = ψα:

∂L∂ψα

=∑

β iγµαβ∂µψβ −mδαβψβ

∂L∂(∂µψα)

= 0

β

iγµαβ∂µψβ −mψα = 0

1one has to use anticommutator in case of fermions as in the previous chapter.

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66 A brief view of Quantum Electrodynamic

Since this equation is valid for the 4 components α, we do recover the Dirac’s equation: iγµ∂µψ−mψ = 0. Now, applying the Euler-Lagrange equation for ψβ:

∂L∂ψβ

=∑

α−mδαβψα∂L

∂(∂µψβ) =∑

α ψαiγµαβ

−mψβ −

α

∂µψαiγµαβ = 0

where putting together the 4 equations (1 per component), we find i∂µψγµ+mψ = 0, the adjoint

equation 2.25.

3.1.3 Free photon Lagrangian

As seen in the previous chapter, the free photon obeys the wave equation (2.72) namely Aµ −∂µ(∂νA

ν) = 0 (before exploiting the gauge invariance). The corresponding Lagrangian is then:

Lγ = −1

4FµνF

µν (3.8)

where Fµν is the usual electromagnetic tensor. Let us check this by applying the Euler-Lagrangeequation. Rewriting first the Lagrangian:

Lγ = −1

4(∂µAν − ∂νAµ)Fµν = −1

4(∂µAνF

µν + ∂νAµFνµ) = −1

2∂µAνF

µν

= −1

2∂µAν(∂µAν − ∂νAµ)

= −1

2gµαgνβ∂µAν(∂αAβ − ∂βAα)

where in the first line we used successively Fµν = −F νµ and µ↔ ν. Thus:

∂Lγ∂(∂µAν)

= −1

2gµαgνβ(∂αAβ − ∂βAα)− 1

2gµαgνβ∂µAν

∂(∂αAβ)

∂(∂µAν)− ∂(∂βAα)

∂(∂µAν)

= −1

2(∂µAν − ∂νAµ)− 1

2∂αAβ

δµαδ

νβ − δµβδνα

= −1

2(∂µAν − ∂νAµ)− 1

2∂µAν − ∂νAµ

= −(∂µAν − ∂νAµ)

the δ denoting the Kronecker symbol (δµα = 1 if µ = α and 0 otherwise.).The Euler-Lagrangeequation is then:

∂Lγ∂Aν

− ∂µ∂Lγ

∂(∂µAν)= 0⇒ 0 + ∂µ∂

µAν − ∂µ(∂νAµ) = Aν − ∂µ(∂νAµ) = 0

which is the wave equation of a free photon.

3.1.4 Gauge invariance consequences

In 1917, Emmy Noether published a theorem that states that every continuous symmetry yieldsa conservation law, and conversely every conservation law is a sign of an underlying symmetry.One of the famous consequences of her theorem is the fact that the momentum is conserved fora system which is invariant under translations in space. Similarly, for the conservation of theangular momentum and the invariance under rotations.

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Gauge invariance consequences 67

Global phase invariance: Now, consider the Dirac’s Lagrangian 3.7 and the transformationof the field:

ψ → ψ′ = e−iqχψ (3.9)

where q and χ are simple real numbers. The Lagrangian becomes:

LD → L′ = e+iqχψ(i/∂ −m)e−iqχψ = LD

Thus, the Lagrangian is invariant under this transformation. Such transformation has all theproperties of a symmetry namely 1) the product of 2 transformations is also a transformation, 2)there exists an identity transformation, 3) there exists an inverse transformation and 4) there isassociativity: if R1,2,3 are 3 transformations, R1(R2R3) = (R1R2)R3. The set of transformations3.9 forms a group called U(1) and the properties just mentioned are actually the ones of a group.Moreover, we see that the order of 2 transformations doesn’t matter, or in other words, the U(1)group elements commute. Such group is called Abelian. According to Noether’s theorem, theremust be a conservation law. Consider an U(1) infinitesimal transformation of parameter δχ:

δψ = −iqδχψ

and imposing the Lagrangian to remain unchanged δLδχ = 0

δL = ∂L∂ψ δψ + ∂L

∂(∂µψ)δ(∂µψ) + δψ ∂L∂ψ

+ δ(∂µψ) ∂L∂(∂µψ)

= −∂L∂ψ iqδχψ − ∂L

∂(∂µψ) iqδχ∂µψ + iqδχψ ∂L∂ψ

+ iqδχ∂µψ∂L

∂(∂µψ)

∂Lδχ = −iq

[∂L∂ψψ − ∂µ

(∂L

∂(∂µψ)

)ψ + ∂µ

(∂L

∂(∂µψ)ψ)]−[ψ ∂L∂ψ− ψ∂µ

(∂L

∂(∂µψ)

)+ ∂µ

(ψ ∂L∂(∂µψ)

)]

The first two terms in the two brakets cancel because of the Euler-Lagrange equation appliedto ψ and ψ, so that:

∂Lδχ = ∂µ

−iq

(∂L

∂(∂µψ)ψ − ψ ∂L∂(∂µψ)

)= 0

and hence the current jµ = −iq(

∂L∂(∂µψ)ψ − ψ ∂L

∂(∂µψ)

)is conserved. Using the Dirac’s Lagrangian

3.7, ∂L∂(∂µψ) = iψγµ and ∂L

∂(∂µψ)= 0, thus we finally have

jµ = qψγµψ (3.10)

which is the charge-current we introduced in equation 2.28! The continuity equation ∂µjµ then

means that the charge Q =∫d3x j0 must be conserved.

Local phase invariance: Now, let us assume that the phase χ in 3.9 is a real functiondepending on the spacetime coordinates:

ψ → ψ′ = e−iqχ(x)ψ (3.11)

How is the Dirac’s Lagrangian transformed?

LD → L′ = e+iqχ(x)ψ(i/∂ −m)e−iqχ(x)ψ

= e+iqχ(x)ψiγµ[(−iq∂µχ(x))e−iqχ(x)ψ + e−iqχ(x)∂µψ

]−mψψ

= LD + qψγµψ ∂µχ(x)= LD + jµ∂µχ(x)

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68 A brief view of Quantum Electrodynamic

If we wish to have a Lagrangian invariant under this local transformation, we need to add anextra term Lint to the Dirac Lagrangian:

Lint = −jµAµ = −qψγµψAµ (3.12)

which will compensate the term jµ∂µχ(x). Lint couples the vector Aµ to the charge-current ofthe Dirac’s field jµ. Let us see how the modified Lagrangian now transforms:

LD + Lint → L′ = LD + jµ∂µχ(x)− j′µA′µBut j′µ = qψ′γµψ′ = qψγµψ = jµ, thus:

LD + Lint → L′ = LD + jµ(∂µχ(x)−A′µ

)

If Aµ transforms as:Aµ → A′µ = Aµ + ∂µχ(x) (3.13)

the new Lagrangian LD + Lint would be invariant. We have already encountered the abovetransformation: it is the same as the one allowed by the gauge invariance of the electromagneticfield 2.73!

3.1.5 QED Lagrangian

Let us recap: promoting the free Dirac Lagrangian to be invariant under a local phase trans-formation U(1), requires an additional term that couples a vector field Aµ, consistent with thephoton field (having the same transformations), to the electron charge-current. Furthermore, ifwe use the correspondence principle 2.46 as in the case of the Dirac’s equation in presence of anelectromagnetic field:

∂µ → Dµ = ∂µ + iqAµ (3.14)

the Dirac Lagrangian becomes:

LD → ψ(iγµDµ −m)ψ→ LD − qψγµψAµ→ LD + Lint

(3.15)

where Lint is the same as in 3.12! Dµ is called the covariant derivative. Conclusion: the localgauge invariance requirement, ie the local phase invariance, included in the free electrons theory,has generated a new field, the photon, that couples to the electron, through the Lagrangian Lint.This is an example of the “gauge principle” which elevates a global symmetry (U(1) here), intoa local one. The “gauge principle” requires to replace the normal derivative by the covariantderivative.

We can now specify the full QED Lagrangian:

L = LD + Lγ + Lint = ψ(i/∂ −m)ψ − 1

4FµνF

µν − qψγµψAµ (3.16)

It is the sum of the free electron Lagrangian (ie the Dirac Lagrangian), the free photon La-grangian, and the interaction Lagrangian between electrons2 and photons. The recipe to obtainthe QED Lagrangian is to start from the free Lagrangians and replace the derivative by the co-variant derivative. Doing so, we obtain a QED Lagrangian that is locally U(1) gauge invariant3:the ψ field transformation (3.11) being absorbed by the Aµ transformation (3.13).

2For electrons q = −e has to be used.3The term Fµν being obviously invariant under transformation 3.13.

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Perturbation with interacting fields: the S-matrix 69

3.2 Perturbation with interacting fields: the S-matrix

We saw in the previous section that electron field and photon field interact. We then expectprocesses where for example, the electron will be scattered by a photon. Hence, we need to beable to calculate measurable quantities as the cross-section using time-dependent perturbationtheory as exposed in section 1.6. This section aims at making a connection between the fieldsand the matrix element appearing in formulas such as 1.62.

3.2.1 Schrodinger, Heisenberg and Interaction representations

In the well known Schrodinger reprensentation, the time evolution of a system is given by:

i∂

∂tψs(t) = Hψs(t) (3.17)

where we use the subscript s to clearly remind that we work in the Schrodinger representation.If the hamiltonian H doesn’t depend on time, ψs can be computed for any time t provided thestate is known at another time let us say t = 0 via4:

ψs(t) = US(t, 0)ψs(0) = e−itHψs(0) (3.18)

where US is unitary (U †S = U−1S ) since H is hermitian. Thus, the state vectors carry the entire

time dependance. The field ψ can be expanded in terms of creation and annihilation operators att = 0. However, the problem with this representation, is this particular time: it is not manifestlyLorentz invariant.

Now, note that the matrix element of an operator A between state ψs and ψ′s can be estimatedat the time t with:

〈ψ′s|A|ψs〉 =

∫ψ′s∗(t)Aψs(t) =

∫ψ′s∗(0)US(t, 0)†AUS(t, 0)ψs(0)

Now, this expression can be seen differently, with an operator AH depending on time:

AH(t) = US(t, 0)†AUS(t, 0)

and states ψH = ψs(0), ψ′H = ψ′s(0) without time dependance, such as:

〈ψ′s|A|ψs〉 = 〈ψ′H |AH |ψH〉

This new representation is the Heisenberg one (thus the subscript H) where operators dependon time and states do not. The time-dependency is entirely transferred to operators via theHeisenberg equation:

id

dtAH(t) = [AH(t), H]

With this representation, the evolution of the creation and annihilation operators appearingin the field expansion, is given by the previous equality, which requires the usage of the fullhamiltonian. When this hamiltonian is the one of free particles, we know how to do it. For morecomplicated cases, the full hamiltonian is not necessarily known. Note that the Hamiltonian H

4The exponential means e−itH =∑∞k=0(−itH)k/k!

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70 A brief view of Quantum Electrodynamic

is identical in the 2 representations since it commutes with itself. Also, note that ψH can bewritten as:

ψH = US(t, 0)†ψs(t) = eitHψs(t)

Finally, there is the Interaction representation which is a hybrid of the two previous repre-sentations. It is well adapted for the perturbative calculations. In this representation, both thestates and the operators have a time dependency. Consider an hamiltonian:

H = H0 +Hint

where H0 describes a free system and Hint the perturbation source of interactions. In theInteraction picture, the state ψI is defined as:

ψI(t) = eitH0ψs(t) (3.19)

Notice the H0 instead of H as in the Heisenberg representation. Since ψ(t) obeys to 3.17, it iseasy to conclude that ψI(t) obeys to:

i ∂∂tψs(t) = Hψs(t) ⇒ i ∂∂t(e−itH0ψI(t)) = (H0 +Hint)e

−itH0ψI(t)

⇒ i ddtψI(t) = eitH0 Hint e−itH0ψI(t) = HI int(t)ψI(t)

(3.20)

with:

HI int(t) = eitH0 Hint e−itH0

Similarly as for the Hamiltonian, any operator in the Interaction picture can be expressed fromthe corresponding operator in the Schrodinger picture as:

AI(t) = eitH0Ae−itH0 (3.21)

which means that its time evolution satisfies the Heisenberg equation:

id

dtAI(t) = [AI(t), H0] (3.22)

with H replaced by H0. The field operators satisfy to this equation which is valid even in thepresence of an interaction. In other words, the field can still be expressed as a linear combinationof the free field solution.

3.2.2 Dyson expansion

We wish to find the evolution operator U which allows to calculate the state at any time tknowing the state at a given time t0:

ψI(t) = U(t, t0)ψI(t0) (3.23)

Thus:d

dtψI(t) =

dU(t, t0)

dtψI(t0)

but ψI also satisfies:

id

dtψI(t) = HI int(t)ψI(t) (3.24)

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Dyson expansion 71

so that:

idU(t, t0)

dtψI(t0) = HI int(t)ψI(t) = HI int(t)U(t, t0)ψI(t0)

and hence:

id

dtU(t, t0) = HI int(t)U(t, t0) (3.25)

From now on, we are going to use reduced notation by forgetting the subscript I referring tothe Interaction representation. Equation 3.25 can be solved by an iterative procedure:

U(t, t0) = U(t0, t0)− i∫ t

t0

dt1Hint(t1)U(t1, t0)

But U(t0, t0) is by definition 1 (see 3.23), so that:

U(t, t0) = 1− i∫ tt0dt1Hint(t1)U(t1, t0)

= 1− i∫ tt0dt1Hint(t1)

[1− i

∫ t1t0dt2Hint(t2)U(t2, t0)

]

= 1− i∫ tt0dt1Hint(t1) + (−i)2

∫ tt0dt1∫ t1t0dt2Hint(t1)Hint(t2)U(t2, t0)

Pursuing the same approach at higher order, we finally get:

U(t, t0) =

∞∑

n=0

(−i)n∫ t

t0

dt1

∫ t1

t0

dt2 · · ·∫ tn−1

t0

dtnHint(t1)Hint(t2) · · · Hint(tn)

Now, we can re-arrange the integrals to have the same range of integration [t0, t]. Consider theintegrals of the second order term:

∫ t

t0

dt1

∫ t1

t0

dt2Hint(t1)Hint(t2) =

∫ t

t0

dt1

∫ t

t0

dt2 θ(t1 − t2)Hint(t1)Hint(t2)

where θ is the usual step function, so the implicit condition t1 > t2 is imposed. Now, splittingin 2 and changing the order of the integrals in the second term, it comes:∫ tt0dt1∫ t1t0dt2Hint(t1)Hint(t2)

= 12

∫ tt0dt1∫ tt0dt2 θ(t1 − t2)Hint(t1)Hint(t2) + 1

2

∫ tt0dt2∫ tt0dt1 θ(t1 − t2)Hint(t1)Hint(t2)

= 12

∫ tt0dt1∫ tt0dt2 θ(t1 − t2)Hint(t1)Hint(t2) + 1

2

∫ tt0dt1∫ tt0dt2 θ(t2 − t1)Hint(t2)Hint(t1)

= 12

∫ tt0dt1∫ tt0dt2 θ(t1 − t2)Hint(t1)Hint(t2) + θ(t2 − t1)Hint(t2)Hint(t1)

(3.26)where in the last line, the integration labels were interchanged. The trick now is to introducethe T -product (or time-ordered product) of operators defined as:

T (Hint(t1)Hint(t2)) = Hint(t1)Hint(t2) if t1 > t2Hint(t2)Hint(t1) if t2 > t1

= Hint(t1)Hint(t2)θ(t1 − t2) +Hint(t2)Hint(t1)θ(t2 − t1)(3.27)

which for the product of n operators A reads:

T (A(t1)A(t2) · · ·A(tn)) = A(ti1)A(ti2) · · ·A(tin) with ti1 > ti2 > · · · > tin (3.28)

Hence, 3.26 becomes:∫ t

t0

dt1

∫ t1

t0

dt2Hint(t1)Hint(t2) =1

2

∫ t

t0

dt1

∫ t

t0

dt2 T (Hint(t1)Hint(t2))

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72 A brief view of Quantum Electrodynamic

For the higher order terms, one has to take into account the number of permutations in theT -product, and the final formula is:

U(t, t0) =∞∑

n=0

(−i)nn!

∫ t

t0

dt1

∫ t

t0

dt2 · · ·∫ t

t0

T (Hint(t1)Hint(t2) · · · Hint(tn)) (3.29)

which can be simply written in the symbolic form:

U(t, t0) = T e−i∫ tt0dτHint(τ)

(3.30)

where again, Hint is understood here in the interaction representation. Now let us consider thefollowing scattering process: long before the interaction, say at t→ −∞, Hint is negligible andaccording to 3.24, ψ is a steady state in the Interaction representation. That’s one of the maininterest of this representation where the kets evolve with time only when there is an interaction.Let us denote this initial state |i〉 = |ψ(−∞)〉. Experimentally, |i〉 is typically the 2 particlesthat are supposed to collide in an accelerator. As time goes by, the particles described by |i〉may scatter and Hint is not negligible anymore. The time evolution of ψ is then given by:

ψ(t) = U(t,−∞)ψ(−∞)

Long after the interaction, for the same reason, the system potentially constituted of manyparticles resulting of the collision, is in another steady state |ψ(+∞)〉. The amplitude of theprobability for finding a given state |f〉 is then:

Sfi = 〈f |ψ(+∞)〉 = 〈f |U(+∞,−∞)ψ(−∞)〉 = 〈f |U(+∞,−∞)|i〉

The matrix:S = U(+∞,−∞) = T e−i

∫+∞−∞ dτHint(τ) (3.31)

is called the scattering matrix (S-matrix). Its element Sfi gives the amplitude probability forhaving a transition i→ f . According to 3.29 the S-matrix expression is thus:

S =∞∑

n=0

S[n] (3.32)

where the nth order term is:

S[n] =(−i)nn!

∫· · ·∫dt1 · · · dtn T (Hint(t1) · · · Hint(tn)) (3.33)

3.2.3 Connection with the general formula of transition rate

We consider an initial state |i〉 made of several particles with definite momenta and similarlyfor the final state |f〉. |i〉 and |f〉 are the states corresponding to the asymptotic conditions att = −∞ and t = +∞. The probability of a transition from |i〉 to |f〉 is then given by:

Pi→f =| 〈f |S|i〉 |2〈i|i〉 〈f |f〉 (3.34)

If there is no reaction at all, we must have S = l1 (the identity operator), and i = f and so we dohave P = 1. Therefore, we expect S to be written as S = l1+ · · · where the dots are an expression

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Connection with the general formula of transition rate 73

of the operators for a non trivial interaction. We do not need to have an explicit formulation ofthese operators (which could be horribly complicated), only the matrix element 〈f |S|i〉 for theinitial and final states envisaged is required. How can we write it? It is convenient to factor outa delta function implementing the 4-momentum conservation5 to define the Lorentz invariantscattering amplitude, Mfi:

〈f |S|i〉 = 〈f |i〉+ (2π)4δ(4)(Pf − Pi) iMfi

where Pf and Pi are respectively the final and initial total momenta. The factor (2π)4 isconveniently factorized so that iMfi will be the result of a Feynman diagram calculation. Sincewe are interested in non trivial interactions, we have:

〈f |S|i〉 = (2π)4δ(4)(Pf − Pi) iMfi , i 6= f (3.35)

Now let us make explicit the terms of (3.34). Using the commutation rules for bosons 2.14 (butwe would find the same result with fermions), we notice that:

〈p′|p〉 = 〈0|ap′a†p|0〉 = 〈0|a†pap′ |0〉+ (2π)32Ep δ(3)(~p′ − ~p) = (2π)32Ep δ

(3)(~p′ − ~p)

where a† is the creation operator of the corresponding particles (boson or fermion). This resultis known as the relativistic normalization of momenta states. However, that means that if weconsider for simplicity an initial state with just one particle with momentum pi, we have:

〈i|i〉 = (2π)32Epiδ(3)(0)

which is an infinite quantity! If instead of plane waves (with definite momenta) we had consideredlocalized wave packets for the asymptotic initial and final states, the result would have beenfinite. The problem appears here because we implicitly use plane waves in an infinite volume(the whole 3-D space). Indeed, according to formula 1.69, we have:

δ(3)(p) =1

(2π)3

∫eip.xd3x = lim

L→∞∆L(p)

with∆L(p) = 1

(2π)3

∫ L/2−L/2 dx e

ipxx∫ L/2−L/2 dy e

ipyy∫ L/2−L/2 dz e

ipzz

= 1(2π)3L

sin(pxL/2)pxL/2

Lsin(pyL/2)pyL/2

L sin(pzL/2)pzL/2

= V(2π)3

sin(pxL/2)pxL/2

sin(pyL/2)pyL/2

sin(pzL/2)pzL/2

where V = L3 is the volume of 3-space (a box with sides of length L). We conclude that∆L(0) ≡ V/(2π)3 (since limx→0

sinxx = 1) is equivalent to δ(3)(0) in the limit V → ∞ and hence

〈i|i〉 is equivalent to 2EpiV. We can then do all our calculations with a finite volume V, and forphysics calculations, the volume should cancel out so that the infinite volume limit can be takensafely. It is straightforward to generalize to n particles in the initial state:

〈i|i〉 =n∏

k=1

2EkV = Vnn∏

k=1

2Ek

5All interactions are assumed to conserve the 4-momenta.

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74 A brief view of Quantum Electrodynamic

and similarly for n′ particles in the final states:

〈f |f〉 = Vn′n′∏

k=1

2E′k

The last term in 3.34 to be determined is | 〈f |S|i〉 |2. According to 3.35, it is:

| 〈f |S|i〉 |2 = (2π)4δ(4)(PF − PI)(2π)4δ(4)(PF − PI)|Mfi|2

Here, we have to evaluate the square of the delta-function δ(4)(PF − PI). Being applied twice,the first delta-function imposes PF = PI into the second and hence δ(4)(PF − PI)δ

(4)(PF −PI) = δ(4)(PF − PI)δ(4)(0). The question is thus to estimate δ(4)(0) = δ(3)(0)δ(0) = V

(2π)3 δ(0).

Proceeding as for our estimation of δ(3)(0), we have:

δ(E) =1

(2π)

∫eiE.tdt =

1

(2π)limT→∞

∫ T/2

−T/2dt eiEt =

1

(2π)limT→∞

Tsin(ET/2)

ET/2

and hence we conclude δ(0) ≡ T(2π) leading to:

| 〈f |S|i〉 |2 = (2π)4δ(4)(PF − PI)VT |Mfi|2

Putting all the pieces together, we finally establish:

Pi→f = V1−n−n′T (2π)4δ(4)(PF − PI)|Mfi|2n∏

k=1

1

2Ek

n′∏

k=1

1

2E′k

or more appropriately in terms of transition rate (i.e. probability per unit of time):

Γi→f = V1−n−n′ (2π)4δ(4)(PF − PI)|Mfi|2n∏

k=1

1

2Ek

n′∏

k=1

1

2E′k

However, we forgot something: the final states have a continuous energy spectrum. The particlesin the final state have not an infinitely accurate momenta/energies but should rather be describedby a set of final states belonging to a given phase space element (i.e. belonging to px±δpx, py±δpy, pz ± δpz). Let us consider an infinitesimal phase space element so that we can assume that|Mfi| remains constant for these dN states. The infinitesimal transition probability per unittime is then the sum of the probabilities of each individual states:

dΓi→f = V1−n−n′ (2π)4δ(4)(PF − PI)|Mfi|2(

n∏

k=1

1

2Ek

n′∏

k=1

1

2E′k

)dN

As well known from quantum mechanics, in case of a single particle in a box of sides L, thecomponents of its momentum are quantized as: px,y,z = (2π/L) nx,y,z, and therefore the numberof states in d3~p = dpxdpydpz is dN = dnxdnydnz = Vd3~p/(2π)3. Therefore for n′ particles in

the final state, dN becomes dN =∏n′k=1 Vd3 ~p′k/(2π)3 and thus:

dΓi→f = V1−n (2π)4δ(4)(PF − PI)|Mfi|2n∏

k=1

1

2Ek

n′∏

k=1

d3 ~p′k(2π)32E′k

(3.36)

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Feynman rules and diagrams 75

which is exactly the formula 1.62 we used in the first chapter to establish the formulas of thedecay rate and cross-section where, as it should, the volume V cancels out.

We have seen in equation (3.35) that the matrix element Mfi is related to 〈f |S|i〉 definedat a order [n] by (3.33). Hence:

(2π)4δ(4)(Pf − Pi) iM[n]fi =

(−i)nn!

∫dt1 · · · dtn T (〈f |Hint(t1) · · · Hint(tn)|i〉)

In order to make explicit the Lorentz invariance, it is better to use the hamiltonian density H(H =

∫d3xH(x)). The final formula for the matrix elements then becomes:

(2π)4δ(4)(Pf − Pi) iM[n]fi =

(−i)nn!

∫dx1 · · · dxn T (〈f |Hint(x1) · · · Hint(xn)|i〉) (3.37)

where, once again, Hint is the interaction hamiltonian given in the Interaction representation(meaning that the product of fields constituting the hamiltonian are themselves given in theinteraction representation).

3.3 Feynman rules and diagrams

Let us consider the Lagrangian of QED 3.16. We wish to find the expression of Hint knowingthat according to 3.4 and 3.3:

H =∂L

∂φ(xµ)φ(xµ)− L (3.38)

Fields derivative in 3.16 only enters in the two free Lagrangians LD and Lγ . Hence, Hint reducesto:

Hint = −Lint = qψγµψAµ (3.39)

3.3.1 Electron-photon vertex

Consider the first order where q has to be understood as the charge of the particle, not the oneof the antiparticle6:

S[1] = −i∫d4xHint(x) = −iq

∫d4x ψγµψAµ

Now, inject the 3 fields development for ψ (2.67), ψ (2.68), and Aµ (2.76):

S[1] = −iq∫d4x

∫ d3~p(2π)32Ep

d3~p′

(2π)32Ep′d3~q

(2π)32Eq

∑r=1,2

∑r′=1,2

∑3λ=0[

c†~prurpe+ipx + d~prvrpe

−ipx]γµ[c~p′r′ur′p′e

−ip′x + d†~p′r′vr′p′e+ip′x

][εµqλα~qλe

−iq.x + ε∗µqλα†~qλe

+iq.x]

6For electron-positron, q = −|e|.

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76 A brief view of Quantum Electrodynamic

Integrating over x (using 1.69) and rearranging, we finally obtain:

S[1] = (2π)4∫ d3~p

(2π)32Ep

d3~p′

(2π)32Ep′d3~q

(2π)32Eq

∑r=1,2

∑r′=1,2

∑3λ=0

c†~prc~p′r′α~qλ (urp (−iqγµ)ur′p′)εµqλ δ(4)(p− p′ − q)+

c†~prd†~p′r′α~qλ (urp (−iqγµ) vr′p′)εµqλ δ

(4)(p+ p′ − q)+d~prc~p′r′α~qλ (vrp (−iqγµ)ur′p′)εµqλ δ

(4)(p+ p′ + q)+

d~prd†~p′r′α~qλ (vrp (−iqγµ) vr′p′)εµqλ δ

(4)(p′ − p− q)+c†~prc~p′r′α

†~qλ (urp (−iqγµ)ur′p′)ε

∗µqλ δ

(4)(p− p′ + q)+

c†~prd†~p′r′α

†~qλ (urp (−iqγµ) vr′p′)ε

∗µqλ δ

(4)(p+ p′ + q)+

d~prc~p′r′α†~qλ (vrp (−iqγµ)ur′p′)ε

∗µqλ δ

(4)(q − p− p′)+d~prd

†~p′r′α

†~qλ (vrp (−iqγµ) vr′p′)ε

∗µqλ δ

(4)(p′ − p+ q)

(3.40)

Thus, S[1] is constituted of 8 terms7 having a similar form: product of 3 operators, a charge-current (product of an adjoint-spinor and a spinor sandwiching a γµ matrix), a photon polarisa-tion vector and a delta function. Looking at the operators, one notices that the total charge isalways zero: for example the first line creates an electron, annihilates an electron and annihilatesa photon, while the second creates an electron and a positron and annihilates a photon.

Let us detailed the first line as an example. Clearly, it gives a contribution only if it is sand-wiched between an initial state containing a photon and an electron |ei, γ〉 (that are annihilated)and a final state containing an electron |ef 〉 (that is created):

〈ef | c†~prc~p′r′α~qλ |ei, γ〉 = 〈0| c~pf rf c†~prc~p′r′α~qλc

†~piri

α†~qγλγ |0〉= 〈0| c~pf rf c

†~prc~p′r′c

†~piri

α~qλα†~qγλγ|0〉

= 〈0|(−c†~prc~pf rf + (2π)32Efδ

(3)(~pf − ~p)δrf r)

(−c†~piric~p′r′ + (2π)32Eiδ

(3)(~p′i − ~p′)δrir′)

(α†~qγλγα~qλ − gλλγ (2π)32Eγδ

(3)(~q − ~qγ))|0〉

= −(2π)32Efδ(3)(~pf − ~p)δrf r(2π)32Eiδ

(3)(~p′i − ~p′)δrir′gλλγ (2π)32Eγδ

(3)(~q − ~qγ) 〈0|0〉

Inserting this result into the S[1] expression and performing the integrations and summations,we get a term8

(2π)4δ(4)(pf − pi − qγ)(urfpf (−iqγµ)uripi

)εµqγλγ (3.41)

Comparing with 3.37 we conclude in that case:

iM =(urfpf (−iqγµ)uripi

)εµqγλγ (3.42)

The creation of the final electron is thus associated to an adjoint spinor u while the annihilationof the initial electron is associated a spinor u. Similarly, the annihilation of the photon isassociated to a vector polarization. This yields our first Feynman diagram shown on figure 3.1.The arrow of time on the horizontal axis is going from the left to the right while the spacialdimension is vertical. The photon is symbolized by a wavy line, and the electron (actually anyfermion) by a solid line. The intersection of the lines is called a vertex. The direction of the

73 fields with 2 operators give 23 = 8 combinations.8Real photons have only transverse polarisation, ie λγ = 1 or 2. Thus gλγλγ = −1.

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Electron-photon vertex 77

Figure 3.1: A basic vertex in QED.

arrow on the electron line has a meaning: for electrons in the initial state, it goes toward thevertex while for electrons in the final state it points away from the vertex. In other words, thearrow of a fermion is always oriented from the past to the future.

The development we made to the first order of perturbation leads to diagrams with only 1vertex. The number of vertices represents the order of the perturbation development. The 7other terms of S[1] development yield similar kind of diagrams. They are all shown on the figure3.2. Positrons (or any antifermions) are represented by the same line as electrons. What differs

Figure 3.2: The vertices of QED at first order of perturbation.

is only the direction of the arrow. For example, the second diagram on the first row, depicts aphoton that annihilates in 1 electron (line going up) and 1 positron. Note that for the outgoingpositron, the arrow “goes back in time”. Similarly, in the fourth diagram of the first row, theincoming particle is a positron, the arrow going back in time. On the photon, there is no arrowsince it is its own antiparticle. According to 3.40, following the same procedure that yields 3.41,we see that antifermions in the initial state are associated to adjoint antispinor v and antispinorv in the final state. For photons, a polarisation vector εµ is associated to an incoming photonsand ε∗µ for an outgoing one.

Looking back to formula 3.40 and 3.42 we see that all terms contain a −iqγµ contributionsandwiched by 2 (anti-)spinors. This factor is logically associated to the vertex itself in thediagram. The charge q = −|e| is the coupling between electrons or positrons and photons.Development at nth order would produce n vertices, leading to a factor |e|n in the amplitude.The convergence of the perturbation series depends on |e|, which finally represents the “strength”of the perturbation. In natural units |e| '

√4π/137 ' 0.3.

Finally, There is a last term that we haven’t exploited yet in 3.41: the delta function. Itclearly imposes the energy-momentum conservation at the vertex. Well, it appears that none

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78 A brief view of Quantum Electrodynamic

of these 8 terms are physically possible with free particles! An obvious one: the third onthe first row of figure 3.2. The energy conservation imposed by δ(4)(p+ p′ + q) means thatEe− + Eγ + Ee+ = 0 ⇒ Ee− = Eγ = Ee+ = 0 which is impossible (there is at least the restenergy since electrons are massive). Even, more conventional diagrams as the first one, areimpossible. Both energy conservation and momentum conservation can’t be satisfied at thesame time (because the photon mass is 0). The only possibility would be to envisage that oneof these particles does not have its normal mass value (or more appropriately that it does notrespect the relativistic relation E2 = p2 +m2). Such a particle is called a virtual particle and isallowed by the uncertainty inherent in quantum mechanics:

∆E ≥ ~/T

The relation can be interpreted as follow: the greater the energy (or mass) is shifted from itsnominal value, the shorter the particle will live. In other words, virtual particles (thus notsatisfying E2 = m2 +p2 with m the usual mass), must live a short time: they have to be quicklyreabsorbed (annihilated) by another process9. Conclusion: another vertex is needed and onehas to go to the second order of the perturbation development.

3.3.2 Photon and electron propagators

Consider the second order of perturbation:

S[2] =(−i)2

2q2

∫d4x d4y T (ψ(x)γµψ(x)Aµ(x)ψ(y)γνψ(y)Aν(y)) (3.43)

If we develop all quantities as in the previous section we have 82 = 64 terms, and thus 64diagrams. Two such diagrams are shown in figure 3.3. In the first diagram, we see that a

Figure 3.3: Example of QED diagrams at the second order.

(virtual) photon is created at the spacetime point x (assuming tx < ty) and annihilated aty. Similarly, the diagram on the right-hand side shows a virtual electron being exchanged.We can de-couple the initial and final states and consider that a virtual particle must be ableto be created from the vacuum and annihilated. Looking at the formula 3.43, we see that,for the photon, the product Aµ(x)Aν(y) has to be involved. Indeed, developed in terms of

creation-annihilation operators, the combination α~qλ(x)α†~q′λ′(y) appears, which will give a non-zero contribution when sandwiched with the vacuum |0〉. Similarly, for the fermions, the only

product giving non-zero term involved c~pr(x)c†~p′r′(y) or d~pr(x)d†~p′r′(y) which originates only from

9In fact, all particles can be considered as virtual: for instance a photon emitted by a star is detected on earthby a photomultiplier which annihilates the photon. Its “degree of virtuality” is however ridiculously tiny to beable to live during million years!

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Summary of QED Feynman rules 79

ψ(x)ψ(y). Conclusion: the two quantities in 3.43 that can describe the propagation of a virtualparticle from a point x to a point y (ie its creation followed by its annihilation) are:

〈0|T (Aµ(x)Aν(y)) |0〉 , 〈0|T (ψ(x)ψ(y)) |0〉 (3.44)

They are called propagators (respectively of the photon and a fermion). The calculation (notso complicated) can be found in any quantum field theory book (see for instance [8, p. 62]).We simply give here the results in the momentum space10 (i.e. after having integrated on allpossible position for x and y):

Photon propagator:−igµνp2 + iε

(3.45)

where p is the 4-momentum of the (virtual) photon and ε is an infinitesimal real positive numberto avoid singularities when p → 0. (In practice, ε is dropped). This formula corresponds to aparticular choice of gauge called Feynman gauge. For the electron, we have:

spin 1/2 propagator: i/p+m

p2 −m2 + iε(3.46)

where again, p is the 4-momentum of the (virtual) fermion and m its mass (ie the nominal massused in E2 = |~p|2 +m2). Remember that /p = γµpµ is a matrix. Hence, diagrams, as the first offigure 3.43 correspond mathematically to an amplitude:

iM =(vre+pe+ (−iqγµ)ure−pe−

) −igµνp2 + iε

(ur′

e−p′e−

(−iqγν) vr′e+p′e+

)(3.47)

where the first charge current jµ = vre+pe+ (qγµ)ure−pe− corresponds to the first perturbationterm between the electron, positron and the virtual photon, −igµν/(p2 + iε) corresponds to thepropagation of the virtual photon, and the second charge-current jν = ur′

e−p′e−

(qγν) vr′e+p′e+

to the second perturbation term between the electron, positron and the virtual photon, whichannihilates the virtual photon.

Before concluding this section, let us examine the left diagram of figure 3.4. At first sight, itseems hard to interpret in term of a particle being created and then absorbed since both verticesare at the same time (time goes from left to right here). But, consider the 2 other diagrams in3.4. When t1 < t2, in the centre diagram, a virtual positron (= an electron moving backward intime on the diagram) is first created and then absorbed whereas when t2 > t1, in the right handside diagram, a virtual electron is first created and then absorbed. Since, in the definition of thepropagators 3.44, there is a T-product of field operators, the Feynman diagram on the left-handside, includes the 2 possibilities depicted by the centre and right pictures! Both time-orderingsare always included in each propagator line! It makes perfectly sense, since there is no way todistinguish experimentally the two modalities.

3.3.3 Summary of QED Feynman rules

In this section, we wish to summarise the recipe to calculate the amplitude of a process.

1. Draw, at a given order, all diagrams describing the transition between an initial state (inpractice, 1 or 2 particles) and a final state (can contain many particles). A QED vertexmust always connect 2 (anti-)fermions and 1 photon.For each vertex:

10Generally, the positions of particles in a reaction are not known but their momenta are.

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80 A brief view of Quantum Electrodynamic

1

e!"

e!" e!" e!"

e!"e!" γ"

γ"γ"

γ"γ"

γ"

1"

2"

e+" e!"

Figure 3.4: Left: e−γ → e−γ Feynman diagram. Centre and right: the corresponding 2 time-ordereddiagrams.

• Energy-momentum is conserved.

• Charge of the incoming particle(s) is always equal to the charge of the outgoingparticle(s). QED can be seen as a flow of charges (that’s why the charge-current isinvolved).

2. Calculate iM of each diagram by using the factors given in table 3.1. Incoming and out-going particles are connected to vertices exchanging virtual particles. The correspondingfactors in 3.1 are respectively the ones of the “External lines”, “Vertex” and “Propagators”.

3. Combine the amplitudes of all diagrams to get the total amplitude of the process.

• If 2 diagrams differ only by the exchange of 2 external identical fermions (2 outgoing,or 2 incoming, or 1 incoming fermion and 1 outgoing anti-fermion, or 1 incominganti-fermion and 1 outgoing fermion), subtract the amplitudes11.

• add all other amplitudes.

For completeness, we also give the factors for massive spin 0 boson interacting with photons.The procedure is similar to the fermion case: from the wave equation, the free Lagrangianis deduced, allowing the quantization of fields. The propagator can then be calculated. Theinteraction term is then obtained, by replacing in the Lagrangian, the derivatives by covariantderivatives 3.14. The vertex factors and external lines factors are determined accordingly.

3.4 QED and helicity/chirality

We have seen that QED interactions involve the charge current. So terms as quγµu, qvγµv,quγµv, qvγµu are expected. Now, we saw in section 2.3.4.3, that any spinor can be decomposedin left-handed and right-handed components of chirality through the projectors 2.61. Hence, thecurrent can be decomposed:

quγµu = q(uL + uR)γµ(uL + uR)= quLγ

µuL + quRγµuR + quLγ

µuR + quRγµuL

11This can be understood as a consequence of the fact that when 2 identical fermions are interchanged, thewave function gets a minus sign. It is clearly related to the spin-statistic theorem.

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QED and helicity/chirality 81

External lines

Spin 0 Boson incoming 1Boson outgoing 1anti Boson incoming 1anti Boson outgoing 1

Spin 12 Fermion incoming u spinor

Fermion outgoing u spinoranti Fermion incoming v spinoranti Fermion outgoing v spinor

Spin 1 Photon incoming εµPhoton outgoing ε∗µ

Propagators

Spin 0 Boson ip2−m2+iε

Spin 12 Fermion i /p+m

p2−m2+iε

Spin 1 Photon −igµνp2+iε

Vertex

Spin 0 −iq(pµ + p′µ)

Spin 1

2 −iqγµ

q = −|e| for e±

Table 3.1: Feynman rules in QED.

But:

uLγµuR = u† 1

2(1− γ5†)γ0γµ 12(1 + γ5)u

= 14u†(1− γ5)γ0γµ(1 + γ5)u

= 14u†γ0(1 + γ5)(1− γ5)γµu

= 0

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82 A brief view of Quantum Electrodynamic

And similarly:uLγ

µuR = uRγµuL = vLγ

µvR = vRγµvL = 0

uLγµvL = uRγ

µvR = vLγµuL = vRγ

µuR = 0

(remember that vL = PRv). Now, at high energy in the ultra-relativistic regime where E m,the helicity states and chirality states are the same (see section 2.3.4.3).We can then conclude that:

In scattering processes (t-channel): only the combinations:

uLγµuL + uRγ

µuR or vLγµvL + vRγ

µvR

give non-zero contribution. The helicity of the particle after the scattering is the same as theone before. Helicity is conserved in the ultra-relativistic limit by QED (and chirality is alwaysconserved).

In annihilation/pair creation processes (s-channel): only the combinations:

uLγµvR + uRγ

µvL or vLγµuR + vRγ

µuL

give non-zero contribution. The helicity of the particle is the opposite of the one of the anti-particle. In our jargon, we’re still talking about helicity conservation in the sense that onlyspecific helicity configurations are possible. Hence, in the centre of mass frame of the pair, thespins of the particle and antiparticle must be aligned, leading to Jz = ±1. The following figuresummarize the situation.

Figure 3.5: The conservation of Helicity in QED.

3.5 Simple examples of graph calculation

3.5.1 Spin and polarisation summations: traces theorems

Consider the simplest QED graph e−µ− → e−µ− shown on figure 3.6.

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Spin and polarisation summations: traces theorems 83

µ−

e−

µ−

e−

Figure 3.6: Lowest order diagram e−µ− → e−µ−

Let us denote, p and k the 4-momenta of respectively e− and µ− in the initial state. And samenotation with a prime (’) for the final state. The photon has a momentum q = p− p′ = k′ − k.Following Feynman rules 3.1, the amplitude is:

iM = (uk′,r′(i|e|γµ)uk,r)−igµνq2

(up′,s′(i|e|γν)up,s) = i|e|2q2

(uk′,r′γµuk,r)(up′,s′γµup,s)

The cross section formula is proportional to the probability of the process and thus to |M|2.Generally, the initial state is not polarized and the spin of the final state is not measured. Hence,the measured cross-section is an average over the spins of the initial state and a sum over thespins of the final state12. Let us denote:

|M|2 =1

2

s

1

2

r

s′

r′|M|2 =

|e|4q4Lµν(µ−)Lµν(e−) (3.48)

with:Lµν(µ−) = 1

2

∑r,r′[uk′,r′γ

µuk,r]∗ [

uk′,r′γνuk,r

]

Lµν(e−) = 12

∑s,s′[up′,s′γµup,s

]∗ [up′,s′γνup,s

] (3.49)

Now, noting that:

(uk′,r′γµuk,r)

∗ = u†k,r㵆u†k′,r′ = u†k,rγ

0γµγ0(u†k′,r′γ0)† = uk,rγ

µuk′,r′

Lµν can be rewritten:

Lµν(µ−) = 12

∑r,r′[uk,rγ

µuk′,r′] [uk′,r′γ

νuk,r]

= 12

∑r uk,rγ

µ(∑

r′ uk′,r′ uk′,r′)γνuk,r

= 12

∑r uk,rγ

µ(/k′+m′

)γνuk,r

where we used the completeness relation 2.66. Now, we write explicitly the matrix element ofLµν with label α, β:

Lµν = 12

∑α,β

∑r u

(α)k,r

[γµ(/k′+m′

)γν](αβ)

u(β)k,r

= 12

∑α,β

[∑r u

(β)k,r u

(α)k,r

] [γµ(/k′+m′

)γν](αβ)

= 12

∑α,β [/k +m]

(βα) [γµ(/k′+m′

)γν](αβ)

= 12Tr

((/k +m)γµ

(/k′+m′

)γν)

12One may wonder why we don’t sum over the amplitudes of the different spin states and take the modulus-squared of the sum. The key here is that a particle with spin-up or spin-down defines 2 different spin states thatare in principle distinguishable (even if spin is not measured).

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84 A brief view of Quantum Electrodynamic

Tr denoting the trace13. We obtained these useful equalities:

Lµνuu = 12

∑r,r′[uk′,r′γ

µuk,r]∗ [

uk′,r′γνuk,r

]= 1

2Tr((/k +m)γµ

(/k′+m′

)γν)

Lµνvv = 12

∑r,r′[vk′,r′γ

µvk,r]∗ [

vk′,r′γνvk,r

]= 1

2Tr((/k −m)γµ

(/k′ −m′

)γν) (3.50)

where the result for anti-spinor was obtained similarly. Following the same approach, one canalso show that:

Lµνvu = 12

∑r,r′[vk′,r′γ

µuk,r]∗ [

vk′,r′γνuk,r

]= 1

2Tr((/k +m)γµ

(/k′ −m′

)γν)

Lµνuv = 12

∑r,r′[uk′,r′γ

µvk,r]∗ [

uk′,r′γνvk,r

]= 1

2Tr((/k −m)γµ

(/k′+m′

)γν) (3.51)

Now recall that:

Tr(αA+ βB) = αTr(A) + βTr(B) , Tr(AB) = Tr(BA)

so that when m = m′ as in our case:

Lµνuu(m = m′) =1

2

(kρk′ηTr(γργµγηγν) +mkρTr(γργµγν) +mk′ηTr(γµγηγν) +m2Tr(γµγν)

)

(3.52)Therefore, we need to evaluate traces of products of γ matrices. Since, the γ’s satisfies theClifford algebra 2.17 (γµ, γν = γµγν + γνγµ = 2gµν l1), we conclude the following rules:

Tr(γµγν) = gµνTr(l1) = 4gµν (3.53)

In passing, we thus have:

Tr(/p/k) = 4p.k (3.54)

Now recall γ5 properties 2.55 and 2.58 (ie (γ5)2 = 1, γ5γµ = −γµγ5)

Tr(γµ1 · · · γµn) = Tr(γµ1 · · · γµnγ5γ5)= Tr(γ5γµ1 · · · γµnγ5)= (−1)nTr(γµ1 · · · γµnγ5γ5)= (−1)nTr(γµ1 · · · γµn)

Hence:

Tr(odd nb of γ) = 0 (3.55)

Now, for 4 matrices:

Tr(γργµγηγν) = Tr(γµγηγνγρ)= −Tr(γµγηγργν) + 2gρνTr(γµγη)= −Tr(γµγηγργν) + 8gρνgµη

= Tr(γµγργηγν)− 8gηρgµν + 8gρνgµη

= −Tr(γργµγηγν) + 8gρµgην − 8gηρgµν + 8gρνgµη

Thus:

Tr(γργµγηγν) = 4(gρµgην − gρηgµν + gρνgµη) (3.56)

13If A and B are 2 matrices, the product C = AB has elements Cij =∑k AikBkj and thus Tr(C) =

∑i Cii =∑

i,k AikBki.

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Electron-muon scattering: e−µ− → e−µ− 85

In passing, we thus have:

Tr( /p1 /p2 /p3 /p4) = 4(p1.p2 p3.p4 − p1.p3 p2.p4 + p1.p4 p2.p3) (3.57)

Hence, coming back to 3.50

Lµνuu(m = m′) =1

2(4kρk

′η(g

ρµgην − gρηgµν + gρνgµη) + 4m2gµν)

so that:Lµνuu(m = m′) = Lµνvv (m = m′) = 2(kµk′ν + kνk′µ + (m2 − k.k′)gµν) (3.58)

For completeness, we give other useful formulas:

γµγµ = 4γµγνγµ = −2γν

γµγνγργµ = 4gνρ

γµγνγργηγµ = −2γηγργν

(3.59)

Tr(γ5γµγν) = 0Tr(γ5γµγνγργη) = −4iεµνρη

(3.60)

with εµνρη = 1 for µ, ν, ρ, η an even permutation of 1, 2, 3, 4, −1 for odd permutation, 0 otherwise.One can deduce the following properties of εµνρη:

εµνρη = −ενµρη = ενρµη = −ενρηµεµνρη = −ενρηµ = ερηµν = −εηµνρεµνρηεµναβ = −2(δραδ

ηβ − δ

ρβδηα)

(3.61)

Another equality which is worth mentioning is that for any 4-vector p, we have /p2 = pµpνγµγν =

pµpν(2gµν − γνγµ) = 2p2 − /p2 and thus:

/p2 = p2 (3.62)

Polarization summation: for photons, there is a similar summation as for fermions (ie∑r urur = /k +m). The result is [8, p. 159]:

4∑

λ=1

ε∗µλεν λ = −gµν (3.63)

3.5.2 Electron-muon scattering: e−µ− → e−µ−

We can now complete the calculation of the amplitude of e−µ− → e−µ−. Coming back to 3.48and inserting 3.58, we have:

|M|2 =|e|4q4Lµν(µ−)Lµν(e−)

Lµν(µ−)Lµν(e−) = 2(kµk′ν + kνk′µ + (m2µ − k.k′)gµν)2(pµp

′ν + pνp

′µ + (m2

e − pp′)gµν)

= 4(2k.p k′.p′ + 2k.p′ k′.p+ 2(m2

e − p.p′)k.k′ + 2(m2µ − k.k′)p.p′+

(m2e − p.p′)(m2

µ − k.k′)4)

= 8(k.p k′.p′ + k.p′ k′.p−m2

ek.k′ −m2

µp.p′ + 2m2

em2µ

)

(3.64)

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86 A brief view of Quantum Electrodynamic

Let us consider the ultra-relativistic limit neglecting terms with me and mµ. We can evaluatethe matrix element in the centre of mass frame and using the Mandelstam variables:

s = (k + p)2 = (k′ + p′)2 ' 2k.p ' 2k.′p′

t = (p− p′)2 = (k′ − k)2 = q2

u = (p− k′)2 = (p′ − k) ' −2p.k′ ' −2p′.k

so that:

|M|2 =2|e|4t2

(s2 + u2) (3.65)

3.5.3 Electron-positron annihilation: e−e+ → µ+µ−

The graph for e−(p) e+(k)→ µ+(k′) µ−(p′) is shown in figure 3.7. Of course, we can follow the

e+

e−

µ−

µ+

Figure 3.7: Lowest order diagram e−e+ → µ+µ−

same kind of calculation as in the previous example, but we can do better using the crossingsymmetry (and neglecting the masses). Comparing graph 3.7 and 3.6 we see that the t-channelof e−µ− → e−µ− is just the s-channel of e−e+ → µ+µ− as illustrated in the figure 3.8.

(2) µ−

(1) e−

(4) µ−

(3) e−

(3) e+

(1) e−

(4) µ−

(2) µ+

(3) e+

(1) e−

(4) µ−

(2) µ+

Figure 3.8: Left: the reaction e−µ− → e−µ− considered as the s-channel i.e. s = (p1 + p2)2, t =(p1 − p3)2 u = (p1 − p4)2. Centre: in order to obtain the t-channel of the left diagram, one has to takethe opposite of p2 and p3 (thus s↔ t) and interpret the result as anti-particles. Right: same diagram asthe centre but redraw in a more conventional way (after “horizontal stretching”).

Hence, we just have to replace t↔ s in the previous matrix element! That’s a nice example ofcrossing. We get:

|M|2e−e−→µ+µ− =2|e|4s2

(t2 + u2) (3.66)

Let us go further by computing the differential cross-section. Using 1.86, we have:

dΩ∗=

1

64π2s2|e|4 (t2 + u2)

s2

|~pf ||~pi|

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Compton scattering: e−γ → e−γ 87

Inserting the fine structure constant14:

α =|e|24π

(3.67)

and developing the Mandelstam variables in the center of mass frame where |~pi| = |~pf | = p∗

(because the masses are neglected):

s = (k + p)2 = 4p∗2

t = (p− k′)2 = −2p∗2(1− cos θ)

u = (p− p′)2 = −2p∗2(1 + cos θ)

where θ is the angle (in the center of mass frame) between in incident e− and the outgoing µ+.we have:

dσdΩ∗ = α2

2s(4p∗4(1−cosθ)2+4p∗4(1+cosθ)2

16p∗4

= α2

4s (1 + cos2 θ)

Integrating over the φ angle, we conclude:

d(cos θ)=α2π

2s(1 + cos2 θ)

Experimentally, the angle which is used is generally the one between the 2 particles e−, µ−

(which is equal to the one between e+, µ+). Thus θ → θ − π leaving however unchanged theformula. Is this prediction in agreement with the experimental data? Look at figure 3.9 on thetop which shows the s dσ

d cos θ distribution for the reaction e+e− → µ+µ−. The data were collectedat an e−e+ collider at

√s = 29 GeV (PEP collider at SLAC in the 1980’). The plot on the

top is for e+e− → µ+µ− and the dashed line describes QED predictions but at higher order(but largely dominated by the cos2 θ behaviour). Comparing to the data point, the agreementis good but the solid line gives a better fit. It corresponds to QED and additional electroweakcorrections where e+e− → µ+µ− can occur via Z boson exchange. We will see this in the nextchapters.

To conclude this calculations, we can determine the total cross-section by integration overcos θ:

σ =

∫ 1

−1d(cos θ)

α2π

2s(1 + cos2 θ) =

4πα2

3s(3.68)

which is in agreement with the data at the % level.

3.5.4 Compton scattering: e−γ → e−γ

We are going to calculate the cross-section of the Compton scattering:

e−(p) γ(k)→ e−(p′) γ(k′)

There are 2 diagrams, shown in figure 3.10, contributing to this process at the lowest order:s-channel (left-hand side of 3.10) and u-channel15 (right-hand side). Since photons are bosons,

14It’s a dimensionless measure of the strength of the electromagnetic interaction15It may not be obvious that the 2 contributions are s and u-channels. Let us assign to the particles in

e−(p) γ(k)→ e−(p′) γ(k′) the labels (1) + (2)→ (3) + (4). Then the u-channel is obtained by swapping (2) and(4) (namely the 2 photons) in the left-hand side diagram of figure 3.10. It gives the diagram on the right after“stretching” in the vertical direction.

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88 A brief view of Quantum Electrodynamic

VOLUME 517 NUMBER 21 PHYSICAL REVIEW LETTERS 21 NOVEMBER 198$

60

50

40

30— (a)

50

40

30— (b)

8Vl

ob 00

b

I.05—

1.00

0.95(c)

0.90-0.75 0.750

cosaFIG. 1, The differential cross sections for (a) Reac-

tion (2) and (b) Beaction (3), normalized to the QEDprediction. (c) The ratio of the differential cross sec-tion for Reaction (1) to the g(~3) QED prediction. Theexperimental cross section in (c) is normalized to theelectroweak cross section taken from a simultaneousfit to all three reactions. In (a)—(c) the dashed linecorresponds to the O(0. ) QED cross section and thesolid line to the fitted cross section.

es due to the limited azimuthal coverage of theLA system, event acceptances at cos0 =0.0 are77%%u~, 80%%ug, and 60%%uo for Reactions (1), (2), and (3),respectively, and fall to 73%, 77%, and 52/0 atIcosa' =0.6.The acceptance- corrected, background- sub-

tracted, differential cross sections are shownin Figs. 1(a)-1(c) and are compared to the O(n')QED prediction. Third-order QED processes cancreate angular asymmetries and normalizationchanges which are comparable to the weak-inter-action effects. Asymmetries in the angular dis-tribution arise from the interference between dia-grams of opposite charge-conjugation parity. 'The QED prediction is derived from a MonteCarlo calculation" that includes contributionsfrom initial- and final-state radiation, vertexcorrections, and leptonic and hadronic vacuumpolarization. Within the experimental acceptancethe QED forward-backward asymmetry is calcu-lated to be A„=1.0% in Reaction (2) and A,o=0.5/o in Reaction (3).Systematic errors in the relative normalization

of Reactions (2) and (3) to Reaction (1) arise from

three major sources: (1) errors in the back-ground estimates, 0.9% for e'e —g'y, and 1.7'%for e "e —~'7, (2) errors in detector simula-tion, 0.6/o in each process, and (3) uncertainty inthe decay modes of the tau lepton. The error(1.8/o) due to uncertainties in tau branching frac-tions is determined by folding the sensitivity toindividual decay modes with the errors in themeasurements of the branching fractions. ' Wehave neglected higher-order corrections to theO(n') QED cross section. These contributions(= 1~1%) are estimated from a leading-log calcu-lation" and are included in the systematic errorestimate. Uncertainties in the hadronic vacuum-polarization corrections introduce an additionalerror in the relative normalizations of roughly0.5/o. Combining the errors from all sourceslisted in quadrature, we find a 1.6%%uo normaliza-tion error on the ratio c»/o'„and a 2.8/o erroron the ratio cr«/c„.Reactions (1)-(3) are fitted by the O(a') QED

cross section plus the weak contributions. ' Frommaximum likelihood fits to the corrected cos0distributions, setting g„'= 0, me find g, 'g, &

=0.32 + 0.07+ 0.02 and g,'g, = 0.19+ 0.09+ 0.02,where the first error is statistical and the secondsystematic. ' The fits yield acceptance-, back-ground-, and QED-corrected forward-backwardelectroweak asymmetries, extrapolated to thefull cosG interval, of A„"" =—7.1/o+ 1.7/o andA, """=—4.2/o+ 2.0% where the standard modelpredicts" A " =—5.7%%uo. The fitted cross sec-tions are superimposed on the data in Figs. 1(a)and 1(b). The p, -~ universality of the axial-vectorcurrent is tested, and we find g, '/g, "=0.6+ 0.3.If we assume e -p -7 universality, then g,' =0.27+ 0.06+ 0.02 from Reactions (2) and (3). From asimultaneous fit to all three reactions which takesinto account the angular distributions and relativecross sections, we find g, ' =0.23 + 0.05 + 0.02 andg„' =0.03+ 0.03+ 0.03 where systematic errorshave been explicitly included in the fit, and thefitted cross section is superimposed on the datain Fig. 1(c). These results are in good agree-ment with the standard model and with other ex-periments. '" The probability that QED alonewould have led to the results presented here is6x10 '. The weak couplings above have beencomputed in the limit ~,—~. Alternatively, thevalues quoted above may be considered to meas-ure the product g'[M, '/(M, ' —s)]. Also g, ' andg„' only correspond to the couplings of the lowest-order &' exchange diagram. Inclusion of themass dependence and radiative corrections" to

1943

Figure 3.9: s dσd cos θ vs cos θ for e+e− → µ+µ− in (a), e+e− → τ+τ− in (b). In (c) the ratio of dσ

d cos θwith the expected QED contribution for the reaction e+e− → e+e−. Dashed line: QED contribution [10]

e−

γ

γ

e−

e−

γ

γ

e−

Figure 3.10: The 2 lowest order compton diagrams

the total amplitude is the sum of the two. Following then Feynman rules 3.1, the amplitude ofthe first diagram is:

iM1 = up′(−iqγµ)ε∗µk′ i/p+ /k +m

(p+ k)2 −m2(−iqγν)εν kup

There is no ambiguity in the position of the different terms in the equation: since the amplitudeis a simple complex number, the adjoint spinor has to be on the left, and hence the second vertex

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Compton scattering: e−γ → e−γ 89

terms are on the left. Similarly, the second amplitude is:

iM2 = up′(−iqγµ)εµk i/p− /k′ +m

(p− k′)2 −m2(−iqγν)ε∗ν k′up

We can simplify a bit these expressions by using the fact that incoming/outgoing particles areon shell so that we can use: p2 = p′2 = m2 and k2 = k′2 = 0. Hence, the amplitudes becomesusing q = −|e|:

M1 = −|e|2ε∗µk′εν k up′γµ/p+/k+m

2p.k γνup

M2 = |e|2ε∗µk′εν k up′γν/p−/k′+m

2p.k′ γµup

where we have changed in M2: µ↔ ν. Now, noting that /pγν = γρpργν = pρ(−γνγρ + 2gρν) =

−γν/p+ 2pν and similarly /pγµ = −γµ/p+ 2pµ, it comes for the total amplitude:

M = −|e|2ε∗µk′εν k up′(γµ−γν/p+2pν+/kγν+γνm

2p.k − γν −γµ/p+2pµ−/kγµ+γµm

2p.k′

)up

= −|e|2ε∗µk′εν k up′(γµ 2pν+/kγν

2p.k − γν 2pµ−/kγµ2p.k′

)up

= −|e|2ε∗µk′εν k up′(γµ 2pν+/kγν

s−m2 + γν 2pµ−/kγµu−m2

)up

where the identity (/p − m)up = 0 has been used in the second line for on-shell fermions ands = (p+ k)2 = m2 + 2p.k ⇒ p.k = (s−m2)/2 and p.k′ = −(u−m2)/2 has been used in the lastline.We wish to calculate the un-polarized cross-section. Thus we have to average over the 2 trans-verse polarizations of the incoming real photon, and the 2 spins of the electron. And we haveto sum over the final polarization states. Hence:

|M|2 =1

2

λ=1,2

1

2

s

λ′=1,2

s′|M|2 = |M1|2 + |M2|2 +M1M∗2 +

(M1M∗2

)∗

with:

|M1|2 = |e|44(s−m2)2

∑[ε∗µk′εν k up′γ

µ(2pν + /kγν)up

] [ε∗ρ k′εη k up′γ

ρ(2pη + /kγη)up

]∗

|M2|2 = |e|44(u−m2)2

∑[ε∗µk′εν k up′γ

ν(2pµ + /kγµ)up

] [ε∗ρ k′εη k up′γ

η(2pρ + /kγρ)up

]∗

M1M∗2 = |e|44(s−m2)2(u−m2)2

∑[ε∗µk′εν k up′γ

µ(2pν + /kγν)up

] [ε∗ρ k′εη k up′γ

η(2pρ + /kγρ)up

]∗

We are just going to detail the first calculation. No need to calculate |M2|2 since the secondgraph is just the u-channel of the first: replacing s↔ u will give the answer. For the interferenceterm, there is no other solution than doing the calculation (which is very similar to the firstterm).

|M1|2 =|e|4

4(s−m2)2

λ

εν kε∗η k

λ′ε∗µk′ερ k′

s,s′

[up′γ

µ(2pν + /kγν)up] [up′γ

ρ(2pη + /kγη)up]∗

According to 3.63,∑

λ εν kε∗η k = −gνη and

∑λ′ ε∗µk′ερ k′ = −gµρ and:

[up′γ

ρ(2pη + /kγη)up]∗

= u†p(2pη + γη†kαγα†)γρ†γ0†up′ = u†p(2pη + γ0γηγ0kαγ0γαγ0)γ0γργ0γ0up′

= u†p(2pη + γ0γη/kγ0)γ0γρup′ = up(2pη + γη/k)γρup′

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90 A brief view of Quantum Electrodynamic

Hence:

|M1|2 = |e|44(s−m2)2 gνηgµρ

∑s,s′[up′γ

µ(2pν + /kγν)up] [up(2p

η + γη/k)γρup′]

= |e|44(s−m2)2

∑s,s′[up′γ

µ(2pν + /kγν)up] [up(2pν + γν/k)γµup′

]

= |e|44(s−m2)2 Tr

((/p′ +m)γµ(2pν + /kγν)(/p+m)(2pν + γν/k)γµ

)

= |e|44(s−m2)2 Tr

(/p′γµ(2pν + /kγν)/p(2pν + γν/k)γµ

)

+m2Tr (γµ(2pν + /kγν)(2pν + γν/k)γµ)

(using 3.55)

Let us consider the first trace:

Tr(/p′γµ(2pν + /kγν)/p(2pν + γν/k)γµ

)= 4p2Tr

(/p′γµ/pγµ

)+ 2Tr

(/p′γµ/k/p2γµ

)+ 2Tr

(/p′γµ/p2/kγµ

)

+Tr(/p′γµ/kγν/pγν/kγµ

)

= 4m2(Tr(/p′γµ/pγµ

)+ Tr

(/p′γµ/kγµ

))+ Tr

(/p′γµ/kγν/pγν/kγµ

)

= 4m2(−2Tr

(/p′γν

)(pν + kν)

)− 2pρTr

(/p′γµ/kγρ/kγµ

)

= −8m2 (4p′ν(pν + kν)) + 4pρTr(/p′/kγρ/k

)

= −32m2(p′.p+ p′.k) + 4Tr(/p′/k/p/k

)

= −32m2(p′.p+ p′.k) + 16(p′.k p.k − p′.p k2 + p′.k k.p)= 32(−m2p′.p+ p′.k(p.k −m2))

where we have used successively the properties /p2 = p2 = m2, (3.59), (3.53), (3.57) and k2 = 0.Using a similar approach, one finds for the second trace:

Tr (γµ(2pν + /kγν)(2pν + γν/k)γµ) = 4p2Tr(γµγµ) + 2Tr(γµ/p/kγµ) + 2Tr(γµ/k/pγµ) + Tr(γµ/kγνγν/kγµ)

= 16m2Tr( l1) + 8p.kTr( l1) + 8k.pTr( l1) + 4Tr(γµ/k2γµ)

= 64m2 + 64p.k

and thus, injecting the results of the two traces:

|M1|2 =4|e|4

(s−m2)2

(−2m2p′.p+ 2p′.k p.k − 2m2p′.k + 4m4 + 4m2p.k

)

Now expressing the expression as function of the 2 Mandelstam variables s and u (rememberingthat s+ t+ u = 2m2):

s = (p+ k)2 = m2 + 2p.k ⇒ p.k = s−m2

2t = (p′ − p)2 = 2m2 − 2p′.p⇒ p′.p = m2 − t

2 = s+u2

u = (k′ − p)2 = (k − p′)2 = m2 − 2p′.k ⇒ p′.k = m2−u2

It finally comes:

|M1|2 = 4|e|4[

2m4

(s−m2)2+

m2

s−m2− 1

2

u−m2

s−m2

]

The expression for |M2|2 is obtained by swapping s and u:

|M2|2 = 4|e|4[

2m4

(u−m2)2+

m2

u−m2− 1

2

s−m2

u−m2

]

and for the interference (after a similar boring calculation):

M1M∗2 = 2|e|4m2

[1

u−m2+

1

s−m2+

4m2

(s−m2)(u−m2)

]

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Compton scattering: e−γ → e−γ 91

so that finally |M|2 = |M1|2 + |M2|2 +M1M∗2 + (M1M∗2)∗ reads:

|M|2 = −2|e|4[u−m2

s−m2+s−m2

u−m2− 4

(m2

s−m2+

m2

u−m2+

(m2

s−m2+

m2

u−m2

)2)]

Let us denote p = (m,~0), k = (ω,~k), p′ = (E′, ~p′), k′ = (ω′,~k′) and θ the angle between ~k and~k′. We notice that:

u = (p− k′)2 = m2 − 2mω′ ⇒ u−m2 = −2mω′

s = (p+ k)2 ⇒ s−m2 = 2mωm2 = p′2 = (p+ k − k′)2 = m2 − 2k.k′ + 2p.k − 2p.k′ = m2 − 2ωω′(1− cosθ) + 2mω − 2mω′

⇒ 1ω′ − 1

ω = 1m(1− cos θ)

and hence:

|M|2 = 2|e|4[ω′ω + ω

ω′ + 4(m2ω − m

2ω′ + ( m2ω − m2ω′ )

2)]

= 2|e|4[ω′ω + ω

ω′ + 2(−(1− cos θ) + 1

2(1− cos θ)2)]

= 2|e|4[ω′ω + ω

ω′ − sin2 θ]

We can then express the cross-section using formula 1.76:

dσ = 1

4√

(p.k)2−0(2π)4δ(4)(p′ + k′ − p− k)|M|2 d3~p′

(2π)32E′d3 ~k′

(2π)32ω′

= 18mω(2π)2 δ

(4)(p′ + k′ − p− k)|e|4[ω′ω + ω

ω′ − sin2 θ]

d3~p′E′

d3 ~k′ω′ ←

∫d3~p′

= 18mω(2π)2 δ(E

′ + ω′ −m− ω)|e|4[ω′ω + ω

ω′ − sin2 θ]

ω′2dω′dΩE′ω′ ←

∫dφ

= 18mω2π δ(E

′ + ω′ −m− ω)|e|4[ω′ω + ω

ω′ − sin2 θ]

ω′dω′d(cos θ)E′

But:

E′ =√m2 + |~p′|2 =

√m2 + (~k − ~k′)2 =

√m2 + ω2 + ω′2 − 2ωω′ cos θ

And thus:

dσd(cos θ) = 1

16mωπ |e|4[ω′ω + ω

ω′ − sin2 θ]

ω′dω′E′

δ(√m2 + ω2 + ω′2 − 2ωω′ cos θ + ω′ −m− ω)

= 116mωπ |e|4

[ω′ω + ω

ω′ − sin2 θ]

ω′E′

12ω′−2ω cos θ

2E′ +1

= 116mωπ |e|4

[ω′ω + ω

ω′ − sin2 θ]

ω′ω′−ω cos θ+E′ ← E′ = m+ ω − ω′

= 116mωπ |e|4

[ω′ω + ω

ω′ − sin2 θ]

ω′m+ω(1−cos θ) ← 1− cos θ = m( 1

ω′ − 1ω )

= 116m2ωπ

|e|4[ω′ω + ω

ω′ − sin2 θ]

ω′2ω

where we used the delta property∫ +∞−∞ δ(g(x)) dx =

∑i

1

| ∂g∂x (xi)| to integrate over ω′. Inserting

e2 = 4πα, we obtain the formula:

d(cos θ)=πα2

m2

[ω′

ω+ω

ω′− sin2 θ

] (ω′

ω

)2

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92 A brief view of Quantum Electrodynamic

which is known as the Klein-Nishina formula. At low energy, when ω → 0 (and thus ωω′ =

1 + ωm(1− cos θ)→ 1)

d(cos θ)' πα2

m2

[2− sin2 θ

]=πα2

m2

[1 + cos2 θ

]⇒ σ =

∫ 1

−1

πα2

m2

[1 + cos2 θ

]=

8πα2

3m2

In this expression, we recognize the classical radius of the electron16 re = α/m =√

3σ/(8π) '3 10−13 cm (with σ ' 6 10−25 cm2).

3.6 Few words about the renormalization

Let us consider again the e−µ− scattering 3.6:

e− (p1) µ−(p2)→ e− (p′1) µ−(p′2)

for which the amplitude is:

iM1 = up′1i|e|γµup1

−igµνq2

up′2i|e|γνup2 = 4πα up′1γ

µup1

igµνq2

up′2γνup2

with α = |e|2/4π is the usual electromagnetic fine structure constant. Suppose you want tomeasure α with this process. Let us imagine, we have a beam of e− and µ− and we detect thescattered e− and µ−. By playing with the beam energy and the scattering angle, we can countthe number of recorded events and so measure the cross-section at a given momentum transferq2. From the measurement, we would then get the value of αmeas = α.

Now, there are higher order corrections to this process. One is shown on figure 3.11. The

µ−

e−

µ−

e−

Figure 3.11: A virtual photon fluctuating in e−e+ pair.

virtual photon splits into a pair of virtual electron-positron (the fermion loop). If the photonhas a momentum q, then one member of the pair carries a momentum p and the other q − p,so that the energy-momentum is conserved. however, there is no constraint on the value of p!Therefore, all possible values have to be taken into account. The amplitude of this diagram canbe calculated but it requires additional rules with respect to the one given in section 3.1, namelya factor −1 for the closed fermion loop17, and the trace of the associated photon vertex must beused. Justification for the trace can be found in [8, p. 120], but a naive “explanation” is that,

16The classical radius of the electron is the size the electron that is determined assuming that only the electro-static potential energy contributes to its mass. Namely, in natural unit: E = m = |e|2/(4πre)⇒ re = α/m.

17The electron and positron in the loop can be interchanged, thus the −1.

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Few words about the renormalization 93

basically we have a γ → γ process which cannot depend on spinor indices. The amplitude isthen given by:

iM2 = up′1i|e|γµup1

−igµµ′q2

∫ d4p(2π)4 (−1)Tr

[i|e|γµ′i /p+m

p2−m i|e|γν′i /p−/q+m(p−q)2−m

]−igν′νq2

×

up′2i|e|γνup2

= up′1i|e|γµup1

−1q4

∫ d4p(2π)4 (−1)(i|e|)2Tr

[γµi

/p+m

p2−m γνi/p−/q+m

(p−q)2−m

up′2i|e|γνup2

Adding the 2 amplitudes, we finally have:

iM = up′1i|e|γµup1

−igµνq2 +

−iΠ[2]µν(q2)q4

up′2i|e|γ

νup2

M = up′1 |e|γµup1

gµνq2 +

Π[2]µν(q2)q4

up′2 |e|γ

νup2

where:

iΠ[2]µν(q2) =

∫d4p

(2π)4(−1)(i|e|)2Tr

[γµi

/p+m

p2 −m γνi/p− /q +m

(p− q)2 −m

]

Hence, we see that adding this contribution can be interpreted as a modification of the initialphoton propagator as:

−igµνq2

→ −igµνq2

+−iΠ[2]

µν(q2)

q4

We can consider this new photon propagator in the graph calculation, but there is a problemsince the integral in Πµν diverges! More precisely, one can show that Πµν can be reduced to theonly contributing term:

Π[2]µν(q2) = −q2gµν

α

([∫ ∞

m2

dp2

p2

]− f(q2)

)+ · · ·+O(α)

with f(q2) a finite contribution function given by:

f(q2) = 6

∫ 1

0dz z(1− z) log

(1− q2

m2z(1− z)

)(3.69)

Clearly∫∞m2

dp2

p2 diverges as a logarithm. Let us introduce a cut-off Λ2, namely a maximum valuefor the upper bound of the integral. Is it shocking to truncate the integral this way? Well, anytheory must have its domain of validity. We can see Λ as a parametrization of our ignorance,

above which new physics must appear. Then, with this prescription, Π[2]µν becomes:

Π[2]µν(q2) = −q2gµν

α

(log

(Λ2

m2

)− f(q2)

)+ · · ·+O(α)

and hence, the amplitude becomes (rewritten in terms of α):

M = up′1γµup1

gµνq2

4πα

1− α

3πlog

(Λ2

m2

)+

α

3πf(q2)

up′2γ

νup2

Now in order to avoid confusion, we are going to rename the α above as α0:

M = up′1γµup1

gµνq2

4πα0

1− α0

3πlog

(Λ2

m2

)+α0

3πf(q2)

up′2γ

νup2

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94 A brief view of Quantum Electrodynamic

The label 0 refers to the fact that this is the α that is used in the case of 0 loop as at thebeginning of this section. α0 is called the “bare” constant. The equation now is how can weinterpret the measurement of the cross section we have made? The actual scattering amplitudeM is not supposed to depend on an arbitrary parameter Λ. If Λ is changed, we don’t wantthat our predicted cross-section (that we’re going to compare to the measured one) changes. Inother words, we would have to change α0 itself in order to compensate the change of Λ ! α0,the constant of the QED lagrangian is not a physical parameter! The key of the renormalizationidea is to express what we measure with physical (measured) parameters. Here, with our secondorder development, if our experiment is made at q2 = µ2, we would identify:

αmeas = α0

(1− α0

3πlog

(Λ2

m2

)+α0

3πf(µ2)

)= α0 −

α20

(log

(Λ2

m2

)− f(µ2)

)+O(α2

0) (3.70)

Our measurement was performed at q2 = µ2. To clearly point this out, I now use the nameαmeas(µ

2). Let’s revert the previous equation:

α0 = αmeas(µ2) +

α20

(log(

Λ2

m2

)− f(µ2)

)+O(α2

0)

= αmeas(µ2) + α2

meas(µ2)

(log(

Λ2

m2

)− f(µ2)

)+O(α2

meas(µ2))

The second equality is equivalent to the first one, to the second order used here. But what ifinstead of µ2, we would have chosen a arbitrary value q2? We would have:

αmeas(q2) = α0 − α2

03π

(log(

Λ2

m2

)− f(q2)

)+O(α2

0)

= αmeas(µ2) + α2

meas(µ2)

(log(

Λ2

m2

)− f(µ2)

)

−α2meas(µ

2)3π

(log(

Λ2

m2

)− f(q2)

)+O(α2

meas(µ2))

= αmeas(µ2) + α2

meas(µ2)

(f(q2)− f(µ2)

)+O(α2

meas(µ2))

The good news now, is that the arbitrary Λ doesn’t appear anymore! We can express any processat any energy if we have measured αmeas at a given reference point. This is the miracle of therenormalization. But it has a price to pay: what we called a coupling constant is not a constant!It depends on the energy scale. We manage to eliminate Λ at the second order of perturbation.The renormalization works if we can eliminate it at any order. In QED, it turns out to be thecase.

Coming back to f(q2), when the momentum transfer is large with respect to m (which is

the case in high energy physics experiments), by developing f(q2) at order O( m2

−q2 ), it is not

too difficult to show that: f(q2) → − log(m2

−q2

)− 5/3 = log

(−q2

e5/3m2

). Note that q2 < 0 with

scattering processes. We can define the positive quantity Q2 = −q2. In this approximation,αmeas(Q

2) that we now simply denote by α(Q2) becomes18:

α(Q2) = α(µ2)

[1 +

α(µ2)

3πlog

(Q2

µ2

)]+O(α2)

Now, this result is valid at order α2. We can go at higher order by adding more loops. Thedependency of α with Q2 then becomes:

α(Q2) =α(µ2)

1− α(µ2)3π log

(Q2

µ2

) (3.71)

18µ2 is now understood as a positive quantity

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A major test of QED: g-2 95

Let us interpret this formula valid only in the regime Q2 m2: it gives the value of thefine structure constant at a given energy, if it has been measured at another energy scale. Itsvariation is only logarithmic with the scale. In Coulomb scattering, at Q2 ' 0, α = 1/137 whileat the Z0 pole, it is about 1/128. The variation of the charge (or α) as function of the scale is areal physical effect. It is due to the vacuum polarization. Imagine you want to test the charge ofan electron by approaching a probe-charge. In the vicinity of the electron, virtual charged pairscan be created for a short amount of time ∆t ∼ ~/mc2. They can spread apart at a maximumdistance c∆t, creating a dipole. The virtual positrons tend to be closer to the electron thanthe virtual electrons. If the probe-charge doesn’t approach enough (ie d c∆t), it will “see”mainly the surrounding clouds of positron, yielding to a measurement underestimating the realcharge of the electron. In other words, at small Q2 (ie high distance), the charge seems smaller.This phenomena is usually called a “screening”. On the contrary, at high Q2, the probe chargepenetrates the virtual cloud and can see the whole electron charge.19

In summary, the renormalization is a procedure that allows to absorb all infinite quantitiesby a redefinition of a finite number of parameters. In QED, they are the charge of the electron(or equivalently the fine structure constant), the mass of the electron (yes, the mass!) and thenormalization of the fields (thus the name renormalization).

3.7 A major test of QED: g-2

There are many tests that validate QED, both at high energy (in accelerator) and low energy.One of the most impressing test is probably the prediction of the gyromagnetic ratio g whichenters in the magnetic moment of the electron.

3.7.1 Prediction with Dirac’s equation

Non relativistic prediction without spin: Let us recall first, the classical formula forthe magnetic moment of charged particle moving in a circular orbit. Classically, the magneticmoment is defined as:

µ = current× area

For a circular orbit or radius r, the area is πr2. And the current is the charge per unit time,namely the charge times the frequency of rotation (the velocity divided by the circumference ofthe orbit). Hence:

µ = qv

2πr× πr2 = q

v

2r

Expressing as function of the angular momentum ~L = ~r × ~p = ~r ×m~v, we have:

~µ = gq

2m~L g = 1

g = 1 in the classical case is called the Lande g-factor or gyromagnetic ratio. Electrons placedin a magnetic field ~B would then have an additional potential energy due to the field:

EB = −~µ. ~B = −g q

2m~L. ~B

19Well, it’s not true: virtual pairs of higher mass particles can be created. There is always a virtual cloud seenby the probe.

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96 A brief view of Quantum Electrodynamic

which can be confirmed by considering the non-relativistic hamiltonian appearing in the Schrodingerequation is

H =p2

2m

In the presence of a magnetic field, we make as usual the substitution p→ p− q ~A, and then:

H =p2

2m− q

m~A.p+

q2A2

2m' p2

2m− q

m~A.p

where the term in q2 is neglected. The additional energy is then given by:

EB = − q

m~A.p

But, ~B = ~∇× ~A and ~A can be chosen satisfying the gauge ~∇. ~A = 0. Thus, ~A = 12B×~r and EB

becomes:

EB = − q

2m( ~B × ~r).p = − q

2m(~r × p). ~B = − q

2m~L. ~B = −~µ. ~B

where we have applied a scalar triple product property. We finally find the expected expressionfor ~µ. Since electrons have an intrinsic angular momentum, the spin ~S, we would expect anadditional energy:

EB = −g q

2m~S. ~B with g = 1

However, at the time of Dirac, the measurements for g excluded the classical value 1, and wascompatible with 2.

Non relativistic prediction with spin: we can follow the same procedure, starting fromDirac’s equation instead of Schrodinger one. Hence, starting from the momentum space equation2.30:

(/p−m)u = (γµpµ −m)u = 0pµ→pµ−qAµ−−−−−−−−→ (γµpµ − qγµAµ −m)u = 0

(γ0E − qγ0φ− ~γ.~p+ q~γ. ~A−m)u = 0

In terms of 2-components, this equation reads:

((E − qφ−m) −~σ.~p+ q~σ. ~A

~σ.~p− q~σ. ~A −(E − qφ+m)

)(uaub

)= 0

(~σ.~p− q~σ. ~A)ub = (E − qφ−m)ua(~σ.~p− q~σ. ~A)ua = (E − qφ+m)ub

Now, to be able to easily identify the magnetic momentum as in the case of the Schrodingerequation, we wish to find a relation satisfied by the spinors in the non-relativistic limit of theDirac’s equation. In this limit, E ' m, but E−m = T and T m where T is the kinetic energy.For usual electromagnetic field, the field energy is negligible with respect to mc2. Therefore,m qφ. and thus:

(~σ.~p− q~σ. ~A)ub = (T − qφ)ua(~σ.~p− q~σ. ~A)ua ' 2m ub

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Higher order corrections 97

So that by multiplying the first equation by 2m and inserting the second:

(~σ.~p− q~σ. ~A)(~σ.~p− q~σ. ~A)ua = 2m(T − qφ)ua[(~σ.~p)2 + q2(~σ. ~A)2 − q

((~σ.~p)(~σ. ~A) + (~σ. ~A)(~σ.~p)

)]ua = 2m(T − qφ)ua

(1)[|~p|2 + q2| ~A|2 − q

(~p. ~A+ i~σ.(~p× ~A) + ~A.~p+ i~σ.( ~A× ~p)

)]ua = 2m(T − qφ)ua[

(~p− q ~A)2 − iq~σ.(~p× ~A+ ~A× ~p

)]ua = 2m(T − qφ)ua

where in (1), we have used the equality (~σ.~a)(~σ.~b) = ~a.~b+ i~σ.(~a×~b). Going back to the languageof wave functions (and thus replacing ua by ψ = uae

−ip.x and ~p by −i~∇), we have:

(−i~∇−q ~A)2

2m ψ − q2m~σ.

(~∇× ( ~Aψ) + ~A× ~∇ψ

)= (T − qφ)ψ

(−i~∇−q ~A)2

2m ψ − q2m~σ.

(~∇× ~A

)ψ = (T − qφ)ψ[

(−i~∇−q ~A)2

2m − q2m~σ.

~B + qφ]ψ = T ψ

The quantity in the bracket:

HE'm =(~p− q ~A)2

2m− q

2m~σ. ~B + qφ

corresponds to the non-relativistic hamiltonian of the Dirac’s equation. Recalling that the spinoperator for 2-components spinors is ~S = 1

2~σ, we identify:

~µ =q

2m~σ = 2

q

2m~S

meaning that:g = 2

Hence, the non-relativistic limit of the Dirac’s equation does imply a gyromagnetic ratio of 2.It was one of the successes of Dirac’s prediction.

3.7.2 Higher order corrections

Consider the electron charge current involved in the basic vertex of the left diagram of figure3.12. It can be decomposed in 2 parts (as known as Gordon decomposition):

jµ = −eψfγµψi= − e

2 [ufγµui + ufγ

µui] e−i(pi−pf ).x ← ψ = ue−ip.x

= − e2m [ufpfνγ

νγµui + ufγµγνpiνui] e

−i(pi−pf ).x ← uf (/pf −m) = 0 , (/pi −m)ui = 0

= − e2m uf [pfν (2gµν − γµγν) + γµγνpiν ]ui e

−i(pi−pf ).x ← γµγν + γνγµ = 2gµν

= − e2m uf

[2pµf + γµγν(piν − pfν)

]ui e

−i(pi−pf ).x

Now, using the matrix:

σµν =i

2(γµγν − γνγµ)⇒ γµγν = gµν − iσµν

so that we get:

jµ = − e2m uf

[2pµf + gµν(piν − pfν)− iσµν(piν − pfν)

]ui e

−i(pi−pf ).x

=[− e

2m uf (pµf + pµi )ui + iufe

2mσµν(piν − pfν)ui

]e−i(pi−pf ).x

(3.72)

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98 A brief view of Quantum Electrodynamic

Referring to the Feynman rules of table 3.1, we see that the first term in the bracket − e2m uf (pµf +

pµi )ui is analogous to the interaction of a charged spin 0 particle. We then expect the secondterm:

jµs = iufe

2mσµν(piν − pfν)ui e

−i(pi−pf ).x

to be related to the interaction involving the spin of the electron. To check this assumption,consider, to the first order of perturbation, the amplitude given by the interaction of such currentwith an electromagnetic field Aµ:

S[1] = −i∫d4x jµsAµ =

∫d4x uf

e

2mσµν(piν − pfν)ui e

−i(pi−pf ).xAµ

Let us consider a static field (coulomb scattering) for simplification. Then Aµ does not dependon time. We can perform the integration over the time:

S[1] = 2π δ(Ei − Ef )

∫d3x uf

e

2mσµν(piν − pfν)ui e

−i(pi−pf ).xAµ

Because of the Dirac-delta function, the time-like component of the 4-momenta in the integralhas to be equal, and hence cannot contribute to the amplitude. Only the spatial-like componentsmatter, and after re-arrangement, it reads:

S[1] = 2πi δ(Ei − Ef )

∫d3x uf

e

2mσµjui ∂j

(e−i(pi−pf ).x

)Aµ

In the expression above, only the exponential and Aµ depend on space coordinates (the spinorsdepend only on 4-momenta). Hence, after integration by part (the field Aµ vanishing at infinity):

S[1] = −2πi δ(Ei − Ef )

∫d3x uf

e

2mσµjui e

−i(pi−pf ).x ∂jAµ

!"Figure 3.12: Left: simplest QED vertex. Right: a correction to the vertex.

As in the previous section, in order to see clearly the spin contribution, we are going to considerthe low energy limit, where the mass dominates over the 3-momentum. Then, only the uppercomponent of the spinors matters since in the lower component, there is the term ~σ.~p/(E +m)(see formula 2.40). In other words, spinors simplify to:

u =

(ua

~σ.~pE+mua

)→(ua0

)

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Higher order corrections 99

In these conditions, the term ufσµjuiAµ = ufσ

0juiA0 + ufσkjuiAk ≈ ufσ

kjuiAk. Indeed,according to the Dirac’s representation of the γ matrices:

σ0j = i2

[(l1 0000 −l1

)(00 σj

−σj 00

)−(

00 σj

−σj 00

)(l1 0000 −l1

)]= i

(00 σj

σj 00

)

and hence ufσ0jui ≈ 0. Thus, only ufσ

kjui gives a significant contribution to the amplitude:

S[1] = −2πi δ(Ei − Ef )

∫d3x uf

e

2mσkjui e

−i(pi−pf ).x ∂jAk

Let us express σkj :

σkj = i2

[(00 σk

−σk 00

)(00 σj

−σj 00

)−(

00 σj

−σj 00

)(00 σk

−σk 00

)]

= i2

(−(σkσj − σjσk) 00

00 −(σkσj − σjσk)

)

= i2

(−2i

∑l εkjlσ

l 0000 −2i

∑l εkjlσ

l

)

=∑

l εkjl

(σl 0000 σl

)

where εkjl is the usual antisymmetric tensor (εkjl = 1 for cyclic permutation of 123, = −1 foranticyclic permutation, 0 otherwise). Hence:

ufσkjui =

l

εkjl(u†af, 0)

(l1 0000 − l1

)(σl 0000 σl

)(uai0

)=∑

l

εkjl u†afσluai

Now, noticing for example that:

ufσ1jui∂j = ufσ

11∂1ui + ufσ12∂2ui + ufσ

13∂3ui=∑

l ε11l u†afσ

l∂1uai +∑

l ε12l u†afσ

l∂2uai +∑

l ε13l u†afσ

l∂3uai= u†af

(∑l ε11lσ

l∂1 +∑

l ε12lσl∂2 +

∑l ε13lσ

l∂3

)uai

= u†af(σ3∂2 − σ2∂3

)uai

= −u†af[~σ × ~∂

]1uai

and similarly for the other components. Thus, we conclude:

ufσkjui ∂jAk = −u†af

[~σ × ~∂

]kAk uai = −u†af

[~σ × ~∂

]. ~A uai = −u†af

[~∂ × ~A

].~σ uai

Hence, since ~B = ~∂ × ~A, we finally have:

S[1] = −2πi δ(Ei − Ef )

∫d3xu†af

(− e

2m

)~B.~σ uai e

−i(pi−pf ).x

where we identify the magnetic moment of the electron ~µ = − e2m~σ. Conclusion, as announced

at the beginning of this section, the current:

jµs = iufe

2mσµνqνui e

−i(pi−pf ).x

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100 A brief view of Quantum Electrodynamic

with q = pi − pf , yields naturally the gyromagnetic ratio.

Now, consider the second diagram on the right hand side of figure 3.12. The computa-tion of the correction leads to a modification of the current20 in the low momentum transferapproximation [9, p. 160] (dropping the exponential):

jµ = uf

− e

2m(pµf + pµi ) + i e2mσ

µνqν

ui

→ uf

−eγµ

[1 + α

3πq2

m2

(log( m

mγ− 3

8

)]+[α2π i

e2mσ

µνqν]ui

where mγ is a cut-off to avoid an infinite term due to the so called infra-red divergence. We canimmediately deduce that adding the 2 diagrams will yield a modification of the gyromagneticratio:

~µ = − e

2m~σ → − e

2m

(1 +

α

)~σ ⇒ ae =

g − 2

2=

α

where ae, the deviation with respect to the value g = 2, is called the anomalous magneticmoment. Numerically, we obtain ae ' 0.001 16, to be compared to the most precise experimentalvalue obtained so far [11]:

aexpe = 1 159 652 180.73 (0.28)× 10−12 (3.73)

the digits in parentheses denoting measurement uncertainty in the last two digits at one standarddeviation. Pretty close! Such experimental accuracy pushes very far the theory and requires acalculation at least at the 8th order (ie α4). The 10th order, which represents 12672 vertex-typeFeynman diagrams (!!), is already partially evaluated [12]. Hadronic (vacuum polarization),electroweak effects and small QED contributions from virtual muon and tau-lepton loops con-tributions have also to be taken into account. Fortunately, the diagrams are now evaluatednumerically and a recent computed value of ae is [12]:

athe = 1 159 652 181.13 (0.11)(0.37)(0.02)(0.77)× 10−12

where the first, second, third, and fourth uncertainties come from the calculated eighth-orderQED term, the crude tenth-order estimate, the hadronic and electroweak contributions, and thefine structure constant, respectively. Both theory and experiment are in very good agreementsince:

aexpe − athe = −0.40 (0.88)× 10−12

The largest source of uncertainty of athe is now the experimental value used for the fine structureconstant (obtained from Cesium or Rubidium atom experiments), and not anymore an uncer-tainty coming from the calculation itself! Therefore, it makes sense now to obtain α from thetheory and the measured value of ae instead of the contrary.

The accuracy of such computation of the g-factor is probably one of the most impressivetriumphs of the theory of quantum electrodynamics.

3.7.3 Measurement of the g-factor

We have already seen that an electron in a magnetic field gets extra contribution to its energywith a term −~µ. ~B = −g q

2m~S. ~B. A field on the Z-axis will yield −g −e2m~msBz = hνsms with

20after renormalization.

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Measurement of the g-factor 101

ms = ±12 and the spin frequency νs is related to the cyclotron frequency νc with νs = g

2νc and

νc = eBz2πm . The principle of the measurement is thus to measure both νc and νs or more precisely

νc and νa = νs − νc, the “anomaly” frequency since:

g

2=νsνc

= 1 +νs − νcνc

= 1 +νaνc

The experimental set-up [11] is based on a Penning-trap which suspends a single electronthanks to a strong magnetic field (5.36 T) along the z-axis and an electrostatic quadrupolepotential as shown on figure 3.13. The effect of the electric field is to confine the electronin a potential well in which it makes small vertical oscillation. The Penning trap is used toartificially bind the electron in an orbital state, as if it was an electron of an atom. Actually, theelectrostatic potential shifts the cyclotron frequency from νc to νc and thus νa to νa = νs − νc.The lowest energy level including the leading relativistic correction can be approximated by [11]:

En,ms = hνsms +

(n+

1

2

)hνc −

1

2hδ(n+

1

2+ms)

2

where the quantity δ is of the order δ ≈ 10−9νc. According to this formula giving the energylevels, the transitions |n,ms〉 = |1, 1

2〉 → |0, 12〉 → |1,−1

2〉 → |0,−12〉:

E1, 12− E0, 1

2= h(νc − 3

2δ)

E0, 12− E1,− 1

2= hνa

E1,− 12− E0,− 1

2= h(νc − 1

2δ)

allow to measure νc, δ and νa and hence g/2. It turns out that νc ≈ 150 GHz while νa ≈ 174MHz. Since νc ' νc is proportional to B, a high magnetic field is necessary to increase thespacing between the cyclotron energy levels. The cavity of the Penning-trap is maintained at a

Figure 3.13: Schema of a Penning trap [13]. The constant electric field (blue) is generated by aquadrupole (a and b). The superposed constant and homogeneous magnetic field (red) is generated by asurrounding cylinder magnet (c). A particle, indicated in red (here positive) is stored in between caps ofthe same polarity. The particle is trapped inside a vacuum chamber.

very low temperature (100 mK) to avoid transitions from the ground state to other states due

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102 A brief view of Quantum Electrodynamic

to blackbody photons. The electron is initially prepared in the state |0, 12〉. The spin-up state is

obtained by playing with the electrostatic potential (which induced an additional magnetic field).The higher cyclotron state are artificially excited by injecting microwaves into the trap cavity.Then, the measurement detects the anomaly transition |0, 1

2〉 → |1,−12〉 followed by the decay

to the ground state |1,−12〉 → |0,−1

2〉. Four measurements are performed at slightly differentcyclotron frequencies (by varying the B field) in order to estimate the frequency shift due to thecavity trap (the shape of the trap cavity modifies the density of states of the radiation modes offree space). They are presented on figure 3.14, from which the measured value of equation 3.73is deduced.

Figure 3.14: Four measurements of g/2 without (open) and with (filled) cavity-shift corrections. Thelight gray uncertainty band shows the average of the corrected data. From [11].

3.8 Exercises

Exercise 3.1 Draw the diagrams at the minimal order for the following reactions and writedown the corresponding probability amplitudes.

1. e+µ− → e+µ−. Use the 4-momenta: e+(k) µ−(p)→ e+(k′) µ−(p′).

2. e+µ+ → e+µ+. Use the 4-momenta: e+(k) µ+(p)→ e+(k′) µ+(p′).

3. e+e− → µ+µ−. Use the 4-momenta: e+(k) e−(p)→ µ+(k′) µ−(p′).

What about e+µ− → e−µ+?

Exercise 3.2 Draw all diagrams at the first order (in αe) for the reactions below. If there areseveral diagrams per reaction, precise whether you have to add or subtract the diagrams.

1. e−e− → e−e− (Moller scattering)

2. e+e− → e+e− (Bhabba scattering)

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Exercises 103

3. e−γ → e−γ (Compton scattering)

4. e−e+ → γγ (pair annihilation)

Exercise 3.3 Electron-positron annihilation into a muon pair, e−e+ → µ−µ+. We consideronly the ultra-relativistic regime and use the notation for the 4-momenta: e−(p)e+(k)→ µ−(p′)µ+(k′).

1. List the helicity combinations that contribute to the process.

2. For each of them, compute the probability amplitude M. (Use the formulas 2.59 for thehelicity states spinors).

3. In order to determine the average probability < |Mtot|2 > of the process, do you have toadd first the different amplitudes and square the result or do you have to add the individualsquared amplitudes? Justify your answer.

4. Using the Mandelstam variables, show that < |Mtot|2 > previously determined can be

written 2e4

s2(t2 + u2).

5. Redo the calculation of |Mtot|2 by using the spin summations and trace theorems. Checkthe consistency of your result.

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104 A brief view of Quantum Electrodynamic

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Chapter 4

From hadrons to partons

Few references:F. Halzen and D. Martin, “Quarks & Leptons”, Chapters 8 and 9, Wiley (1984)D.H. Perkins, “Introduction to High Energy Physics”, Chapter 7, Addison-Wesley (1982)

In this chapter, the notion of partons is introduced. The evidences of thesubstructure of the nucleon are given and the formalism of the deep inelasticscattering is exposed.

4.1 Electron-proton scattering

In this section, we wish to probe the internal structure of the proton. The proton is not anelementary particle (in first approximation made of quarks as we will see) which can be easilyrevealed by the measurement of its magnetic moment (M is the proton mass):

~µp = gp|e|

2M~S = gp~S µN

which gives in unit of Nuclear magneton µN = |e|2M (knowing that a proton has a spin 1/2):

|~µp| =gp2

= 2.79

Thus gp = 5.59 differs significantly from the value 2 expected for the gyromagnetic ratio of ele-mentary fermions (see previous chapter). Probing the proton structure is based on experimentssimilar in their spirit to the famous Rutherford experiment. In the proton case, one has touse a probe which will ease the interpretation of the scattering experiment. Such probe is forexample an electron which interacts only via the electromagnetic interaction with the proton1.Therefore, the reaction considered in this section is:

e−(k) + p(p)→ e−(k′) +X(p′)

where X is a proton if the scattering remains elastic or a set of multi-hadrons in the case ofa deep inelastic scattering (see later). The energy of the incident or outgoing electron will be

1We consider the energy of the electron not high enough, say below 20 GeV, so that weak interaction can beneglected

105

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106 From hadrons to partons

denoted respectively E and E′. The 4-momentum transfer is denoted by

q = k − k′ = p′ − p (4.1)

4.1.1 Elastic scattering with a point-like proton

As a starting point, let us consider the simple case where the proton is considered point-like andthe 4-momentum transfer of the scattering is low enough to yield an elastic scattering, ie X isa single proton. When the momentum transfer between the electron and the proton is small,and hence the wavelength large, the proton is seen as a point. Assuming the proton to be anelementary particle, we can follow the calculation made in section 3.5.2 for the electron-muonscattering yielding:

|M|2 = |e|4q4 L

µν(p)Lµν(e−) = 16π2α2

q4 Lµν(p)Lµν(e−)

dσ = 1

4√

(p.k)2−(mM)2(2π)4δ(4)(p′ + k′ − p− k)|M|2 d3~p′

(2π)32p′0d3 ~k′

(2π)32E′

where we used e2 = 4πα. We are interested here only in the unpolarized cross-section. Let usassume the proton in the target to be at rest and the electron to be moderately relativistic sothat:

k = (E,~k) (m = 0) , p = (M, 0)

(M is the proton mass). Writing d3~k′ = E′2dE′dΩ we have:

dσdE′dΩ = 1

4EM(2π)2 δ(4)(p′ + k′ − p− k)|M|2E′2

d3~p′2p′0

= 18M(2π)2 δ

(4)(p′ + k′ − p− k)|M|2E′Ed3~p′2p′0

Now, we are going to integrate over d4p′ instead of d3~p′ using the relation 1.68:

∫ +∞

−∞d4p′ δ(p′2 −M2)θ(p′0) =

∫ +∞

−∞

d3~p′

2p′0

and hence:

dσdE′dΩ =

∫d4p′ δ(p′2 −M2)θ(p′0) 1

8M(2π)2 δ(4)(p′ + k′ − p− k)|M|2E′E

= 18M(2π)2 δ((p+ q)2 −M2)|M|2E′E

= 18M(2π)2 δ(q

2 + 2p.q)|M|2E′E= 1

8M(2π)2 δ(q2 + 2M(E − E′))|M|2E′E

And finally, using a property of the δ-functions:

dE′dΩ=

1

16M2(2π)2δ(

q2

2M+ E − E′)|M|2E

E(4.2)

where we have to keep in mind that p′ = p+ k− k′ = p+ q and p′0 ≥ 0. Note that until now, itwas not necessary to know the dynamic of the process. Formula 4.2 is general regardless of theform of |M|2 as soon as the electron is moderately relativistic and the target proton at rest.

Now according to 3.64:

Lµν(p)Lµν(e−) = 8(k.p k′.p′ + k.p′ k′.p−M2k.k′

)

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Elastic scattering with a point-like proton 107

with:

k.p = EM

k′.p′ = k′.(k − k′ + p) = E′E(1− cos θ) + 0 + E′M = 2EE′ sin2 θ2 + E′M

k.p′ = k(k − k′ + p) = 0− EE′(1− cos θ) + EM = −2EE′ sin2 θ2 + EM

k′.p = E′Mk.k′ = 2EE′ sin2 θ

2

q2 = (k − k′)2 = −4EE′ sin2 θ2

Thus:

Lµν(p)Lµν(e−) = 8(2ME2E′ sin2 θ

2 +M2EE′ − 2MEE′2 sin2 θ2 +M2EE′ − 2M2EE′ sin2 θ

2

)

= 16MEE′(E sin2 θ

2 +M − E′ sin2 θ2 −M sin2 θ

2

)

= 16M2EE′(

cos2 θ2 + E−E′

M sin2 θ2

)

Putting everything together, we have:

dσdE′dΩ = α2

4M216E2E′2 sin4 θ2

δ( q2

2M + E − E′)16M2EE′(

cos2 θ2 + E−E′

M sin2 θ2

)E′E

= α2

4E2 sin4 θ2

δ( q2

2M + E − E′) cos2 θ2

(1 + E−E′

M tan2 θ2

)

Finally, replacing E − E′ by −q2/2M because of the δ:

dE′dΩ=

α2

4E2 sin4 θ2

δ(q2

2M+ E − E′) cos2 θ

2

(1− q2

2M2tan2 θ

2

)(4.3)

We can go a bit further, by integrating over E’, noticing that:

δ(q2

2M + E − E′)

= δ(−2EE′

M sin2 θ2 + E − E′

)= δ

(−(

2EM sin2 θ

2 + 1)E′ + E

)

= 11+ 2E

Msin2 θ

2

δ

(E′ − E

1+ 2EM

sin2 θ2

)

so that the integration over E′ gives:

dΩ=

α2

4E2 sin4 θ2

1

1 + 2EM sin2 θ

2

cos2 θ

2

(1− q2

2M2tan2 θ

2

)

Now since:

E − E′ = − q2

2M=

2EE′

Msin2 θ

2⇒ E = E′

(1 +

2E

Msin2 θ

2

)⇒ 1

1 + 2EM sin2 θ

2

=E′

E

we conclude:

dΩ=

α2

4E2 sin4 θ2

E′

Ecos2 θ

2

(1− q2

2M2tan2 θ

2

)=

(dσ

)

Mott

(1− q2

2M2tan2 θ

2

)(4.4)

The term: (dσ

)

Mott

=α2

4E2 sin4 θ2

E′

Ecos2 θ

2(4.5)

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108 From hadrons to partons

is the Mott cross-section, describing a scattering of spin 1/2 with spin 0: this is what wewould have obtained by considering that the proton is spinless (this can be checked by usingthe appropriate Feynman rules). Therefore, only the charge of the proton matters: the Mott

scattering is due to the charge. The additional term : − q2

2M2 tan2 θ2 is thus due to the spin of the

proton and hence to its magnetic moment. Now, recall that the tensor Lµν is defined in 3.50 as:

Lµν =1

2

r,r′

[uk′,r′γ

µuk,r]∗ [

uk′,r′γνuk,r

](4.6)

The terms in bracket being the current (the charge missing) that we have decomposed in 3.72as:

jµ = ufγµui = uf

(1

2M(pµf + pµi ) + i

1

2Mσµν(pfν − piν)

)ui

We showed in the previous chapter that the last term gives rise to the spin-spin interaction.

This is thus this term that is responsible for the factor q2

2M2 tan2 θ2 .

4.1.2 Elastic scattering with a non-point like proton: nucleon form-factor

Knowing that the proton is not point-like, how should we modify the scattering cross-section or,in other words, how the proton current must be modified assuming the interaction still proceedvia a single photon exchange? The answer is [9]:

ufγµui → uf

[F1(q2)γµ + i

κ

2MF2(q2)σµν(pfν − piν)

]ui (4.7)

where F1 and F2 are 2 form factors (scalar functions) depending on q2 = (pf − pi)2. We seethat κF2(q2) will give an extra contribution to the spin-spin interaction. Indeed:

uf

[F1(q2)γµ + i

κ

2MF2(q2)σµν(pfν − piν)

]ui =

uf

(F1(q2)

2M(pµf + pµi ) + i

F1(q2) + κF2(q2)

2Mσµν(pfν − piν)

)ui

In the limit q2 → 0, the proton must be seen as point-like, and hence F1(0) = F2(0) = 1. Thevalue of κ ' 1.79 explains why gp = 2(1 + κ) ' 5.59. In the case of the neutron, since it isneutral, one has F1(0) = 0 and F2(0) = 1 with κ = −1.91.

We can re-do the previous calculation of the cross-section. This time, the tensor Lµν (4.6)of the proton has to be replaced according to 4.7 by:

Lµν → Kµν = 12

∑r,r′

[up′,r′

(γµ(F1(q2) + κF2(q2)− pµ+p′µ

2M κF2(q2))up,r

]∗[up′,r′

(γµ(F1(q2) + κF2(q2)− pµ+p′µ

2M κF2(q2))up,r

]

where we used the Gordon decomposition 3.72 to replace uf i1

2M σµν(pfν − piν)ui by ufγ

µui −uf

12M (pµf + pµi )ui. The cross-section will then be similar to 4.2:

dE′dΩ=

α2

4M2q4δ(

q2

2M+ E − E′)KµνLµν(e−)

E′

E(4.8)

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Inelastic electron-proton scattering 109

The rest of the calculation is very similar to the previous case (but lengthy) and the final resultis:

dE′dΩ=

α2

4E2 sin4 θ2

δ(q2

2M+ E − E′) cos2 θ

2

(F2

1 −κ2q2

4M2F2

2 −q2

2M2(F1 + κF2)2 tan2 θ

2

)(4.9)

and after integration over E′:

dΩ=

(dσ

)

Mott

(F2

1 −κ2q2

4M2F2

2 −q2

2M2(F1 + κF2)2 tan2 θ

2

)(4.10)

This formula is known as the Rosenbluth cross-section. By measuring it with scattering experi-ment, the form factor F1,2(q2) can then be determined.

What is the physical significance of the form factors? To better appreciate, consider thesimple case of an electron scattered by a static Coulomb potential with a spatial extension (i.e.not point-like). Let us denote by ρ(~r) the charge density (

∫ρ(~r)d~r = 1) and Q the total charge.

The potential felt by the electron at ~r is thus :

V (~r) =

∫d3~r′

Qρ(~r′)

4π|~r − ~r′|

using natural units. The scattering of the electron by the potential yields the matrix amplitudeat first order of perturbation (treated classically):

Mfi = 〈ψf |V (~r)|ψi〉=∫d3~r e−i~pf .~r

∫d3~r′ Qρ(~r′)

4π|~r−~r′|

ei~pi.~r

=∫ ∫

d3~r d3~r′ ei~q.~r Qρ(~r′)4π|~r−~r′| ← ~q = ~pi − ~pf

=∫d3 ~R ei~q.

~R Q

4π|~R|∫d3~r′ ei~q.~r

′ρ(~r′) ← ~r = ~R+ ~r′

=Mpoint−likefi F (~q)

with F (~q) =∫d3~r′ ei~q.~r

′ρ(~r′). In this simple example, the form-factor F is interpreted as the

Fourier transform of the charge distribution. The cross-section will logically be:

dΩ=

(dσ

)

Point−like|F (~q)|2

Now, coming back to the nucleon case, it can be shown that the linear combinations, calledSachs form factors:

GE(q2) = F1(q2) +κq2

4M2F2(q2) , GM (q2) = F1(q2) + κF2(q2)

are related to the spatial distributions in the nucleon of the charge (GE) and magnetic moment(GM ) . Both distributions are consistent with a proton of size r ' 0.8 fm (see slides or [14,p. 288]).

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110 From hadrons to partons

Figure 4.1: Diagram of e−p deep inelastic scattering. From [6].

4.1.3 Inelastic electron-proton scattering

When the momentum transfer Q2 = −q2 increases, the proton can be excited into a higher massresonance such as e−p → e−∆+ which then decays in e−pπ0. We are in the inelastic regime.If Q2 is very large, the proton breaks up and many hadrons are produced: e−p → e−X, assymbolised by the figure 4.1. This is the deep inelastic scattering. The figure 4.2 shows thedifferential cross-section of e−p scattering as function of the invariant mass of the proton decayproducts, W . When W = mp = 938 MeV, we are in the elastic regime. The other peaks aredue to the resonance states of the excited proton (∆+, N+ etc). For W & 2 GeV, the largemultiparticles state yields a smooth spectrum.

Figure 4.2: Differencial cross-section of e−p scattering as function of the invariant mass of the protondecay products W . The electron energy is 10 GeV and the detector was placed at 6 with respect to theelectron beam axis. The elastic scattering peak at W = mp has been scaled down by a factor 8.5. From[15].

How can we write the inclusive cross-section e−(k)p(p) → e−(k′)X(p′) corresponding to thediagram 4.1? Since, the electron current remains untouched, we still have the electron tensorLµν(e−). The matrix amplitude being a Lorentz scalar, the Lµν(e−) has to be necessarilycontracted by another tensor Kµν(X) describing the decay of the proton in X. So that theunpolarized matrix element for a given Xi final state is:

|M|2ep→eXi =|e|4q4Lµν(e−)Kµν(Xi)

If the Xi states contains ni particles of 4-momentum pi1 · · · pini with p′ =∑ni

n pin, the cross-section

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Recapitulation 111

should be:

dσ =|e|4

4√

(p.k)2 − (mM)2q4(2π)4δ(4)(p′ + k′ − p− k)Lµν(e−)Kµν(Xi)

d3~k′

(2π)32E′

ni∏

n=1

d3 ~pin

(2π)32pin0

Since we are interested in the inclusive cross-section for any Xi final state, we have to sum overall possible Xi and integrate over the corresponding 4-momenta. Let us denote by

4πM Wµν =∑

i

∫· · ·∫

(2π)4δ(4)(p′ + k′ − p− k)Kµν(Xi)

ni∏

n=1

d3 ~pin

(2π)32pin0

where the factor 4πM is there to stick with the usual convention (M is still the proton mass).Thus:

dσ =|e|4

4√

(p.k)2 − (mM)2q4Lµν(e−) 4πM Wµν d3~k′

(2π)32E′

Neglecting the electron mass and calculating the cross-section in the proton rest-frame (p.k =ME), it comes:

dσ =α2

q4Lµν(e−)Wµν d

3~k′

EE′=α2

q4Lµν(e−)WµνE

EdE′dΩ (4.11)

where, as usual, we use e2 = 4πα. The tensor Wµν seems horribly complicated. However,since we are interesting in the inclusive cross-section, it must depend only on the initial proton4-momentum p and the 4-momentum transfer q (remember that Wµν is already the result of theintegration over the decay products of the proton). Other arguments like the necessary symmetryof the tensor and the conservation of the proton 4-current that implies qµWµν = qνWµν = 0 (seefor example [9]) leads to the structure:

Wµν(p, q) = W1(q2, ν)

(−gµν +

qµqν

q2

)+W2(q2, ν)

M2

(pµ +

2

)(pν +

2

)(4.12)

The functions W1,2 are called structure functions. Contrary to the elastic case where the equiv-alent W1,2 depends only on q2 (via F1,2(q2)), this time, the function W1 and W2 depends on 2scalar-Lorentz:

q2 or equivalently Q2 = −q2 > 0 and ν =p.q

M(4.13)

In the proton rest frame (so in the lab if the proton is a fixed target), the variable ν representsthe energy transfer from the electron to the hadronic state. Indeed: ν = M(E−E′)/M = E−E′.The rest of the calculation is very similar to the elastic case. Thus, if we inject 4.12 into 4.11,knowing 3.58, we finally get:

dE′dΩ=

α2

4E2 sin4 θ2

cos2 θ

2

[W2(q2, ν) + 2W1(q2, ν) tan2 θ

2

](4.14)

4.1.4 Recapitulation

Let us recap. Assuming a point-like proton, we found the elastic cross-section 4.3 (replacingE − E′ by ν):

dE′dΩ=

α2

4E2 sin4 θ2

δ(q2

2M+ ν) cos2 θ

2

(1− q2

2M2tan2 θ

2

)

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112 From hadrons to partons

And for non-point-like proton (eq 4.9):

dE′dΩ=

α2

4E2 sin4 θ2

δ(q2

2M+ν) cos2 θ

2

(F2

1 (q2)− κ2q2

4M2F2

2 (q2)− q2

2M2(F1(q2) + κF2(q2))2 tan2 θ

2

)

For deep inelastic scattering (eq 4.14):

dE′dΩ=

α2

4E2 sin4 θ2

cos2 θ

2

[W2(q2, ν) + 2W1(q2, ν) tan2 θ

2

]

We see manifestly a similar structure, where moving from deep inelastic scattering to elasticscattering, the structure functions would have to be replaced by:

2W1(q2, ν)elastic−−−−−→ − q2

2M2 (F1(q2) + κF2(q2))2δ( q2

2M + ν)point-like−−−−−−−→ − q2

2M2 δ(q2

2M + ν)

W2(q2)elastic−−−−−→ (F2

1 (q2)− κ2q2

4M2F22 (q2)δ( q2

2M + ν)point-like−−−−−−−→ δ( q2

2M + ν)

(4.15)

4.2 The parton model

4.2.1 Bjorken scaling

In the late sixties-early seventies, at Stanford Linear Accelerator Center (SLAC), several ex-periments led by J. Friedman, H. Kendall and R. Taylor have been of essential importance forthe development of the quark model thanks to their measurements of the structure functionsW1,2. They received the Nobel prize in 1990 for their discoveries. They measured the scatteringcross-section for various angles θ of the scattered electrons of a given energy E′. Both θ and E′

determines the values of Q2 = 4EE′ sin2 θ2 , ν = E − E′ and the mass of the hadronic system

W 2 = M2 −Q2 + 2Mν. The figure 4.3 shows one of these measurements. Firstly, one sees thatelastic cross-section falls down rapidly (∝ q−6) at high Q2, so that the inelastic cross-sectiondominates for Q2 & 1 GeV. Secondly, the Q2 dependance of the inelastic scattering is ratherweak, and become almost independent for high masses of the hadronic system. High masses ofthe hadronic system mean necessarily high energy transfer ν. These experimental facts corrobo-rated a prediction made by Bjorken (based on theoretical considerations concerning the currentalgebra) who claimed that at high Q2 and ν, the structure functions are only function of theirratio (thus the term scaling) via the dimensionless x-Bjorken variable:

x =Q2

2Mν(4.16)

Note that since Q2 > 0 ⇒ x > 0 and because of the conservation of the baryon number in theelectromagnetic interaction, the final state contains at least one baryon, and hence W ≥ M .Thus, W 2 = M2 −Q2 + 2Mν ≥M2 ⇒ x ≤ 1. The case x = 1 corresponds to elastic scatteringwhile for the inelastic scattering, 0 < x < 1.More, precisely, Bjorken scaling prediction was:

MW1(Q2, ν)Q2→∞−−−−→ν→∞

F1(x) , νW2(Q2, ν)Q2→∞−−−−→ν→∞

F2(x) (4.17)

The figure 4.4 confirms the validity of Bjorken prediction. It shows νW2 versus Q2 for a fixedx. The behaviour is compatible with no dependence with Q2 as predicted by Bjorken.

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The partons 113

lO-3

\ \

\ ,

\

\ - .ELASTIC \ SCATTERING

\ -\

\,

\ \

\ \

Id4 0 I 2 3 4 5 6 7

. -W=2 GeV x ---- W = 3 GeV

q2 (GeV/d2

Fig. 1 Figure 4.3: Differencial cross-section of e−p scattering normalized to the Mott cross-section as functionof the momentum transfer Q2 for various mass of the hadronic system. From [16].

Figure 4.4: The νW2 structure function measured as function of Q2 for a fixed x = 0.25 value [17].

4.2.2 The partons

How this scaling is interpreted? According to equations 4.15, we expect:

MW1(q2, ν)point-like p−−−−−−−−−→ 1

2−q2

2M δ( q2

2M + ν) = 12Q2

2M δ(ν −Q2

2M ) = 12Q2

2Mν δ(1−Q2

2Mν ) = x2 δ(1− x)

νW2(q2)point-like p−−−−−−−−−→ νδ( q2

2M + ν) = νδ(ν − Q2

2M ) = δ(1− Q2

2Mν ) = δ(1− x)

We see that the point-like proton structure functions do correspond to functions depending onlyon x. But the proton cannot be point-like since we had to add form factors to fit the data in theelastic scattering. The logical interpretation is then that when Q2 and ν are large enough, theprobe electron, or more precisely the virtual photon, seems to see point-like fermions2 objects

2They must be fermions since we used cross-section formula based on Feynman rules with spinors.

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114 From hadrons to partons

inside the proton, called partons (for historical reason). Let us examine this assumption. Letbe the charge of a parton i: qi|e|. This parton should carry only a fraction zi of the proton4-momentum3 and we assume that the parton momentum is aligned with the proton one. Inother word, we neglect any transverse motion. Hence:

Ei = ziEp , ~pi = zi~p ⇒ mi = ziM

Of course, a parton with a variable mass is meaningless. In fact, we have to consider the massas negligible, namely we work in a frame where the proton energy is much larger than its massso that both M and thus mi can be considered as equal to zero. Moreover, in such a frame (socalled infinite momentum frame), the time dilation is very large and thus the interactions amongpartons are strongly slowed down, so that we can assume the partons inside the proton as freeparticles4 only interacting with the virtual photon. The scattering cross-section of a parton ofcharge qi is then the elastic one 4.3 replacing M by ziM and one α by q2

i α:

dE′dΩ=

α2q2i

4E2 sin4 θ2

δ(ν − Q2

2ziM) cos2 θ

2

(1 +

Q2

2z2iM

2tan2 θ

2

)

to be compared to:

dE′dΩ=

α2

4E2 sin4 θ2

cos2 θ

2

[W2(Q2, ν) + 2W1(Q2, ν) tan2 θ

2

]

Hence, we would have for this particular parton:

MW1(q2, ν)(Q2,ν)→∞−−−−−−−→ q2

i2

Q2

2z2iM

δ(ν − Q2

2ziM) =

q2i x

2z2iδ(1− x

zi) =

q2i x

2ziδ(zi − x)

νW2(q2)(Q2,ν)→∞−−−−−−−→ νq2

i δ(ν − Q2

2ziM) = q2

i δ(1− xzi

) = q2i ziδ(zi − x)

Now, there is a probability fi(zi) that the parton carries a fraction zi of the proton momentum.Thus, we have to integrate over zi, so that for this parton i, we expect:

MW1(q2, ν)(Q2,ν)→∞−−−−−−−→

∫ 10 dzi fi(zi)

q2i x

2ziδ(zi − x) = 1

2q2i fi(x)

νW2(q2)(Q2,ν)→∞−−−−−−−→

∫ 10 dzi fi(zi)q

2i ziδ(zi − x) = q2

i xfi(x)

However, there might be other partons in the proton. Hence, we have to sum-up over all partons:

MW1(q2, ν)(Q2,ν)→∞−−−−−−−→ F1(x) = 1

2

∑i q

2i fi(x)

νW2(q2)(Q2,ν)→∞−−−−−−−→ F2(x) = x

∑i q

2i fi(x)

(4.18)

This scenario seems consistent with Bjorken prediction 4.17. Moreover, 4.18 predicts the socalled Callan-Gross formula of the parton model:

2xF1(x) = F2(x) (4.19)

Is it verified experimentally? Yes! See figure 4.5. The ratio 2xF1(x)/F2(x) is compatible with

3It means that internal motions of partons inside the proton are neglected considering the large proton 4-momentum.

4This strong assumption will be justified in the next chapter.

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Parton distribution functions 115

Figure 4.5: Measurement of the ratio 2xF1/F2 vs x. Open circles: 1.5 < Q2 < 4 GeV2, solid circles5 < Q2 < 11, stars 12 < Q2 < 16.. From [14].

1 as expected for fermions (since we used the elastic scattering cross-section computed withfermions vertices).

What is the physical meaning of the variable x-Bjorken? In a frame where the proton hasan energy Ep M and moves for example along the z-axis, its four momentum is thus p =(Ep, 0, 0, Ep). The parton inside the proton carries a fraction z of the proton 4-momentum: anytransverse motion is neglected and the parton mass m = zM is also negligible since M has beenneglected. After the elastic scattering with the virtual photon of 4-momentum q, the parton hasa 4-momentum zp+ q. Then:

m2 ≈ 0⇒ (zp+ q)2 = 0⇒ z2M2 + q2 + 2zp.q ≈ q2 + 2zp.q = 0⇒ z = − q2

2p.q= x

where we used 4.16 and 4.13. Hence, x-Bjorken is nothing more than the fraction of longitudinalmomentum carried by the parton.

Let us recap: the scattering experiments have shown that when the 4-momentum exchangedis large enough (and so the wave length short enough), we “see” inside the proton point-likefermions, that have been called partons. Anticipating on the next chapter, we assume thesepartons to be quarks (at first approximation).

4.3 Parton distribution functions

The function fi(x) appearing in 4.18 is called the Parton Distribution Function (PDF) andcorresponds to the Probability Density Function (PDF) of the fraction of momentum of theproton carried by partons/quarks of type i. In other words, fi(x)dx is the number of quarksof type i within the proton carrying a momentum fraction between x and x + dx. There isno tool (theory) on the market that can predict the values of the PDFs. Hence, they rely onexperimental measurements based on the deep inelastic scattering cross-sections that give accessto F1,2(x) and thus fi(x).

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116 From hadrons to partons

If the proton were made of only the 3 valence quarks |uud〉 responsible for its quantumnumber5, since they have equivalent masses and assuming that they do not interact amongthem, we would expect that they all share the same fraction of proton’s momentum. In otherwords,

u(x) ≡ fu(x) = 2δ(x− 1/3) , d(x) ≡ fd(x) = δ(x− 1/3)

However, this naive picture does not match the experimental observations. Indeed, any of thesequarks may radiate a gluon which may create a pair of quarks-antiquark qq (during a shorttime). Obviously, the heavier the quark is, the less likely this process occurs. So we should alsointroduce u(x), d(x), s(x) etc. To distinguish valence quarks from the others, we are going touse the subscript s for the latters: us(x) etc, where, s stands for sea, meaning all the (virtual)quarks present in the proton. Let us first neglect all quarks heavier than u and d. According to4.18, F2(x) for a e−p scattering is:

F ep2 (x) = x

[4

9(up(x) + ups(x)) +

1

9(dp(x) + dps(x))

](4.20)

where we have added the exponent p to clearly states that we deal with PDFs in proton. However,the same kind of deep inelastic scattering experiment can be performed with neutrons (|udd〉)giving:

F en2 (x) = x

[4

9(un(x) + uns (x)) +

1

9(dn(x) + dns (x))

]

Now, if we assume the isospin symmetry (see next chapter), we expect that u↔ d exchange doesnot change the experimental results and hence we can replace un → dp and dn → up yielding:

F en2 (x) = x

[4

9(dp(x) + dps(x)) +

1

9(up(x) + ups(x))

](4.21)

What can we learn by comparing the expected PDFs 4.20 and 4.21 to experimental data? Thefigure 4.6 shows the experimental data for F ep2 (x).

Figure 4.6: Measurement of F2 vs x for 2 < Q2 < 18GeV 2. From [18].

5The reader is supposed to have attended the course of the first semester about the “introduction in particlephysics”. See the beginning of the next chapter for a very brief summary.

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What is a proton? 117

The area corresponds to:

Aep =

∫ 1

0dx F ep2 (x) =

4

9

∫ 1

0dx x(u(x) + us(x)) +

1

9

∫ 1

0dx x(d(x) + ds(x)) =

4

9fu +

1

9fd ' 0.18

where 0.18 is the area measured on the figure. We have dropped the index p and fu, fdcorrespond respectively to the fraction of the proton momentum carried by the u, u and d, dquarks. We can do the same with the neutron data and based on similar measurements, it isfound:

Aep = 49fu + 1

9fd ' 0.18Aen = 4

9fd + 19fu ' 0.12

leading to:

fu ' 0.36 , fd ' 0.18

Conclusion: we do see that in the proton, on average the u quarks carry twice more momentumthan the d quarks! That’s expected for a proton made of |uud〉. However, we learn somethingnew: fu + fd ' 0.54. There are 46% missing! One could argue that this is due to the heavierquarks we neglected. But it’s not, their mass being really too large to compensate the 46%missing. Since the scattering with the electrons probes only the charged component of thenucleon, we must conclude that the missing fraction is due to neutral partons in the proton: themost natural candidates in the QCD framework (see next chapter) are the gluons.

4.4 What is a proton?

4.4.1 PDFs of the proton

We clearly realize that the proton (or any hadron) has a more complicated structure thanexpected: it is made of its valence quarks responsible for its quantum number, the sea quarksand anti-quarks, and finally the gluons from the sea! Ultimately, the PDFs are obtained from a fitto all data (including neutrino scattering, hadron collisions etc) using complicated models withmany parameters. Several groups in the world provides these PDFs. An example is shown onfigure 4.7 for Q2 = 10 GeV. According to the figure, at low x, the proton is mainly dominated bygluons and partially by quarks from the sea. For x ' 1/3, the valence quarks play the dominantrole matching more or less the naive expectations with uv(x) ' 2dv(x) (twice more u quarksthan d quarks). For x very close to one, the situation is less clear as we will see at the end ofthis section.

In order to obtain the PDFs as the ones shown on figure 4.7, constraints called sum rulesare applied so that the quantum numbers of the proton (charge, isospin etc) are recovered. Weshould find that there are a total of 2 u-quarks, 1 d and no other quarks. Since a quark and thecorresponding antiquark have opposite quantum number, we deduce the following sum rules:

∫ 10 dx [uv(x) + us(x)− us(x)] =

∫ 10 dx uv(x) = 2∫ 1

0 dx[dv(x) + ds(x)− ds(x)

]=∫ 1

0 dx dv(x) = 1∫ 10 dx [qs(x)− qs(x)] = 0 (q = s, t, b, t)

(4.22)

where the subscript v denotes the valence quarks and in the first 2 lines we have assumedus(x) = us(x) and ds(x) = ds(x) because qq are produced via a gluon splitting g → qq (equivalent

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118 From hadrons to partons

Figure 4.7: Distributions of several PDFs xf(x) for f = uv, dv, the valence quarks, f = u, d, s, c, thesea (anti-)quarks, and f = g the gluons from the sea where its contribution is divided by 10. From [19].

to γ → e−e+ in QED). Moreover, since u and d quarks have similar mass, it seems reasonableto assume:

us(x) = ds(x) = us(x) = ds(x) (4.23)

Hence the function structures 4.20 for the proton becomes:

F ep2 (x) = x[

49(uv(x) + us(x) + us(x)) + 1

9(dv(x) + ds(x) + ds(x))]

= x[

49uv(x) + 1

9dv(x) + 109 us(x)

]

while for the neutron 4.21 is now:

F en2 (x) = x[

49(dp(x) + dps(x)) + 1

9(up(x) + uns (x))]

= x[

49dv(x) + 1

9uv(x) + 109 us(x)

]

and thus:F en2 (x)

F ep2 (x)=

4dv(x) + uv(x) + 10us(x)

4uv(x) + dv(x) + 10us(x)

At very low x, the contribution of sea quarks must dominate (in other words, the valence quarksmust always carry a significant fraction of the nucleon momentum) and hence:

limx→0

F en2 (x)

F ep2 (x)' 1

This prediction is verified by the experimental data. On the contrary, when x ' 1, the quarksfrom the sea must be negligible, and thus:

limx→1

F en2 (x)

F ep2 (x)'

4 dv(x)uv(x) + 1

dv(x)uv(x) + 4

∈ [1

4, 1]

The experimental observations give a value of about 1/4 which would mean dv(1)/uv(1) ' 0.The reason for this value is not really understood: naively, we would expect uv(x) = 2dv(x)⇒limx→1 F

en2 (x)/F ep2 (x) ' 2/3.

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Proton spin 119

4.4.2 Proton spin

If the proton were a non-relativistic bound system, the spin of the ground state (proton beingthe lightest baryon) where all orbital angular momenta between its constituents are zero, wouldbe only due to the addition of the spins of its 3 valence quarks. This is the naive static modelwhere it is assumed that two of the quarks have opposite spins and the spin of the third quark isparallel to the proton spin. However, the proton is clearly a relativistic system: the constituentquark masses (below 10 MeV) are negligible compared with the proton mass6. In addition, thesize of the proton is about 1 fm, and because of Heisenberg uncertainties, the quarks must havea typical momenta of the order of 197 MeV. Knowing the small mass of u and d quarks, theyare clearly relativistic!

Since the late 80s, several experiments have measured the contribution of spin of the valencequarks to the spin of the proton using deep inelastic scattering with polarized beams. The veryfirst measurements made by the European Muon Collaboration (EMC) at CERN in 1987 gavethe surprising result that none of the proton’s spin was due to its quarks constituent! Today,it is estimated that about 30% of the proton’s spin is due to quarks spins contribution (valencequarks and quark-anti-quark pairs from the sea). This quantity is usually refereed as Σ inthe literature. Where are the other 70% coming from? Well, there are quarks orbital angularmomentum Lq, gluons spins ∆G (denoting the contributions of gluons7 aligned in the samedirection as the proton spin minus the gluons aligned in the opposite direction), and the gluonorbital angular momentum LG. Imagine for instance a proton polarized in the spin eigenstate+1/2 along z. The previous terms must then satisfy:

1

2=

1

2Σz + Lzq + ∆z

G + LzG (4.24)

The contribution ∆G is estimated to be below 0.2, the others contributions Lq and LG have neverbeen measured. The decomposition in 4.24 is not unambiguous (related to gauge invariance andthe sharing between ∆G and LG) and is a subject of debate in the theoretical community [20].

4.5 Scaling violation

With the constant rise of the energy of the accelerators, one can access to larger Q2 and henceprobe smaller scale. If the quarks were finally composite particles, we would expect its elasticcross-section and hence the proton function structure to fall very rapidly at a large Q2 as what wehave seen for the elastic cross-section of the proton. In the 1990s until the end of the last decade,the HERA collider at DESY8 laboratory in Germany, strongly contributed to the measurementsof structure functions with large Q2 and a wide range of x values, especially small values. HERAcollided electrons or positrons of 27.5 GeV with protons of 820 GeV. Two experiments operatedat HERA: H1 and ZEUS. A summary of the measurements of F2 is provided in figure 4.8 wherethe HERA’s measurements (black circles) are presented in addition to previous measurementsat lower Q2.Firstly, we see that at high Q2, there is no evidence of a rapid decrease of the cross-section.Therefore, a limit on the quark’s radius can be set giving about:

Rquark < 10−18 m (4.25)

6For hadrons made of heavy quarks as b or c quarks, the static approximation remains however valid since themass of those hadrons is dominated by the mass the heavy quarks.

7In the next chapter, we will see that gluons are spin 1 bosons.8Deutsches Elektronen-SYnchrotron.

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120 From hadrons to partons

16. Structure functions 21

NOTE: THE FIGURES IN THIS SECTION ARE INTENDED TO SHOW THE REPRESENTATIVE DATA.

THEY ARE NOT MEANT TO BE COMPLETE COMPILATIONS OF ALL THE WORLD’S RELIABLE DATA.

Q2 (GeV2)

F 2(x,Q

2 ) * 2

i x

H1+ZEUSBCDMSE665NMCSLAC

10-3

10-2

10-1

1

10

10 2

10 3

10 4

10 5

10 6

10 7

10 -1 1 10 10 2 10 3 10 4 10 5 10 6

Figure 16.8: The proton structure function Fp2 measured in electromagnetic scattering of electrons and positrons on protons (collider

experiments H1 and ZEUS for Q2 ! 2 GeV2), in the kinematic domain of the HERA data (see Fig. 16.10 for data at smaller x and Q2),and for electrons (SLAC) and muons (BCDMS, E665, NMC) on a fixed target. Statistical and systematic errors added in quadrature areshown. The data are plotted as a function of Q2 in bins of fixed x. Some points have been slightly o!set in Q2 for clarity. The H1+ZEUScombined binning in x is used in this plot; all other data are rebinned to the x values of these data. For the purpose of plotting, F

p2 has

been multiplied by 2ix , where ix is the number of the x bin, ranging from ix = 1 (x = 0.85) to ix = 24 (x = 0.00005). References: H1 andZEUS—F.D. Aaron et al., JHEP 1001, 109 (2010); BCDMS—A.C. Benvenuti et al., Phys. Lett. B223, 485 (1989) (as given in [66]) ;E665—M.R. Adams et al., Phys. Rev. D54, 3006 (1996); NMC—M. Arneodo et al., Nucl. Phys. B483, 3 (1997); SLAC—L.W. Whitlowet al., Phys. Lett. B282, 475 (1992).

Figure 4.8: The proton function structure F2 for different x versus Q2. F2 has been multipled by anumber depending on the x value to ease the reading of the plot. From [19].

Secondly, we see that at very small x (typically x < 0.02), there is a clear dependence with Q2:the Bjorken scaling is not true anymore and F2(x) must be replaced by F2(x,Q2):

F2(x)x→0−−−→ F2(x,Q2)

The reason for the violation of Bjorken scaling is the following: at very low x, the gluon radiationdominates, producing many qq pair. In QCD, the transverse momentum of these gluons withrespect to the proton motion can be large, contrary to our initial assumption with the partonmodel. And this is precisely this assumption which led us to the Bjorken scaling property. Wewill come back on this aspect in the next chapter.

4.6 Exercises

Exercise 4.1 The Rutherford cross-section as a spinless scattering.We consider the elastic electron-proton scattering e−(k) + p(p) → e−(k′) + p(p′) assuming thatboth the electron and the proton are spinless. We assume that the proton is at rest and neglectits recoil. The electron mass is also neglected.

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Exercises 121

1. show that the amplitude is:

M =4παM

E sin2 θ2

where M is the proton mass, θ the scattering angle of the electron and E the energy of theincoming electron.

2. Conclude that the differential cross-section (Hint: use formula 1.88) can be expressed whenthe proton recoil is neglected as:

dΩ=

α2

4E2 sin4 θ2

which is the well-known Rutherford formula, that is usually derived using a spinless particle(historically alpha) in a static Coulomb field.

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122 From hadrons to partons

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Chapter 5

Quantum Chromodynamic

Few references:I.J.R Aitchison & A.J.G. Hey,“Gauge theories in particle physics”, vol2, Chapter 14 & 15, IoP2003.

In this chapter, the Quantum ChromoDynamic (QCD) theory is introduced.We try to follow more or less the historical developments. The existence of thegluons, briefly introduced in the last chapter will be justified and the notion ofcolors introduced.

5.1 Quark model

The whole section is essentially a summary of the first semester courses “Physique des particuleselementaires” by Michel Gonin. Hence, I won’t give so much details.

5.1.1 Baryon number

In the late forties, beginning of fifties, the panorama of “elementary” particles changed radicallywith the emergence of the accelerators. In addition to the well known electron, proton, neutron,positron, new particles were observed π0 (π± had been observed in cosmic ray data), K0,±,Λ, ∆, Σ etc. It was noticed that the heavy particles (p, n,Λ,∆,Σ) were conserved: decays liken→ p e−(+ν), Λ→ p π− were observed but not n→ e+e−, Λ→ π+π−. A new quantum numberwas then invented: the Baryon number (B) supposed to be conserved by the interactions, sothat the heavy particles have B 6= 0 while the lightest one (m < mp) have B = 0. In addition,it gave an “explanation” for the stability of the proton, the lightest of the heavy particles. Thelight particles appearing in strong interaction were called Mesons, the heavy ones, Baryons.

5.1.2 Isospin and SU(2)

The masses of the proton and neutron are so close (938.3 MeV and 939.5 MeV), that Heisenbergin 1932 suggested to interpret them, as far as nuclear interactions are concerned, as 2 states of asame object, the nucleon. Since the masses are (almost) equal, the 2 states would be degenerated.The situation is then similar to the one of the spin 1

2 systems, for which the 2 states sz = ±12

are degenerated in absence of magnetic field. The nucleon state |N〉 is then a superposition

123

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124 Quantum Chromodynamic

of the 2 degenerated states |p〉 isospin up and |n〉 isospin down1 with a corresponding isopinor2-components notation:

|N〉 =

(αβ

)= α

(10

)+ β

(01

)= α |p〉+ β |n〉

α and β being respectively the amplitude for the nucleon to have isospin up and down. Since |p〉and |n〉 are degenerated, any linear transformation should give the same physical results. Hencethe state:

|N ′〉 = U |N〉with U a 2× 2 matrix should be equivalent to |N〉. This imposes constraints on the possible Umatrices similar to the spin case, where we know that they have to be special unitary matrices(U †U = 1, det U = 1). This set of matrices forms a group called SU(2) (for Special Unitary ofrank 2), a group which is a Lie group for which a representation of its elements is:

U = ei~α.~I (5.1)

αi being numbers and Ii the generators satisfying:

[Ii, Ij ] = iεijkIk

where εijk = 1 for cyclic permutation of 123, -1 for anticyclic and 0 otherwise. A matrixrepresentation of the generators of rank 2 × 1

2 + 1 = 2 is given by the 3 Pauli matrices (equ.1.42):

Ii =1

2σi with σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

)(5.2)

With this approach, proton and neutron belong to the same iso-doublet with I = 12 . By con-

vention, the most positively charged particle has the largest value of I3. There is a subtle pointconcerning the definition of the antinucleon state from the nucleon one. Consider a infinitesimaltransformation from 5.1:(p′

n′

)= ei~ε.

~σ2

(pn

)'(

1 + i[ε1σ1

2+ ε2

σ2

2+ ε3

σ3

2

])( pn

)=

(1 + i ε32 i ε1−iε22

i ε1+iε22 1− i ε32

)(pn

)

⇒p′ = (1 + i ε32 )p+ i ε1−iε22 n

n′ = (1− i ε32 )n+ i ε1+iε22 p

Now, we wish the antinucleon to transform consistently with respect to the nucleon. Naively,we would expect (putting n with I3 = 1

2 since its charge is greater than the one of p):

(n′

p′

)=(ei~ε.

~σ2

)∗( np

)'(

1− i ε32 −i ε1+iε22

−i ε1−iε22 1 + i ε32

)(np

)⇒p′ = (1 + i ε32 )p− i ε1−iε22 n

n′ = (1− i ε32 )n− i ε1+iε22 p

which is clearly different than the nucleon transformation. The solution is to define the antistateas:

|N〉 =

(−np

)

1Unlike the spin, the isospin is a simple number. It has no units (i.e. there is no ~).

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Strangeness and SU(3) flavours 125

Indeed, we obtain:

(−n′p′

)'(

1− i ε32 −i ε1+iε22

−i ε1−iε22 1 + i ε32

)(−np

)⇒p′ = (1 + i ε32 )p+ i ε1−iε22 n

n′ = (1− i ε32 )n+ i ε1+iε22 p

Similarly to the nucleon doublet, the masses of the 3 pions π−, π0, π+ are so close that theycorrespond to an iso-triplet with I = 1. The four ∆−,0,+,++ particles (and the four ∆)2 haveI = 3

2 . It is worth mentioning that there is no iso-multiplet with I > 32 .

A rotation of an isospin vector (via formula 5.1) does not affect the way it couples to thestrong force. Therefore, there are necessarily quantities that are conserved (Noether’s theorem):I3 and I (analogous to conservation of J3 and J for angular momentum).

5.1.3 Strangeness and SU(3) flavours

The discovery of the kaon, Λ, Σ and the possibilities to study them at accelerator in the fifties,led to the observation that they were copiously produced but their lifetime was much longerthan for other particles (∆ etc). The hypothesis of a production mechanism different than thedecay one emerged: they are produced by strong interaction and decay by weak interaction.But what prevents them to decay via strong interaction? In 1953, Gell-Mann and Nishijimaintroduced a new additive quantum number: the strangeness which is conserved in strong inter-action (explaining why the strange particle always appear by pair) but not in weak interaction.Eight years later, in 1961, Gell-Mann realised that particles having similar properties can bearranged into geometrical patterns, mesons and baryons of spin 1

2 in octet and baryons of spin32 in decuplet (see figure 5.1) according to their charge Q and strangeness S or equivalently totheir hypercharge Y = B + S (as defined in the 1960s) and I3 with:

Q = I3 +Y

2(5.3)

This formula is known as the Gell-Mann-Nishijima formula.

Figure 5.1: Classification of baryons in octet and decuplet in 1961 (the Ω− was not known at that time).Hadron having same charge q are on the downward sloping diagonal lines of the pattern. The strangenessS is on the horizontal.

2Warning ∆+ ≡ |uud〉 is not the ∆− ≡ |ddd〉!

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126 Quantum Chromodynamic

Why such patterns?The fact that there was no hadrons with I3 >

32 , which conditioned the shape of the patterns,

led Gell-Mann and Zweig to postulate that hadrons are made of quarks (remember, it was beforethe deep inelastic scattering experiments revealing the substructure of nucleons). Two quarks,u and d, were members of an isospin doublet with S = 0, and one quark, s, was supposed to bea S = −1 isosinglet (the unfortunate sign is historical):

(ud

):I3 = +1

2 , S = 0I3 = −1

2 , S = 0(s) : I3 = 0, S = −1 (5.4)

Note that when 2 spins 12 are summed we get:

2⊗ 2 =

(↑↓

)⊗(↑↓

)= 3

↑↑1√2(↑↓ + ↓↑)↓↓

⊕ 1

1√2

(↑↓ − ↓↑)

Similarly when combining the isodoublet 5.4 and its anti-isodoublet:

2⊗ 2 =

(ud

)⊗(−du

)= 3

−ud1√2(uu− dd)

du

⊕ 1

1√2

(uu+ dd)

The isotriplet is then π+, π0 and π−, the isosinglet being a η.

The 3 quarks flavours u, d, s were thought as 3 degrees of freedom of the symmetry groupSU(3)f , f standing for flavour. The symmetry is approximative since members of a multipletcan have a significant dispersion of masses, reflecting the non equality of the 3 masses of thequarks. Baryons are constituted of 3 quarks while mesons are made of a pair quark-antiquark.Quarks are assumed to share democratically the baryon number, meaning that they have B = 1

3 .Consequently, according to formula 5.3, they have a charge qu = 2

3 and qd = qs = −13 .

In this model, the 3 quarks flavours belong to a SU(3) triplet. So let us define:

|u〉 =

100

, |d〉 =

010

, |s〉 =

001

and hence, a general quark wave function is:

q =

q1

q2

q3

The Lie group SU(3) has 32 − 1 = 8 generators Ti. Similarly to the (iso-)spin case, the wavefunction is rotated under a SU(3) transformation in:

q′ = Uq with U = ei~α.~T

The generators Ti take the following matrix representation:

Ti =λi2

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Strangeness and SU(3) flavours 127

the λ’s being 3 × 3 matrices known as the Gell-Mann matrices. They can be viewed as ageneralization of the Pauli matrices where we had Ii = σi

2 . To identify the matrix representation,we can consider just the first 2 flavours u and d, for which we should recover the SU(2) generators.Hence:

λ1 =

σ1

00

0 0 0

λ2 =

σ2

00

0 0 0

λ3 =

σ3

00

0 0 0

namely:

λ1 =

0 1 01 0 00 0 0

λ2 =

0 −i 0i 0 00 0 0

λ3 =

1 0 00 −1 00 0 0

We see that λ3 is the matrix related to the I3 values:

I3 =1

2λ3 ⇒ I3u = +

1

2u , I3d = −1

2d , I3s = 0 s

In other words, u, d, s are eigenstates of I3 generator. For the other flavours pairs (u, s) and(d, s), we should recover the SU(2) generators but “shifted” or “splitted”, namely, for (u, s):

λ4 =

0 0 10 0 01 0 0

λ5 =

0 0 −i0 0 0i 0 0

λ′3 =

1 0 00 0 00 0 −1

while (d, s) gives:

λ6 =

0 0 00 0 10 1 0

λ7 =

0 0 00 0 −i0 i 0

λ3” =

0 0 00 1 00 0 −1

We notice that 2 matrices are diagonal: λ′3 and λ3”. Hence, there is another quantum numberbeing conserved. Noticing that λ3, λ

′3 and λ3” are not linearly independent since λ3−λ′3+λ3” = 0,

we can define a linear combination of λ′3, λ3” such as the eigenvalues will give the hyperchargeY . Hence, defining the eighth matrix3 with:

λ8 =1√3

[λ′3 + λ3”

]=

1√3

1 0 00 1 00 0 −2

we have:

Y =1√3λ8 ⇒ Y u =

1

3u , Y d =

1

3d , Y s = −2

3s

which matches the values obtained with Y = B + S formula (u : 13 + 0, d : 1

3 + 0, s : 13 − 1).

With the two conserve numbers I3 and Y , we can specify any states with its eigenvalues: |i3, y〉.u = |12 , 1

3〉 , d = |−12 ,

13〉 , s = |0,−2

3〉. Similarly the antiquarks have opposite quantum numbers.Geometrically, it gives a representation in the plan I3,Y as shown in figure 5.2. The combinationof these basic patterns allows to deduce the mesons content qq generated by the representation

3The factor 1/√

3 is such that any 8 matrices verifies Tr(λaλb) = 2δab. This property will be used later.

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128 Quantum Chromodynamic

Figure 5.2: SU(3) quark (3-representation) and anti-quark (3-representation) multiplets in the Y − I3plan. From [9].

3 ⊗ 3 as shown on figure 5.3. It suffices to superimpose the centre of gravity of the antiquarkmultiplet on top of every site of the quark multiplet. We see that we obtain:

3⊗ 3 = 8⊕ 1

When several combinations give the same quantum number |i3, y〉, they have to be combinedto give appropriate orthogonal states in the correct multiplet. For example, there are 3 com-binations giving |0, 0〉: uu, dd and ss. One combination, say A in figure 5.3 should give theπ0 = 1√

2(uu−dd) derived previously and completing the isotriplet made with π+ = −ud = |1, 0〉

and π− = du = |−1, 0〉. The SU(3) singlet combination C must be invariant (modulo a possiblephase) for any interchange u ↔ d ↔ s and hence is necessarily 1√

3(uu + dd + ss). The last

combination at |0, 0〉 uu + αdd + βss must be orthogonal to the 2 other states at |0, 0〉 andthus 〈uu− dd|uu+ αdd+ βss〉 = 0 ⇒ α = 1 and 〈uu+ dd+ ss|uu+ dd+ βss〉 = 0 ⇒ β = −2giving the state 1√

6(uu+ dd− 2ss).

Figure 5.3: The quark content of the meson resulting from 3 ⊗ 3 = 8 ⊕ 1. See text for A,B, and Ccontent. From [9].

Similarly one can obtain all the states for the baryons4:

3⊗ 3⊗ 3 = (6⊕ 3)⊗ 3 = (6⊗ 3)⊕ (3⊗ 3) = 10⊕ 8⊕ 8⊕ 1

Note that there are 2 octets appearing in the decomposition of 3⊗ 3⊗ 3. These are 2 equivalentrepresentations of the same particles. They differ in their symmetry properties under permuta-tion of the three triplets. One octet is symmetric under a permutation of the first 2 quarks while

4Please, refer to the group theory courses PHY575 “Groupe de symetrie en physique” for a more rigorousderivation of these combinations.

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Need for colors 129

the other is anti-symmetric. For example, in the case of the proton the flavor wave function are

ψ(8)s = 1√

6[(ud+du)u− 2uud] (symmetric in interchange of the first 2 quarks) while in the other

it is ψ(8)a = 1√

2[(ud−du)u] (anti-symmetric in interchange of the first 2 quarks). Of course, these

flavor wave functions have to be combined with the spin ones ψ(2)s (symmetric in first 2 spins)

and ψ(2)a (anti-symmetric in first 2 spins). For a spin-up baryon they are:

ψ(2)s (↑) =

1√6

[(↑↓ + ↓↑) ↑ −2 ↑↑↓], ψ(2)a (↑) =

1√2

[(↑↓ − ↓↑) ↑]

(the corresponding spin down is obtained by doing the replacement ↑↔↓). The flavor+spin

wave function of the proton is then ψ(8)+(2) = 1√2(ψ

(8)s ψ

(2)s +ψ

(8)a ψ

(2)a ). We have made a globally

symmetric wave function for a reason that will be explained in the next section. After expansion,a spin-up proton reads5:

ψ(8)+(2)(proton) = 1√18

( 2u ↑ d ↓ u ↑ +2d ↓ u ↑ u ↑ +2u ↑ u ↑ d ↓−u ↓ d ↑ u ↑ −d ↑ u ↑ u ↓ −u ↑ u ↓ d ↑−u ↑ d ↑ u ↓ −d ↑ u ↓ u ↑ −u ↓ u ↑ d ↑ )

The original quark model as proposed by Gell-Mann and Zweig was remarkably successful inexplaining the observed hadrons in terms of basic constituents and in particular in calculatingbaryon magnetic moments. Gell-Mann even predicted the existence of a particle with quantumnumber S = −3 and Q = −1 to complete the decuplet 3 years before it was actually observed:the Ω− finally discovered in 1964.

5.1.4 Need for colors

Consider the Ω− in the right-hand side figure 5.1. It has S = −3 and hence is made of 3 quarkss, while being a member of the decuplet, its spin is 3

2 . Quarks being fermions, it means that the3 quarks must be in the up-spin state. Furthermore, because of the Pauli’s exclusion principle, abound state of 3 identical quarks must have a wave function anti-symmetric with the interchangeof any 2 quarks:

ψq1q2q3 = −ψq2q1q3 = −ψq3q2q1 = −ψq1q3q2The wave function of the Ω− is a priori:

|Ω−〉 = ψflavour ⊗ ψspin ⊗ ψspatial = |sss〉 ⊗ |↑↑↑〉 ⊗ ψl=0

Obviously, ψflavour and ψspin are symmetric. Since Ω− is in the ground state l = 0, its spatialwave function is symmetric as well. We then have a symmetric global wave function in contra-diction with Pauli’s principle! In 1964, Greenberg proposed to add a new quantum number toquarks, named colours. Each quark is postulated to come with 3 possible colours, red, green,blue (for example), so that the wave function of a baryon made of 3 identical quarks can beantisymmetric if the 3 quarks have 3 different colours:

|Ω−〉 = ψcolour ⊗ ψflavour ⊗ ψspin ⊗ ψspatial5Keep in mind, that such wave function corresponds to the static model where the proton spin is entirely due

to the spin of its valence quarks. We know it is not the case as explained in section 4.4.2.

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130 Quantum Chromodynamic

where

ψcolour(q1q2q3) = N

∣∣∣∣∣∣

r1 r2 r3

g1 g2 g3

b1 b2 b3

∣∣∣∣∣∣= N [r1(g2b3 − b2g3)− g1(r2b3 − b2r3) + b1(r2g3 − g2r3)]

Choosing the norm N so that |ψcolour|2 = 1,

ψcolour =1√6

[rgb− rbg + gbr − grb+ brg − bgr]

where the index was dropped (the first colour applies to first quark etc). The color wavefunction is antisymmetric with the interchange of any 2 colors. In addition, it is postulated thatall hadrons must have zero colour, namely be colour singlet. Consequently all baryons have thesame colour wave function6

ψBaryonscolour =1√6

[rgb− rbg + gbr − grb+ brg − bgr] (5.5)

and the complex conjugate for the anti-baryon. Since by construction 5.5 is antisymmetric, wehave for all baryons

Baryons: ψflavour ⊗ ψspin ⊗ ψspatial ⇒ symmetric (5.6)

Mesons are made with a pair quark-antiquark. The only colour singlet qq state is:

ψMesonscolour =

1√3

[rr + gg + bb

](5.7)

This wave function is symmetric. Its form is obviously the same as the single meson in SU(3)flavour by replacing the flavour by colours (u → r, d → g, b → s), as we will see in the nextsection. It is important to realize that a state rr is not colorless (in the sense that it is not apure color-singlet)! Only the state given in 5.7 is colorless because there is the same amount ofr, g and b for the quark and the same amount of r, g and b for the antiquark.

The rule forbidding hadrons having a color, imposes that states made with 1, 2 or 4 quarks forinstance, are not physical states. So far, this rule has never been contradicted.

5.1.5 Summary of the quark model

Between 1974 and 1995, three others quarks flavours have been discovered: the charm c, beautyb (sometimes called bottom) and top t (sometimes called truth). The corresponding additivequantum numbers have been created with the convention that the flavour quantum number hasthe same sign as the charge of the quark. The hypercharge number, still satisfying to 5.3 hasbeen changed accordingly:

Y = B + S + C + B + T (5.8)

The table 5.1 summarises the quarks quantum numbers relevant for the strong interaction andtheir mass (from [19]). Concerning the parity, since quarks are fermions with spin 1/2, they

6You might argue that the other wave function 1√6

[rgb+ rbg + gbr + grb+ brg + bgr] would be fine as well.In fact, no. This wave-function belongs to the decuplet resulting from the decomposition of 3 ⊗ 3 ⊗ 3 and thuscarries some color. By analogy with the spin, it would be the same mistake as confusing the state |1, 0〉 (whichhas a spin 1 but a projection 0) with |0, 0〉 with is really spinless.

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The colour as a gauge theory 131

u d s c b t

Mass (GeV/c2) 1.7-3.1 10−3 4.1-5.7 10−3 0.1 1.29 4.19 172.9

Charge 23 −1

3 −13

23 −1

323

Spin 12

12

12

12

12

12

Isospin |I, I3〉 |12 , 12〉 |12 ,−1

2〉 |0, 0〉 |0, 0〉 |0, 0〉 |0, 0〉Baryon nb B 1

313

13

13

13

13

strangeness S 0 0 -1 0 0 0

charm C 0 0 0 1 0 0

beauty B 0 0 0 0 -1 0

truth T 0 0 0 0 0 1

Hypercharge Y 13

13 −2

343 −2

343

Parity ηP 1 1 1 1 1 1

Table 5.1: Quantum numbers and mass of quarks. These quantum numbers are conserved by the stronginteraction.

satisfy the Dirac’s equation and thus, according to the equality 2.53, their parity is 1. Lookingat the masses of the quarks in the table 5.1, we see that the isospin is an excellent symmetry:u and d quarks have almost the same mass, and thus hadrons members of an isospin multipletdiffer in mass by at most 3%, well within the effects induced by the electromagnetism (since theydon’t have the same charge). This has practical consequences as it can be seen in an exercise atthe end of this chapter. Members of multiplets made with a strange quark can have a significantdifference of masses (up to 40%), so the flavor symmetry SU(3) is approximative. With quarksof larger mass, the differences are much more pronounced breaking frankly the flavor symmetry.

5.2 The colour as a gauge theory

5.2.1 Experimental evidences of the colour

One may wonder if the quarks colour is a pure mathematical artifice to “anti-symmetrize” thebaryon wave function or if it has real physical fundaments, with experimental observations. Infact, there are. For example, consider the production of hadrons at e+e− collider. Quarks beingcharged, they can be produced via the s-channel of the e+e− annihilation shown in figure 5.4. Assoon as

√s mZ , we can neglect electroweak interaction (see next chapter) and consider only

the virtual photon exchange. What is the cross-section of this process? It is easy to calculatesince it is strictly similar to the process e−e+ → µ−µ+ seen in section 3.5.3 for which we found:

σe−e+→µ−µ+

=4πα2

3s

The quarks flavour i having a charge Qi, we would expect:

σe−e+→qiqi =

4πα2Q2i

3s

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132 Quantum Chromodynamic

e−

e+

q

q

Figure 5.4: Lowest order diagram e−e+ → qq

However, since it can have 3 different colors, we should then have7:

σe−e+→qiqi = 3

4πα2Q2i

3s

Now if√s is large enough, several quarks flavour can be produced, so that:

σe−e+→qq = 3

4πα2

3s

mqi≤√s

2∑

i

Q2i

Hence the ratio quarks production / muon production is:

Rhad =σe−e+→qq

σe−e+→µ−µ+ = 3

mqi≤√s

2∑

i

Q2i (5.9)

Since the quarks have colors, they cannot be seen as free particle in the detector according to thezero-colour postulate. As the qq move apart, they will “dress” in zero-colour hadron, a processcalled hadronization. We will see in the next sections how. But let us assume, that to firstorder, the formula 5.9 remains valid for the hadrons final state. Experimentally, these hadronsare detected thanks to calorimeters detectors. Our prediction of Rhad can then be comparedto experimental measurements. The measured ratio is shown on figure 5.5. Before the charmthreshold (

√s ' 2.6 GeV), or beauty threshold (

√s ' 9 GeV)

Rudshad = 3

[(2

3

)2

+ 2

(1

3

)2]

= 2 , Rudschad = 3

[2

(2

3

)2

+ 2

(1

3

)2]

=10

3

while without the colour factor 3, we would have found Rudshad = 2/3 and Rudschad = 10/9. Com-parisons to the data rule out the no-colour scenario. The ratio (more or less guided by theyellow band) appears close to 2 below the charm threshold and pretty close to 10/3 above asexpected in a model with 3 colours. This ratio has to be compared in the continuum namelyoutside of resonances peaks (where we should have added a form factor to describe the modifiedcross-section).

There are many other evidences of the quarks colours: the π0 lifetime, leptonic τ branchingratio etc.

7Have a look to the exercise 5.4 to better appreciate this factor 3.

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The symmetry group SU(3)c 133

Zhengguo Zhao R Values in Low Energy e+e! Annihilation

Place Ring Detector Ecm Points Year(GeV)

Beijing BEPC BES II 2.0–5.0 85 1998–1999Novosibirsk VEPP-2M CMD-2 0.6–1.4 128 1997–1999

SNDVEPP-2 Olya,ND CMD 0.3–1.4

SLAC Spear MarkI 2.8–7.8 78 1982Frascati Adone !!2,MEA 1.42–3.09 31 1978

Boson,BCFOrsay DCI M3N,DM1,DM2 1.35–2.13 33 1978

Hamburg Doris DASP 3.1–5.2 64 1979PLUTO 3.6–4.8 27 1977

Table 1: Measurements of R at low energy by di!erent laboratories.

!s in GeV

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8 9 10

Bacci et al.

Cosme et al.

Mark I

Pluto

Cornell,DORIS

Crystal Ball

MD-1 VEPP-4

VEPP-2M ND

DM2

Rhad

",#,$ %'s &'s

15 % 15 % 6 % 3 %

relative

error in

continuum

Burkhardt, Pietrzyk '95

Figure 1: Experimental R values in the energy region below 10 GeV, from [10].The relative errors on R in the continuum is given in numbers at the bottom ofthe figure.

light quark massesmu andmd as well as the inherent non-perturbative nature ofthe problem at small energy scales, where the free quark loops are strongly mod-ified by strong interactions at low energy. An ingenious way to handle this [19]is to relate !"had from the quark loop diagram to R, making use of unitarity and

380

Figure 5.5: Rhad (see text) as function of√s in various experiments. From [21]

5.2.2 The symmetry group SU(3)c

We wish to detail the way in which the strong force between quarks depend on their colours.Let us consider these remarks:

• The colour symmetry is assumed to be exact: the transformation corresponding to thatsymmetry must keep the physical states unchanged.

• Experimentally, there are 3 colours per quark.

• Only hadrons without colour and hence belonging to a colour singlet are found in nature.This is the hypothesis of colour confinement.

• Quarks colours and anti-quarks colours are different otherwise we would have found stateswith qq since states qq exist. Consequently ψ∗c 6= ψc ⇒ ψc not real.

Based (partially) on these considerations, Han and Nambu have proposed in 1965 SU(3) as thegroup symmetry of the colours. In fact, it is the simplest one. The 3 colours belong to a SU(3)ctriplet. So let us define:

|r〉 =

100

, |g〉 =

010

, |b〉 =

001

and hence, a general quark field of flavour f is:

qf =

f1

f2

f3

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134 Quantum Chromodynamic

Where f1,2,3 are respectively the quark fields for red, green and blue colour. The Lie groupSU(3) has 32 − 1 = 8 generators Ti. The formalism is strictly the same as with SU(3) flavour.The field is transformed under a SU(3)c transformation in:

q′f = Uqf with U = e−i~α.~T

where the minus sign is here a matter of convention. The generators Ti take the following matrixrepresentation:

Ti =λi2

the λ’s being the 8 Gell-Mann matrices already introduced and reproduced again below:

λ1 =

0 1 01 0 00 0 0

λ2 =

0 −i 0i 0 00 0 0

λ3 =

1 0 00 −1 00 0 0

λ4 =

0 0 10 0 01 0 0

λ5 =

0 0 −i0 0 0i 0 0

λ6 =

0 0 00 0 10 1 0

λ7 =

0 0 00 0 −i0 i 0

λ8 =

1√3

0 0

0 1√3

0

0 0 −2√3

(5.10)

The norm of these matrices has been chosen so that for the eight λi we have:

Tr(λaλb) = 2δab

The matrices λ3 and λ8 are diagonal, meaning that as for SU(3)flavor, there are 2 quantumnumbers that are conserved. Let us call them the color-Isopsin I3c and the color-hyperchargeYc by analogy with the previous case. A color eigenstate can then be represented by these 2numbers |i3c, yc〉 and a graphical representation of the color or anti-color states is given in figure5.6.

+2/3

-2/3

+1/2 -1/2

Yc

I3c

b

g r

+2/3

-2/3 +1/2 -1/2

Yc

I3cg r

b Figure 5.6: SU(3)c representation for color (3-representation) and anti-color (3-representation).

The generators of SU(3)c obeys to the commutation rule:

[Ta, Tb] = ifabcTc ⇒[λa2,λb2

]= ifabc

λc2

(5.11)

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Gauge invariance consequences 135

where ifabc are the structure constants of SU(3)c with8:

f123 = 1f147 = f246 = f257 = f345 = f516 = f637 = 1

2

f458 = f678 =√

32

fijk = fjki = fkij = −fjik = −fikjfiij = fijj = 0fijk = 0 for other permutations

(5.12)

But a fundamental question remains: in this theory, what prevent mesons, or baryons tohave colours? Indeed, we know that the combination of a colour and an anti-colour, or thecombination of 3 colours gives:

3⊗ 3 = 8⊕ 13⊗ 3⊗ 3 = 10⊕ 8⊕ 8⊕ 1

(5.13)

What forces the mesons and baryons to be in their respective colour singlet and not in an octetor decuplet? One can show (see section 5.4.1 for basic ideas or [6, p. 90]-[8, p. 782]) that thepotential between a quark and an anti-quark is attractive if qq is in the singlet state and repulsiveotherwise. The potential between 2 quarks is always repulsive explaining why there are not suchbound state. In summary, only the colour singlet states have an attractive potential, justifyingthe colour confinement postulate9.

5.2.3 Gauge invariance consequences

In 1954, Yang and Mills generalized the idea of a local phase symmetry to a symmetry grouplarger than U(1) (the symmetry group of QED) implying unitary rotations of several fields.Such theory is called a non-abelian gauge theory, because the transformations of the fields donot commute.

Quarks being spin 1/2 fermions, they satisfy the Dirac’s equation and hence the free fieldLagrangian is given by the Dirac Lagrangian:

L = qf (iγµ∂µ −m)qf =3∑

i=1

fi(iγµ∂µ −m)fi with qf =

f1

f2

f3

where for simplicity, we just consider one quark flavour for the moment. Now consider theSU(3)c global transformation in the colour space:

q′f = Uqf = e−igs~α.~λ2 qf = e−igsα

a λa2 qf

where we have factorized a common real gs, and αa is a set of 8 reals. U belongs to the SU(3)cgroup: special unitary properties implies UU † = U †U = 1 and detU = 1. Note that repeatedindices are, as usual, summed over. Since U acts on the colour space while γµ acts on the spinorspace, they both commute and:

L global−−−−→ L′ = qfU†(iγµ∂µ −m)Uqf = L

8Please, see the group theory courses PHY575, for a derivation of these numbers.9Strictly speaking, the repulsive potential should be infinite to justify the impossibility to observe colored

states.

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136 Quantum Chromodynamic

The Lagrangian is thus invariant under a global transformation. But what about a local trans-formation in which U depends on space-time through αa(x)? We know than in the QED case,it gave rise to the photon field.

L local−−−→ L′ = qfU†(iγµ∂µ −m)Uqf

= qfU†(iγµ∂µ(Uqf )−mUqf )

= qfU†(iγµ(∂µU)qf + iγµU∂µqf −mUqf )

= L+ qf iγµ(U †∂µU)qf

(5.14)

The Lagrangian is manifestly not invariant anymore because of the additional term:

U †∂µU = U †(−igs∂µαa(x)λa2

)U (5.15)

Let us proceed as in QED and introduce 8 new fields Gµa , the gluons fields via the covariantderivative:

Dµ = ∂µ + igsλa2Gaµ (5.16)

The gluons field will have to compensate the additional term 5.15 to maintain the local invari-ance. Hence, the new Lagrangian we consider is now:

L = qf (iγµDµ −m)qf = qf (iγµ∂µ −m)qf − jµa Gaµ (5.17)

with the 8 (conserved) currents given by:

jµa = gs qfγµλa

2qf (5.18)

These Gaµ fields must transform under a SU(3)c transformation so that they will compensatethe additional term in 5.14. The transformation of the Lagrangian 5.17 gives:

L local−−−→ L′ = qf (iγµ∂µ −m)qf + qf iγµ(U †∂µU)qf − j′µa G′aµ

= L+ jµa Gaµ + qf iγµ(U †∂µU)qf − j′µa G′aµ

The Lagrangian 5.17 is thus invariant if

jµa Gaµ + qf iγ

µ(U †∂µU)qf − j′µa G′aµ = 0 (5.19)

Now, let us denoteG′aµ = Gaµ + δGaµ

where we assume |δGaµ| |Gaµ|. The condition 5.19 becomes:

j′µa δGaµ = (jµa − j′

µa)Gaµ + qf iγ

µ(U †∂µU)qf (5.20)

According to 5.18,

j′µa = gsqfU†γµ

λa2Uqf

and hence 5.20 is:

gsqfγµU †

λa2U δGaµ qf = gsqfγ

µ

(λa2− U †λa

2U

)qf G

aµ + qf iγ

µ(U †∂µU)qf

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Gauge invariance consequences 137

and since it must be true for any spinor, it simplifies after multiplying with (γµ)−1 to:

gsU† λa

2 U δGaµ = gs(λa2 − U † λa2 U

)Gaµ + i(U †∂µU)

⇒ λa2 U δGaµ =

(U λa

2 − λa2 U)Gaµ + i

gs∂µU

where we have used the property UU † = 1. For simplicity, we will consider the infinitesimallocal transformation:

U = 1− igsαb(x)λb2

and we are going to neglect second order terms as α2 or α δG:

λa2 δG

aµ =

((1− igsαb(x)λb2 )λa2 − λa

2 (1− igsαb(x)λb2 ))Gaµ + ∂µα

b(x)λb2

= igsαb(x)

[λa2 ,

λb2

]Gaµ + ∂µα

b(x)λb2

= −gsαb(x)fabcλc2 Gaµ + ∂µα

b(x)λb2

where we have used 5.11. Note that in this expression, there are implicit summation over indicesa, b. We can simplify the expression by changing the labels of the indices: in the first term ofthe right-hand side, we do a↔ c and in the second term b↔ a. Hence:

λa2δGaµ = −gsαb(x)fcba

λa2Gcµ + ∂µα

a(x)λa2

so that we can remove the λa2 (because the expression above must be satisfied for any αa(x), αb(x)

). Now using the property 5.12 fcba = fbac = −fabc, we finally obtain the transformation rule10

for the gluon fields:

Gaµlocal−−−→ G′aµ = Gaµ + gsα

b(x)fabcGcµ + ∂µα

a(x) (5.21)

where there is an implicit summation over b and c.

Let us recap: the invariance of the Lagrangian 5.17 when the quarks field are transformedunder a SU(3)c transformation led us to introduce 8 gluons fields via the covariant derivative5.16 that transform as in 5.21.

Can we go further to interpret the 8 gluons fields? The fact that there are 8 such objects(all a priori equivalent with respect to the strong interaction) suggests that they belong to anoctet of colour. Recalling that:

3⊗ 3 = 8⊕ 1

we notice that the single state can be built (modulo a factor) with:

1 ≡ c†.c = (r, g, b).

rgb

= rr + gg + bb

(here r∗ and r are equivalent notation). Similarly, 8 linearly independent states can be builtwith:

8 ≡ ~w =

w1

w2...w8

with wa = c†

λa2c = (r, g, b)

λa2

rgb

(a = 1, 8)

10In some text book, the sign is reversed because the initial transformation is e+igsαa λa

2 . In others, gs is notincluded into the exponent but only in the covariant derivative leading to the final term αb(x)fabcG

cµ+ 1

gs∂µα

a(x).

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138 Quantum Chromodynamic

If these 8 states correspond to a eight-dimension representation of SU(3)c (as we claim), theirtransformation under SU(3)c must give specific linear combination of themselves, so that thetransformed fields are still members of the same multiplet (by definition of a group). Let uscheck using an infinitesimal transformation:

w′a = c†(e−igsα

b(x)λb2

)†λa2 e−igsαb(x)

λb2 c

' c†(

1 + igsαb(x)λb2

)λa2

(1− igsαb(x)λb2

)c

' c†λac− igsαb(x)c†(λa2λb2 −

λb2λa2

)c

= wa − igsαb(x)ifabc c† λc

2 c= wa + gsα

b(x)fabc wc

(5.22)

where we have used the property λ† = λ, the commutation properties of λ and kept only firstorder terms. As announced, the transformed w′a results from a linear combination of themselves.For a finite transformation, we must “exponentiate” the last result giving:

wa − igsαb(x)ifabc wc = wa − igsαb(x)(−ifbac) wc ⇒ ~w′ = e−igs~α(x). ~G(8)~w

where ~G(8) is a vector of eight 8× 8 matrices G(8)1 · · ·G

(8)8 having elements

(G

(8)b

)ij

= −ifbij (5.23)

Moreover, one can show that the G(8) matrices verify:[G(8)a , G

(8)b

]= ifabcG

(8)c (5.24)

Hence, these matrices constitute a representation of SU(3)c. This is a property of the Lie groupfor which, such representation, called adjoint representation always exists and has the matrixelements given by minus the structure constants.

Comparing the transformation of the gluon fields 5.21 and the one of wa in 5.22, we see thatthe gluons fields belong to the adjoint representation11 of SU(3)c. Hence, they are members ofthe colour octet and carry both colour and anti-colour charge. For example, w1 = c† λ1

2 c = rg+gr.

In conclusion, promoting the SU(3)c colour symmetry as a local gauge symmetry gives riseto 8 fields, the gluons, that are members of a SU(3)c octet. The gluons carry a colour and ananti-colour charge. This is a fundamental difference with respect to the QED theory where thephoton does not carry an electric charge. As for the photon, the gluons must be massless sincea term m2GµaGaµ would violate the gauge invariance. Moreover, the gluons are spin 1 particlesince their fields are vector-fields. In addition, looking at Lagrangian 5.17, we see that the gluonfield is coupled to a quark (spin 1/2) and antiquark field (spin 1/2) via the current 5.18.

5.2.4 QCD Lagrangian

The generalization of the Lagrangian 5.17 to the 6 flavours of quarks (denoted by f below) isstraightforward:

L =∑

f

qf (iγµ∂µ −m)qf − gs qfγµλa2qf G

aµ (5.25)

11If the transformation was global, the identification of Gµa to wa would be perfect, since the term ∂µαa in 5.21

would be zero (α not depending on x for global transformation).

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QCD Lagrangian 139

the term:

Lint =∑

f

−gs qfγµλa2qf G

aµ (5.26)

describing the coupling (ie the interaction) between the gluons and the (anti-)quarks. It isimportant to realize that 5.25 is a compact way of writing:

L =∑

f

i,j

fi(iγµ∂µ −m)δij fj − gs fiγµ

1

2(λa)ij fj G

where i, j are the colour indices (from 1 to 3). As in QED, we must add a term describing thepropagation of the gluons, namely the kinetic energy term. In QED, this term is:

L = −1

4FµνFµν with Fµν = ∂µAν − ∂νAµ

It would be tempting to propose:

L = −1

4Gµνa Gaµν with Gaµν = ∂µG

aν − ∂νGaµ

However, this term is not gauge invariant. Indeed:

∂µGaν → ∂µG

′aν = ∂µG

aν + gs(∂µα

b)fabcGcν + gsα

bfabc ∂µGcν + ∂µ∂να

a

∂νGaµ → ∂νG

′aµ = ∂νG

aµ + gs(∂να

b)fabcGcµ + gsα

bfabc ∂νGcµ + ∂ν∂µα

a

∂µG′aν − ∂νG′aµ = ∂µG

aν − ∂νGaµ + gsfabc

[(∂µα

b)Gcν − (∂ναb)Gcµ

]+ gsfabcα

b[∂µG

cν − ∂νGcµ

]

⇒ δ(∂µGaν − ∂νGaµ) = gsfabc

[(∂µα

b)Gcν − (∂ναb)Gcµ

]+ gsfabcα

b[∂µG

cν − ∂νGcµ

]

In fact, in order to compensate these additional terms, one should use instead:

Gaµν = ∂µGaν − ∂νGaµ − gsfabcGbµGcν (5.27)

It is important to realize that in addition to the purely kinetic terms ∂µGaν − ∂νGaµ, the term

−gsfabcGbµGcν adds an interaction between 2 gluon fields. We will stress this point in the nextsection. Let us first check the gauge invariance:

δGaµν = δ(∂µGaν − ∂νGaµ)− gsfabcδ(GbµGcν)

but (keeping only first order terms):

G′bµG′cν = (Gbµ + gsα

dfbdeGeµ + ∂µα

b)(Gcν + gsαdfcdeG

eν + ∂να

c)

⇒ δ(GbµGcν) = gsα

d(fcdeGeνG

bµ + fbdeG

eµG

cν) +Gbµ∂να

c +Gcν∂µαb

and thus:

δGaµν = gsfabc[(∂µα

b)Gcν − (∂ναb)Gcµ

]+ gsfabcα

b[∂µG

cν − ∂νGcµ

]

−g2sα

d(fabcfcdeGeνG

bµ + fabcfbdeG

eµG

cν)− gsfabc(Gbµ∂ναc +Gcν∂µα

b)

Changing the dummy labels in −gsfabcGbµ∂ναc → −gsfacbGcµ∂ναb = gsfabcGcµ∂να

b, and in

fabcfcdeGeνG

bµ → fabefedcG

cνG

bµ → faebfbdcG

cνG

eµ = −faebfdbcGcνGeµ , it remains:

δGaµν = gsfabcαb[∂µG

cν − ∂νGcµ

]− g2

sαd(−faebfdbc + fabcfbde)G

eµG

= gsfabcαb[∂µG

cν − ∂νGcµ

]− g2

sαd(−faebfdbc + fabcfdeb)G

eµG

= gsfabcαb[∂µG

cν − ∂νGcµ

]− g2

sαd(fdebfabc − faebfdbc)GeµGcν

= gsfabcαb[∂µG

cν − ∂νGcµ

]− g2

sαd(i(G

(8)d

)ebi(G

(8)a

)bc− i(G

(8)a

)ebi(G

(8)d

)bc

)GeµG

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140 Quantum Chromodynamic

where we have used 5.23. Now, one has to keep in mind that colour indices that are repeatedmean a summation regardless of the indices position12 (in other words fabc = fabc = fabc etc). Ifyou’re not convinced, it suffices to notice that the left-hand side only depends on the a index.Thus, the right-hand side cannot depend on the other indices. Hence:

δGaµν = gsfabcαb[∂µG

cν − ∂νGcµ

]+ g2

sαd((G

(8)d G

(8)a

)ec−(G

(8)a G

(8)d

)ec

)GeµG

= gsfabcαb[∂µG

cν − ∂νGcµ

]+ ig2

sαd fdaf

(G

(8)f

)ecGeµG

cν relation 5.24

= gsfabcαb[∂µG

cν − ∂νGcµ

]+ g2

sαb fbafffecG

eµG

cν d→ b

= gsfabcαb[∂µG

cν − ∂νGcµ

]+ g2

sαb fbacfcekG

eµG

kν c→ k and f → c

= gsfabcαb[∂µG

cν − ∂νGcµ − gsfcekGeµGkν

]

= gsfabcαb Gcµν

Finally, the variation of the kinetic energy gives:

δ(Gµνa Gaµν) = 2Gµνa δGaµν = 2gsfabcαbGµνa Gcµν = gsfabcα

bGµνa Gcµν + gsfabcαbGµνa Gcµν

= gsfabcαbGµνa Gcµν + gsfcbaα

bGµνc Gaµν = gsfabcαbGµνa Gcµν − gsfabcαbGµνc Gaµν

= gsfabcαbGµνa Gcµν − gsfabcαbGcµνGµνa

= 0

Thus, the kinetic energy term is indeed gauge invariant. Hence, the whole QCD Lagrangian is:

LQCD =∑

f

qf (iγµ∂µ −m)qf − gs qfγµλa2qf G

aµ −

1

4Gµνa Gaµν (5.28)

the tensor Gµνa being given by 5.27.

5.3 QCD Feynman rules

We are going to derive the QCD Feynman rules from the Lagrangian 5.28. Only the first order(no loop) will be shown. The comparison with QED Lagrangian will help us in this task (avoidingto describe quantization of field). Let us recall it here:

LQED = ψ(iγµ∂µ −m)ψ − qψγµψAµ −1

4FµνFµν

where ψ is the field of a spin 1/2 fermion. The first term:

ψ(iγµ∂µ −m)ψ

describes a free fermion. In QCD we have the equivalent term:

qf (iγµ∂µ −m)qf ≡∑

αβ

fα(iγµ∂µ −m)δαβ fβ

where the colour indices are noted α and β. The delta is there to be sure that the propagationof a free quark of flavour f does not change the colour it carries. Note that we can rewrite theprevious expression differently using the color vector c:

c1 =

100

for red, c2 =

010

for green, c3 =

001

for blue

12This is not the case for space-time indices!

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QCD Feynman rules 141

Hence, the free lagrangian is:

qf (iγµ∂µ −m)qf ≡∑

αβ

f c†α(iγµ∂µ −m)cβf

with of course: c†αcβ = δαβ.

The propagator of a quark is the same as the one of any massive fermion namely:

= i /p+m

p2−m2+iε

Now, consider the second term in QED:

−qψγµψAµ

It gave rise to the interaction between a charged fermion and a photon with a vertex factor−iqγµ. The interaction between a quark and a gluon is given equivalently by:

−gs qfγµλa2qf G

aµ ≡

αβ

−gs fαγµ(λa)αβ

2fβ G

aµ =

αβ

−gs f c†αγµλa2cβf G

where the summation over the 8 gluons fields (a running from 1 to 8) is implicit. The vertexfactor for qqg is straightforward to extrapolate from QED:

qβ qα−igsγµ

(λa)αβ2

Gaµor equivalently

cβq c†αq−igsγµ λa2

Gaµ

Now, let us focus on the kinetic energy term:

−14G

µνa Gaµν = −1

4(∂µGνa − ∂νGµa − gsfabcGµbGνc )(∂µGaν − ∂νGaµ − gsfadeGdµGeν)

= −14(∂µGνa − ∂νGµa)(∂µG

aν − ∂νGaµ)

+12gsfabc(∂

µGνa − ∂νGµa)GbµGcν

−14g

2sfabcfadeG

µbG

νcG

dµG

The first term is the pure kinetic energy term similar to −14F

µνFµν in QED. It corresponds tothe gluon propagator, almost similar to the photon propagator since it is a massless particle ofspin 1:

= −igµνδabp2+iε

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142 Quantum Chromodynamic

where a, b runs over the 8 gluon colour indices 1 · · · 8. This formula is valid in a gauge calledthe Feynman gauge. The gluon being a massless particle of spin 1, it must be described as forthe photon by a polarization vector εµ for incoming real gluon and εµ∗ for outgoing real gluon.Its massless nature also imposes kµεµ = 0 where kµ is the gluon 4-momentum (transversality).The color/anti-color of the gluon defines the λ matrix that has to be used at the vertex (in caseof vertex qqg). Concretely, a gluon Ga with a being within [1,8], is necessarily connected to avertex with the λa matrix (in case of vertex qqg).

The 2 others terms are specific to QCD: they describe respectively 3 gluons interactionsand 4 gluons interactions. They are allowed since the gluons carry colours (and anti-colours)contrary to the photon which has a zero charge. One can show [22, p. 280] that it yields thefollowing vertices factors:

Gbν(q2) Gcρ(q3)

Gaµ(q1)

−gsfabc [gµν(q1 − q2)ρ + gνρ(q2 − q3)µ + gρµ(q3 − q1)ν ]q1 + q2 + q3 = 0

and the 4 gluons vertex factor:

Gaµ

Gbν

Gcρ

Gdη

−ig2s

[fabef cde(gµρgνη − gµηgνρ) + fadef bce(gµνgρη − gµρgνη) + facef bde(gµηgνρ − gµνgρη)

]

e : 1→ 8 and∑4

i=1 qi = 0

where e is running from 1 to 8.A final remark about these Feynman’s rules. It is important to notice that for any QCD

interactions (qqg, ggg, gggg), the same coupling constant gs is involved.

5.4 QCD colour factors

5.4.1 Quark-antiquark interaction

Let us consider the scattering between a u quark and an antiquark d, as it often occurs atprotons colliders:

ud→ ud

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Quark-antiquark interaction 143

We assume here that we can use the perturbation approach of Feynman’s diagram, namely thatthe coupling constant of the strong interaction is weak enough. The corresponding graph is:

dρ(p2)

uα(p1)

dη(p′2)

uβ(p′1)

Gaµ

Gbµ

Applying the QCD Feynman rules of the previous section we find:

iM =

[uu(p′1)

(−igsγµ

(λa)βα2

)uu(p1)

](−igµνδ

ab

q2

)[vd(p2)

(−igsγν

(λb)ρη2

)vd(p

′2)

]

where q = p1 − p′1. Simplifying with the δ and gµν , it comes:

M =g2s

4(λa)βα(λa)ρη

1

q2

[uu(p′1)γµuu(p1)

] [vd(p2)γµvd(p

′2)]

(5.29)

where there is an implicit summation over a (and over µ of course). The amplitude can becompared to an equivalent scattering in QED e−µ+ → e−µ+:

M = |e2| 1q2

[ue(p

′1)γµue(p1)

] [vµ(p2)γµvµ(p′2)

](5.30)

We have obviously the change of coupling constant |e| → gs or equivalently α to:

αs =g2s

4π(5.31)

and in addition, a colour factor:

Cαρ→βη =1

4(λa)βα(λa)ρη =

1

4

8∑

a=1

(λa)βα(λa)ρη (5.32)

Hence the difference between the QED process and the QCD one can be summarised with:

QED: α QCD: C αs

The calculation of a QCD graph is thus identical to a QED one knowing these prescriptions. Assoon as you know how to compute a colour factor, there is no difference. We are going to seeseveral examples in what follows.

We wish now to address several consideration about the QCD potential felt by the system qq.For the QED processes with opposite charges, we know that an attractive potential is describedby:

V (r) = −αr

Notice the negative sign: the energy of the system made of opposite charges decreases. In otherwords, the presence of the opposite charges has lowered the energy, hence the attractive nature

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144 Quantum Chromodynamic

of the interaction. We can then proceed by analogy and since both QED and QCD amplitudeshave a similar form, we can suppose that a QCD potential (at high energy) would be:

V (r) = −Cαsr

Hence, if C is positive, the potential would be attractive and repulsive otherwise. Let us calculateC assuming the u and d form a meson bound state (π+ for example). We claimed that suchbound state must be a colour singlet meaning that it is

ψ =1√3

(rr + gg + bb)

Thus, the potential felt by a singlet is:

〈V singlet

ud〉 = 〈ψ|Vud|ψ〉

= 13

(∑c=r,g,b 〈cc|Vud|cc〉+

∑c′ 6=cc,c′=r,g,b 〈cc|Vud|c′c′〉

)

= 13

(∑c=r,g,b−Ccc→cc αsr +

∑c′ 6=cc,c′=r,g,b−Ccc→c′c′ αsr

)

= −Csinglet αsrwith the colour factor for a singlet state:

Csinglet =1

3

c=r,g,b

Ccc→cc +

c′ 6=c∑

c,c′=r,g,b

Ccc→c′c′

According to 5.32 and the Gell-Mann matrices 5.10, for example:

Crr→rr =1

4

8∑

a=1

(λa)11(λa)11 =1

4[(λ3)11(λ3)11 + (λ8)11(λ8)11] =

1

4

[1 +

1

3

]=

1

3

The colour factors for gg → gg or bb → bb are identical because of the gauge symmetry. For

Cc′ 6=ccc→c′c′ , we can compute Crr→gg where the colour of the u-quark moves from red to green, so

α = 1 and β = 2 and the colour of d from r to g, so ρ = 1 and η = 2. Hence:

Cc′ 6=ccc→c′c′ = Crr→gg =

1

4

8∑

a=1

(λa)21(λa)12 =1

4[(λ1)21(λ1)12 + (λ2)21(λ2)12] =

1

4[1 + i×−i] =

1

2

And hence:

Csinglet =1

3

c=r,g,b

1

3+

c′ 6=c∑

c,c′=r,g,b

1

2

=

1

3(3× 1

3+ 6× 1

2) =

4

3> 0

The meson potential is thus:

V qqsinglet(r) = −4

3

αsr

(5.33)

which is attractive as expected for a bound state. What about a qq state member of the coloroctet? The eight states are:

rg, gr, rb, br, gb, bg,1√2

(rr − gg),1√6

(rr + gg − 2bb) (5.34)

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Quark-quark interaction 145

(see figure 5.3 and replace u→ r, d→ g and s→ b). The octet potential is then:

〈V udoctet〉 = 〈rg|Vud|rg〉

(or any states of the octet) and therefore, we must evaluate u : r → r ⇒ α = β = 1 andd : g → g ⇒ ρ = η = 2 :

Coctet = Crg→rg =1

4[(λ3)11(λ3)22 + (λ8)11(λ8)22] =

1

4

[−1 +

1

3

]= −1

6

We would have found the same factor for any of the 8 members of the octet in virtue of thegauge symmetry. And hence:

V qqoctet(r) =

1

6

αsr

(5.35)

which is repulsive.

It is worth mentioning that there is an alternative recipe to determine the magnitude ofcolour factors. The recipe is to compute a factor at each vertex c1 and c2 knowing the gluonsthat have to be exchanged and the global colour factor is then:

C =∑

gluons

1

2c1c2 (5.36)

Let us take the example of Crg→rg. The u quark vertex is r → r and for d, g → g. Among the 8gluons 5.34, only the one having rr and gg can be exchanged ie 1√

2(rr− gg), 1√

6(rr+ gg− 2bb).

With the first, we have:

c1 =1√2, c2 = − 1√

2⇒ c1c2 = −1

2

For the second gluon:

c1 =1√6, c2 =

1√6⇒ c1c2 =

1

6

And thus:

C =1

2

(−1

2+

1

6

)= −1

6

as we found before.

5.4.2 Quark-quark interaction

We can do the same exercise with a pair quark-quark instead of qq.

dρ(p2)

uα(p1)

dη(p′2)

uβ(p′1)

Gaµ

Gbµ

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146 Quantum Chromodynamic

According to the graph above, the colour factor is:

Cαρ→βη =1

4

8∑

a=1

(λa)βα(λa)ηρ

Note the inversion of the η and ρ indices with respect to the previous case (equ. 5.32): thedirection of the arrow in the graph indicates the order of indices. We know that 3⊗ 3 = 6⊕ 3.We would have found that the colour factor for states in 3 yields an attractive potential and arepulsive one in the sextet (see an exercise at the end of this chapter), namely for 3:

V qq3

(r) = −2

3

αsr

However, we have never seen a bound state of qq (which would be coloured). Firstly, one shouldnotice that it is twice less attractive than for a qq singlet (-2/3 instead of -4/3). Secondly,if combined to a third quark to form a baryon, the singlet state of qqq yields again a moreattractive potential.

Of course, what was done in this section is not a proof of the colour confinement: we shouldhave found an infinitely repulsive potential for any coloured state. It just gives strong indicationswhy bound states are always found in singlet configuration. Bound states are characterized bygluon exchange at low Q2 for which the coupling constant αs becomes large. Perturbation theorycannot be applied reliably. One should use other approaches, such as Lattice QCD based onnumerical calculations. Using the lattice QCD approach, Wilson showed that QCD exhibitsconfinement of colour, with the only finite energy states are those that are singlets of colourSU(3)c.

5.5 QCD properties

5.5.1 Asymptotic freedom

There is an apparent paradox: we showed how to calculate QCD amplitude, based on pertur-bation theory, meaning that the strong coupling constant must be weak enough. However, weadmitted and it has never been contradicted by the experimental facts, that QCD should be verystrong to bind (anti-)quarks together so that no coloured objects can be seen as a free particle innature. Furthermore, in the previous chapter, where the parton model was presented, we alwaysassumed for deep inelastic scattering that partons (namely quarks and gluons) inside the hadrondo not interact together as if they were free. And this assumption was valid since the partonmodel, in general, compares well with experimental data. The resolution of this paradox, relieson a QCD property related to the non-abelian gauge, called asymptotic freedom and was foundby Gross, Politzer and Wilczek in 1973 who won the Nobel price in 2004 for their discovery.

In QED, we saw that the renormalization procedure leads to the concept of running couplingconstant (and mass). In QCD, we have the same situation. The QCD vacuum polarizationreceives contribution from diagram such as the ones below.

These 2 contributions have opposite effect as one can imagine since in the first diagram, afermion loop runs while in the second it is a boson loop. The latter is possible in QCD becausegluons carry colour charges. The global effect of all diagrams yields the following behaviour ofthe coupling constant [8, p. 541]:

αs(Q2) =

αs(µ2)

1 + αs(µ2)(33−2Nf )

12π log(Q2

µ2

) (5.37)

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Non perturbative QCD 147

q

q

q

q

q

q

q

q

Figure 5.7: The 1 loop contribution to QCD vacuum polarization.

where Nf is the number of flavours, and µ2 the renormalization scale. In QED, we found asimilar equation 3.71:

α(Q2) =α(µ2)

1− α(µ2)3π log

(Q2

µ2

)

But there is a major difference: since Nf = 6, the term in front of the logarithm is positive whileit is negative in QED. We see that when Q2 becomes larger and larger, in QED, α increases,while in QCD it decreases: as soon as the energy transfer is large enough, αs becomes very smallleading to a quasi-free behaviour of the partons: this is the asymptotic freedom. Hence, in deepinelastic scattering we can safely consider the partons inside the hadrons as free.

According to 5.37, one can find an energy scale ΛQCD for which αs(Λ2) becomes infinite:

1 + αs(µ2)

(33− 2Nf )

12πlog

(Λ2QCD

µ2

)= 0⇒ αs(Q

2) =12π

(33− 2Nf ) log

(Q2

Λ2QCD

)

This expression is valid only for Q2 > Λ2QCD. The scale ΛQCD gives an idea where the strong

interaction becomes really strong. It is estimated to be of the order of 200 MeV, so a distanceof 1 fm which is the typical size of hadrons.

5.5.2 Non perturbative QCD

The particles that are detected in the experiments are not the quarks or gluons, but hadronswhich are colourless objects. What is the process that yields those hadrons? The idea is thefollowing: imagine a pair qq produced by e+e− annihilation in a collider. Because of the initialboost, they separate. When the distance increases or in other words the Q2 decreases, theircolour interaction becomes stronger because of the way αs depends on Q2. The potential energybetween them becomes so strong that it exceeds the energy needed to create another pair qqthrough gluon emission. If there is still enough kinetic energy to separate the quarks, theywill create other pairs until all pairs of qq have not enough kinetic energy to separate anymoreand hence remain bound in the created hadrons. This process, highly non-perturbative is calledhadronization. The hadrons produced from the initial qq pair move roughly in the same directionas the initial quark or antiquark, creating what is called jets (of hadrons). A sketch of the processis shown in figure 5.8. Experimentally, a quark of 10 GeV produces about 7 hadrons, while at100 GeV, it yields about 15 hadrons.

For distance of the order of 1 fm (thus in the non perturbative QCD regime), the so-calledlattice QCD in which the space–time continuum is represented as a discrete lattice of points,

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148 Quantum Chromodynamic

Figure 5.8: Production of jets of hadrons due to colour confinement. From [23]

allows to do some numerical prediction. For example, the QCD static potential is found to be[7, p. 187]:

V (r) = −4

3

αsr

+ λr (5.38)

which describes well the data with λ ' 0.85 GeV.fm−1. This is obviously the λr part which isresponsible for the colour confinement. In usual units, λ is of the order of 105 N!!!

5.6 Experimental evidences

5.6.1 Discovery of the gluon

In QCD, high energetic quarks can radiate a gluon via for example the process below:

e−

e+

q

q

Figure 5.9: The first order graph for e+e− → qqg.

If the energy of the gluon is large enough, we expect it to produce a jet well separated from thequark jet. At e+e− collider, the topology of such event would be characterized by the presenceof 3 jets. Moreover, one of these jets must have a total zero electric charge if originated from agluon (which is neutral). In 1979, at the PETRA e+e− storage ring, the first 3 jets events withthese properties were observed for the first time by the TASSO experiment. A display of one ofthis event is shown in figure 5.10. The energy in the centre of mass was about 30 GeV. The 3jets were coplanar as expected for e+e− → qqg. The analysis of the angular distribution showedthat the gluon is a spin-1 particle.

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Measurement of αs 149

The radiation of energetic gluons o! high energetic quarks was predicted from QCD, and hence theemergence of a third hadron jet was expected in e+e! annihilation at higher energies [17]. In 1979, 3-jetstructures were observed at the PETRA e+e! collider at Ecm ! 30 GeV [18], see figure 2. The 3rd jetcould be attributed to the emission of a third parton with zero electric charge and spin 1 [19] – thegluon was discovered and explicitly seen!

Figure 2: The first 3-jet event, observed by the TASSO experiment at the PETRA e+e! storage ring[18].

After an almost parameter-free description of hadronic event shapes [20, 21] had served to establishgluon radiation beyond any doubts, first determinations of the coupling strength !s were performed.More detailed studies of the jet structure and the shape of hadronic events provided insights into thecolor structure of the gluon, like for instance the “string e!ect” [22] which was expected in QCD dueto the higher color charge of the gluon [23].

2.7 Status in the late 1980’s

Towards the end of the 1980’s, after successful exploitation of the experimental programs of the e+e!

storage rings PETRA at DESY and PEP at SLAC, the proton-antiproton collider at CERN and manydeep inelastic lepton-nucleon experiments at CERN, at Fermilab and at SLAC, “the experimental sup-port for QCD is quite solid and quantitative” [24], however convincing proofs of the key features ofQCD, of the gluon-selfcoupling and/or of asymptotic freedom, were still not available. As an example,the summary of measurements of !s at di!erent energy scales in 1989, as reproduced in figure 3 [24],

7

Figure 5.10: The first 3 jets event, evidence for a gluon emission. From the review [24]

5.6.2 Measurement of αs

Consider again the graph of figure 5.9. The gluon coupling involves αs. Hence the measurementof the cross-section of 3 jets events, or more generally the ratio:

Rhad =σe+e−→hadronsσe+e−→µ+µ−

allows the measurement of the strong coupling constant. Many experiments contribute to theαs measurement. Since they usually probe different Q2 regions, one has to use equation 5.37(or more precisely a version of 5.37 which involves more loops that the ones shown in diagrams5.7) to compare the measurement at the same Q2 which is usually chosen at the Z0 mass. Asummary is shown in figure 5.11. The data are well reproduced by the theory visualized withthe yellow band. Note that for Q = mZ = 91 GeV, αs ≈ 0.12, which is still 15 times larger thanαem at that mass (αem(mZ) ≈ 1/128).

5.6.3 Measurement of colour factors

The colour factors we introduced in the previous section result directly from the SU(3) gaugegroup of QCD. It is then fundamental to measure them and compare to the theory. We canenvisage 3 specific colour factors involved in the following processes:

• gluon radiation, q → qg, colour factor denoted traditionally by CF

• triple gluon vertex g → gg, denoted by CA

• gluon splitting g → qq denoted by TF .

One can show that they are given by:

1

4

8∑

a=1

3∑

β=1

λaαβλaβρ = δαρCF ,

8∑

b,c=1

fabcfdbc = CAδad ,1

4

3∑

α,β=1

λaαβλbβα = TF δab

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150 Quantum Chromodynamic

threshold matching at the heavy quark pole masses Mc = 1.5 GeV and Mb = 4.7 GeV. Results fromdata in ranges of energies are only given for Q = MZ0 . Where available, the table also contains thecontributions of experimental and theoretical uncertainties to the total errors in !s(MZ0).

Finally, in the last two columns of table 1, the underlying theoretical calculation for each mea-surement and a reference to this result are given, where NLO stands for next-to-leading order, NNLOfor next-next-to-leading-order of perturbation theory, “resum” stands for resummend NLO calculationswhich include NLO plus resummation of all leading und next-to-leading logarithms to all orders (see[39] and [32]), and “LGT” indicates lattice gauge theory.

Figure 17: . Summary of measurements of !s(Q) as a function of the respective energy scale Q, fromtable 1. Open symbols indicate (resummed) NLO, and filled symbols NNLO QCD calculations used inthe respective analysis. The curves are the QCD predictions for the combined world average value of!s(MZ0), in 4-loop approximation and using 3-loop threshold matching at the heavy quark pole massesMc = 1.5 GeV and Mb = 4.7 GeV.

In figure 17, all results of !s(Q) given in table 1 are graphically displayed, as a function of theenergy scale Q. Those results obtained in ranges of Q and given, in table 1, as !s(MZ0) only, are notincluded in this figure - with one exception: the results from jet production in deep inelastic scatteringare represented in table 1 by one line, averaging over a range in Q from 6 to 100 GeV, while in figure 17combined results for fixed values of Q as presented in [67] are displayed.

28

Figure 5.11: Summary of αs measurements as function of the energy scale. From [24]

but we can easily calculate them by hand using the alternative method shown at the end ofthe previous section. For instance, in the process q → qg, the initial quark can have 3 differentcolours while there are 8 possible outgoing gluons. Hence the average colour factor is:

CF =1

2

8

3=

4

3

where the 1/2 is coming from the convention (see in 5.36). Similarly, for g → qq, 8 initial gluonsand 8 possible qq states (since the gluon has a colour the system qq must carry a colour excludingthe singlet state):

TF =1

2

8

8=

1

2

And finally for g → gg, a given initial gluon carries a colour and an anticolour that must bepresent in the 2 final gluons. And since the colour is conserved, any new colour in one finalgluon must be compensated by the corresponding anti-colour in the other. Hence, there are only3 possibilities:

CA = 3

Experimentally, several approaches exist. For example, consider the 4 jets production at e+e−

colliders. The relevant graphs are:

e−

e+

q

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Back to the parton model 151

The first graph involves CF , the second TF and CF and the last one CA and CF . Moreover,the angular correlations between the 4 jets differ in these 3 graphs: q → qg implies spin 1/2 tospin 1/2-spin 1 objects, while g → gg implies spin 1 to two spin 1 objects. Hence, it allows todetermine the different contributions and thus to measure CF , CA and TF .

The figure 5.12 shows an example of the colour factor measurement (yellow ellipse) obtainedat the LEP collider e+e− at CERN. Different models of gauge group are compared and as canbeen seen only SU(3) for which CF = 4/3 and CA = 3 is compatible with the data.

0

0.5

1

1.5

2

2.5

0 1 2 3 4 5 6

U(1)3

SU(1)

SU(2)

SU(4)

SU(5)

Combined result

SU(3) QCD

ALEPH 4-jet

OPAL 4-jet

Event Shape

OPAL Ngg

DELPHI FF

CF

CA

86% CL error ellipses

Figure 11: Measurements and combination of the QCD colour factors CA and CF [55].

and based on many di!erent methods and observables. While an overall summary of !s measurements,as presented in section 6, gives a very distinct signature for asymptotic freedom, the demonstrationof the running of !s from single experiments, minimising point-to-point systematic uncertainties, addsextra confidence in the overall conclusion.

One of the most recent developments in determining !s in e+e! annihilation is the precise extractionof !s from di!erential 4-jet distributions, i.e. the 4-jet production rate as a function of the jet resolutionycut [56, 57, 58]. QCD predictions in NLO are available [59], which is in O(!3

s ) perturbation theory.4-jet final states appear only at O(!2

s ) as their leading order, since they require at least two radiationor splitting processes o! the primary quark-antiquark pair. Although “only” in NLO, the theoreticaluncertainties for 4-jet observables appear to be rather small, smaller than for typical 3-jet event measuresor event shapes. Since 4-jet observables are proportional to !2

s , they provide very sensitive measurementsof !s.

As another new development, the data of the JADE experiment, at the previous PETRA colliderwhich was shut down in 1986, are currently re-analysed, with the experimental and theoretical experienceand tools of today. This adds important new information especially at lower energies, seen from theperspective of the LEP experiments. Moreover, the JADE experiment can be regarded as the smallerbrother or prototype of the OPAL detector at LEP, since both experiments were based on similardetector techniques.

A LEP-like study of !s from 4-jet production from JADE is available [60], and is summarised infigure 12, together with the corresponding results from OPAL and the other LEP experiments. In orderto appreciate the high significance for the running of !s, one should keep in mind that only the inner(experimental) error bars need to be considered in a relative comparison of these data.

21

Figure 5.12: Combined result of the measurement of the colour factor at the LEP collider. From [24]

Can we distinguish the nature of the jets? In other words, can we know whether a given jet wasgenerated by a gluon or by a quark? Since CA > CF , we conclude that there is more radiationof soft gluons in a gluon jet than in a quark jet. Consequently:

• the gluons jets have a larger multiplicity than the quarks jets and thus the hadrons ingluon jets tend to have a softer energy spectrum.

• the gluons jets are less collimated that the quarks jet.

It is indeed verified experimentally and this property is used to distinguish statistically bothkinds of jets.

5.7 Back to the parton model

5.7.1 QCD corrections

At the end of the previous chapter, we noticed that at very small x-Bjorken (fraction of nucleonmomentum carried by a quark inside the proton), the Bjorken scaling is violated (see figure

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152 Quantum Chromodynamic

4.8). We can now give at least qualitative explanations.13 So far, the basic diagram for the deepinelastic scattering of a proton with an electron is the one below:

p

e−

q

e−

=⇒ q

γ∗

But now, we know that quarks can radiate gluons, and hence, to first order αs, one should addthese diagrams:

q(pi = yp)

γ∗(q)

g

q(zpi = yzp = xp)

q

γ∗

g

q

We are now in a situation where the initial quark carries a fraction y of the proton momentumie pi = yp, radiates a gluon reducing the quark momentum to pf = zpi = zyp = xp, the differentvariables being related by:

Q2 = −q2 , x =Q2

2p.q, x = zy

Since the quark radiates a gluon, its transverse momentum pT with respect to the proton direc-tion is not anymore necessarily zero (the momentum is not aligned with the proton direction).One can then show, that these first order diagrams induce a modification of the structure func-tions F1(x) and F2(x) = 2xF1(x):

F2(x) = x∑

i

q2i fi(x)→ F2(x,Q2) = x

i

q2i fi(x,Q

2) (5.39)

with the quark density of flavour i given by:

fi(x,Q2) = fi(x) +

αs2π

log

(Q2

µ2

)∫ 1

x

dy

yfi(y)Pqq

(x

y

)(5.40)

where µ2 is a cut-off to avoid divergences when pT → 0 (due to very soft gluon radiation). Thefunction Pqq, called splitting function represents the probability of a quark emitting a gluon andso becoming a quark with momentum reduced by a fraction z = x

y :

Pqq(z) = CF

(1 + z2

1− z

)(5.41)

13I won’t provide a demonstration of the formulas that will be written in this section. It is beyond the scope ofthis course. Please consult for example chapter 10 of [9].

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QCD corrections 153

The scaling violation is clearly visible in 5.40 since F2 or equivalently fi depends now on both xand Q2 via the logQ2 term. To first order in αs, the way it varies with logQ2 is from 5.40:

∂ logQ2fi(x,Q

2) =αs2π

∫ 1

x

dy

yfi(y,Q

2)Pqq

(x

y

)(5.42)

This equation is called the DGLAP evolution equation, DGLAP standing for the five physicistswho established it: Dokshitzer, Gribov, Lipatov, Altarelli and Parisi. The physical meaningof the equation is the following: a quark with momentum xp, could have come from a parentquark with a larger momentum yp which has radiated a gluon with a probability αs

2πPqq(x/y).This probability is of course proportional to the colour factor relevant for q → qg. The fractionof momentum of the parent quark carried by the final quark is x/y. According to the formula5.41, the probability is larger when x/y is close to 1. For large x close to 1, since 1 ≥ y ≥ x, theratio x/y is necessarily close to the unity. Hence, the probability to radiate a gluon when x ' 1is very high. Since x = Q2/(2p.q), when Q2 increases and x ' 1, x gets closer to one. Thus, asQ2 increases, we expect the quark to radiate more and more gluons in large x regime. Since itproduces more gluons inside the nucleon, the quark density function and thus F2 should thendecrease at large x with increasing Q2. This is what is observed in figure 4.8 for x > 0.2. Onthe contrary, we expect the opposite behaviour at small x, ie an larger quark density when Q2

increases. The prediction of DGLAP describes well the experimental data in figure 4.8.If we know/measure the quark structure function at a given Q2

0, we can compute it at anyvalue Q2 thanks to the DGLAP equation.

Finally, since we have considered processes at order αs, we should also include the followingprocesses occurring via a gluon producing a pair qq:

g

γ∗

q

q

g

γ∗

q

q

Such processes are possible since we know that by quantum fluctuation, gluons (from the sea)are present in the nucleon. The consequences is a modification of the DGLAP equation 5.42 to:,

∂ logQ2fi(x,Q

2) =αs2π

∫ 1

x

dy

y

[fi(y,Q

2)Pqq

(x

y

)+ g(y,Q2)Pqg

(x

y

)](5.43)

where g(y,Q2) is the gluon density and αs2πPqg represents the probability that a gluon creates a

qq pair, the quark carrying a fraction z = x/y of the gluon momentum (which itself carries afraction y of the proton momentum). Pqg is given by:

Pqg(z) = TF (z2 + (1− z)2) (5.44)

The gluon density g(y,Q2) satisfies as well a DGLAP evolution equations of the kind 5.43.

In summary, we have learnt that when the nucleon structure is probed at larger and largerQ2 i.e. finer and finer scales, more and more partons, namely gluons and qq pairs, appear insuch a way that it can be predicted by QCD theory.

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154 Quantum Chromodynamic

5.7.2 Application to hadrons collider

The parton distribution functions provide a good test of QCD theory and they are essentialat hadron collider since at high energy, the hard scattering occur between partons inside the2 incident hadrons. Let us consider the case of a proton collider like the LHC. The collisionprocess can be summarised by the schema below:

ij

p

p

σ

where i and j denote the 2 partons involved in the hard scattering and σ is the cross sectionof the reaction considered ij → X. This cross-section can be calculated with the perturbationtheory using Feynman diagrams. But, what collides are the 2 protons, and hence, this is thecross-section that the experiment measures. The 2 cross-sections are related:

σpp→X =

∫ 1

0dx1dx2

ij

fi(x1, Q2)fj(x2, Q

2)σij→X (5.45)

where fi,j are the parton distribution function of the partons i, j and Q2 is the momentum scaleat which they are evaluated (via the appropriate DGLAP equation). We have to sum over alltype of partons which could contribute to the final state X. And of course, we have to considerall the possible fraction of momentum x1,2 ∈ [0, 1]. The Q2 should be the scale representativeof the hard process: squared mass of a resonance (if X is produced via the resonance), p2

T etc.

As an example, let us take the Drell-Yan production where i = q, j = q and X = µ−µ+.Let be

√s, the energy in the center of mass of the 2 protons and p1 and p2 their 4-momenta.

Neglecting their mass:s = (p1 + p2)2 ' 2p1p2

Now denoting by pq and pq, the partons 4-momenta, the actual energy√s available for the

partons interaction is (still neglecting the masses):

s = (pq + pq)2 ' 2pqpq = 2x1x2 p1p2 ⇒ s ' x1x2s

The cross-section σqq→µ−µ+ is easy to guess from the one of the QED process e−e+ → µ−µ+.The latter was already calculated in 3.68:

σe−e+→µ−µ+ =4πα2

3s

If the quark charge is Qq, α2 has to be changed to α2Q2

q . What about the additional colourfactor? Since the final state has zero colour and colour is conserved, the quark and antiquarkmust have opposite colour (and obviously opposite charge). Hence, there are only 3 possibilities.Since q and q can come in 3 different (anti-)colours, the colour averaging cross-section is:

σqq→µ−µ+ =1

3× 1

3× 3×

4πα2Q2q

3s= σ0Q

2q with σ0 =

4πα2

9s

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Application to hadrons collider 155

According to 5.45, the double differential cross-section for the protons collision is then:

d

dx1dx2σpp→µ

−µ+= σ0

q

Q2q

[fq(x1,M

2)fq(x2,M2) + fq(x2,M

2)fq(x1,M2)]

where M denote the mass of the pair µ+µ−. The natural scale for Q2 is M2 in this particularcase. We can go a bit further by using the rapidity y (see equ. 1.33) of the muon pair (orequivalently of the parton pair). In the centre of mass of the 2 protons, the component ofparton 4-momenta are:

pq =

√s

2(x1, 0, 0, x1) , pq =

√s

2(x2, 0, 0,−x2)

where as usual, the 2 protons beams are moving along the z-axis. Hence,

E =

√s

2(x1 + x2) , pL =

√s

2(x1 − x2)

Thus:

y =1

2ln

(E + pLE − pL

)=

1

2lnx1

x2

and combine with the relation:s = x1x2s

it gives:

x1 =

√s

sey , x2 =

√s

se−y (5.46)

namely:

x1,2 =M√se±y (5.47)

We can then change the variables x1,2 for y and M that can be measured in the detector. Now:

dx1dx2 =

∣∣∣∣∣∂x1∂y

∂x1∂M

∂x2∂y

∂x2∂M

∣∣∣∣∣ dydM =

∣∣∣∣∣M√sey 1√

sey

−M√se−y 1√

se−y

∣∣∣∣∣ dydM =2M

sdydM

Hence:

d

dydMσpp→µ

−µ+=

2M

sσ0

q

Q2q

[fq(x1,M

2)fq(x2,M2) + fq(x2,M

2)fq(x1,M2)]

and injecting the formula of σ0 with s = M2, we finally obtain:

d

dydMσpp→µ

−µ+=

8πα2

9Ms

q

Q2q

[fq(x1,M

2)fq(x2,M2) + fq(x2,M

2)fq(x1,M2)]

Hence, different values of M and y probe the parton distribution functions at different values ofx-Bjorken. The formula relating x1,2 to y and M is given in 5.47. It is obviously general and notonly valid in the Drell-Yan process. The domain in (x,Q2) plan accessible for different valuesof the measured variables y and M is shown in figure 5.13, for the LHC collider assuming thenominal energy

√s = 14 TeV. For a given rapidity y there are two (dashed) lines, corresponding

to the values of x1 and x2. Since both x1 and x2 must be within the range [0,1], it restrictedthe range of possible rapidities y. As can be seen, the larger the mass of the produced system,the more central (ie y → 0) it is produced.

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156 Quantum Chromodynamic

Hard Interactions of Quarks and Gluons: a Primer for LHC Physics 9

Figure 3. Graphical representation of the relationship between parton (x, Q2)

variables and the kinematic variables corresponding to a final state of mass M produced

with rapidity y at the LHC collider with!

s = 14 TeV.

The double–di!erential cross section is therefore

d!

dM2dy=

!0

Ns

! "

k

Q2k(qk(x1, M

2)qk(x2, M2) + [1 " 2])

#. (11)

with x1 and x2 given by (10). Thus di!erent values of M and y probe di!erent values

of the parton x of the colliding beams. The formulae relating x1 and x2 to M and y

of course also apply to the production of any final state with this mass and rapidity.

Assuming the factorization scale (Q) is equal to M , the mass of the final state, the

relationship between the parton (x, Q2) values and the kinematic variables M and y is

illustrated pictorially in Figure 3, for the LHC collision energy!

s = 14 TeV. For a

given rapidity y there are two (dashed) lines, corresponding to the values of x1 and x2.

For y = 0, x1 = x2 = M/!

s.

In analogy with the Drell–Yan cross section derived above, the subprocess cross

sections for (on–shell) W and Z production are readily calculated to be

!qq!!W ="

3

!2GFM2

W |Vqq!|2#(s # M2W ),

!qq!Z ="

3

!2GFM2

Z(v2q + a2

q)#(s # M2Z), (12)

Figure 5.13: Relationship between parton (x,Q2) variables and the kinematic variables correspondingto a final state of mass M produced with rapidity y. From [25].

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Exercises 157

5.8 Exercises

Exercise 5.1 Applications of the isospin symmetry.We recall that the deuteron d is a bind state of one proton and one neutron (nucleus of theDeuterium) and is an isopsin singlet.

1. The reaction p + p → π+ + d is easily seen. Explain why it excludes that the pion is anisospin singlet or a member of a doublet.

2. We now assume that pions are members of an isotriplet. Using the isospin invariance,show that:

σ(p+ p→ π+ + d) = σ(n+ n→ π− + d)

3. What do you expect for σ(n+ p→ π0 + d) with respect to the 2 previous cross-sections?

Exercise 5.2 Color factors for baryons.

1. Show that the combination of the color triplet and anti-triplet gives: 3⊗ 3 = 6⊕ 3

2. Show that a member of sextet is for example |rr〉 and a member of the triplet 1√2|rg − gr〉.

3. Compute the corresponding color factors and conclude about their repulsive or attractivebehavior.

4. The baryons are based on 3⊗ 3⊗ 3. Based on the previous conclusion, can you guess ifthe baryon multiplet color results from 6⊗ 3 or 3⊗ 3?

Exercise 5.3 Quarks annihilation into gluons.

1. Draw all diagrams at the lowest order for the process qq → gg.

2. How these diagrams must be combined?

3. Write down the amplitudes and express them when appropriate as function of the corre-sponding amplitude in QED for qq → γγ. Use the following notations q(p1, color α)q(p2, color β)→g(p3, color a)g(p4, color b).

Exercise 5.4 e+e− → qqWe recall some tensorial products relevant for QCD: 3⊗ 3 = 8⊕1, 8⊗8 = 27⊕10⊕8⊕1⊕10⊕8.

1. What is the “color hypercharge” quantum number and the “color isospin” quantum numberof the pair qq? Justify the answer.

2. Deduce which states among the octet and the singlet fulfill the condition of the first question.

3. Show that with respect to the cross-section σ0 of the reaction e+e− → µ+µ−, the cross-section of e+e− → qq can be written as: σ = σ0Q

2(Noctet + Nsinglet) with Noctet = 2 andNsinglet = 1 and Q is the electric charge of the quark.

4. We assume the qq was produced in the octet states. Explain why additional gluons areexpected in the event? Imagine that the qq pair finally forms a meson, what is the minimalnumber of gluons expected?

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158 Quantum Chromodynamic

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Chapter 6

Weak interaction

Few references:F. Halzen and D. Martin, “Quarks & Leptons”, Wiley, (1984) chapter 12.Carlo Giunti and Chung W. Kim“Fundamentals of Neutrino Physics and Astrophysics”, OxfordUniversity press (2007) chapter 7 and 8.

So, far we have studied the electromagnetic and the strong interaction as 2 ex-amples of gauge theories. This chapter presents the last interaction describedby the standard model of particle physics, the weak interaction. We will followan historical approach, trying to explain the evolution of its theoretical descrip-tion. The gauge theory of the weak interaction will be described later, in thenext chapter, within the context of the unification of the electromagnetism andthe weak interaction.

6.1 The Fermi’s theory

6.1.1 Birth of a new interaction

In 1934, Fermi proposed his theory to describe the so-called β decay:

n→ p e−ν

where the neutrino, not yet discovered, had been previously postulated by Pauli to explain theobserved continuous energy spectrum of the electron. By analogy with the nuclear γ decaywhere a photon is emitted

N → Nγ

Fermi assumed that the electron and neutrino, were created in the process leading to the graph:

np

e−

ν

159

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160 Weak interaction

This graph is an example of a so-called four-fermion contact interaction (no propagation ofbosons). While in QED, the Lagrangian for p→ pγ (the proton being in a nucleus) is:

L = e(upγµup)Aµ

Fermi replaced the proton current by a current with the transition n → p and the photon Aµby a current expressing the creation of the electron and neutrino:

L = GF (upγµun)(ueγµuν)

the charge e being replaced by a constant, the Fermi’s constant [19]:

GF = 1.6637 10−5 GeV−2 (6.1)

consistent with the β decay rate. A new coupling constant meant that a new force was born.The Fermi coupling constant is so small that the probability of interaction is necessarily small,leading to naming this new interaction: the weak force. Moreover, the currents in the latterLagrangian are charged while in QED they are always neutral. So far, so good. Perturbationtheory based on this Lagrangian described pretty well the β-decay.

6.1.2 The θ+ and τ+ mystery

Many experiments were made to better understand the property of the weak interaction. Atthat time (in the 1950s), two particles called θ+ and τ+ (the latter was not the now well-knownlepton of the third family) had very similar properties: same mass and same lifetime (withinexperimental errors). However they decayed via the weak interaction differently:

θ+ → π+π0

τ+ → π+π+π−

They were considered as two different particles because according to the decays above, they haveopposite parities (see section 2.3.4.2) which were assumed to be conserved for all interactions.Indeed, knowing that the intrinsic parity of pions is -1, the parity of θ+ is:

η(θ+) = η(π+)η(π0)η(spatial) = (−1)(−1)(−1)l=0 = +1

where l = 0 because the θ+ had spin-0 as well as the pions and thus by angular momentumconservation (J = L+ S), l is necessarily 0. In the case of τ+, for the spatial parity one has toconsider the orbital momentum l1 between the first 2 π’s and the orbital momentum l2 betweenthe third π and the barycenter of the first 2 π’s. Since the τ+ had also a spin 0, necessarily0 = l1 + l2 ⇒ l1 = l2 and thus η(spatial) = (−1)2l1 = +1. Hence:

η(η+) = η(π+)η(π+)η(π−)η(spatial) = (−1)(−1)(−1)(+1) = −1

In 1956, two physicists, Yang and Lee suggested that the θ+ and τ+ were the same particle, theK+ in our modern nomenclature, and that parity is not conserved by weak interactions. A realrevolution in physics! Nature distinguishes left and right1!

1Indeed, your left hand appears as a right hand in a mirror. The non-conservation of parity induces that thetwo situations lead to different physics measurements (with particles not your hands!).

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The Wu’s experiment: parity is violated 161

6.1.3 The Wu’s experiment: parity is violated

After such hypothesis, the challenge was to prove it experimentally. Yang and Lee published apaper2 suggesting several experiments. The first experiment was led by Wu using a source ofCobalt polarized thanks to an external magnetic field. The Cobalt has the following β decay:

60Co→ 60Ni∗ e− ν

and the total angular momenta are respectively J(Co) = 5 and J(Ni∗) = 4. The field was largeenough and the temperature low enough (about 0.01 K!) to be sure that the spin of the Cobaltwas properly aligned. Now recall that the angular momentum is a pseudovector (or equivalentlynamed axial vector) meaning that contrary to usual vectors as the position or the momentum, itremains unchanged under a parity transformation. Indeed, for the orbital angular momentum:

~L = ~r × ~p parity−−−−→ ~L′ = −~r ×−~p = ~L

The spin S, the intrinsic angular momentum, is thus a pseudovector as well as the total J = L+S.Now, if parity were conserved, we should find the same rate of electrons emitted with an angleα and an angle π − α, the angle being measured with respect to the Nuclei angular momentum(or the B field) axis (see figure 6.1). However Wu and her team measured an asymmetry whichindicated a clear sign of parity violation! Electrons were more likely to be emitted at an angleα = π than α = 0.

Figure 6.1: Schema of the Wu’s experiment. From [26].

With J(Co) = 5 and J(Ni∗) = 4, we have ∆J = 1, the projection of the spin of the system e−νon an axis is thus necessarily 1 and hence both the electron and the neutrino have Sz = +1/2.Since they are emitted back to back (the Nitrogen stays (almost) at rest as was the Cobalt),they have opposite helicity. The electron being more likely emitted in the opposite direction ofits spin, left handed helicity is favored, while for the anti-neutrino, we expect the right-handedhelicity. Recall that when the masses are neglected, both helicity and chirality are equivalentwhich is reasonably the case here.

6.2 Modification of Fermi’s theory: the V − A current

6.2.1 Parity conservation in QED and QCD

The parity violation signed the death of Fermi’s theory. Indeed, the current involved in theLagrangian has the same structure as the one in QED (by construction):

jµ = eu1γµu2 = eu†1γ

0γµu2

2Physical Review 104:254

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162 Weak interaction

We saw in section 2.3.4.2, that the parity operator is P = γ0 when applied on spinors. Hence,under a parity transformation the current becomes:

jµ = e(γ0u1)†γ0γµγ0u2 = eu†1γµγ0u2

and thus:

j0 parity−−−−→ eu†1γ0γ0u2 = eu1γ

0u2 = j0

jiparity−−−−→ eu†1γ

iγ0u2 = −eu†1γ0γiu2 = −eu1γiu2 = −ji

We have just checked that jµ is a vector (polar vector more precisely), namely the temporalcomponent is not affected by parity transformation while the spatial components are inverted.Therefore, the amplitude being made with the product of 2 currents jµ1 and jµ2 will give:

M = jµ1 j2µ = j01j20 + ji1j2i

parity−−−−→ j01j20 + (−ji1)(−j2i) = jµ1 j2µ

We do have invariance of the QED amplitude under parity transformation. Similarly, in QCDthere is just an additional Gell-Mann matrix which remains unchanged under parity transfor-mation. QCD is thus parity invariant.

6.2.2 The V − A current

How should we modify the QED current to include parity violation? We have just seen thatelectrons with left-handed helicity were favored in the Wu’s experiment. We could think thatinserting an helicity projector would be enough. However, in section 2.3.3.5, we saw that helicityis not Lorentz invariant. We cannot have a weak interaction that would depend on the frame!Relativistic fermions are described by spinors, so a covariant current is necessarily of the form:

j = ψOψ

where O is an operator having a 4 × 4 matrix representation. The constraints of Lorentzinvariance and hermiticity reduce the possibilities of O to these 5 combinations:

O = 1, γ5, γµ, γµγ5, γµγν − γνγµ

Let us consider the case O = γµγ5. A parity transformation of such current gives:

jµ = u1γµγ5u2

parity−−−−→ (γ0u1)†γ0γµγ5γ0u2 = −u†1γµγ0γ5u2

where we have used the anticommutation property 2.58. The time and spatial components read:

j0 parity−−−−→ −u1γ0γ5u2 = −j0

jiparity−−−−→ +u1γ

µγ5u2 = ji

The spatial components are not affected meaning that jµ is an axial vector (pseudo). Thisbehavior is opposite to that of the (polar) vector. Clearly, an amplitude based on the productsof 2 axial vectors remains invariant under parity transformation. However, the parity violationis obtained by a mixture of vector current jµV and axial vector jµA:

jµV = u1γµu2, jµA = u1γ

µγ5u2, jµ = cV jµV − cAj

µA = u1γ

µ(cV − cAγ5)u2

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The V −A current 163

where the minus sign is purely conventional (it will be more clear in few lines). Let us see howan amplitude based on the product of 2 currents of this type is transformed by parity:

M =(cV j

µV − cAj

µA

) (c′V j

′V µ − c′Aj′Aµ

)

= cV c′V j

µV j′V µ + cAc

′Aj

µAj′Aµ − cV c′Aj

µV j′Aµ − cAc′V j

µAj′V µ

M parity−−−−→M′ = cV c′V j

0V j′V 0 + cAc

′A(−j0

A)(−j′A0)− cV c′Aj0V (−j′A0)− cAc′V (−j0

A)j′V 0

cV c′V (−jiV )(−j′V i) + cAc

′Aj

iAj′Ai − cV c′A(−jiV )j′Ai − cAc′V jiA(−j′V i)

=(cV j

µV + cAj

µA

) (c′V j

′V µ + c′Aj

′Aµ

)

6= M if cA or c′A 6= 0 and cV 6= 0 or c′V 6= 0

Conclusion: a weak current based on the difference (or addition) of a vector current and an axialvector current, thus the name “V −A” does permit parity violation.

Let us come back to the β decay: n → p e− ν. We then expect a current between the(outgoing) electron and the anti-neutrino of this kind:

jµeν = ueγµ(cV − cAγ5)vν (6.2)

Wu’s experiment showed that the electron was mainly produced with a left-handed helicity,which in the low mass limit, is equivalent to a left-handed chirality state. But, in section 2.3.4.3,we saw that the left-handed chirality projector for fermions (equ. 2.61) is:

PL =1

2(1− γ5)

Hence the left-handed electron reads:

ueL = 12(1− γ5)ue ⇒ ueL = 1

2(1− γ5)ue =(

12(1− γ5)ue

)†γ0 = u†e

(12(1− γ5)

)γ0

= ue12(1 + γ5)

where we used the γ5 properties of anticommutation and (γ5)† = γ5. Hence if only left(right)handed components of (anti)fermions are involved in weak interaction3, a current like:

jµeν = PLueγµPLvν = ue

1

2(1 + γ5)γµ

1

2(1− γ5)vν = ueγ

µ 1

2(1− γ5)

1

2(1− γ5)vν = ueγ

µ 1

2(1− γ5)vν

(using P 2L = PL) has precisely the form required for parity violation: see equ. 6.2 with cV =

cA = 12 . This is precisely what was proposed in 1958 by Feynman, Gell-Mann, Sudarshan and

Marshak. The amplitude of the β decay4 then reads:

M =4√2GF

[upγ

µ 1

2(1− γ5)un

] [ueγµ

1

2(1− γ5)vν

]

where the factor 4 and√

2 are purely conventional to keep the original value of the Fermi’sconstant.

3Recall that PL for antifermions projects on the right component!4Historically a factor 1.26 was put in front of the γ5 of the n→ p current to match the experimental rate. We

know now that it comes from the composite nature of the neutron. There is no such factor when the β decay istransposed to quark level: d→ ue−ν. Hence we have omitted it here.

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164 Weak interaction

6.2.3 Evidence for V − A current

Consider the decays of the charged pion. Since it is the lightest meson, it cannot decay via thestrong interaction. And since its quarks content is ud of respective charge -2/3 and 1/3, these2 quarks cannot annihilate into a photon (the photon would carry an electric charge). Thus, itmust decay via the weak interaction into lighter objects namely:

π− → e−νe , π− → µ−νµ

Since the phase space available in the electron channel is much larger than in the muon channel,we would expect naively a branching ratio larger for the former than the latter. However,experimentally, the measurement gives:

Γ(π− → e−νe)Γ(π− → µ−νµ)

= 1.23 10−4

The opposite of our naive expectation! The explanation is the following: the pion has a spin0, thus the charged lepton and the anti-neutrino have an opposite spin projection. In the pionrest frame, they fly back to back and hence, they have necessarily the same helicity. Now, weknow that the neutrino has a tiny mass (because of the neutrino oscillation phenomenon). Thus,we can safely consider that its helicity state corresponds to its chirality state. But, because ofthe V −A structure of the weak interaction, we know that only right-handed anti-neutrino caninteract. Consequently, its helicity state is also right-handed, meaning that the charged leptonstate has also a right-handed helicity state. However, V −A predicts that only the left-handedchirality state of charged lepton interacts. So we have a situation in which only the left-handedchirality projection of a right handed helicity state is involved. Now consider the right-handedhelicity state (equation 2.45 for fermion) and project it on left-handed chirality state:

12(1− γ5)ψ+ 1

2= 1

2

1 0 −1 00 1 0 −1−1 0 1 00 −1 0 1

√E +m

cos θ2eiφ sin θ

2|~p|

E+m cos θ2|~p|

E+meiφ sin θ

2

= 12

√E +m

(1− |~p|

E+m

)

cos θ2eiφ sin θ

2

− cos θ2−eiφ sin θ

2

We see clearly that if the mass of the charged lepton was 0 (and thus |~p| = E), the projectionwould give 0, and hence the decay of the pion would not occur. Since me or mµ are howeververy small, the decay is naturally suppressed explaining the long lifetime of the charged pion(2.6 10−8 s). And since me mµ, the decay must be much more suppressed in the electronchannel than in the muon one explaining the ratio of the decay widths above.

6.2.4 The four-fermion contact interaction failure

The four-fermion contact interaction with V − A currents described well the weak interactionprocesses known in the 1950s-1960s, namely at low energy. Rapidly, it was realized that thistheory was only an approximation, ie an effective theory in the low energy range. For example,consider the following neutrino electron interaction νµ e

− → νe µ−:

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The four-fermion contact interaction failure 165

e−(p)

νµ(k)

νe(p′)

µ−(k′)

The matrix element is then:

M =4√2GF

[uνeγ

µ 1

2(1− γ5)ue

] [uµγµ

1

2(1− γ5)uνµ

]=GF√

2

[uνeγ

µ(1− γ5)ue] [uµγµ(1− γ5)uνµ

]

(6.3)We wish to calculate the cross-section of this process. The calculation is very similar to e−µ− →e−µ−. But instead of γµ, we need to replace it by γµ(1−γ5). The spin average requires attention.The neutrino mass is so tiny that we will consider that only one helicity state (left-handed) ispossible. Hence, only the spins of incoming electron has to be averaged. Using equality 3.50replacing γµ → γµ(1− γ5) and neglecting all masses, it comes:

|M|2 =

(G2F

2

)1

2Tr(/pγ

µ(1− γ5)/p′γν(1− γ5)

)Tr(/kγµ(1− γ5)/k

′γν(1− γ5)

)

Let us evaluate the traces:

Tr(/pγµ(1− γ5)/p′γν(1− γ5)

)= Tr

(/pγµ/p′γν

)−

Tr(/pγµ/p′γνγ5

)−

Tr(/pγµγ5/p′γν

)+

Tr(/pγµγ5/p′γνγ5

)

= 2× pρp′η4(gρµgην − gρηgµν + gρνgµη)−pρp′η ×−4iερµην−

pρp′η ×−4iεηνρµ

where the following properties were used: 3.56, 3.60, the anticommutation of γ5, and (γ5)2 = 1.Hence, using the cyclic permutation εηνρµ = −ενρµη = ερµην :

Tr(/pγµ(1− γ5)/p′γν(1− γ5)

)= 8(pµp′ν − p.p′gµν + pνp′µ + ipρp

′ηερµην)

Tr(/kγµ(1− γ5)/k

′γν(1− γ5)

)= 8(kµk

′ν − k.k′gµν + kνk

′µ + ikρk′ηερµην)

(6.4)

Now we can make the product of the 2 traces, denoted in what follows by Tr1 × Tr2. Noticethat ερµηνgµν = 0 since gµν 6= 0 only when µ = ν for which ερµην = 0.

Tr1 × Tr2 = 64(2 p.k p′.k′ + 2 p.k′ p′.k+ipρp

′ηkµk

′νερµην + ipρp

′ηkνk

′µερµην + ipµp′νkρk′ηερµην + ipνp′µkρk′ηερµην

−pρp′ηkαk′βερµηνεαµβν)

In the first term of the second line, we can change the dummy indices µ↔ ν. Doing so, we haveερνηµ = −ερνµη = ερµνη = −ερµην . Hence the first 2 terms of the second line will cancel. Andsimilarly with the third and fourth terms of the second line. Now, using properties 3.61:

ερµηνεαµβν = εµνρηεµναβ = −2(δραδηβ − δ

ρβδηα)

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166 Weak interaction

And thus:

Tr1 × Tr2 = 64(

2p.k p′.k′ + 2 p.k′ p′.k + 2pρp′ηkαk′β(δραδ

ηβ − δ

ρβδηα))

= 256 p.k p′.k′(6.5)

Because of 4-momentum conservation, p + k = p′ + k′ ⇒ p.k = p′.k′ where the masses areneglected. The square of the center of mass energy is s = (p + k)2 ' 2p.k. Hence, the matrixelement is simply reduced to:

|M|2 =

(G2F

2

)1

2256

s2

4= 16G2

F s2 (6.6)

Now using formula 1.86 for the differential cross-section and 1.54 for the momentum in the

rest-frame ~p∗2 = ~p′∗2

= s/2:

dΩ∗=

1

64π2s|M|2 |

~p′∗||~p∗| =

G2F s

4π2⇒ σ =

G2F s

π

After this lengthy calculus, we see that there is clearly a problem: the cross section is proportionalto the square of the center of mass energy! Hence, it is divergent for s → ∞ ! In our jargon,we say that the unitarity is violated, meaning that the probability of observing this reaction isgreater than 1! Clearly the theory cannot be used at high energy: it is an effective theory, i.e.an approximation of a more general theory, only valid in the low energy regime.

There is another (related) problem with the four-fermion contact interaction: it can be shownthat it not renormalizable. Thus any calculation to higher order diverges.

6.3 A more modern weak interaction

6.3.1 The W boson

In our modern interpretation of the interactions, the forces are mediated by vector bosons. Sincein the β decay, we have currents that are vector-like (for Lorentz transformation), the W bosonresponsible for the weak interaction, must be a spin 1 boson as the photon. However, there is anobvious difference: the weak current is charged! A charge is transferred between the neutrinoand the electron for instance. Therefore, the W boson must be charged: The W− and its chargeconjugate W+ are the intermediate vector bosons of the weak interaction with charged current.The β decay now corresponds to the graph below:

W−

d

e−

ν

u

At each vertex, only left(right)-handed chirality for (anti)fermions is involved. It would betempting to imagine that the W ’s are just charged photons. If it would be the case, we shouldbe able to see this kind of Compton process producing W bosons as easily as photons:

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The charged current of leptons 167

e−

γ

ν

W−

It is not the case. However, nothing forbids this process. Why don’t we detect weak chargedbosons as easily as photon in detectors? The logical explanation is that they simply have a largemass suppressing their production at low energy. A natural consequence of their mass is theshort range of the weak interaction.

Experimentally, the W± was discovered at CERN in 1983 in the UA1 experiment, and fewdays later by the UA2 experiment, both operating at the SPS, a proton-anti-proton colliderwith

√s = 540 GeV (at that time). The channel pp→ W−X → e−νX (and charge conjugate)

was used. Nowadays, the W property is well known and it mass has been measured with agood accuracy thanks to the Tevatron collider (near Chicago) and more recently to the LHC atCERN. Its mass is [19]:

MW = 80.399± 0.023 GeV/c2 (6.7)

How can we change the photon propagator to accommodate for the W mass? It can beshown that it is [7, p. 250]:

Wpropagator = i−gµν + qµqν/M2

W

q2 −M2W

(6.8)

which can be compared to the usual photon propagator −igµν/q2. Well, setting MW → 0 doesnot recover the photon propagator. The reason is related to the fact that with a massive spin1 boson, the 3 polarizations are allowed contrary to the photon which has only 2 polarizations.And this is the longitudinal polarization that is responsible for the term qµqν/M2

W .

6.3.2 The charged current of leptons

Looking closely at the amplitude squared 6.6 which is dimensionless, we conclude that GF hasthe dimension GeV−2. It is not dimensionless as its analog α in QED. In complete analogywith QED, let us define gw

5 the universal weak coupling constant, the analog of e in QED. Thecharged current for leptons is then straightforward:

jµe−→νe =

gw√2uνeγ

µ 1

2(1− γ5)ue (6.9)

where for convenience, the factor√

2 is used so that it will match the electroweak definition thatwe will see later. The reaction considered in the section 6.2.4, νµ e

− → νe µ− is now described

by the graph:

5the w of gw stands for weak, not for W boson.

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168 Weak interaction

e−(p)

νµ(k)

νe(p′)

µ−(k′)

for which the amplitude is:

iM = jµe−→νe × i

−gµν+qµqν/M2W

q2−M2W

× jννµ→µ−=[gw√

2uνeγ

µ 12(1− γ5)ue

]i−gµν+qµqν/M2

W

q2−M2W

[gw√

2uµγ

ν 12(1− γ5)uνµ

] (6.10)

When q2 MW , we should recover the Fermi’s effective theory. In this approximation, theamplitude becomes:

M =[gw√

2uνeγ

µ 12(1− γ5)ue

]gµνM2W

[gw√

2uµγ

ν 12(1− γ5)uνµ

]

= g2w

8M2W

[uνeγ

µ(1− γ5)ue] [uµγµ(1− γ5)uνµ

]

Comparing this expression to 6.3, we can immediately identify:

GF√2

=g2w

8M2W

(6.11)

Now, let us come back to the general case without assuming q2 MW . Neglecting the massesof the fermions, and applying the Dirac’s equations in the momentum space to e− and νe gives:

(/p−m)ue ' /pue = 0 , uνe(/p′ −m) ' uνe /p′ = 0

Since q = p−p′, the term qµqν due to the W propagator in 6.10 contracted with jµe−→νe becomes:

[gw√

2uνeγ

µ 12(1− γ5)ue

]qµqν =

[gw√

2uνe(/p− /p′)1

2(1− γ5)ue

]qν

=[gw√

2uνe/p

12(1− γ5)ue

]qν

=[gw√

2uνe

12(1 + γ5)/pue

]qν

= 0

Hence, the amplitude is simply reduced to:

M =[gw√

2uνeγ

µ 12(1− γ5)ue

]gµν

M2W−q2

[gw√

2uµγ

ν 12(1− γ5)uνµ

]

= g2w

8(M2W−q2)

[uνeγ

µ(1− γ5)ue] [uµγµ(1− γ5)uνµ

]

Compared to 6.3, we see that now, the amplitude does not diverge anymore thanks to thequadratic term 1/q2. (The calculation of the cross-section is very similar to the previous case).

If gw is universal, we expect by definition the same coupling for the 3 families of leptons(νee−

),

(νµµ−

),

(νττ−

)and the W± namely:

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The charged current of leptons 169

e−

νe

W−−i gw√2γµ 1

2(1− γ5) µ−

νµ

W−−i gw√2γµ 1

2(1− γ5) τ−

ντ

W−−i gw√2γµ 1

2(1− γ5)

The vertex factor has been simply deduced by analogy with QED. Indeed in QED, the currentis jµ = quγµu with a vertex factor −iqγµ. For weak interaction, jµ = gw√

2uγµ 1

2(1 − γ5)u and

hence a vertex factor −i gw√2γµ 1

2(1 − γ5). Can we test that gw = gew = gµw = gτw or equivalently

GF =√

2g2w

8M2W

= GeF = GµF = GτF ? Consider for example the muon decay µ− → e−νeνµ. One can

show that [9, p. 263] (neglecting masses of the final state):

1

τµ= Γµ−→e−νeνµ =

G2Fm

192π3

Assuming non universal constants, one would have found:

1

τµ= Γµ−→e−νeνµ =

GeFGµFm

192π3

Similarly for the τ− decays:

Γτ−→e−νeντ =GeFG

τFm

192π3Γτ−→µ−νµντ =

GµFGτFm

192π3

But the partial decay-width is related to the branching-ratio and the lifetime via:

Γτ−→e−νeντ = Br(τ− → e−νeντ )× Γ =Br(τ− → e−νeντ )

ττ

Hence combining the muon and tau lifetime:

τµττ

=192π3

GeFGµFm

GeFGτFm

192π3

Br(τ− → e−νeντ )=GτFGµF

(mτ

)5 1

Br(τ− → e−νeντ )

Now, these quantities are well known with a good accuracy [19]:

τµ = 2.197034(21)× 10−6 s mµ = 105.658367(4) GeVττ = (290.6± 1.0)× 10−15 s mτ = 1776.82± 0.16 GeV Br(τ− → e−νeντ ) = (17.82± 0.04)%

so that:GτFGµF

= 1.0024± 0.0033

A value compatible with 1. A similar combination yields a ratio of coupling constants elec-tron/muon compatible with 1 as well. Hence, the weak coupling constant is universal for allleptons.

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170 Weak interaction

6.3.3 The charged current of quarks

6.3.3.1 The Cabibbo angle

It seems logical to assume that the quarks current is identical to the lepton one, or in otherwords, that the coupling constants are the same. However, compare the values of the Fermiconstant measured in β decay and obtained from muon lifetime measurement:

GβF = 1.136(3) 10−5 GeV−2 GµF = 1.16637(1) 10−5 GeV−2

There is a slight, but statistically significant, discrepancy between the two. In addition, in theearly time of the weak interaction, there was another indication that the quark sector behavesdifferently: the decay K+ → µ+νµ was observed:

u

s

µ+

νµ

meaning that a transition u → s was possible. Compared to the corresponding pion decay (soud→ µ+νµ) with ∆S = 0, the kaon decay ∆S = 1 is suppressed by a factor 20. Do we have togive up the universality of weak interaction? In 1963, the physicist Cabibbo postulated that thestrength GF of the weak interaction was shared between ∆S = 0 transition and ∆S = 1. Thesharing was described by an angle θc, the Cabibbo angle, so that:

Γ∆S=0 ∝ G2F cos2 θc and Γ∆S=1 ∝ G2

F sin2 θc

Empirically, the factor 20 required an angle θc ' 13 or sin θc ' 0.23. After the discovery ofquarks, the reinterpretation of the Cabibbo angle was the following: the quark that couples tothe u-quark in the weak interaction is a mixture of d and s quarks:

|d′〉 = cos θc |d〉+ sin θc |s〉

d′ is the weak eigenstates whereas d and s are the mass eigenstates (the ones that propagate).This is the mass state that matters for the QCD interaction. In terms of vertex factor:

d

u

W−−i cos θcgw√

2γµ 1

2(1− γ5) s

u

W−−i sin θcgw√

2γµ 1

2(1− γ5)

6.3.3.2 The GIM mechanism

In 1970, the physics community was puzzled by the very small branching ratio of the decayK0 → µ+µ−, about 9× 10−9 which was supposed to be described by a so-called box diagram:

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The charged current of quarks 171

u νµ

d

s

µ+

µ−

sin θc

cos θc

By symmetry with the leptonic sector where 2 families were already known, Glashow, Iliopoulosand Maıani postulated the existence of a fourth quark, the charm so that another box diagramwas possible with a quark c instead of a u in the box. Since the c quark can couple to the s or dquark, one can use the linear combination of d and s orthogonal to the d′ to define the s′ stateso that: (

d′

s′

)=

(cos θc sin θc− sin θc cos θc

)(ds

)(6.12)

and hence the W boson couples c quark to s′ and u quark to d′:

(ud′

) (cs′

)

The u and d′ quarks are respectively the counterpart of the νe and e− in the leptonic sector(and c, s′ play the role of νµ, µ

−).

The new diagram is thus:

c νµ

d

s

µ+

µ−

cos θc

− sin θc

The amplitude of the first diagram is thus proportional to g4w cos θ sin θ while the second is

proportional to −g4w cos θ sin θ (the power 4 is coming from the 4 vertices including the muons

ones). Hence, if the u and c quarks had the same mass, these 2 diagrams would perfectly cancel(one has to add the amplitudes since we have the same initial/final states). Since mc mu, theoverall contribution is simply strongly suppressed with respect to just considering the originaldiagram above. That’s the so-called GIM mechanism, GIM standing for Glashow, Iliopoulos andMaıani. The three physicists predicted that the c quark had to have a mass between 1 to 3 GeVto explain the measured rate of the K0 → µ+µ− decay. Four years later, in 1974, the charmwas discovered as a constituent of the J/ψ meson made of a pair cc. Its mass is mc ' 1.3 GeV !

6.3.3.3 The CKM matrix

Nowadays, 6 quarks have been discovered. The Cabibbo-Kobayashi-Maskawa (CKM) matrix,VCKM is an extension of GIM matrix 6.12 to the 3 families:

d′

s′

b′

=

Vud Vus VubVcd Vcs VcbVtd Vts Vtb

dsb

(6.13)

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172 Weak interaction

where d′, s′, b′ are the weak eigenstates, and d, s, b the mass eigenstates. The CKM matrix is aunitary matrix (V †V = 1) and hence the following equalities hold:

k

|Vik|2 =∑

i

|Vik|2 = 1 ,∑

k

VikV∗jk =

k

VkiV∗kj = 0 (i 6= j) (6.14)

The charge rising current with the 3 generations is then:

jµcc+ =gw√

2(u, c, t)γµ

1

2(1− γ5)VCKM

dsb

(6.15)

with a VCKM element being for example with d→ u: Vud. The charge lowering weak current isobtained with the hermitician conjugate:

jµcc− = (jµcc+)† =gw√

2(d, s, b)V †CKMγ

µ 1

2(1− γ5)

uct

(6.16)

with:

V †CKM =

V ∗ud V ∗cd V ∗tdV ∗us V ∗cs V ∗tsV ∗ub V ∗cb V ∗tb

This time the matrix element is for example with u→ d: V ∗ud. The vertex factor is then:

W+

di

uj−i gw√2γµ 1

2(1− γ5)Vujdi W−

ui

dj−i gw√2γµ 1

2(1− γ5)V ∗uidj

where ui is an up-type quark (charge +2/3) ui=1,2,3 = u, c, t and di a down-type quark (charge-1/3) di=1,2,3 = d, s, b.

What is the number of free parameters in the CKM unitary matrix? Any 3×3 unitary matrixcan be decomposed as the product of 3 rotations (3 real parameters, the angles θij=12,13,23) plusone complex parameter interpreted as a phase δ. The common choice is then:

VCKM =

1 0 00 c23 s23

0 −s23 c23

c13 0 s13e−iδ

0 1 0−s13e

iδ 0 c13

c12 s12 0−s12 c12 0

0 0 1

=

c12c13 s12c13 s13e−iδ

−s12c23 − c12s23s13eiδ c12c23 − s12s23s13e

iδ s23c13

s12s23 − c12c23s13eiδ −c12s23 − s12c23s13e

iδ c23c13

(6.17)

where cij = cos θij and sij = sin θij . If the third generation of quarks did not mix with the firstone, the angle θ12 would be exactly θc. The mixing is however very small, and hence θ12 ' θc.These parameters have to be measured, they are not predicted by the theory. The CKM matrix

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The neutral current 173

elements can be more precisely determined by a global fit that uses all available experimentalmeasurements and imposes constraints as assuming three generations and the unitarity of thematrix (i.e. |Vub|2 + |Vcb|2 + |Vtb|2 = 1) [19]:

∣∣V expCKM

∣∣ =

0.97428(15) 0.2253(7) 3.47(16)× 10−3

0.2252(7) 0.97345(16) 41(1)× 10−3

8.62(26)× 10−3 40.3(1.1)× 10−3 0.999152(45)

(6.18)

As expected, the dominant decay is always within a given family u ↔ d, c ↔ s and t ↔ b, thematrix being almost diagonal. Mixing between the first and third family is almost zero.

6.3.4 The neutral current

In the mid-sixtees, Glashow, Salam and Weinberg developed a theory to unify the weak inter-action and the electromagnetism. This theory, called electroweak will be described in the nextchapter, but it predicted the existence of a neutral boson the Z0 that mediates the weak inter-action via neutral current (and hence no exchange of electric charge). Hence reactions of thiskind were predicted:

νµ e− → νµ e

− , νµ N → νµ X (6.19)

the former occurring via the diagram:

Z0

e−

νµ

e−

νµ

Notice that a W cannot be exchanged, this diagram:

W−

e−

νµ

e−

νµ

being forbidden because of a violation of lepton number (or in other words because the W couplesonly to leptons of the same family). In 1973, the processes 6.19 were observed at CERN in theGargamelle experiment (a big bubble chamber that can be seen now in the CERN garden!).The coupling to the Z0 respects the lepton number, meaning that in:

Z0 → ff

the fermion f is necessarily produced with its anti-fermion. Such coupling Z0 → e−µ− inducinga Flavor Changing Neutral Current (FCNC) are thus forbidden (but actively looked for as a signof new physics). Moreover, the measured cross section of:

νµ N → νµ X

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174 Weak interaction

related at the parton level to reaction νµ q → νµ q was about a third larger than the correspond-ing pure charged current reaction:

νµ N → µ+ X

suggesting, as the electroweak theory predicts (see next chapter) that the structure of the neutralcurrent is different than the one of the charged current. In fact, the experimental data ruled outa pure V −A structure [30], and justified a neutral current:

jµNC ∝ qγµ1

2(cV − cAγ5)q

with cV 6= cA 6= 1, meaning that not only the left handed chirality state is involved in the neutralcurrent but also the right handed chirality state!

The Z0 boson has been observed for the first time at CERN in 1983, still in the UA1 andUA2 experiments few months after the discovery of the W±. It was observed in its leptonicdecay:

pp→ Z0X → l+l−X

where l = e or µ. Later on, it has been extensively studied at the LEP e+e− collider. Its massis now very well known [19]:

MZ = 91.1876± 0.0021 GeV/c2 (6.20)

explaining again the weakness of the weak interaction. In the next chapter, more details will begiven.

6.3.5 Strength of the weak interaction

The very small value of GF ' 1.6 10−5 GeV−2 compared to α = 1/137 explains why the weakinteraction is qualified as “weak” at low energy (where GF is valid). However, injecting the Wmass value 6.7 into 6.11 allows to estimate gw:

gw =

√8M2

WGF√2

= 0.65⇒ αw =g2w

4π' 1

29

Finally αw is larger than α! Intrinsically, the weak force is thus stronger than the electromag-netism. Its weakness, observed at low energy, is simply due to the large mass of the W boson.At high energy, the weak interaction becomes stronger than the electromagnetism. At LHC forinstance, it is more likely to interact via the weak interaction than the electromagnetism.6

6.3.6 Failure of the model

There is a major problem with the weak theory described so far: it is not renormalizable! Suchhigh order (O(g4

w)) process νµνµ → νµνµ corresponding to the diagram:

6At very large energy, the distinction between the 2 interactions become irrelevant. See the electroweak modelin the next chapter.

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The neutrinos case 175

µ−

W+

W−

µ+

νµ

νµ

νµ

νµ

is divergent! Even if this process cannot be detected, it is a problem for the theory! In fact, onecan show that this problem is related to the fact that the W or Z boson in the theory exposedin this chapter are not quanta of a gauge field. What is missing here is a gauge invariance tomake the theory renormalizable. The next chapter is dedicated to this topic.

6.4 The neutrinos case

6.4.1 The PMNS matrix

Until the 1990s, the neutrinos were assumed to be massless. In these conditions, the solutionsof the Dirac’s equation simplify. Looking back to the 2 coupled equations 2.33, setting m = 0and being interested by the E = +|~p| solution gives:

|~p|ua − ~σ.~p ub = 0~σ.~p ua − |~p|ub = 0

where ua and ub are 2-components spinors of the wave function:

ψ(x) =

(uaub

)e−ip.x

Let us define the following linear combination:

uL = ua − ubuR = ua + ub

We see that they satisfy two independent equations:

~σ.~p2|~p|uL = −1

2uL~σ.~p2|~p|uR = 1

2uR

Since ~σ.~p2|~p| is the helicity projector (equivalent to chirality in the massless assumption), we see

that uL, uR are respectively the left-handed and right-handed spinors. Let us define the 2 wavefunctions χL/R(x) = uL/Re

−ip.x so that the original wave function ψ(x) can be expressed as:

ψ(x) =

(uR+uL2

uR−uL2

)e−ip.x =

1

2

(χR(x) + χL(x)χR(x)− χL(x)

)

Since ψ(x) satisfies the Dirac equation iγµ∂µψ(x) = 0, it comes with χL/R(x):

i(∂0χR(x) + ∂0χL(x) + σi∂iχR(x)− σi∂iχL(x)

)= 0

i(−∂0χR(x) + ∂0χL(x)− σi∂iχR(x)− σi∂iχL(x)

)= 0

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176 Weak interaction

which decouples into:

i(∂0 − σi∂i)

)χL(x) = 0

i(∂0 + σi∂i)

)χR(x) = 0

These two equations known as the Weyl equations7 describe independently the evolution ofa massless left-handed particle and a massless right-handed particle. Thus, the massless left-handed and the right-handed particles can be considered as two different particles. This appliesto νL and νR. Since only νL is sensitive to the interactions, there was no way (even in principle)to detect a νR. Conclusion: the νR did not have any physical meaning and was then absentfrom the Standard Model.

However, in the late 1990s, several experimental facts suggested that neutrinos are massive.As we will see in the next section, massive neutrinos lead to a phenomena called neutrinooscillations, where neutrinos can change their flavour while propagating. The very first hints ofthe oscillations came from a large deficit (50% to 70%) in the number of νe coming from thesun (produced by the nuclear reaction 4p + 2e− → 4He + 2νe) with respect to the standardsolar model predictions. It turns out that because of their oscillation, the fraction of νe reachingthe earth is only the third of the total flux of neutrinos (assuming 3 flavors). These last years,neutrino oscillations have been confirmed by observing directly the change of flavor of neutrinoscoming from a nuclear reactor or neutrino beam. Such oscillations (i.e flavour changing) areexplained by assuming that neutrinos eigenstates of the weak interaction differ from the masseigenstates (the eigenstates of the free-propagation of the neutrinos). This scenario is thensimilar to the quark case of the previous section. After the quark mixing, we now have theneutrino mixing! However, there is an historical difference: quarks were discovered first throughtheir strong interaction. The usual labels u, d etc names the strong or equivalently the masseigenstates8. Only later, it was realized that weak interaction acts on other states. In the caseof neutrinos, the reverse occurred: since neutrinos are only sensitive to the weak interaction andare not directly observed, the usual neutrinos νe, νµ, ντ are by definition the ones associated tothe charged leptons that are used to detect them: those neutrinos are thus weak eigenstates.Only recently, we consider that they differ from the mass eigenstates. But contrary to the quarkscase, there is no way to detect directly the mass eigenstates.

Anyway, as for the quarks, there is a unitary matrix similar to CKM, called PMNS forPontecorvo-Maki-Nakagawa-Sakata that relates the mass eigenstates ν1, ν2, ν3 to the weak eigen-states νe, νµ, ντ :

νeνµντ

= VPMNS

ν1

ν2

ν3

=

Ve1 Ve2 Ve3Vµ1 Vµ2 Vµ3

Vτ1 Vτ2 Vτ3

ν1

ν2

ν3

(6.21)

with VPMNS having a similar parametrization 6.17 in terms of mixing angles θ12, θ13, θ23 anda phase δ as the CKM matrix. However, the components of the neutrino matrix is known witha very poor accuracy. The phase is totally unconstrained and a crude estimation (combining

7We would have found these equations more simply by using another representation of the Dirac matricescalled the chiral representation.

8The quarks cannot be seen as free particles, they always feel the strong interactions. Thus the propagationof free quarks (pure mass eigenstates) does not really make sense. Therefore, what is called mass eigenstate is bydefinition the eigenstate of the kinetic and strong interaction parts of the Hamiltonian.

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Neutrino oscillations formalism 177

various sources of data) of the absolute value of the components of the matrix gives [27]:

|VPMNS | =

0.78− 0.86 0.51− 0.61 0.00− 0.180.19− 0.57 0.39− 0.73 0.61− 0.800.20− 0.57 0.40− 0.74 0.59− 0.79

The difference of this matrix with respect to CKM is striking: it is not diagonal at all, allelements having pretty large values9.

One may wonder why the mixing is applied to the neutrinos and not to the charged leptonswhich are considered with a definite mass. All charged leptons have the same quantum numbers,the only difference being their masses. Thus, the flavor of a charged lepton is determined by itsmass. However, the mass difference between charged leptons is very large. Consequently, themass measurement uncertainties is small enough to discriminate the 3 masses and moreover, thelarge difference of masses yields very distinct experimental signature (very fast decay of the τ ,Bremsstrahlung of the light e− etc.). Therefore, there is no ambiguity in the assignment of thenature of the particle. Hence, charged leptons with a definite flavor are, by definition, particleswith definite mass. On the other hand, neutrinos are never detected directly: they are identifiedvia the charged leptons produced by the weak interaction which define the flavor of the neutrino.Therefore, flavor neutrinos are not required to have a definite mass and the mixing implies thatthey are superpositions of neutrinos with definite masses.

6.4.2 Neutrino oscillations formalism

Let us denote by να=e,µ,τ the neutrinos with a given flavor α and νk=1,2,3, the neutrinos masseigenstates. The charged current transition producing a neutrino να from a charged lepton lα(and coupled to the W field, see equ. 6.9) can be expressed as function of the νk:

jµlα→να = gw√2uναγ

µ 12(1− γ5)ulα

= gw√2

∑3k=1 Vαkuνkγ

µ 12(1− γ5)ulα

= gw√2

∑3k=1 V

∗αkuνkγ

µ 12(1− γ5)ulα

Consequently, the created neutrino with flavor α is described by the state:

|να〉 =

3∑

k=1

V ∗αk |νk〉 (6.22)

the states being normalized: 〈νβ|να〉 = δβα, 〈νi|νk〉 = δik. The flavor state is expressed as alinear combination of the mass eigenstates. Let us consider that a neutrino νk created in thespace-time point (0,~0) has an energy Ek and a momentum ~p with:

E2k = |~p|2 +m2

k (6.23)

The evolution of the state νk in space-time is simply:

|νk(x, t)〉 = e−i(Ekt−~p.~x) |νk〉9Recently, thanks to neutrino oscillation experiments, the angle θ13 has been measured with a good accuracy

giving sin2(2θ13) = 0.092± 0.017 [28].

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178 Weak interaction

νk being an eigenstate of the free hamiltonian. Let us assume that the 3 mass eigenstatespropagate in the same direction, for example ~p = p~ex where ~ex is a unit vector in the x-axisdirection. Then, the state να at a time t and a distance L along the x-axis is given by:

|να(L, t)〉 =∑

k

V ∗αk e−i(Ekt−pL) |νk〉 (6.24)

Now, because of equ. 6.22, we have:

|νe〉|νµ〉|ντ 〉

= V ∗PMNS

|ν1〉|ν2〉|ν3〉

and since VPMNS is a unitary matrix:

V †PMNSVPMNS = (V ∗PMNS)t VPMNS = 1⇒ (VPMNS)t V ∗PMNS = 1⇒ (V ∗PMNS)−1 = (VPMNS)t

and hence: |ν1〉|ν2〉|ν3〉

=

Ve1 Vµ1 Vτ1

Ve2 Vµ2 Vτ2

Ve3 Vµ3 Vτ3

|νe〉|νµ〉|ντ 〉

so that |νk〉 =∑

β Vβk |νβ〉. Conclusion, equ. 6.24 can be re-written:

|να(L, t)〉 =∑

β

k

V ∗αk e−i(Ekt−pL)Vβk |νβ〉

where only flavor neutrinos appear. The probability to observe the transition να → νβ is thus:

Pνα→νβ (L, t) = | 〈νβ|να(L, t)〉 |2 =∣∣∑

k V∗αk e

−i(Ekt−pL)Vβk∣∣2

=∑

k,j V∗αkVβkVαjV

∗βj e−i(Ekt−pL) ei(Ejt−pL)

For ultra-relativistic neutrinos, t ≈ L and Ek ≈ Ej = E ≈ p thus

Ek,jt− pL = (Ek,j − p)L =E2k,j − p2

Ek,j + pL =

m2k,j

2EL

Hence we conclude:

Pνα→νβ (L) =∑

k,j

V ∗αkVβkVαjV∗βj e−i(

∆m2kj

2E

)L

(6.25)

where:∆m2

kj = m2k −m2

j (6.26)

Clearly, Pνα→νβ (0) = δαβ since the unitarity of the PMNS matrix imposes∑

k V∗αkVβk =∑

j VαjV∗βj = δαβ. In other words, there is no change of flavor without propagation. For very

short distances, close to the point of creation of the neutrino, the probability converges towardsδαβ. In order to observe a change of flavor between the creation of neutrinos and their detectionin an experiment located at L, three conditions have to be fulfilled:

• the corresponding PMNS matrix elements must not be 0,

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Neutrino oscillations formalism 179

• neutrinos must be massive.

• their masses must not be degenerated.

It is convenient to introduce the oscillation length

Losckj =4πE

∆m2kj

(6.27)

and decompose formula 6.25 as:

Pνα→νβ (L) =∑

k

|Vαk|2|Vβk|2 + 2Re∑

k>j

V ∗αkVβkVαjV∗βj exp

(−i 2π

L

Losckj

)(6.28)

The first term does not depend on L. Very far from the production point of the neutrinowhere L Losckj , the probability will converge toward this constant value because of the finiteresolution in the measurement of E and L [27]. Even if there is a transition from one flavor toanother (mixing), it is not an oscillation. The second term depending on L is the oscillation: theprobability of a transition from one flavor to another does depend on the distance (or equivalentlyon the time). Using standard units, the oscillation length reads:

Losckj (km) = 247E(GeV)

∆m2kj(10−2 eV2)

(6.29)

Global constraints on various sources of data (solar neutrinos, atmospheric neutrinos etc.) givean estimation of ∆m2

21 ≈ O(10−4) eV2 and |∆m231| ≈ |∆m2

32| ≈ O(10−3) eV2. Typical neutrinosexperiments trying to observe the neutrinos oscillations use either (anti-) neutrinos (νe) fromnuclear plants (with typical energy of the order of few MeV) or neutrinos from accelerators(νµ with few GeV produced by π decay). Consequently, the oscillation length is necessarilylarge (macroscopic), ranging typically from one kilometer (Double-CHOOZ, Daya Bay, Renoexperiments etc) to several hundreds of kilometers (T2K, OPERA etc.). This is the typicaldistance between the source of neutrinos and their detection to be able to observe the oscillationphenomena. Notice that the sign of ∆m2

31 and ∆m232 is still unknown. The mass hierarchy could

be m1 < m2 < m3 (so called normal hierarchy) or m3 < m1 < m2 (inverted hierarchy).Coming back to the oscillation formula 6.28, we see that when α 6= β, Pνα→νβ can be different

from Pνβ→να . Indeed,

Re[V ∗αkVβkVαjV

∗βj exp

(−i 2π

L

Losckj

)]6= Re

[V ∗βkVαkVβjV

∗αj exp

(−i 2π

L

Losckj

)]

as soon as the elements of the PMNS are complex. In addition, if we assume that the CPTsymmetry is conserved by the weak interaction, the CPT transformation of the reaction νβ → ναbeing να → νβ

10 , we would expect Pνβ→να = Pνα→νβ and thus a complex PMNS matrix impliesPνα→νβ 6= Pνβ→να ⇒ Pνα→νβ 6= Pνα→νβ . But the reaction να → νβ is just the CP transformationof να → νβ (remember the anti-neutrinos are only right handed and neutrinos left-handed).Conclusion, the complexity of the PMNS matrix implies that CP is violated in the neutrinosector.

10The only interacting (anti-)neutrinos are left (right) handed so the CPT transformation of νβ → να ≡ νβL →

ναLC=⇒ νβL → ναL

P=⇒ νβR → ναR

T=⇒ ναR → νβR ≡ να → νβ

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180 Weak interaction

6.4.3 Considerations on the measurement of neutrinos masses and the oscil-lations phenomena

We have seen that the νk neutrinos have by definition a definite mass. What about the να?Consider the 4-momentum operator Pµ. By definition, we have:

PµPµ |νk〉 = m2k |νk〉

And the average value of the mass squared is as expected 〈νk|PµPµ|νk〉 = m2k. Hence, according

to equ. 6.22:

PµPµ |να〉 =∑

k

m2kV∗αk |νk〉

Showing that να is not a mass eigenstate as announced. The average mass squared is however:

〈m2α〉 = 〈να|PµPµ|να〉 =

k

m2kV∗αk 〈να|νk〉 =

k

m2k|Vαk|2

The same result would have been obtained by using the development of |να(x, t)〉 (from 6.24).On average, we would measure 〈m2

α〉 which does not depend on t or L (and so on the distanceof the detector), but event by event we would measure m2

k with a probability |Vαk|2. Themeasurement supposes that the mass can be measured directly. Let us take for instance thepion decay π+ → µ+νµ. Assuming an ideal detector, able to measure perfectly the 4-momentumof the µ+, according to formula 1.52:

m2k = m2

π +m2µ − 2mπ

√m2µ + |~p∗µ|

where ~p∗µ is the muon momentum in the pion rest frame. If the 4-momentum of the pion is alsoperfectly measured, then one can determine ~p∗µ from the 4-momentum of the µ+ measured inthe lab. But, beware that the measurement forces the superposition to collapse on the massiveneutrino whose mass has been measured. In other words, there is no more oscillation. Indeedafter the collapse, the state is |νk(x, t)〉 = |νk〉 e−i(Ekt−~p.~x) and the probability to detect the state|νβ〉 (via weak interaction) is:

Pνα→νk→νβ = |〈νβ|νk(x, t)〉|2 |〈νk|να〉|2 = |Vβk|2|Vαk|2

which does not depend anymore on t or L. As soon as the measurement is accurate enoughto determine m2

k with an error less than ∆m2ij and thus determine unambiguously which νk

was involved, oscillation cannot exist anymore. What is the reason for the destruction of theoscillation pattern? It relies on the uncertainty principle. Coming back to our example of thepion decay, the more accurate the pion momentum is measured, the more uncertain the pointwhere the neutrino has been created would be. Thus the uncertainty on L can even exceedthe oscillation length Losckj , and so washing out the oscillation pattern. A rigorous proof of thisexplanation, would require to use a more complicated formalism than the one we used based onplane-wave approximation. It is necessary to apply wave packets treatment11 which introducesthe coherence length Lcoh [27] related to the size of the neutrinos wave-packets. When this sizeis larger that Losckj , or when L is larger than Lcoh, the oscillation is suppressed.

11By the way, the wave-packet treatment eliminates idealizing assumptions like assuming that all νk have thesame momentum during the propagation.

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Lepton number: conserved or not conserved? 181

6.4.4 Lepton number: conserved or not conserved?

As a conclusion for this section about neutrinos, let us give few considerations about individuallepton number conservation. Because of neutrinos oscillations, a neutrino produced with agiven flavor can be detected with another flavor. Hence, there is clearly no conservation of theindividual lepton number for neutrinos. But what about charged leptons? Consider for instancethe following diagram describing the muon decay µ− → e−γ (the photon is here only to conservethe 4-momentum):

νµW−

νeµ−

e−

γ

Because of the neutrino mixing (symbolized by the cross), this reaction is possible, leadingto muon and electron lepton numbers violation. However, the probability for such reaction isproportional to (∆m2/M2

W )2 [29]. Since ∆m2 . 10−3 eV2, this decay is virtually impossible toobserve: a branching ratio of the order of 10−56 is expected! Hence, even if strictly speakingthe individual lepton number for charged lepton is also violated, it remains in all practical casesconserved. Beware that at the vertex level, the lepton number is always conserved. Such diagramdoes not exist in the standard model:

µ−

νe

W−

The source of the individual lepton number violation must always be due to a neutrino oscillation.However, the total lepton number remains always conserved: if a lepton is present in the initialstate (of any flavor), there must be a lepton in the final state (but not necessarily with the sameflavor).

6.5 What is violated by weak interaction?

6.5.1 Charge conjugation C

We already know that the weak interaction violates the parity. What about the charge conju-gation? Consider the charged pion decay:

π− → µ−νµ

Since the pion has a spin 0, the spin of µ− and νµ is opposite and since in the rest-frame ofthe pion, they are produced back-to-back, they necessarily have the same helicity. Consideringthe neutrino massless, its chirality state coincides to its helicity state and thus because of weak

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182 Weak interaction

interaction property, νµ is necessarily in the right-handed helicity (=chirality) state νµR . There-fore, µ− is also in the right-handed helicity state µ−R. Now let us transform every particles in itsantiparticle (and vice-versa):

π− → µ−RνµRcharge conjugation−−−−−−−−−−−−−−−→ π+ → µ+

RνµR

Is this last reaction possible? No obviously since νµ only interacts as a left-handed particle!Hence, weak interaction violates both P and C transformations.

6.5.2 CP transformation

Continuing with the previous example, if we apply in addition a parity transformation:

π− → µ−RνµRcharge conjugation−−−−−−−−−−−−−−−→ π+ → µ+

RνµRparity−−−−→ π+ → µ+

LνµL

which is perfectly possible and gives a similar rate as the original reaction. So weak interactionseems to preserve the CP symmetry. Actually not, but only in rare cases.

The first case where it was observed is in the K0−K0 system in 1956. The K0 = (ds), K0 =(ds) are very easily produced by strong interaction. For instance:

K+ + n→ K0 + p , K− + p→ K0 + n

Notice that the strangeness is conserved (strong interaction) since K+ = (us). In order todecay to lightest mesons (pions), they have to decay via the weak interaction, since this is theonly interaction that does not conserve the quark flavour (and hence the strangeness). But theK0’s, eigenstates of the strong interaction, have a peculiarity: they can transform spontaneouslyto their anti-particle and vice versa: K0 ↔ K0 thanks to weak couplings. We say that theyoscillate via the two box diagrams:

u; c; t

W−

W+

u; c; t

s

d

d

s

W+

u; c; t

W−

u; c; ts

d

d

s

Since K0 and K0 are permanently oscillating, they do not have a definitive lifetime, and thephysical states that propagates are a linear combination of the two. Those states are eigenstatesof the total hamiltonian which includes the strong hamiltonian and the weak hamiltonian (sincethe original K0 are sensitive to both interactions). These states are called KL and KS wherethe L and S stand for “Long” and “Short”. This appellation emphasizes the large differencebetween their (definite) lifetimes:

τKL ' 5× 10−8 s , τKS ' 9× 10−11 s

The masses of the KL and KS are almost equal, about 498 MeV/c2 but differ however by only3.5× 10−12 MeV/c2. What is the relationship with CP violation? Consider the K0 parity:

ηK0 = ηdηs(−1)l = (1)(−1)(−1)0 = −1⇒ P |K0〉 = − |K0〉

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CP transformation 183

For the same reason, P |K0〉 = − |K0〉. Concerning the charge conjugation, it is arbitrarilypositive (the negative value would not change what follows). Therefore:

C |K0〉 = |K0〉 C |K0〉 = |K0〉

Hence, combining both C and P gives:

CP |K0〉 = − |K0〉 CP |K0〉 = − |K0〉

Thus, K0 and |K0〉 are not eigenstates of CP . Instead, the linear combinations:

|K1〉 =1√2

(|K0〉 − |K0〉

)|K2〉 =

1√2

(|K0〉+ |K0〉

)

are CP eigenstates. Indeed:

CP |K1〉 = 1√2

(CP |K0〉 − CP |K0〉

)= 1√

2

(− |K0〉+ |K0〉

)= + |K1〉

CP |K2〉 = 1√2

(CP |K0〉+ CP |K0〉

)= 1√

2

(− |K0〉 − |K0〉

)= − |K2〉

The kaons can easily decay in pions because of their large difference of masses (' 498 MeV/c2

against ' 140 MeV/c2). Some decays in 2 pions (π+π− or π0π0), others in 3 pions (π+π−π0

or π0π0π0). We have already seen in section 6.1.2 that the 2 and 3 pions coming from a spin 0particle have parities:

P |2π〉 = +1 |2π〉 , P |3π〉 = −1 |3π〉For the charge conjugation, it is a bit more complicated. Consider a state made of a particle x itsantiparticle x. Before and after C transformation, the system has the same content in terms ofparticle and antiparticle. Hence it is necessarily an eigenstate of C with an eigenvalue12 ηC = ±1.Now, if the particle x has a momentum ~p (in the center of mass frame of the system), a chargec, its antiparticle x has a momentum −~p and charge −c. If the particle x is a spin 0 boson,then the system (as π+π−) is simply described by the wave function |xx〉 ≡ ψ(~p, c ;−~p,−c), i.e.the “first” x has (~p, c) and the “second” (−~p,−c). Since they are indistinguishable particles,the wave function ψ′′(−~p,−c ; ~p, c) describes obviously the same system and because of thespin-statistc theorem, in case of bosons, ψ = +ψ′′. If x carries a non-zero spin, this additionalquantum number has to be taken into account too13. Now, one can transform ψ into ψ′′ via 2successive transformations: a C transformation followed by parity:

ψ(~p, c ;−~p,−c) C−−→ηC

ψ′(~p,−c ;−~p, c) parity−−−−→(−1)l

ψ′′(−~p,−c ; ~p, c)

Hence, we have14:ψ′′ = +ψ ⇒ ηC × (−1)l = +1⇒ ηC(xx) = (−1)l (6.30)

For the system π+π−, since both the kaons and the pions are spin 0 particles, the angularmomentum conservation imposes 0 = 0 + 0 + l⇒ l = 0 and thus:

ηC(π+π−) = +1

12ηC is necessarily ±1 because after 2 successive C-transformations, the initial state is recovered.13When constructing the wave function of a system under the strong interaction, one must also take into account

the isospin (relevant only for hadrons made of u and d quarks).14For spin 1/2 particles, because of the additional interchange of the spin, the result is ηC = (−1)l+s where s

is the total spin of the system.

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184 Weak interaction

The case of |π0π0〉 is simpler since π0 is an eigenstate of C. Indeed, since π0 → γγ via theelectromagnetic interaction which conserves C, and knowing that ηC(γ) = −115, we have:

ηC(π0) = η2C(γ) = (−1)2 = +1⇒ ηC(π0π0) = η2

C(π0) = +1

Conclusion: both final states π0π0 and π+π− have CP number:

ηCP (2π) = ηC × ηP = 1× 1 = 1

In the case of the 3 pions final state, we have to take into account the angular momentum between2 pions l1 and the one between the third pion and the system of the 2 pions l2. Because of angularmomentum conservation we have 0 = 0 + 0 + 0 + l1 ⊕ l2 ⇒ l1 = l2 and thus (−1)l1(−1)l2 = 1.This argument led us to conclude in section 6.1.2 that ηP (3π) = −1. The charge conjugation ofthe 3π0 state is straightforward knowing that ηC(π0) = 1 and hence:

ηCP (3π0) = 1×−1 = −1

For π+π−π0, we have ηC(π+π−π0) = ηC(π+π−)× ηC(π0) = (−1)l1 × 1 = (−1)l1 where we haveused equ. 6.30 and l1 is the angular momentum number between the 2 charged pions. Therefore:

ηCP (π+π−π0) = ηC(π+π−π0)× ηP (π+π−π0) = (−1)l1 ×−1 = (−1)l1+1

The difference of mass between K0 and 3 pions is only ' 80 MeV. Hence, l1 = 0 is muchmore likely to occur that l1 > 0 (only tunnel quantum effect would allow the latter case). Inconclusion, the 3 pions are almost always in ηCP = −1 state.

Let us recap: we have defined 3 kinds of kaons:

• K0 and K0, produced by the strong interaction.

• KS and KL, the physical states that propagate and decay by weak interaction.

• K1 and K2, the CP eigenstates with ηCP (K1) = 1 and ηCP (K2) = −1.

If CP is conserved, the 2 pions final state (CP even) must result from a CP -even state decayi.e. K1 → 2π. And similarly K2 → 3π. Since the phase space in the 2 pions final states is muchlarger, we expect K1 to have a much smaller lifetime. Hence we can identify the physical state:

If CP conserved: KS = K1 → 2π , KL = K2 → 3π

Now, in 1964, Cronin and Fitch made the following experiment: By dumping a proton beam(30 GeV) into a target, they produced a K0 beam mixture of the 2 physical states KL and KS .About 18 m after the production point, they placed a detector able to identify and measure themomentum of the charged pions, decay products of the KL,S (the π0 contribution was deducedfrom energy-momentum conservation). Since the KS is a short-lived meson (cτ = 2.7 cm), itdecays very early in the beam and hence, the only mesons arriving at the detector level are theKL. Among the 22700 events they analyzed, 45 corresponded to a 2 π’s decay: the KL coulddecay in 0.2% of the cases in 2 π’s and in 99.8% in 3 π’s. Conclusion: CP is violated by theweak interaction at the level of 0.2% in the Kaon system! There can be 2 origins of this CPviolation:

15This comes from the fact that the hamiltonian of a charged particle in an electromagnetic field is H =p2

2m− q

m~A.p

C−→ p2

2m− −q

m~A.p⇒ ~A

C−→ − ~A so that H is conserved under C.

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Summary 185

1. direct violation: KL is a K2 CP eigenstate, but CP is violated directly in the decay processitself: K2 → 2π (and 3π)

2. indirect violation: KL is a mixture of K2 → 3π and K1 → 2π i.e.

|KL〉 =1√

1 + |ε|2(|K2〉+ ε |K1〉)

Both sources of violation exist but in the Kaon system the latter dominates (about 600 timeslarger than the former). Thanks to the B factories (SLAC, BaBar experiment in USA, KEKB,Belle experiment in Japon), the CP violation has been observed in B mesons and recently (2007)in the charmed mesons.

So CP is violated by the weak interaction. It is finally not so surprising: we live in a worldmade of particles, not anti-particles. Thus CP must be violated to explain this asymmetry.Note however, that the CP violating mechanisms described in this section are, by several ordersof magnitude, too rare to accommodate the observed asymmetry. Something else must be atwork, but so far, we do not have a clear explanation of its origin.

6.5.3 Summary

The interactions are all described by quantum field theory. A general theorem states that theproduct of the parity P, charge conjugation C, and time reversal T namely CPT is conservedby quantum field theory. A consequence of CPT symmetry is the equality of the masses as wellas the lifetimes of a particle and its anti-particle. This prediction has been tested for severalparticles. One of the stringent tests is [19]:

∣∣∣m(K0)−m(K0)∣∣∣ < 4× 10−19 GeV/c2

Since CP is violated by weak interaction, T must also be violated.

The table 6.1 summarises the quantum numbers that are conserved or violated by the dif-ferent interactions. Note that one may define a fermion number defined as F = B +L where Bis the baryon number and L the lepton number. The fermion number is always conserved sinceboth B and L are individually conserved.

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186 Weak interaction

QED QCD Weak

Baryon nb yes yes yes

Lepton nb yes yes yes

Total isospin I no† yes no

Isopsin third component I3 yes yes no

Lepton flavour (le, lµ, lτ ) yes yes yes and no*

quark flavour (strangeness, charm number, etc) yes yes no

Parity P yes yes no

Charge conjugation C yes yes no

CP yes yes no

Time reversal T yes yes no

CPT yes yes yes

Table 6.1: Quantities conserved (“yes”) or violated (“no”) by the different interactions.

†obvious in π0 → γγ. Note however that I3 is always conserved by QED since QED current never changesthe flavor of the fermions .*As soon as a neutrino is involved, mixing is possible. See section 6.4.4.

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Exercises 187

6.6 Exercises

Exercise 6.1 Allowed and forbidden reactions.Indicate whether the reactions are allowed or not. If not, specify the reason. In any cases, whichinteraction would be involved? For convenience, we give the main quantum number of particles:I(JP ): p = 1

2(12

+), n = 1

2(12

+), π± = 1(0−); I(JPC): π0 = 1(0−+), γ = 1−−.

1. n→ p+ e+ + νe

2. π0 → γγγ

3. p+ π− → n+ π0

4. p→ n+ e+ + νe

5. π0 → γγ

6. p→ e+ + νe

7. π0 → γ

Exercise 6.2 Electron-neutrino scattering.We suppose that the energy of particles are large enough to neglect their mass but remain negli-gible with respect to the weak bosons mass. The two following reactions are considered:

(1) νµ + e− → νe + µ−, (2) νe + e− → νµ + µ−

1. Draw the Feynman diagrams and deduce the two amplitudes. Use the labels: νµ(k) +e−(p)→ νe(k

′) + µ−(p′) and νe(k) + e−(p)→ νµ(k′) + µ−(p′).

2. With the help of formulas 6.4 and 6.5, show that:

|M1|2 = 64G2F

s2

4and |M2|2 = 64G2

F

t2

4

3. Conclude that:σ(νµ + e− → νe + µ−)

σ(νe + e− → νµ + µ−)= 3

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188 Weak interaction

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Chapter 7

The Standard Model

Few references:I.J.R Aitchison & A.J.G. Hey,“Gauge theories in particle physics”, vol2, Chapter 19 & 22, IoP2003.M.E. Peskin & D.V. Schroeder, “An introduction to Quantum Field Theory”, Chapter 20, West-view Press Inc 1995

At this step of the courses, we have all the pieces to build the Standard Model ofparticle physics. In this chapter, we will see how gauge theories can underlie allelementary interactions (except gravity). The notion of spontaneous symmetrybreaking will be introduced as well, leading to the prediction of a new particle,the Higgs boson.

7.1 The Electroweak or Glashow-Salam-Weinberg theory

7.1.1 Weak isospin

In the previous chapter, we learned that the charged current couples a left handed electron to aleft-handed neutrino. The doublet made of the two fields:

(νeLeL

)≡(νee

)

L

remains thus unchanged by the charged weak interaction. And similarly for the other leptonsand quarks families: (

νµµ

)

L

(νττ

)

L

(ud′

)

L

(cs′

)

L

(tb′

)

L

Notice that the quarks here are the ones eigenstates of the weak interaction. This structurereminds the early days of the strong interaction where neutron and proton were considered ina strong isospin doublet (see section 5.1.2). Like SU(2) was the symmetry group of the strongisospin, it is then tempting to consider that SU(2)L is the symmetry group describing the weakinteractions. We add the subscript L to emphasize that this group acts only on the chiral leftcomponent of the fields. Hence, we can define a weak isospin denoted with the letter T insteadof the letter I to clearly distinguish the weak isospin T from the strong isospin I. Therefore,

189

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190 The Standard Model

the particles carry now a new quantum number, the weak isospin T and its projection T3:

T =1

2

T3 = +12

T3 = −12

(νee

)

L

(νµµ

)

L

(νττ

)

L

(ud′

)

L

(cs′

)

L

(tb′

)

L

(7.1)

Since the right handed components are not sensitive to the (charged current) weak interaction,they must belong to a weak isospin singlet1:

T = 0 νeR, eR, νµR, µR, ντR, τR, uR, dR, cR, sR, tR, bR (7.2)

For convenience, let us denote the weak isospin doublets of leptons and quarks by:

liL =

(νi

ei

)

L

with: νi=1,2,3 = νe, νµ, ντ and ei=1,2,3 = e, µ, τ (7.3)

qiL =

(ui

d′i

)

L

with: ui=1,2,3 = u, c, t and d′i=1,2,3 = d′, s′, b′ (7.4)

and similarly the weak isospin singlets:

νiR, eiR, u

iR, d

′iR (7.5)

For simplicity we can denote by

Li =

(ψiLψ′iL

)ψiR ψ′iR (7.6)

any weak isodoublet (leptons or quark) and the 2 corresponding weak isosinglets. The index icorresponds to the generation of fermions. We would expect the free Lagrangian of the theoryfor the matter fields to be based on the usual Dirac’s Lagrangian:

L =∑

q,l

3∑

i=1

Li(i/∂)Li + ψiR(i/∂)ψiR + ψ′iR(i/∂)ψ

′iR + mass terms

where for a Dirac field ψ, a mass term can be decomposed as:

−mψψ = −mψ(PL + PR)(PL + PR)ψ = −mψ(PLPL + PRPR)ψ = −mψ(PLψL + PRψR)

= −m(ψRψL + ψLψR)

and the usual property of the chirality projectors PLPR = 0 having been used. The Lagrangianwould then be with the fields of the standard model:

L =∑

q,l

3∑

i=1

Li(i/∂)Li + ψiR(i/∂)ψiR + ψ′iR(i/∂)ψ

′iR −miψiLψ

iR −m′iψ

′iLψ

i′R + h.c.

However, we see immediately that there is a problem: under a SU(2)L gauge transformation,only the left fields are transformed. Hence, this lagrangian with a mass term combining both the

1I warn the reader that I consider also the right-handed component of the neutrinos. In the original StandardModel, neutrinos were assumed to be massless which turns out to be contradicted by experimental facts. Therefore,νR remains an open possibility (neutrino would be a Dirac particle different from its anti-particle) even if notsupported by experimental evidence. However, if they do exist (which is not proven), they cannot be sensitive tointeractions except gravity as we will see later.

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Weak isospin 191

left-handed fields and the right handed ones, cannot be invariant since the right-handed cannotcompensate the transformation of the left handed. This lagrangian is not gauge invariant! Thus,either we have to give up the isodoublet model or we have to find another mechanism to generatethe masses of particles. The latter scenario will be the correct one thanks to the so-called Higgsmechanism that we shall see later in this chapter. Therefore, at this stage we consider allfermions massless. The Lagrangian of the free theory is then:

LEWfree =∑

q,l

3∑

i=1

Lii/∂Li + ψiRi/∂ψiR + ψ

′iRi/∂ψ

′iR

In order to simplify the notation we now replace the double summation by a simple summationover fermions f and dump the index of the generation. Thus:

LEWfree =∑

f

Li/∂L+ ψRi/∂ψR + ψ′Ri/∂ψ′R (7.7)

Following what was done in section 5.1.2, a doublet is rotated in the weak isospin spaceaccording to:

L′ = e−i~α.~TL (7.8)

The three generators Ti of the SU(2)L group are related to the Pauli matrices:

Ti =1

2σi with σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

)(7.9)

exactly in the same was as with the strong isospin (and the usual spin). The Lagrangian 7.7is invariant under a global SU(2)L transformation (the singlet components being unchanged).Now let us promote the invariance at the local level i.e. in every points of the space-time whenα in 7.8 depends on x:

L′ = e−i~α(x).~σ2L (7.10)

The formalism developed in section 5.2.3 in the context of SU(3)c group of QCD is straightfor-ward to transpose to SU(2)L. Imposing the local gauge invariance generates gauge bosons viathe covariant derivative analogous to equation 5.16:

Dµ = ∂µ + igwσa2W aµ (7.11)

where gw is the weak coupling constant (associated to the SU(2)L group) and the summationon a = 1 to 3 is implicit. There are as many gauge bosons (denoted above by W1, W2 and W3)as the number of generators (for the colour group SU(3)c in section 5.2.3, we did find 8 gluons,namely as many gluons as the number of Gell-Mann matrices). Moreover, these gauge bosonsbelong to the so-called adjoint representation (8 gluons in the 8 representation). Hence, in caseof SU(2)L, we expect 3 weak bosons to belong to the adjoint representation, i.e a triplet of weakisospin. Those gauge bosons give rise to the interaction term:

Lint =∑

f

−gw Lγµσa2LW a

µ =∑

f

−gw jµaW aµ (7.12)

The 3 conserved currents for each generation of fermions (quark or lepton) being:

jµa = Lγµσa2L (7.13)

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192 The Standard Model

As an example, let us take L =

(νeLeL

):

jµa = (νeL, eL)γµσa2

(νeLeL

)

so that, using the explicite matrices 7.9:

jµ1 = (νeL, eL)γµ 12

(eLνeL

)= 1

2 (νeLγµeL + eLγ

µνeL)

jµ2 = (νeL, eL)γµ 12

(−ieLiνeL

)= 1

2 (−iνeLγµeL + ieLγµνeL)

jµ3 = (νeL, eL)γµ 12

(νeL−eL

)= 1

2 (νeLγµνeL − eLγµeL)

(7.14)

Since the (charged current) weak interaction involves charge raising current transforming anelectron into a neutrino:

jµcc+ = νeγµ 1

2(1− γ5)e = νeLγ

µeL

or a charge decreasing current doing the reverse transformation (conjugate of above):

jµcc− = eLγµνeL

we see that we have to combine jµ1 and jµ2 in order to recover jµcc+ and jµcc−. The following linearcombination does the job:

jµcc+ = jµ1 + ijµ2 = Lγµ σ+

2 Ljµcc− = jµ1 − ijµ2 = Lγµ σ−2 L

(7.15)

where we have introduced:σ± = σ1 ± iσ2 (7.16)

This combination is valid for any doublet, hence the interaction term 7.12 then becomes:

Lint =∑

f −gw(jµ1W

1µ + jµ2W

2µ + jµ3W

)

=∑

f −gw(

12 [jµcc+ + jµcc−]W 1

µ − i12 [jµcc+ − jµcc−]W 2

µ + jµ3W3µ

)

=∑

f −gw(

12 [W 1

µ − iW 2µ ]jµcc+ + 1

2 [W 1µ + iW 2

µ ]jµcc− + jµ3W3µ

)

Consequently the bosons W1 and W2 must mix2 to create the physical charged bosons W±.Looking at the coupling gw√

2involved with a W± (see equ 6.9 and the Feynman rules in the same

section), we identify:

Lint =∑

f − gw√2

(1√2[W 1

µ − iW 2µ ]jµcc+ + 1√

2[W 1

µ + iW 2µ ]jµcc−

)+∑

f −gw jµ3W

=∑

f − gw√2

(W+µ j

µcc+ +W−µ j

µcc−)

+∑

f −gw jµ3W

meaning that the charged weak bosons are:

W±µ =1√2

[W 1µ ∓ iW 2

µ

](7.17)

2It is not surprising: we have already seen in section 2.2.3 that charged scalar bosons are necessarily a su-perposition of 2 scalar fields. Here, charged bosons of spin 1 are necessarily a superposition of 2 real vectorfields.

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The weak hypercharge 193

In addition, we know that weak interaction also involves a neutral current with transitions ofthe kind of those described by jµ3 in equation 7.14. It is then very tempting to identify the bosonW3 to the Z0. However, it does not work. The reason is that, since in this model the 3 bosonswould be members of the weak isospin triplet, the neutral current must have exactly the sameV-A structure as the charged currents. But we saw in the previous chapter (section 6.3.4) thatit is not the case. Moreover, the three bosons would have to have the same mass which is nottrue. Conclusion: SU(2)L cannot be the gauge group of the weak interaction.

7.1.2 The weak hypercharge

In 1961, much before the weak currents were observed experimentally, Glashow and later Salamand Ward (1964), suggested to enlarge the gauge group of the weak interaction from SU(2)Lto SU(2)L × U(1)Y . Their initial motivations were to unify the weak interaction and the elec-tromagnetism as we shall see. The subscript Y in U(1)Y stands for weak hypercharge which isnot the strong hypercharge we introduced in section 5.1.3. The names weak isospin and weakhypercharge are obviously coming from the analogy with the strong interaction where we had asimilar mathematical structure: SU(2) for the strong isospin, and even if we did not mention itin the QCD chapter, the conservation of the strong hypercharge (a simple number) is necessarilyrelated to the symmetry group U(1). How did Glashow et al. manage to unify the electromag-netism and the weak interaction which seem so different (infinite range and independence withrespect to the chirality for the former, very short range and chirality dependent for the latter)?Consider first the electromagnetic current of an electron:

jµem = qeγµe =1

2(−1)

(eγµ(1− γ5)e+ eγµ(1 + γ5)e

)= −eLγµeL − eRγµeR

Both left and right components are (equally) involved. Now, comparing this expression to theone of jµ3 in 7.14, we see that:

jµem − jµ3 = −1

2(νeLγ

µνeL + eLγµeL)− eRγµeR = −1

2l1Lγ

µl1L − e1Rγ

µe1R + 0× ν1

Rγµν1R (7.18)

where we re-introduced the compact notation of weak iso-doublets and iso-singlets. Hence, thedifference of those two currents can be seen as a third current acting on all fields of the theory(both doublet and singlet) with appropriate coefficients: −1

2 for the doublet, −1 for the electronsinglet and 0 for the neutrino singlet. This is precisely what does the weak hypercharge. Indeedwith a symmetry group U(1), and thus a simple rotation, the two members of a doublet arenecessarily treated on the same footing:

l′iL = e−iα(x)Y2 liL

where we have introduced a factor 1/2 to stick with the convention. Both left-handed electronand neutrino carry the same weak hypercharge quantum number and according to 7.18 it mustbe Y/2 = −1/2⇒ Y = −1. Similarly, for the iso-singlet,

e′iR = e−iα(x)Y2 eiR , ν ′iR = e−iα(x)Y

2 νiR

we identify for eR, Y = −2 and for νR, Y = 0. The new current generated by the weakhypercharge must then satisfy:

jµem − jµ3 =1

2jµY (7.19)

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194 The Standard Model

which can be translated in terms of the generators of each type of current: Q the electrical chargegenerator for the electromagnetism, T3 the generator associated with jµ3 and Y the generator ofU(1)Y :

Q = T3 +Y

2(7.20)

This is exactly the same relation as the Gell-Mann-Nishijima relation 5.3 used with the stronginteraction. Glashow indeed proposed that this relation holds also with the weak quantities.Based on 7.20, 7.1 and 7.2, we can easily determine the quantum numbers of all particlessummarized in table 7.1.

Q T T3 Y

νeL, νµL, ντL 0 1/2 1/2 -1

eL, µL, τL -1 1/2 -1/2 -1

νeR, νµR, ντR 0 0 0 0

eR, µR, τR -1 0 0 -2

uL, cL, tL 2/3 1/2 1/2 1/3

d′L, s′L, b

′L -1/3 1/2 -1/2 1/3

uR, cR, tR 2/3 0 0 4/3

d′R, s′R, b

′R -1/3 0 0 -2/3

Table 7.1: Quantum numbers of weak isospin and hypercharge for quarks and leptons.

A word of caution about neutrinos. As seen in table 7.1, their chirality right handed compo-nent has zero quantum numbers: they have no electric charge, and as we have shown previouslythey have Y = 0 and since they belong to a singlet, they cannot be changed by SU(2)L. Inother words, the right handed neutrinos (or left handed anti-neutrinos), if they exist (assumedhere), interact with nothing except the gravity since we know now that neutrinos have a mass3

(even if tiny). They are said to be sterile (no interaction).

7.1.3 The electroweak unification

The gauge group is now enlarged to SU(2)L × U(1)Y . We wish to clearly show that it willdescribe both the weak interaction and the electromagnetism. The generators of the respectivegroups are σi

2 and Y2 . The local transformation of the weak iso-doublets and the iso-singlets is

then:

(L)′ = e−i~α(x).~σ2−iβ(x)Y

2 L , (ψR)′ = e−iβ(x)Y2 ψR ,

(ψ′R)′

= e−iβ(x)Y2 ψ′R (7.21)

where L is any doublet field (liL or qiL) and R any singlet field (eiR, νiR, uiR, d′iR).As for QCD, the theory is invariant under a gauge local transformation if we add an inter-

action term generated by the covariant derivative to the free Lagrangian 7.7 :

L : Dµ = ∂µ + igwσi2 W

iµ + ig Y2 Bµ

ψR, ψ′R : Dµ = ∂µ + ig Y2 Bµ

(7.22)

3being massive, they might also interact with the Higgs field as we shall see later.

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The electroweak unification 195

where W iµ are the 3 gauge bosons of SU(2)L, Bµ the gauge boson of U(1)Y and gw and g the

corresponding coupling constants. Note that the field Bµ is not the electromagnetic field. Itdoes not couple to the electric charge but rather to another charge: the weak hypercharge.Introducing the covariant derivative into the free Lagrangian then yields:

Lf =∑

f Li /DL+ ψRi /DψR + ψ′Ri /Dψ′R

= LEWfree + Lint(7.23)

the interaction terms between the fermions and the gauge bosons being:

Lint = −∑

f

gw Lγµσi

2LW i

µ + g LγµY

2LBµ + g ψRγ

µY

2ψRBµ + g ψ′Rγ

µY

2ψ′RBµ

We saw in the previous section that W 1µ and W 2

µ mix together to give the 2 charged W± bosons.Using the relations 7.16 and 7.17, we then have:

Lint = −∑fgw√

2Lγµ σ+

2 LW+µ + gw√

2Lγµ σ−2 LW

−µ +

gw Lγµ σ3

2 LW3µ + g Lγµ Y2 LBµ+

g ψRγµ Y

2 ψRBµ + g ψ′Rγµ Y

2 ψ′RBµ

(7.24)

Let us focus on the neutral components described by the second line of the Lagrangian above.Following Glashow, Salam4 and Weinberg, we want to introduce the electromagnetism field. Wealready concluded that it cannot be W 3

µ (it couples to neutrino which is neutral). For the samereason it cannot be the field Bµ since the hypercharge couples similarly to the 2 components ofthe doublet and hence to the neutrino. But what about a linear combination of the 2 fields?Let us assume that the photon field Aµ is a mixing of the 2 neutral fields W 3

µ and Bµ. Theorthogonal combination of Aµ is also a neutral field. Hence, this new field must contribute tothe other interaction described by the theory, that is the weak interaction. At the time wherethe model was built, the weak neutral current had not been discovered. It was then a strongprediction. Let us write:

(AµZµ

)=

(cos θw sin θw− sin θw cos θw

)(BµW 3µ

)(7.25)

where θw is the Weinberg angle. Inverting 7.25, the second line of 7.24 then reads:

gw Lγµ σ3

2 LW3µ + g Lγµ Y2 LBµ = Lγµ

(gw cos θw

σ32 − g sin θw

Y2

)LZµ +

Lγµ(gw sin θw

σ32 + g cos θw

Y2

)LAµ

(7.26)

while the last line of 7.24 reads:

g ψRγµ Y

2 ψRBµ + g ψ′Rγµ Y

2 ψ′RBµ = −g sin θw

(ψRγ

µ Y2 ψR + ψ′Rγ

µ Y2 ψ′R

)Zµ +

g cos θw

(ψRγ

µ Y2 ψR + ψ′Rγ

µ Y2 ψ′R

)Aµ

(7.27)

The terms in 7.26 and 7.27 coupled to the electromagnetic field are:

Lγµ(gw sin θw

σ32 + g cos θw

Y2

)L+ g cos θw

(ψRγ

µ Y2 ψR + ψ′Rγ

µ Y2 ψ′R

)=

gw sin θwLγµT3L+ g cos θw

(Lγµ Y2 L+ ψRγ

µ Y2 ψR + ψ′Rγ

µ Y2 ψ′R

) (7.28)

4and Ward who was “forgotten” for the Nobel prize about the electroweak.

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196 The Standard Model

where we have used T3 = σ3/2. Let us examine the term in bracket. Inserting Y/2 = Q − T3,and noticing that by definition the singlet components have a zero weak isospin, it reads:

Lγµ Y2 L+ ψRγµ Y

2 ψR + ψ′Rγµ Y

2 ψ′R = LγµQL− LγµT3L+ ψRγ

µQψR + ψ′RγµQψ′R

= −LγµT3L+

ψLγµQψL + ψ′Lγ

µQψ′L + ψRγµQψR + ψ′Rγ

µQψ′R

where in the last line, we have decomposed the doublet in each of its components (see 7.6). Byconstruction, ψL = PLψ and ψR = PRψ have the same charge (PL and PR being the chiralityprojector). Hence:

ψLγµQψL + ψRγ

µQψR = ψγµQ(PL + PR)ψ = ψγµQψ

and similarly for ψ′. Therefore 7.28 simplifies to:

Lγµ(gw sin θw

σ32 + g cos θw

Y2

)L+ g cos θw

(ψRγ

µ Y2 ψR + ψ′Rγ

µ Y2 ψ′R

)=

(gw sin θw − g cos θw)LγµT3L+ g cos θw(ψγµQψ + ψ′γµQψ′

) (7.29)

This term couples to the electromagnetic field. Hence, we wish to associate it to the electro-magnetic current:

ejµem = e(ψγµQψ + ψ′γµQψ′

)

(the operator Q giving the charge in unit of e). Thus, comparing 7.29 to the expression above,we deduce the 2 equalities:

gw sin θw − g cos θw = 0 and e = g cos θw = gw sin θw (7.30)

This fundamental relation combined with the constraint cos2 θw + sin2 θw = 1 implies:

e2

g2w

+e2

g2= 1⇒ e =

gwg√g2w + g2

(7.31)

meaning that

cos θw =gw√g2w + g2

sin θw =g√

g2w + g2

(7.32)

The equation 7.31 is a major step toward the unification of both the electromagnetism and theweak interaction. The coupling constant of the electromagnetism e is now expressed as functionof the 2 coupling constants gw and g or equivalently gw and θw needed to describe the weakinteraction!

What about the coupling with the Zµ field? Let us consider the terms in 7.26 and 7.27coupled to the Zµ field:

Lγµ(gw cos θw

σ32 − g sin θw

Y2

)L− g sin θw

(ψRγ

µ Y2 ψR + ψ′Rγ

µ Y2 ψ′R

)

= Lγµ (gw cos θwT3 − g sin θw(Q− T3))L− g sin θw

(ψRγ

µQψR + ψ′RγµQψ′R

)

= Lγµ (gw cos θw + g sin θw)T3 L− g sin θw(ψγµQψ + ψ′γµQψ′

)

= gwcos θw

[LγµT3 L− sin2 θw

(ψγµQψ + ψ′γµQψ′

)]

where again, we have used the fact that T3ψR = T3ψ′R = 0 and similar transformations as

previously with Aµ in addition to 7.31. This term above must correspond to a current associated

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The Feynman rules for fermions-gauge bosons interactions 197

to the Zµ field. Since, the Z boson is neutral, we call it the neutral current and is thus definedas:

jµnc = LγµT3 L− sin2 θw(ψγµQψ + ψ′γµQψ′

)= jµ3 − sin2 θw j

µem (7.33)

the coupling constant to the Z boson being:

gnc =gw

cos θw(7.34)

We can now come back to the Lagrangian 7.24 describing the electroweak interaction. Insertingall the previous steps, it reads in terms of the physical gauge fields:

LEWint = −∑

f

gw√2jµcc+W

+µ +

gw√2jµcc−W

−µ + e jµemAµ +

gwcos θw

jµncZµ (7.35)

namely:

LEWint = −∑fgw√

2LγµT+LW

+µ + gw√

2LγµT−LW−µ +

e(ψγµQψ + ψ′γµQψ′

)Aµ+

gwcos θw

[LγµT3 L− sin2 θw

(ψγµQψ + ψ′γµQψ′

)]Zµ

(7.36)

with:

T+ =σ+

2=σ1 + iσ2

2=

(0 10 0

), T− =

σ−2

=σ1 − iσ2

2=

(0 01 0

), T3 =

σ3

2=

(12 00 −1

2

)

(7.37)and where a given generation of quarks or leptons is defined by the fields:

L =

(ψLψ′L

), ψR , ψ′R with ψ = ψL + ψR , ψ′ = ψ′L + ψ′R (7.38)

with (ψ ,ψ′) = (νe, e−) (νµ, µ

−) (ντ , τ−), (u, d′), (c, s′), (t, b′).

Let us recap: the 4 physical gauge bosons of the electroweak interaction: Aµ, Zµ, W±µ

are now coupled to matter fields in a consistent way with the electromagnetism and the weakinteraction. The couplings to all gauge bosons involved only two free parameters: gw and theangle θw or in terms of measured quantities e and θw instead of more (at least three) withoutunification.

7.1.4 The Feynman rules for fermions-gauge bosons interactions

Looking at the Lagrangian 7.36, we can now give the Feynman rules corresponding to theinteraction between the gauge fields and the matter fields.

7.1.4.1 Interactions with W±

The T+ matrix mixes the 2 components of a weak isospin doublet, inducing the desired chargedcurrent transition: ψL → ψ′L or ψ′L → ψL. Only the left handed chirality fields are involved.The vertex factor is −i gw√

2γµ in terms of left-handed fields or −i gw√

2γµ 1

2(1− γ5) when we include

the projection on the left handed components. The Feynman diagram is then:

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198 The Standard Model

W+

ψ′

ψ−i gw√2γµ 1

2(1− γ5) W−

ψ

ψ′−i gw√2γµ 1

2(1− γ5)

We have recovered what we found in the last chapter. Keep in mind that for down type quarks(ψ′), one should use the VCKM matrix element if the mass eigenstates are used (instead of theweak eigenstates).

7.1.4.2 Interactions with γ

The second line of 7.36 specifies this interaction. As expected, this is the one of the usual QED.Left-handed and right-handed are treated on the same footing. The coupling constant is thecharge of the particle q = eQ. The vertex factor is then:

γ

f

f−iqγµ

f being any fermions (ψ or ψ′).

7.1.4.3 Interactions with Z0

The last line of 7.36 gives the coupling of the matter with the Z0 boson. We notice that bothleft-handed (through the isodoublet) and right-handed (through the field ψ and ψ′) chiralitycomponents are involved. But contrary to the photon, the strength of the coupling differs forleft handed and right handed. Let us develop this last line:

LEWint Z0 = −∑fgw

cos θw[ LγµT3 L− sin2 θw

(ψγµQψ + ψ′γµQψ′

)]Zµ

= −∑fgw

cos θw[ 1

2ψLγµψL − 1

2ψ′Lγ

µψ′L − sin2 θw(ψγµQψ + ψ′γµQψ′

)]Zµ

= −∑fgw

cos θw[ 1

2ψγµ 1

2(1− γ5)ψ − 12ψ′γµ 1

2(1− γ5)ψ′

− sin2 θw(ψγµQψ + ψ′γµQψ′

)]Zµ

= −∑fgw

cos θw[ ψγµ 1

2

(12 − 2 sin2 θwQ− 1

2γ5)ψ +

ψ′γµ 12

(−1

2 − 2 sin2 θwQ+ 12γ

5)ψ′]Zµ

= −∑fgw

cos θw[ ψγµ 1

2

(cfV − c

fAγ

5)ψ + ψ′γµ 1

2

(cf′V − c

f ′A γ

5)ψ′]Zµ

(7.39)

wherecfV = 1

2 − 2 sin2 θwQ cfA = 12

cf′V = −1

2 − 2 sin2 θwQ cf′A = −1

2

(7.40)

These coefficients are given in the table below for each fermion. Notice that the neutrinos area bit special: since they are neutral leptons, they couple only to the Z0 (and W± ) but not tophotons. In addition, looking at the first line of the Lagrangian 7.39, we see that when Q = 0,only the chiral left component contributes to the coupling to the Z0 (this originates from thefact that νR has all his gauge numbers equal to zero Q = T3 = Y = 0, so no coupling to a gauge

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The Feynman rules for fermions-gauge bosons interactions 199

f cV cA

νe, νµ, ντ12

12

e−, µ−, τ− −12 + 2 sin2 θw −1

2

u, c, t 12 − 4

3 sin2 θw12

d, s, b −12 + 2

3 sin2 θw −12

Table 7.2: Coefficients involved in the coupling of fermions with the Z0.

field is possible). In contrast, for charged leptons both left and right components are involved.It is important to realize that there is no flavor changing of quarks (or neutrinos) with the Z0

interaction, contrary to the W± interaction. Indeed, for quarks, flavor changing would occurvia the mixing of the lower components (via the VCKM matrix element), denoted in Lagrangian

7.39 as ψ′. But all lower components have the same coefficients cf′V and cf

′A so that we can write:

L = −∑d′,s′,b′gw

cos θw

[f ′γµ 1

2

(cf′V − c

f ′A γ

5)f ′]Zµ

= −∑k=u,c,tgw

cos θw

[(∑

j=d,s,b Vkjqj)γµ 1

2

(cf′V − c

f ′A γ

5)

(∑

i=d,s,b Vkiqi)]Zµ

= − gwcos θw

∑k=u,c,t

[(∑

j=d,s,b qjV∗kj)γ

µ 12

(cf′V − c

f ′A γ

5)

(∑

i=d,s,b Vkiqi)]Zµ

= − gwcos θw

∑k=u,c,t

∑i,j=d,s,b VkiV

∗kj

[qjγ

µ 12

(cf′V − c

f ′A γ

5)qi)]Zµ

Considering the unitarity of CKM matrix 6.14, we have:

∑k=u,c,t

∑i,j=d,s,b VkiV

∗kj qj · · · qi =

∑k=u,c,t

(∑i 6=j=d,s,b VkiV

∗kj qj · · · qi +

∑i=d,s,b VkiV

∗ki qi · · · qi

)

=∑

i 6=j=d,s,b qj · · · qi∑

k=u,c,t VkiV∗kj +

∑i=d,s,b qi · · · qi

∑k=u,c,t VkiV

∗ki

=∑

i 6=j=d,s,b qj · · · qi × 0 +∑

i=d,s,b qi · · · qi × 1

=∑

i=d,s,b qi · · · qi

Conclusion, the Lagrangian coupling the down quarks to the Z0 can be simply written:

L = −∑

f ′=d,s,b

gwcos θw

[f ′γµ

1

2

(cf′V − c

f ′A γ

5)f ′)]Zµ

meaning that the weak current are diagonal in flavors. The same reasoning with up componentsin the lepton sector5 (since these are the neutrinos that mix) would have yielded a similarconclusion. The Feynman rules are now easy to deduce:

Z0

f

f−i gwcos θw

γµ 12(cV − cAγ5)

f being any fermions (leptons or quarks, up or down type).

5It is unfortunate that due to the order of physics discoveries, we chose a different convention in the quarksector and in the lepton sector. In the former, the weak isospin −1/2 components differ from the mass eigenstatewhile in the latter, it is the +1/2 components.

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200 The Standard Model

7.1.5 The gauge field transformations

After QCD based on the gauge symmetry group SU(3)c, the electroweak theory based on thegauge symmetry group SU(2)L ×U(1)Y is another example of a non abelian theory, also calledYang-Mills theory. Originally in 1954, Frank Yang (the theorist who explained the parity vi-olation in the weak interaction) and Robert Mills, developed the theoretical framework of anon abelian theory in order to describe the strong interaction. However their theory predicteda massless charged boson in conflict with the experimental facts and was soon forgotten. Itis only 10 years later, when theorists like Sheldon Glashow worked on the unification of theelectromagnetism and weak interaction that Yang-Mills theories were fully appreciated.

The field transformations we derived in the QCD chapter is easy to transpose to the SU(2)L×U(1)Y case. Indeed, for QCD we had (equation 5.21):

G′aµ = Gaµ + gsαb(x)fabcG

cµ + ∂µα

a(x)

where fabc are the structure constants verifying:

[λa2,λb2

]= ifabc

λc2

The transformations above guaranty that the QCD free Lagrangian:

L = −1

4GaµνG

µνa

based on the tensor:

Gaµν = ∂µGaν − ∂νGaµ − gsfabcGbµGcν

is gauge invariant.

Now, the generators of SU(2)L obeys to the usual spin commutation:

[Sx, Sy

]= iSz ⇒

[σ1

2,σ2

2

]= i

σ3

2

which can simply be summarized by:

[σa2,σb2

]= iεabc

σc2

ε being the usual antisymmetric tensor. Hence, the transformation of the 3 gauge fields W 1, W 2

and W 3 is:

W ′aµ = W aµ + gwα

b(x)εabcWcµ + ∂µα

a(x) (7.41)

with a = 1, 2, 3 and there is an implicit summation on b and c. The transformations aboveguaranty that the free Lagrangian:

L = −1

4F aµνF

µνa

based on the tensor:

F aµν = W aµν − gwεabcW b

µWcν with W a

µν = ∂µWaν − ∂νW a

µ (7.42)

is gauge invariant under SU(2)L.

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The gauge field transformations 201

For U(1)Y , the situation is much simpler since the group is abelian and mathematicallyequivalent to the usual U(1) of QED. Hence we have:

B′µ = Bµ + ∂µα(x) (7.43)

and:Bµν = ∂µBν − ∂νBµ (7.44)

The free Lagrangian of the electroweak gauge fields is finally:

LEWgauge = −1

4F aµνF

µνa −

1

4BµνB

µν (7.45)

It is invariant by construction under the gauge transformations SU(2)L × U(1)Y knowing thefields transformations 7.41 and 7.43. The next step is to express this Lagrangian in terms of thephysical fields W±µ , Aµ and Zµ.

Evaluation of F 1µνF

1µν + F 2µνF

2µν:Because of the antisymmetric tensor, we have:

F 1µν = W 1

µν − gwε123W2µW

3ν − gwε132W

3µW

2ν = W 1

µν − gw(W 2µW

3ν −W 3

µW2ν )

F 2µν = W 2

µν − gwε213W1µW

3ν − gwε231W

3µW

1ν = W 2

µν + gw(W 1µW

3ν −W 3

µW1ν )

so that after a bit of math (playing with dummy indices), we get:

F 1µνF

1µν + F 2µνF

2µν = W 1µνW

1µν +W 2µνW

2µν + 4gwW3µ(W 1

µνW2ν −W 2

µνW1ν)+

2g2w

[(W 1

µW1µ +W 2

µW2µ)W 3

νW3ν − (W 1

µW1ν +W 2

µW2ν)W 3

νW3µ]

Knowing the definition of W± fields 7.17, it is then straightforward to show that:

W 1µνW

1µν +W 2µνW

2µν = 2W+µνW

−µν

W 1µνW

2ν −W 2µνW

1ν = i(W−µνW+ν −W+

µνW−ν)

W 1µW

1µ +W 2µW

2µ = 2W+µ W

−µ

W 1µW

1ν +W 2µW

2ν = W+µ W

−ν +W−µ W+ν

leading to:

F 1µνF

1µν + F 2µνF

2µν = 2W+µνW

−µν + 4igwW3µ(W−µνW

+ν −W+µνW

−ν)+

2g2w

[2W+

µ W−µW 3

νW3ν − (W+

µ W−ν +W−µ W

+ν)W 3νW

3µ]

Now we have:

W 3µ = cos θwZµ + sin θwA

µ

W 3νW

3ν = cos2 θwZνZν + sin2 θwAνA

ν + 2 cos θw sin θwZνAν

W 3νW

3µ = cos2 θwZνZµ + sin2 θwAνA

µ + cos θw sin θw(ZνAµ +AνZµ)

(W+µ W

−ν +W−µ W+ν)W 3

νW3µ = 2W+

µ W−νW 3

νW3µ

and thus:

F 1µνF

1µν + F 2µνF

2µν = 2W+µνW

−µν + 4igw(cos θwZµ + sin θwA

µ)(W−µνW+ν −W+

µνW−ν)+

2g2w

[2 cos2 θwW

+µ W

−µZνZν + 2 sin2 θwW+µ W

−µAνAν+

4 cos θw sin θwW+µ W

−µZνAν]−

2g2w

[2 cos2 θwW

+µ W

−νZνZµ + 2 sin2 θwW+µ W

−νAνAµ+

2 cos θw sin θwW+µ W

−ν(ZνAµ +AνZ

µ)]

(7.46)

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202 The Standard Model

Evaluation of F 3µνF

3µν +BµνBµν:

F 3µν = W 3

µν − gwε312W1µW

2ν − gwε321W

2µW

1ν = W 3

µν − gw(W 1µW

2ν −W 2

µW1ν )

= W 3µν + igw(W+

µ W−ν −W−µ W+

ν )

leading to:

F 3µνF

3µν = W 3µνW

3µν + 4igwW3µνW

+µW−ν − 2g2w(W+

µ W+µW−ν W

−ν −W+µ W

−µW−ν W+ν)

But we have:

W 3µνW

3µν +BµνBµν = (cos θwZµν + sin θwAµν)(cos θwZ

µν + sin θwAµν)+

(− sin θwZµν + cos θwAµν)(− sin θwZµν + cos θwA

µν)= ZµνZ

µν +AµνAµν

And thus:

F 3µνF

3µν +BµνBµν = ZµνZ

µν +AµνAµν + 4igw(cos θwZµν + sin θwAµν)W+µW−ν−

2g2w(W+

µ W+µW−ν W

−ν −W+µ W

−µW−ν W+ν)

(7.47)

Finally, LEWgauge from 7.45 becomes adding 7.46 and 7.47 and noticing that 2W+µνW

−µν = (W−µν)†W−µν+

(W+µν)†W+µν :

LEWgauge = −14

[(W−µν)†W−µν + (W+

µν)†W+µν + ZµνZµν +AµνA

µν]

−igw[(cos θwZ

µ + sin θwAµ)(W−µνW

+ν −W+µνW

−ν)

+(cos θwZµν + sin θwAµν)W+µW−ν ]

−g2w2

[2 cos2 θw(W+

µ W−µZνZν −W+

µ W−νZνZµ)

+2 sin2 θw(W+µ W

−µAνAν −W+µ W

−νAνAµ)

+2 cos θw sin θw(2W+µ W

−µZνAν −W+µ W

−νZνAµ −W+µ W

−νAνZµ)

−W+µ W

+µW−ν W−ν +W+

µ W−µW−ν W

+ν]

(7.48)

7.1.6 The Feynman rules for the interactions among gauge bosons

The first line of the Lagrangian 7.48 just describes the propagation of the free fields. We shallsee in the next section how it will contribute to the propagator of the massive fields W±µ , Zµand to the massless photon Aµ. The second and third lines describe the interactions between3 gauge fields: they are the trilinear couplings. The last four lines describe the interactionsbetween 4 gauge fields: they are the quadrilinear couplings. Those trilinear and quadrilinearcouplings are possible because of the non-Abelian nature of the theory (it is analogous to the3-gluons and 4-gluons couplings). The Lagrangian 7.48 is then decomposed as:

LEWgauge = LEWgauge free + LEWtrilinear + LEWquadrilinear (7.49)

with:

LEWgauge free = −1

4

[(W−µν)†W−µν + (W+

µν)†W+µν + ZµνZµν +AµνA

µν]

(7.50)

and:LEWtrilinear = −igw

[(cos θwZ

µ + sin θwAµ)(W−µνW

+ν −W+µνW

−ν)

+(cos θwZµν + sin θwAµν)W+µW−ν ](7.51)

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The Feynman rules for the interactions among gauge bosons 203

and:

LEWquadrilinear = −g2w2

[2 cos2 θw(W+

µ W−µZνZν −W+

µ W−νZνZµ)

+2 sin2 θw(W+µ W

−µAνAν −W+µ W

−νAνAµ)

+2 cos θw sin θw(2W+µ W

−µZνAν −W+µ W

−νZνAµ −W+µ W

−νAνZµ)

−W+µ W

+µW−ν W−ν +W+

µ W−µW−ν W

+ν]

(7.52)

7.1.6.1 Gauge fields trilinear couplings

Coupling: Z0W+W−

It is given by the subpart of the Lagrangian 7.51:

LZ0W+W− = −igw cos θw[ZµW−µνW

+ν − ZµW+µνW

−ν + ZµνW+µW−ν

]

Let us denote the 4-momenta of W+, W−, Z0 respectively by k1, k2 and k3. Knowing thati∂µ = kµ, we have:

iW−µν = i(∂µW−ν − ∂νW−µ ) = k2µW

−ν − k2νW

−µ

and similarly for the other tensors, so that LZ0W+W− reads:

LZ0W+W− = −gw cos θw[Zµ(k2µW

−ν − k2νW

−µ )W+ν − Zµ(k1µW

+ν − k1νW

+µ )W−ν)

+(k3µZν − k3νZµ)W+µW−ν ]= gw cos θw [(k1 − k2)µ gλν + (k2 − k3)ν gµλ + (k3 − k1)λ gµν ]ZµW−λW+ν

where we have re-arranged the dummy indices so that the Z has a µ index, the W−, a λ indexand the W+, a ν index. The vertex factor is now straightforward:

W+(ν, k1)

W−(λ, k2)

Z0(µ, k3)igw cos θw [(k1 − k2)µ gλν + (k2 − k3)ν gµλ + (k3 − k1)λ gµν ]

The 3 momenta being understood as going toward the vertex.

Coupling: γW+W−

The procedure is similar to previously.

LγW+W− = −igw sin θw[AµW−µνW

+ν −AµW+µνW

−ν) +AµνW+µW−ν

]

= gw sin θw [(k1 − k2)µ gλν + (k2 − k3)ν gνλ + (k3 − k1)λ gµν ]AµW−λ

Recalling the relation e = gw sin θw, the vertex factor is:

W+(ν, k1)

W−(λ, k2)

γ(µ, k3)ie [(k1 − k2)µ gλν + (k2 − k3)ν gµλ + (k3 − k1)λ gµν ]

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204 The Standard Model

7.1.6.2 Gauge fields quadrilinear couplings

Coupling: Z0Z0W+W−

It is given by the subpart of the Lagrangian 7.52:

LZ0Z0W+W− = −g2w cos2 θw

[W+µ W

−µZνZν −W+µ W

−νZνZµ]

(7.53)

In order to extract the vertex factor, there is a subtlety to be taken into account. Since thereare two undistinguishable Z0 particles, the term W+

µ W−µZνZν can describe both graphs:

W+µ

W−ν

W+µ

W−ν

Hence we have the following equivalence Lagrangian-diagram:

W+µ W

−µZνZν ⇔ (gµνgαβ + gµνgβα)W+µW−νZαZβ = 2gµνgαβW+µW−νZαZβ

and similarly:

W+µ W

−νZνZµ ⇔ (gµβgνα + gµαgνβ)W+µW−νZαZβ

The overall vertex factor corresponding to Lagrangian 7.53 is thus:

W+(µ)

W−(ν)

Z0(β)

Z0(α)

− ig2w cos2 θw [2gµνgαβ − gµβgνα − gµαgνβ ]

Coupling: γγW+W−

It is given by the subpart of the Lagrangian 7.52:

LγγW+W− = −g2w sin2 θw

[W+µ W

−µAνAν −W+µ W

−νAνAµ]

= −e2[W+µ W

−µAνAν −W+µ W

−νAνAµ] (7.54)

Following the same procedure as previously (Z and γ play a symmetric role), we thus obtainthe vertex factor:

W+(µ)

W−(ν)

γ(β)

γ(α)

− ie2 [2gµνgαβ − gµβgνα − gµαgνβ ]

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The Feynman rules for the interactions among gauge bosons 205

Coupling: γZ0W+W−

It is given by the subpart of the Lagrangian 7.52:

LγZ0W+W− = −g2w cos θw sin θw

[2W+

µ W−µZνAν −W+

µ W−νZνAµ −W+

µ W−νAνZµ

]

= −egw cos θw[2W+

µ W−µZνAν −W+

µ W−νZνAµ −W+

µ W−νAνZµ

]

(7.55)leading directly to the vertex factor:

W+(µ)

W−(ν)

Z0(β)

γ(α)

− iegw cos θw [2gµνgαβ − gµβgνα − gµαgνβ ]

Coupling: W+W−W+W−

It is given by the subpart of the Lagrangian 7.52:

LW+W−W+W− = −g2w2

[−W+

µ W+µW−ν W

−ν +W+µ W

−µW−ν W+ν]

= g2w2

[W+µ W

+µW−ν W−ν −W+

µ W−µW−ν W

+ν] (7.56)

Here again, we cannot distinguish the 2 W+ and 2 W−. The term W+µ W

+µW−ν W−ν in the

Lagrangian then leads to 4 possible diagrams:

W+µ

W−ν

W+α

W−β

W+µ

W−β

W+α

W−ν

W+α

W−ν

W+µ

W−β

W+α

W−β

W+µ

W−ν

corresponding to:

W+µ W

+µW−ν W−ν ⇔ [gµαgνβ + gµαgβν + gαµgνβ + gαµgβν ]W+µW−νW+αW−β

And similarly:

W+µ W

−µW−ν W+ν ⇔ [gµβgαν + gµνgαβ + gαβgµν + gανgµβ]W+µW−νW+αW−β

The overall contribution to the vertex factor is then the sum of both:

W+(µ)

W−(ν)

W+(α)

W−(β)

ig2w [2gµαgνβ − gµβgαν − gµνgαβ]

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206 The Standard Model

7.1.7 Recapitulation

The theory defined by the Lagrangian:

L = LEWint + LEWgauge

where LEWint and LEWgauge are given respectively by the equalities 7.36 and 7.49 is able to describethe free fermions of the 3 generations, the free gauge bosons γ, W± and Z0, the interactions of thefermions with the gauge bosons and the interactions between the gauge bosons. It encapsulatesboth QED and the weak interaction. However, there is a very serious issue: none of the particlesinvolved in this theory (fermions and bosons) have a mass. A mass term like

mψψ = mψ(PR + PL)ψ = mψPRψ +mψPLψ = mψPRPRψ +mψPLPLψ = mψLψR +mψRψL

mixes components having different weak isospin and weak hypercharge, and so cannot conserveI3 and Y . It violates the gauge invariance, implying that all particles are massless! We have tosee now how to restore massive particles.

7.2 The Electroweak symmetry breaking

7.2.1 A simple example: spontaneous U(1) symmetry breaking

As a starting point, let us remind the Lagrangian for a massive scalar charged boson (the fieldbeing complex):

L = ∂µφ† ∂µφ− µ2φ†φ

with

φ =1√2

(φ1 + iφ2)

(for a classical scalar field φ† = φ∗). Applying the Euler-Lagrange equation 3.6 with φ† leads to:

∂L∂(φ†)

− ∂µ∂L

∂(∂µ(φ†))= −µ2φ− ∂µ∂µφ = 0⇒ (+ µ2)φ = 0

which is the expected Klein Gordon equation for a massive spin 0 particle of mass µ. Nowconsider the Lagrangian:

L = ∂µφ† ∂µφ− V (φ) (7.57)

where V is a scalar potential having the form6:

V (φ) = µ2φ†φ+ λ(φ†φ)2 (7.58)

When λ = 0 and µ2 > 0, the Klein-Gordon Lagrangian is recovered. We notice that our newLagrangian is invariant under a U(1) transformation:

φ→ φ′ = e−iαφ

6The form is appropriate to generate the spontaneous symmetry breaking and is conditioned by the gaugeinvariance and the renormalizability of the theory.

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A simple example: spontaneous U(1) symmetry breaking 207

The ground state of this theory is the state corresponding to the minimum energy. In the presentcase, the energy is given by the Hamiltonian operator. When only one single field is involved,we saw in equation 3.4 that the Hamiltonian (density) is defined as:

H = πφ− L =∂L∂φ

φ− L

Since here, we have 2 fields (φ1 and φ2 or equivalently φ and φ†), we should use instead:

H =∂L∂φ

φ+ φ†∂L∂φ†− L

which gives with L = ∂µφ† ∂µφ− V (φ) = φ†φ+ ∂iφ

† ∂iφ− V (φ)

H = φ†φ+ φ†φ− (φ†φ+ ∂iφ† ∂iφ− V (φ))

= φ†φ− ∂iφ† ∂iφ+ V (φ)

= φ†φ+ ~∇φ†.~∇φ+ µ2φ†φ+ λ(φ†φ)2

= |φ|2 + |~∇φ|2 + µ2|φ|2 + λ|φ|4

We are going to assume that we can proceed as if φ was a classical field. We notice that thefirst 2 terms of the Hamiltonian are positively defined and vanish when φ is a constant field (notdepending on space-time). Therefore, the minimum of H is reached for a constant field φ0 thatminimizes the last 2 terms i.e. the potential V (φ0). The potential is bound from below if λ ≥ 0(otherwise an infinite value of φ would yield an infinite negative energy). Two situations haveto be considered:

Case µ2 ≥ 0Then, the minimum is reached by the trivial constant field φ0 = 0. The ground state is uniqueand respect the U(1) symmetry as the original Lagrangian which can be re-written as functionof the real scalar fields φ1 and φ2:

L = 12∂µ(φ1 − iφ2)∂µ(φ1 + iφ2)− µ2

2 (φ1 − iφ2)(φ1 + iφ2)− λ4 [(φ1 − iφ2)(φ1 + iφ2)]2

= 12(∂µφ1∂

µφ1 + ∂µφ2∂µφ2)− µ2

2 (φ21 + φ2

2) + λ4 (φ2

1 + φ22)2

This Lagrangian in quantum field theory describes a particle spectrum with 2 bosons of spin 0having the same mass7 µ with quadrilinear interactions via the coupling λ.

Case µ2 < 0The extrema of

V (φ) = µ2|φ|2 + λ|φ|4 =µ2

2(φ2

1 + φ22) +

λ

4(φ2

1 + φ22)2

requires:∂V∂φ1

= µ2φ1 + λ(φ21 + φ2

2)φ1 = 0∂V∂φ2

= µ2φ2 + λ(φ21 + φ2

2)φ2 = 0

7 12∂µφ1∂

µφ1− µ2

2φ2

1 = 12gµµ(∂µφ1)2− µ2

2φ2

1 leads to the Euler-Lagrange equations: (−µ2φ1)− ∂µ(gµµ∂µφ1) =0⇒ (+ µ2)φ1 = 0

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208 The Standard Model

The solution φ1 = φ2 = 0 has V = 0. Since µ2 < 0, it defines a maximum. The other solution is

(φ21 + φ2

2) = 2|φ|2 =−µ2

λ≡ v2 > 0⇒ |φ| =

√−µ2

2λ=

v√2

Hence, all constant fields satisfying |φ| = v/√

2 correspond to the energy minimum of the theory.There is degeneracy of the ground state in infinite ground states lying on a circle in the imaginaryplane (=(φ),<(φ)) of radius v/

√2. This is what is shown in figure 7.1, where the potential 7.58

has the famous shape of a mexican hat. Those ground states respect the initial U(1) symmetry

Figure 7.1: The potential V (φ) 7.58 when µ2 < 0. The red ball symbolizes the degenerated ground states.

in the sense that a change of the phase of fields will not change the circle representing thesolutions. However, as soon as a particular (arbitrary) solution is chosen as the ground state,this ground state is not invariant under U(1) since it will be transformed in another state on thecircle. This situation where the ground state has less symmetry than the original Lagrangian iscalled a spontaneous breaking of the symmetry. This is typically what happens in a ferromagnetwhere above the critical temperature Tc, the spins of the electrons of the ferromagnet formingtiny magnetic dipoles cannot maintain a fixed direction because of the thermal fluctuations.When T < Tc, there are enough tiny dipoles stably oriented in the same direction to create aspontaneous and permanent magnetic field: the material becomes a magnet. Another classicalexample is the buckling of a knitting needle: starting from a state where the needle is verticaland applying a force from the top to the bottom, the initial needle which is symmetric becomesbent in an arbitrary direction because the energy in the bent position is lower than in the verticalposition. This new state state does not respect the cylindrical symmetry: any direction can bespontaneously chosen reaching one of the states minimizing the energy.

Now, let us come back to the quantum field theory case. So far, the field φ was consideredas a classical field and we simply wrote that the ground state corresponds to a field satisfying|φ| = v/

√2. In quantum field theory, we interpret the average values of quantum field as

corresponding to the classical values. When µ2 ≥ 0, we then have:

〈0|φ|0〉 = 0

which is what we have usually met so far when expanding φ in terms of creation and annihilationoperators:

φ ≈∫ake−ik.x + a†ke

+ik.x

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A simple example: spontaneous U(1) symmetry breaking 209

Indeed, we have ak |0〉 = 0 and 〈0| a†k = 0 so that naturally 〈0|φ|0〉 = 0. The quantity 〈0|φ|0〉 iscalled the vacuum expectation value or v.e.v.

When µ2 < 0, our correspondence principle between classical and quantum fields leads to:

〈0|φ|0〉 =v√2

This situation is new: the effect of the vacuum of a spontaneous symmetry breaking theoryon the field is such that the v.e.v. does not vanish. In words, we can say that now thereis on average something in the vacuum8. In quantum field theory, particles are considered asexcitations from the vacuum. In order to have a perturbative quantum field theory we haveto consider excitations around the true vacuum. We cannot have small excitations around theφ = 0 vaccum, since this is not the minimum of the potential anymore. Thus, let us expand φaround the v.e.v using the convenient parametrization:

φ(x) = (v√2

+1√2h(x))eiθ(x) (7.59)

where h(x) and θ(x) are 2 real scalar fields9 (as φ1(x) and φ2(x) were). Here, we assume〈0|h|0〉 = 〈0|θ|0〉 = 0 so that:

〈0|φ(x)|0〉 = v√2〈0|eiθ(x)|0〉+ 1√

2〈0|h(x)eiθ(x)|0〉

= v√2〈0|1 + iθ(x) + . . . |0〉+ 1√

2〈0|h(x)(1 + iθ(x) + . . .)|0〉

= v√2〈0|0〉+ 0

= v√2

Inserting the field 7.59 into the original Lagrangian 7.57, one can show (keeping in mind that

v =√−µ2

λ and expending the exponential) that:

L =1

2∂µh∂

µh− |µ2|h2 +1

2∂µθ∂

µθ + . . .

where θ(x) = vθ(x). The dots correspond to trilinear and quadrilinear interactions between hand θ and a constant term which has no incidence on the equations of motions. We see that thefield h is a massive field with a mass

√2|µ| and θ is a massless field. h corresponds to radial

excitations, orthogonal to the degenerated ground states along the θ angles. The spectrum of thespontaneous symmetry breaking theory (1 massive boson and 1 massless boson) is dramaticallydifferent from that of the normal case (2 massive bosons with same mass). The massless spin 0boson is called a Goldstone10 boson. It appears when the Lagrangian is invariant under a globalgauge transformation (U(1) in this subsection) but the vacuum does not respect that symmetrywith a non-zero vacuum expectation value.

8Please, see the section 7.4.2 where I try to better explain what I mean by something in the vacuum.9h†(x) = h(x) and θ†(x) = θ(x)

10Jeffrey Goldstone is the english physicist who proved a theorem that states that in a theory where there isa spontaneous symmetry breaking, massless spin 0 bosons appears. He gave his name to such bosons and to thetheorem.

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210 The Standard Model

7.2.2 Spontaneous breaking of the global SU(2)L × U(1)Y symmetry

We are going to see the consequences of a spontaneous breaking of the electroweak symmetry.This time, we have to use bosons that are sensitive to the gauge group, thus at minima a weakisospin doublet with non zero weak hypercharge. Weinberg, in 1967, considered the followingdoublet:

φ =

(φ+

φ0

)=

(1√2(φ1 + iφ2)

1√2(φ3 + iφ4)

)(7.60)

where φ1,2,3,4 are real scalar fields. As usual, the most positive charged component is the upperone. Thus, φ+ has T3 = +1

2 while φ0 has T3 = −12 . Note that since φ0 is made of 2 real scalar

fields, the neutral particle described by φ0 would be different than its anti-particle. Assumingthat the Gell-Mann-Nishijima relation 7.20 is satisfied, the hypercharge11 of the fields φ+ andφ0 is Y φ = 2(Q− T3)φ:

yφ = 1 (7.61)

We still consider the Lagrangian:

L = ∂µφ† ∂µφ− V (φ) (7.62)

with V :V (φ) = µ2φ†φ+ λ(φ†φ)2 , µ2 < 0

= µ2(φ+∗φ+ + φ0∗φ0) + λ(φ+∗φ+ + φ0∗φ0)2

= µ2

2 (φ21 + φ2

2 + φ23 + φ2

4) + λ4 (φ2

1 + φ22 + φ2

3 + φ24)2

(7.63)

This time, the potential V must be seen in a 4 dimension space (since there are 4 independentscalar real fields). The figure 7.1 is a cut in 2 of these 4 dimensions.

Note that the Lagrangian 7.62 is invariant under a global transformation SU(2)L × U(1)Y .Indeed φ carrying both T3 and Y numbers:

φ→ φ′ = e−i~σ2.~αe−iα

Y2 φ

∂µφ′† ∂µφ′ = ∂µφ

† ∂µφ and φ′†φ′ = φ†φ

As in the previous section, the ground state with minimal energy is given for the classicalfields minimizing the potential i.e. when:

(φ21 + φ2

2 + φ23 + φ2

4) =−µ2

λ≡ v2 ⇔ φ†φ =

−µ2

2λ≡ v2

2(7.64)

There are again an infinity of ground states. We could choose arbitrarily one of them as theground state. For instance, φ1 = φ2 = φ4 = 0 and φ3 = v. Hence, the doublet field correspondingto that ground state is:

φgroundstate =

(0v√2

)(7.65)

Why this particular choice? First, the component φ+ which carries an electrical charge is notpresent in the ground state. Hence the ground state is electrically neutral. This choice is

11Our argument is a bit sloppy. In fact, the Higgs hypercharge must be 1 in order to keep the Yukawa Lagrangian7.97 invariant under a U(1)Y transformation (see afterward). Consequently, because of the Gell-Mann-Nishijimarelation, the upper and lower components of the doublet are respectively positively charged and neutral, hencethe notation.

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Spontaneous breaking of the global SU(2)L × U(1)Y symmetry 211

consistent with our universe: the vacuum (ground state) is indeed electrically neutral. Second,we shall see later that this choice keeps the U(1)em electromagnetic gauge invariance preserved,and the photon massless. Moreover this choice will allow to explain the mass of the 3 bosonsW+, W−, Z0 (mandatory since the weak interaction has a very short range). Note however,that the ground state violates both SU(2)L and U(1)Y : None of the 4 generators of the group(σi=[1,3] and Y ) keeps the ground state invariant. Indeed if it were invariant, we would have:

e−iαΛφgroundstate = φgroundstate ⇒ Λφgroundstate = 0

where Λ isσi=1,3

2 or Y2 . Instead we have:

σ1φgroundstate =

(0 11 0

)(0v√2

)=

(v√2

0

)6= 0

σ2φgroundstate =

(0 −ii 0

)(0v√2

)=

(−i v√

2

0

)6= 0

σ3φgroundstate =

(1 00 −1

)(0v√2

)=

(0− v√

2

)6= 0

Y φgroundstate =

(1 00 1

)(0v√2

)=

(0v√2

)6= 0

(recall that yφ = 1). However, notice that Q = σ32 + Y

2 keeps invariant the groud state:

(σ3 + Y )φgroundstate =

((1 00 −1

)+

(1 00 1

))(0v√2

)= 0

Conclusion: the ground state does not respect the initial SU(2)L × U(1)Y symmetry of theLagrangian but respects a subgroup of symmetry: U(1)em. The vacuum remains neutral! Theresult of the spontaneous symmetry breaking is thus:

SU(2)L × U(1)Yspontaneous symmetry breaking−−−−−−−−−−−−−−−−−−−−−−−−−→ U(1)em (7.66)

Now, we come back to the quantum field and leave the classical field. The quantum fieldversion of the classical field minimum 7.64 is:

〈0|φ†φ|0〉 =−µ2

2λ≡ v2

2(7.67)

Once again, the vacuum is not empty since the field φ has not a zero v.e.v. In order to describethe excitations from the vacuum (that are interpreted in quantum fields as particles), we aregoing to use a different parametrization of the field φ 7.60 around the classical minimum 7.65:

φ(x) = eiθa(x)σa

(0

1√2(v + h(x))

)

where a runs from 1 to 3 and the 4 fields θ1, θ2, θ3 and h are 4 real scalar fields. For smallexcitations we do check:

φ ' (1+iθaσa)

(0

1√2(v + h)

)=

(1 + iθ3 iθ1 − θ2

iθ1 + θ2 1− iθ3

)(0

1√2(v + h)

)'(

0v√2

)+

(− v√

2θ2 + i v√

2θ1

h√2− i v√

2θ3

)

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212 The Standard Model

(where we have used the Pauli matrices and kept only the first order) that the 2 parametrizationsare equivalent (φ1 = − v√

2θ2 , φ2 = v√

2θ1 etc). If we insert the field above into the Lagrangian

7.62, one can show that:

L =1

2∂µh∂

µh− |µ2|h2 +1

2

(∂µθ

1∂µθ1 + ∂µθ2∂µθ2 + ∂µθ

3∂µθ3)

+ . . .

with θi = vθi. Hence, we have a massive field h of mass√

2|µ| and 3 massless Goldstone bosons.It is worth mentioning that, if the symmetry SU(2)L × U(1)Y were totally broken, we wouldhave 4 massless Goldstone bosons. This is the Goldstone theorem which states that there appearas many Goldstone bosons as the number of broken generators. However, in the present casewe have seen that U(1)em remains unbroken. Therefore, only 4 − 1 = 3 massless Goldstonebosons appear. None of the quanta (i.e. particles) associated to these Goldstone fields havebeen discovered. It is a serious issue since being massless, their production should be easy. Thefield h is called the Higgs field referring to Peter Higgs one of the physicists (with R. Brout, F.Englert and G. S. Guralnik, C. R. Hagen, T. W. B. Kibble) who proposed in 1964 a mechanismto eliminate the 3 (non-physical) massless Goldstone bosons. This is precisely what we are goingto see now.

7.2.3 Spontaneous breaking of the local SU(2)L × U(1)Y symmetry

The standard model is based on the local gauge symmetry SU(2)L×U(1)Y . The new Lagrangianwe introduced in 7.62 with the scalars doublet φ is invariant under the global symmetry SU(2)L×U(1)Y . Imposing the local invariance requires to use the covariant derivative Dµ instead of ∂µ.Since the doublet carries both weak isospin and weak hypercharge, the covariant derivative is:

Dµ = ∂µ + igwσi2W iµ + ig

Y

2Bµ (7.68)

The Lagrangian 7.62 now becomes:

LEWφ = (Dµφ)†Dµφ−(µ2φ†φ+ λ(φ†φ)2

)(µ2 < 0, λ > 0) (7.69)

Notice that this Lagrangian alone is not gauge invariant, the additional contribution of LEWgaugein 7.49 being necessary. As in the global case, the vacuum gives a non zero v.e.v of φ, and φtakes the form around the classical minimum:

φ(x) = eiθa(x)σa

(0

1√2(v + h(x))

)(7.70)

All the other fields involved in the theory, gauge bosons and fermions generically denoted re-spectively by Wµ and ψ below, which are sensitive to a rotation (because of their spin) mustnot be present in the vacuum (on average) because otherwise the vacuum would have a non-zeroangular momentum and invariance of space by rotation would be broken. Hence, they obey to:

〈0|Wµ(x)|0〉 = 〈0|ψ(x)|0〉 = 0 (7.71)

Thus, the only fields that can have a non zero v.e.v are scalar fields. Furthermore, since we believethat the physics is invariant by translation, the v.e.v(s) of the scalar fields cannot depend onspace-time: they must be constant.

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Generation of the bosons masses 213

The whole Lagrangian (i.e. LEWφ + LEWgauge where the latter is defined in 7.49) is still gaugeinvariant under a SU(2)L × U(1)Y transformation, independently of which vacuum we havechosen. However, there is a fundamental difference with respect to the global case: we havethe freedom to perform a local gauge transformation of the scalar field 7.70 that eliminates theGoldstone’s bosons. Indeed, we can do the transformation:

φ(x)→ φ′(x) = e−iαa(x)σa

2 φ(x) = ei(θa(x)−α

a(x)2

)σa

(0

1√2(v + h(x))

)(7.72)

We have the freedom to choose αa(x) = 2θa(x) so that φ is simply:

φ(x) =1√2

(0

v + h(x)

)(7.73)

φ verifying:

〈0|φ†φ|0〉 =−µ2

2λ=v2

2with 〈0|h(x)|0〉 = 0 (7.74)

The lower component of the doublet is such that:

H(x) = v + h(x)⇒ 〈0|H(x)|0〉 = v (7.75)

The field H is constant and present in the vacuum. Its quanta are represented by the excitationof the vacuum with the field h(x) associated to the particle called Higgs boson.

7.2.4 Generation of the bosons masses

Let us inject the field 7.73 into the Lagrangian LEWφ 7.69. We have12:

Dµφ = (∂µ + igwσi2 W

iµ + ig Y2 Bµ) 1√

2

(0

v + h

)

= 1√2

(∂µ + igw2 W

3µ + ig2Bµ igw2 (W 1

µ − iW 2µ)

igw2 (W 1µ + iW 2

µ) ∂µ − igw2 W 3µ + ig2Bµ

)(0

v + h

)

Dµφ =

(igw2 W

+µ (v + h)

1√2

(∂µh− i(v + h)(gw2 W

3µ − g

2Bµ)))

(Dµφ)† =(−igw2 W−µ (v + h), 1√

2

(∂µh+ i(v + h)(gw2 W

3µ − g

2Bµ)))

Where we have used the fact that Y = 1 for the Higgs doublet and the definition of W± fromW 1,2 (equation 7.17). Thus:

(Dµφ)†Dµφ = g2w4 (v + h)2W−µ W

+µ + 12∂µh∂

µh+ 18(v + h)2(gwW

3µ − gBµ)(gwW

3µ − gBµ)

= g2w4 (v + h)2W−µ W

+µ + 12∂µh∂

µh

+18(v + h)2(g2

w + g2)( gw√g2w+g2

W 3µ − g√

g2w+g2

Bµ)( gw√g2w+g2

W 3µ − g√g2w+g2

Bµ)

= g2w4 (v + h)2W−µ W

+µ + 12∂µh∂

µh+ 18(v + h)2(g2

w + g2)ZµZµ

= g2w4 (v + h)2W−µ W

+µ + 12∂µh∂

µh+ 18(v + h)2 g2

wcos2 θw

ZµZµ

12Notice that as for the photon where A†µ = Aµ, we have W 3†µ = W 3

µ and B†µ = Bµ, meaning that they are theirown anti-particle. For the charged field, we have W+†

µ = W−µ .

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214 The Standard Model

where we have used the expression 7.25 of Zµ as function of W 3µ and Bµ and the definition of

the Weinberg angle 7.32. Now:

φ†φ =1

2(v + h)2

so that:µ2φ†φ+ λ(φ†φ)2 = 1

2(v + h)2(µ2 + λ12(v + h)2)

= 12(v + h)2(µ2 − µ2

v212(v + h)2) (using 7.67)

= µ2v2

2 − µ2h2 − µ2

v h3 − µ2

4v2h4

The first term does not depend on any field and has therefore no incidence on the equation ofpropagation. We can omit it. Finally the Lagrangian LEWφ reduces to:

LEWφ = 12(∂µh∂

µh+ 2µ2h2) + g2wv

2

4 W−µ W+µ + g2

wv2

8 cos2 θwZµZ

µ

+g2w4 h

2W−µ W+µ + g2

wv2 hW−µ W

+µ + g2wv

4 cos2 θwhZµZ

µ + g2w

8 cos2 θwh2ZµZ

µ

+µ2

v h3 + µ2

4v2h4

(7.76)

In order to have a Lagrangian invariant under SU(2)L × U(1)Y , we have to take into accountLEWgauge (defined in 7.49):

LEWφ + LEWgauge = LEWφ + LEWgauge free + LEWtrilinear + LEWquadrilinear

More precisely, we are going to focus on LEWφ + LEWgauge free which reads after rearrangement:

LEWφ + LEWgauge free = 12(∂µh∂

µh+ 2µ2h2)

−14(W−µν)†W−µν + 1

2

(gwv2

)2(W−µ )†W−µ

−14(W+

µν)†W+µν + 12

(gwv2

)2(W+

µ )†W+µ

−14ZµνZ

µν + 12

(gwv

2 cos θw

)2ZµZ

µ

−14AµνA

µν

+g2wv2 hW−µ W

+µ + g2w4 h

2W−µ W+µ + g2

wv4 cos2 θw

hZµZµ + g2

w8 cos2 θw

h2ZµZµ

+µ2

v h3 + µ2

4v2h4

(7.77)where we have used W−µ W

+µ = 12 [(W+

µ )†W+µ + (W−µ )†W−µ]. We can now apply the Euler-Lagrange equation to the different fields. Applying it on the Higgs field (first line) leads asbefore to the Klein Gordon equation (recalling µ2 < 0) where we can identify the Higgs mass tobe:.

mH =√

2|µ| =√

2λ v (7.78)

Applying Euler-Lagrange to W−, W+ and Z (respectively second, third and fourth line of 7.77)leads to the Procca equation describing a massive particle of spin 1. For instance, with W− itgives:

(2 +m2W−)W−µ + interactions = 0 , ∂µW

−µ = 0

with:

mW− =gwv

2(7.79)

Similarly we have:

mW+ =gwv

2(7.80)

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Consideration about the choice of the gauge 215

mZ =gwv

2 cos θw=

mW

cos θw(7.81)

As expected, there is no AµAµ term (see fifth line of 7.77), meaning that the photon is indeed

massless:

mγ = 0 (7.82)

Equation 7.79 (or 7.80) gives access to the value of v. Indeed, since gw and mW are related tothe Fermi constant, one has:

GF√2

=g2w

8m2W

⇒ v2 =1√

2GF⇒ v ' 246 GeV (7.83)

The vacuum expectation of the Higgs field is:

〈0|φ†φ|0〉 =v2

2' (174 GeV)2 (7.84)

It sets the scale of the electroweak symmetry breaking. Such energy density corresponds to atemperature of roughly 2× 1015 K! In the big-bang model of the universe, this phase transitionSU(2)L×U(1)Y → U(1)em where the electroweak splits into the electromagnetism and the weakforce occurred at 10−11 s.

We have accomplished a major step! The spontaneous breaking of the local symmetry ofthe gauge group of the Electroweak theory yields massive gauge bosons for the Weak interactionand keeps a massless gauge boson for the electromagnetism. Before the spontaneous symmetrybreaking, we had 4 massless spin 1 bosons (3 W , 1 B) and 4 spin 0 scalar fields coming fromthe Higgs doublet. In term of degrees of freedom, it represents 4 × 2 polarizations + 4 = 12.After the spontaneous symmetry breaking, we have 3 massive spin 1 bosons W+, W− and Z, 1massless spin 1 boson, the photon and 1 massive scalar field, giving a total of 3×3 polarizations(recall that ∂µW

−µ = 0) + 2 polarizations (photon) + 1 = 12. There is no loss of degrees offreedom. Thanks to the gauge transformation, we shifted the 3 massless Goldstone modes intothe 3 gauge fields W+, W− and Z via their longitudinal polarization. What is really remarkableabout this way of giving a mass to the three weak bosons is that at no point we have given upgauge invariance. The symmetry is only hidden.

7.2.5 Consideration about the choice of the gauge

In order to eliminate the Goldstone bosons, we chose an appropriate gauge (called unitary)that eases the interpretation of the particle spectrum of the theory. The fact that the Goldstonebosons can be eliminated just by a judicious choice of gauge shows they are not physical particles.When a theory is invariant under a gauge symmetry, the physics must not depend on a particularchoice of gauge. In other words, we could have decided to not perform the gauge transformation7.72 and keep the Goldstone bosons. In that case, these unphysical fields would have to beincluded in the Feynman diagrams with appropriate propagators. The gauge bosons themselveswould have their propagators modified but overall, all these modification would cancel so thatthe physics would remain the same. The proof of this statement is however not trivial at all.

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216 The Standard Model

7.2.6 Higgs couplings to bosons

The 2 last lines of the Lagrangian 7.77 show that the Higgs boson couples to the massive gaugefield via trilinear and quadrilinear terms. The trilinear ones are:

LEWhWW =g2wv

2hW−µ W

+µ = gwmW hW−µ W+µ (7.85)

and:

LEWhZZ =g2wv

4 cos2 θwhZµZ

µ =1

2

gwcos θw

mZ hZµZµ (7.86)

This latter Lagrangian gives rise to 2 Feynman diagrams because of the interchange of the 2 Zbosons:

1

2

gwcos θw

mZ hZµZµ ⇔ 1

2

gwcos θw

mZ hgµνZνZµ +

1

2

gwcos θw

mZ hgνµZµZν =

gwcos θw

mZ hgµνZνZµ

The vertex factors are then:

W+(ν)

W−(µ)

HigwmW gµν Z(ν)

Z(µ)

Hi gwcos θmZ gµν

Note that the Higgs couplings to gauge bosons can be expressed in a more symmetric way. Forinstance, replacing gw by 2mw/v or equivalently by 2(

√2GF )1/2mW gives the couplings:

cW = gwmw = 2vm

2W = 2(

√2GF )1/2m2

W

cZ = gWcos θW

mZ = 2vm

2Z = 2(

√2GF )1/2m2

Z

we we have used the relation 7.81. The 2 couplings have a similar form

cV=W,Z =2

vm2V = 2(

√2GF )1/2m2

V (7.87)

The other trilinear coupling is the self-coupling of the Higgs given by:

LEWhhh =µ2

vh3 = −|µ

2|vh3 = −m

2H

2vh3 = −gw

m2H

4mWh3 (7.88)

This term corresponds to 3! = 6 possible permutations of the 3 Higgs, and hence, the vertexfactor is 6 times greater:

H

H

H−igw 3m2H

2mW

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Generation of the fermions masses 217

Now, considering the quadrilinear couplings (in 7.77):

LEWhhWW =g2w

4h2W−µ W

+µ (7.89)

LEWhhZZ =g2w

8 cos2 θwh2ZµZ

µ (7.90)

LEWhhhh =µ2

4v2h4 = −|µ

2|4v2

h4 = −g2w

m2H

32m2W

h4 (7.91)

In 7.89, there are 2 possible permutations of H, hence the vertex factor is proportional to 2× g2w4 .

In 7.90 we have 2 possibilities for the Higgs and 2 for the Z, and so a vertex factor proportional to

4× g2w

8 cos2 θw. And finally, in 7.91, there are 4! = 24 permutations of Higgses, giving 24×−g2

wm2H

32m2W

.

In summary, the vertex factors are:

W+(µ)

W−(ν)

H

H

ig2w2 gµν

Z(µ)

Z(ν)

H

H

i g2w

2 cos2 θwgµν

H

H

H

H

− ig2w

3m2H

4m2W

7.2.7 Generation of the fermions masses

The spontaneous symmetry breaking of the local gauge symmetry SU(2)L × U(1)Y gives riseto mass terms of the weak gauge bosons, while keeping the photon massless. This mechanism,known as the Higgs mechanism (but also due to the work of Brout and Englert among others)fixes two issues: it gives mass to gauge bosons and it eliminates the non physical Goldstonescalar bosons. We have seen that technically it was accomplished thanks to the coupling of theHiggs boson doublet to the covariant derivative Dµ. However, such mechanism cannot generatemasses of fermions since the fermions fields do not show up in Dµ. Another mechanism mustbe at work.

7.2.7.1 Masses without fermions mixing

In section 7.1.1, we explained why a mass term is not invariant under the gauge transformation:

−mψψ = −m(ψRψL + ψLψR)

The mass couples the two chirality states. That is precisely the reason why such term is notgauge invariant since the doublets and thus the left-handed states are subject to a differenttransformation than the singlets i.e. the right-handed states. The transformation of the formercan not be compensated by the transformation of the latter. We have to find a way to coupleleft-handed states to right handed ones without explicitly breaking the gauge invariance. Well,the Higgs doublet:

φ(x) =

(φ+(x)φ0(x)

)

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218 The Standard Model

can do this job! For simplicity, we consider for the moment only the first generation of leptons.Let us couple a doublet to a singlet via the Higgs and an arbitrary coupling constant ge (a realnumber):

L = −geLeφ eR = −ge(νeL, eL)

(φ+

φ0

)eR = −ge(νeLφ+ + eLφ

0)eR

We must add the hermetic conjugate:

−gee†R(

(φ+)†(νe†Lγ

0)† + (φ0)†(e†Lγ0)†)

= −geeR(

(φ+)†νeL + (φ0)†eL)

= −geeRφ†Le

so that the Lagrangian is hermitian:

L = −geLeφ eR − geeRφ†Le (7.92)

Let us check the invariance of this Lagrangian under a SU(2)L×U(1)Y transformation (droppingthe hermitian conjugate):

L′ = −ge(e−iα

a(x)σa2−iα(x)Y

2 Le

)(e−iα

a(x)σa2−iα(x)Y

2 φ)(

e−iα(x)Y2 eR

)

= −ge(e−iα

a(x)σa2−iα(x)

yLe2 Le

)(e−iα

a(x)σa2−iα(x)

yφ2 φ)(

e−iα(x)yeR

2 eR

)

= −geLe(e−iα

a(x)σa2

)†e−iα

a(x)σa2 ei

α(x)2

(yLe−yφ−yeR )φ eR

= −geLeφ eR eiα(x)

2(yLe−yφ−yeR )

(capital letter for the hypercharge means the operator while y means the quantum number).The lagrangian would be invariant if:

yLe − yφ − yeR = 0⇒ yφ = yLe − yeRBut looking at table 7.1 where the hypercharge numbers are defined and we see that it imposes:

yφ = −1 + 2 = +1

This is the justification of our choice in the definition 7.61 which guaranties the invariance ofthe Lagrangian13. Notice that the sum of the hypercharges cancels for all combinations L, φand right-handed particles associated to the down component of the weak isospin (denoted byψ′R). We shall address the case of the ψR fields in a moment. Now, after spontaneous symmetrybreaking, the field becomes:

φ =

(φ+

φ0

)s.s.b−−→ φ =

(0

1√2(v + h)

)

and thus the Lagrangian 7.92 now reads:

L = −ge(νeL, eL)

(0

1√2(v + h)

)eR − geeR

(0, 1√

2(v + h)

)(νeLeL

)

= − ge√2

(eLeR(v + h) + eReL(v + h))

= − ge√2ee(v + h)

13And consequently, the upper component of the Higgs doublet is positively charged and the lower one neutral.

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Generation of the fermions masses 219

But we have just created a mass term! It suffices to define the mass of the electron as:

me =ge√

2v

and the Lagrangian after spontaneous symmetry breaking is just:

L = −meee−me

veeh

describing a massive electron (one has to add ei/∂e coming from the free Lagrangian 7.7) in-teracting with the Higgs field with a coupling proportional to the electron mass. A couplingbetween a scalar and 2 spin 1/2 fermions (Dirac field) is called a Yukawa coupling. Historically,this is this coupling that Yukawa used in the 1930s to describe the strong interaction as theexchange of pions (spin 0) between two nucleons (spin 1/2).

We can easily generalize to the other fermions. The Lagrangian is then:

Lyuk =∑

f ′−gf ′Lφψ′R − gf ′ψ′Rφ†L

The spontaneous symmetry breaking then generates the mass terms and the coupling of thefermions to the Higgs:

Lyuk =∑

f ′−mf ′ψ′ψ

′ − mf ′

vψ′ψ′h with mf ′ =

gf ′√2v

Note that only fermions of down type f ′ = e, µ, τ, d′, s′, b′ are described here since only ψ′

appears. We repeat here the notation we introduced in the definition 7.6.

L =

(ψLψ′L

)ψRψ′R

(7.93)

ψR and ψ′R being 2 weak isosinglets. Explicitly:

(νeLeL

)νeReR

,

(νµLµL

)νµRµR

,

(ντLτL

)ντRτR

(uLd′L

)uRd′R

,

(cLs′L

)cRs′R

,

(tLb′L

)tRb′R

How can we generate the mass of the up-type fermions?: Coming back to the first generationof lepton, we see that if we manage to have the following term after spontaneous symmetrybreaking, we may succeed:

L = −gνe(νeL, eL)

(1√2(v + h)

0

)νeR − gνeνeR

(1√2(v + h), 0

)(νeLeL

)

= −gνe√2

(νeLνeR(v + h) + νeRνeL(v + h))

= −gνe√2νeνe(v + h)

The example here is with the neutrino that, we now know, has a (tiny) mass thanks to theoscillation phenomena. In using such Yukawa coupling, we assume implicitly [7, p. 362] that

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220 The Standard Model

the neutrino is a Dirac fermion (as all the other fermions of the Standard Model) and not aMajorana fermion. The nature of the neutrino is still an open question and in the rest of this

document, for simplicity, we will assume it to be of Dirac type14. How can we get

(1√2(v + h)

0

)

from φ? In section 5.1.2, we learned in the context of the strong isospin that the anti nucleondoublet, namely the charge conjugate is:

|N c〉 = |N〉 =

(−np

)⇔(−du

)

where we have specified the modern version using the quarks language. This is what led us todefine the π0 as 1√

2(uu− dd). We could have defined equivalently the conjugate doublet as:

(d−u

)

and the π0 would be 1√2(dd− uu). The factor -1 has no physical consequence. So let us define

the charge conjugate of weak isospin doublet of the Higgs as:

φ =

(φ+

φ0

)⇒ φc =

(φ0

−φ+

)≡(φ0∗

−φ−)

(7.94)

One can easily check that φc is in fact defined by:

φc = iσ2φ∗ (7.95)

which is analogous to ψc = iγ2ψ∗ (equation 2.50) in the 4-spinor world. The quantum numberof φc must be opposite to the ones of φ, and thus we expect:

yφc = −1 (7.96)

Indeed, the charge of the upper component (T3 = +1/2) φ0∗ is 0 and so yφ0∗ = 2(Qφ0∗−T3φ0∗) =

−1. Similarly y−φ− = 2(−1+1/2) = −1. Thanks to this definition, we see that after spontaneoussymmetry breaking, φc reduces to the desired form:

φc =

(1√2(v + h)

0

)

(keeping in mind that h is a scalar real field and v a real) and hence the Lagrangian that givesmass to νe is:

L = −gνe(νeL, eL)

(1√2(v + h)

0

)νeR − gνeνeR

(1√2(v + h), 0

)(νeLeL

)

= −gνeLeφcνeR − gνeνeRφc†LeThe generalization to all fermions is straightforward. The whole Yukawa Lagrangian is then:

LEWyuk =∑

f ′−gf ′Lφψ′R − gf ′ψ′Rφ†L+

f

−gfLφc ψR − gfψRφc†L (7.97)

14According to the Ockham’s razor principle, I don’t see why other kind of fermions would be necessary.

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Generation of the fermions masses 221

which after spontaneous symmetry breaking reads:

LEWyuk =∑

f ′−mf ′ψ′ψ

′ − mf ′

vψ′ψ′h+

f

−mfψψ −mf

vψψh with mf ′ =

gf ′√2v, mf =

gf√2v

(7.98)where f ′ = e, µ, τ, d′, s′, b′ and f = νe, νµ, ντ , u, c, t. Let us stress again that the Lagrangian 7.97respects the SU(2)L×U(1)Y symmetry 15. This is the Higgs field itself which, after spontaneoussymmetry breaking, does not respect the initial symmetry. And this is the coupling of fermions tothe Higgs which generates their mass. Beware, that the masses are not predicted at all! Indeed,the couplings are free parameters, and there are as many couplings as the number of fermions.Furthermore, it is rather unsatisfactory to have couplings with such differences in magnitude.A neutrino of let us say m = 1 eV would have a very weak coupling of gν =

√2mv ' 6× 10−12

while the top quark (175 GeV) would have gt ' 1.According to the Lagrangian 7.98, we see that the Higgs couples to fermions with a constant

−mf

v= −gw

mf

2mW(7.99)

(using 7.80). This yields the Feynman diagram:

f

f

H −igw mf2mW

We see that the coupling to fermions is linear in the mass of the fermion while the coupling togauge boson is quadratic in the mass of the boson (compare 7.99 with 7.87).

7.2.7.2 Masses with fermions mixing

So far, we did not pay attention to the mixing of fermions. Recall that this mixing is due tothe fact that weak eigenstates are not the mass eigenstates (that propagate). Using just onecoupling constant per fermion as before led to the mass of fermions suggesting that the weakeigenstates are the mass eigenstates. To avoid this, we need to use several coupling constantsper weak eigenstate. More precisely, we define the Lagrangian as:

LEWyuk = LEWyuk (q) + LEWyuk (l) (7.100)

where we distinguish the part of the Lagrangian relevant for quarks and the one for leptons.Both have a similar form and can be written:

LEWyuk (q or l) =

3∑

j,k=1

−g′jkLjφψ′kR − gjkLjφc ψkR + h.c. (7.101)

where h.c. stands for hermitian conjugate and j, k correspond to the generation number of quarksor leptons. Developing the fermions doublets, we have:

LEWyuk (q or l) =3∑

j,k=1

−g′jk(ψjL, ψ′jL)φψ′kR − gjk(ψjL, ψ′jL)φc ψkR + h.c.

15we have again yL − yφc − yψR = 0.

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222 The Standard Model

which reads after spontaneous symmetry breaking:

LEWyuk (q or l) =3∑

j,k=1

−g′jk√

2ψ′jLψ

′kR(v + h)− gjk√

2ψjLψkR(v + h) + h.c.

Let us define:

mjk = gjkv√2

, m′jk = g′jkv√2

(7.102)

so that:

LEWyuk (q or l) =3∑

j,k=1

−mjkψjLψkR −m′jkψ′jLψ′kR −

mjk

vψjLψkR h−

m′jkvψ′jLψ

′kR h+ h.c.

We can write this Lagrangian using matrix notation by defining:

Ψ =

ψ1

ψ2

ψ3

, Ψ′ =

ψ′1ψ′2ψ′3

, M =

m11 m12 m13

m21 m22 m23

m31 m32 m33

, M ′ =

m′11 m′12 m′13

m′21 m′22 m′23

m′31 m′32 m′33

(7.103)

so that:

LEWyuk (q or l) = −ΨLMΨR −Ψ′LM ′Ψ′R −1

vΨLMΨR h−

1

vΨ′LM ′Ψ′R h+ h.c. (7.104)

The mass eigenstates denoted thereafter with a tilde are the ones which would be unmixed inthe Lagrangian. So, let us diagonalize the M and M ′ matrices thanks to 4 unitary (U †LUL = 1etc) matrices:

ULMU †R =

m1

m2

m3

, U ′LM

′U ′†R =

m′1

m′2m′3

We define the physical states as:

ΨL =

ψ1L

ψ2L

ψ3L

= UL

ψ1L

ψ2L

ψ3L

, ΨR = URΨR , Ψ′L = U ′LΨ′L , Ψ′R = U ′RΨ′R

In case of quarks we have:

Ψ =

uct

, Ψ′ =

dsb

, ULMU †R =

mu

mc

mt

, U ′LM

′U ′†R =

md

ms

mb

while for leptons:

Ψ =

ν1

ν2

ν3

, Ψ′ =

eµτ

, ULMU †R =

mν1

mν2

mν3

, U ′LM

′U ′†R =

me

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The Standard Model 223

Now coming back to the Lagrangian 7.104, we notice that:

ΨLMΨR = ΨL U†LULM U †RUR ΨR = ΨL U

†L

m1

m2

m3

UR ΨR

= ULΨL

m1

m2

m3

UR ΨR = ΨL

m1

m2

m3

ΨR

and similarly for the other terms. Hence 7.104 reads:

LEWyuk (q or l) = −ΨL

m1

m2

m3

ΨR

(1 +

h

v

)− Ψ′L

m′1

m′2m′3

Ψ′R

(1 +

h

v

)+ h.c.

Thus, we see again that the physical (i.e. mass) eigenstates get their mass via the Higgs mech-anism and interact with the Higgs with the coupling 7.99.

What about the charged current? According to its definition 7.15:

jµcc+ =∑Lγµ σ+

2 L = ΨLγµΨ′L = ΨLU

†LULγ

µU ′†LU′LΨ′L = ULΨLULγ

µU ′†LU′LΨ′L

= ΨLγµULU

′†L Ψ′L

Hence, we identify the mixing matrix to be:

V = ULU′†L (7.105)

For the quarks V = VCKM while for leptons V = VPMNS . A remark about neutrinos: VPMNS

should be used as soon as the mass eigenstates are used in the Feynman diagrams. Usually,this is not what we do at experiments operating on colliders: First, the neutrinos escape fromdetection. Secondly, if by chance a neutrino interacts in the detector, the probability that thephysical state (ν1, ν2 or ν3) that propagates, interacts with a flavor different than the originalflavor at production is virtually 0 just because the size of the detector is not large enough toallow the neutrino to oscillate: significant probabilities of oscillation are encountered only withdistances of several hundred kilometers. Therefore, only dedicated experiments using a beamof neutrinos detected several hundred kilometers further are sensitive to the VPMNS matrixelements.

7.3 The Standard Model

7.3.1 Summary

In this section, we summarize the Standard Model of particle physics exposed in the previoussections and chapters. The gauge group of the Standard Model is:

SU(3)c × SU(2)L × U(1)Y (7.106)

where SU(3)c is the color group of QCD while SU(2)L × U(1)Y are the weak isospin and weakhypercharge groups of the Electroweak theory. After spontaneous symmetry breaking, it isassumed that SU(3)c remains unbroken so that the symmetry breaking induces:

SU(3)c × SU(2)L × U(1)Ys.s.b.−−−→ SU(3)c × U(1)em

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224 The Standard Model

The matter fields (fermions spin 1/2), in terms of SU(2)L representation are :

L =

(ψLψ′L

)ψRψ′R

(νeLeL

)νeReR

,

(νµLµL

)νµRµR

,

(ντLτL

)ντRτR

(uLdL

)uRdR

,

(cLsL

)cRsR

,

(tLbL

)tRbR

(7.107)

where for the neutrinos, we stick to the usual convention to use the weak eigenstates. In addition,we have introduced right-handed neutrinos that were absent from the original formulation of theStandard Model where neutrinos were assumed to be massless. For the SU(3)c representation,the matter fields are:

qf =

f1

f2

f3

u1

u2

u3

d1

d2

d3

s1

s2

s3

c1

c2

c3

b1b2b3

t1t2t3

(7.108)

all the other fermions being singlets of SU(3)c (since not interacting by strong interaction). Theindex 1,2 or 3 refers to the color index.

Concerning the gauge bosons (spin 1), there are the 8 gluons (octet of SU(3)c) , the 3 weakbosons W+, W−, Z0 and the photon γ:

g (Ga=[1,8]µ ), W+ (W+

µ ), W− (W+µ ), Z0 (Zµ), γ (Aµ) (7.109)

and finally, the only scalar of the theory, the Higgs boson, member of a SU(2)L doublet:

φ =

(φ+

φ0

)s.s.b−−→ φ =

(0

1√2(v + h)

)(7.110)

The Lagrangian of the Standard Model is the sum of several Lagrangians:

LSM = LQCD + LEW (7.111)

where:LQCD = LQCDfree + LQCDint + LQCDgauge (7.112)

with (see equation 5.28):

LQCDfree =∑

f=u,d,s,c,b,t

qf (iγµ∂µ −m)qf (7.113)

LQCDint =∑

f=u,d,s,c,b,t

−gs qfγµλa2qf G

aµ (7.114)

LQCDgauge = −1

4Gµνa Gaµν (7.115)

the expression of Gµνa being given in 5.27. The Electroweak Lagrangian is:

LEW = LEWfree + LEWint + LEWgauge + LEWφ + LEWyuk (7.116)

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Strength and weakness of the theory 225

with:LEWfree =

f

Li/∂L+ ψRi/∂ψR + ψ′Ri/∂ψ′R (7.117)

the sum over f being understood as a sum over the representation given in 7.107, i.e. 6 contri-butions (leptons and quarks of 3 generations).

LEWint = −∑fgw√

2ψLγ

µV ψ′LW+µ + gw√

2ψ′Lγ

µV †ψLW−µ +

e(ψγµQψ + ψ′γµQψ′

)Aµ+

gwcos θw

[ψγµ 1

2

(cfV − c

fAγ

5)ψ + ψ′γµ 1

2

(cf′V − c

f ′A γ

5)ψ′]Zµ

(7.118)

where V = VCKM for quarks and is unity for leptons since we use the neutrinos eigenstates ofthe weak interaction as matter field.

LEWgauge = −1

4F aµνF

µνa −

1

4BµνB

µν (7.119)

and an explicit expression as function of the physical vector fields is given in 7.48.

LEWφ = (Dµφ)†Dµφ−(µ2φ†φ+ λ(φ†φ)2

)(µ2 < 0, λ > 0) (7.120)

which is described in details in 7.76. After spontaneous symmetry breaking, the term LEWyuksimply reads:

LEWyuk =∑

f

−mfψψ

(1 +

h

v

)+∑

f ′−mf ′ψ′ψ

′(

1 +h

v

)(7.121)

7.3.2 Strength and weakness of the theory

The major strength of Standard model is its predictive power: Z0 and W± bosons were predictedand found, top quark was predicted and its mass even evaluated before its discovery thanks tothe fits of the electroweak data [31]:

mt = 178± 8 +17−20 GeV/c2

where the main uncertainty was due to the systematics related to the unknown Higgs mass.This prediction has to be compared to the measured [19]:

mt = 172.9± 0.6± 0.9 GeV

It is an extraordinary success. So far, there is no sign of significant (and persistent) deviationwith respect to the numerous predictions. Moreover, the theory is renormalizable (t’Hooft andVeltman showed it) allowing the calculation at any order. Talking about the top quark, it isworth mentioning a particularity: its total decay width is 2 GeV, very large because mt beinggreater than mW , the top can decay into an on mass-shell W : t→W+b. But Γ = 2 GeV meansa lifetime of only 3 × 10−25 s, a maximal distance of only 0.1 fm before decaying. However,we learned that the QCD potential becomes very strong at a scale of typically 1 fm, ten timeslarger than the top mean free path. Conclusion: the top quark is the only quark which decaysbefore “hadronizing”!

In its original formulation where the neutrinos were considered massless, the number of freeparameters of the Standard Model was: 6 masses of quarks, 3 masses of charged leptons, the

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226 The Standard Model

coupling constants gw, g, gs and the Weak (Weinberg) mixing angle θw, the Higgs potentialparameters µ and λ and the 4 independent VCKM constants (3 angles and a phase): total 19free parameters. Now, the neutrinos sector appears richer than expected, so we have the 3masses and the 4 parameters of VPMNS (assuming neutrinos to be of Dirac type), increasing thetotal to 26 parameters. It is generally believed that it is too much for a fundamental theory. TheStandard Model is probably an effective theory, the ultimate theory still being a mystery. Natureseems to perfectly accommodate just the first generation of fermions, so, why 3 generations?

Finally, let us mention that the Higgs potential did not come from any dynamical mechanism.It is a pure add-hoc postulate. In addition, the vacuum energy density coming from the Higgspotential is:

ρH = V (φ) = µ2(0,v√2

)

(0v√2

)+ λ

[(0,

v√2

)

(0v√2

)]2

= λ

(µ2

λ

v2

2+v4

4

)= −λv

4

4= −m

2Hv

2

8

Knowing the Higgs mass (see next chapter) mH = 126 GeV and v = 246 GeV, it gives a densityρH ' −1.2 × 108 GeV4. Since vacuum energy density couples to gravity in general relativity,ρH contributes to the cosmological constant via:

ΛH = κρH with κ =8πGNc2

' 1.86× 10−27cm gr−1

GN being the gravitational constant. Converting ρH in the appropriate units (gr cm−3), onefinds:

ΛH = (1.86× 10−27cm gr−1)−1.2× 108 GeV4

(1.97× 10−14 GeV.cm︸ ︷︷ ︸~c

)3× 1.7827× 10−24

︸ ︷︷ ︸gr GeV−1

= −5.2× 10−2cm−2

which has to be compared to the measured cosmological constant16 :

Λmeas = κ ρcrit Ωtot

= (1.86× 10−27cm gr−1)(1.88× 10−29 × 0.6732g cm−3)× 1= 1.6× 10−56cm−2

In addition to the wrong sign, the cosmological constant due to the Higgs potential is about 54orders of magnitude larger than the measured value! A very serious problem of the theory!!!

There are many other reasons why theorists are not satisfied by the Standard Model and tryto encapsulate it in a larger framework. For instance, why is the electric charge quantified? Whydo quarks and leptons both have spins of one half (particles of matter) if they are so differentwith respect to the interactions?

7.3.3 Tests of the theory

In this section, only tests of the electroweak theory will be addressed. The tests of QCDwere described in the dedicated chapter. The electroweak model has been tested extensively atvarious colliders and is still tested at LHC. So far, there is no significant sign of deviation betweenexperimental measurements and the theoretical predictions. The LEP collider (e+e− collider)

16The numerical values can be found on the particle data group web site.

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Tests of the theory 227

played a particular role in the last decade of the twentieth century thanks to the very highprecision achieved using the Z0 production. Dedicated data taking were acquired around theZ0 peak in order to measure the so-called Z0 lineshape, i.e. the shape of the Z0 resonance usingthe cross-section measurements at various energies around the Z0 mass. The hadronic cross-section measurement obtained at several e+e− colliders is shown in figure 7.2. The Z0 resonance

The ALEPH, DELPHI, L3, OPAL and SLD Collaborations / Physics Reports 427 (2006) 257 –454 261

e+

e!

"

f!

f

e+

e!

Z

f!

f

Fig. 1.1. The lowest-order s-channel Feynman diagrams for e+e! " ff. For e+e! final states, the photon and the Z boson can also be exchangedvia the t-channel. The contribution of Higgs boson exchange diagrams is negligible.

10

10 2

10 3

10 4

10 5

0 20 40 60 80 100 120 140 160 180 200 220

Centre-of-mass energy (GeV)

Cro

ss-s

ecti

on (p

b)

CESRDORIS

PEP

PETRATRISTAN

KEKBPEP-II

SLC

LEP I LEP II

Z

W+W-

e+e!#hadrons

Fig. 1.2. The hadronic cross-section as a function of centre-of-mass energy. The solid line is the prediction of the SM, and the points are theexperimental measurements. Also indicated are the energy ranges of various e+e! accelerators. The cross-sections have been corrected for theeffects of photon radiation.

centre-of-mass energies of approximately 91 GeV, close to the mass of the Z boson.1 Fig. 1.2 illustrates two prominentfeatures of the hadronic cross-section as a function of the centre-of-mass energy. The first is the 1/s fall-off, due tovirtual photon exchange, corresponding to the left-hand diagram in Fig. 1.1, which leads to the peak at low energies.The second is the peak at 91 GeV, due to Z exchange, which corresponds to the right-hand diagram of Fig. 1.1, andallows LEP and SLC to function as “Z factories”.

The LEP accelerator operated from 1989 to 2000, and until 1995, the running was dedicated to the Z boson region.From 1996 to 2000, the centre-of-mass energy was increased to 161 GeV and ultimately to 209 GeV allowing theproduction of pairs of W bosons, e+e! " W+W!, as indicated in Fig. 1.2. Although some results from this laterrunning will be used in this report, the bulk of the data stems from the Z period. When needed, the Z period will bedenoted “LEP-I”, and the period beginning in 1996 “LEP-II”. During the seven years of running at LEP-I, the fourexperiments ALEPH [7], DELPHI [8], L3 [9] and OPAL [10] collected approximately 17 million Z decays in total,distributed over seven centre-of-mass energy points within plus or minus 3 GeV of the Z-pole.

The SLC accelerator started running in 1989 and the Mark-II collaboration published the first observations of Zproduction in e+e! collisions [11]. However, it was not until 1992 that longitudinal polarisation of the SLC electronbeam was established. By then the SLD detector [12,13] had replaced Mark-II. From 1992 until 1998, when theaccelerator was shut down, SLD accumulated approximately 600 thousand Z decays. Although the data set is much

1 In this report h = c = 1.

Figure 7.2: The hadronic cross-section at various e+e− colliders. From [32].

around 91 GeV is clearly seen. When the center of mass energy exceeds the W+W− threshold(2×mW ' 160 GeV), the reaction e+e− → W+W− becomes possible. Note that this reactioninvolves the trilinear couplings γW+W− and Z0W+W−. LEP made very precise measurementsof the cross-section within a few GeV range around the Z0 mass. Such measurements allowedto determine the Z mass, its total decay width and partial decay width when the final state isclearly identified. Figure 7.3 illustrates how the Z0 mass and total decay width are measured.The data are fitted by the green curve which is corrected to take into account the radiativeemission of photons from the initial states (so-called ISR for Initial State Radiation). Indeed,the electron or positron can radiate a photon before colliding. Hence, the energy of the collisionis not necessarily the sum of the energy of the 2 beams. After all corrections, the dashed redcurve is obtained. The position of the Z0 peak is shifted by 100 MeV and the total cross-sectionis increased by 36%. The effect of the correction is therefore large but the accuracy of thecorrection is excellent since it involves pure QED processes which are very well known.

In addition, the Z0 lineshape gives constraints on the number of light neutrinos (below mz/2)as shown in figure 7.4. This number is found to be:

Nν = 2.9840± 0.0082

(very close to 3 at 2 sigmas). This number is obtained by measuring Γinv, the invisible partial

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228 The Standard ModelThe ALEPH, DELPHI, L3, OPAL and SLD Collaborations / Physics Reports 427 (2006) 257 – 454 275

Ecm [GeV]

! had

[nb]

! from fitQED corrected

measurements (error barsincreased by factor 10)

ALEPHDELPHIL3OPAL

!0

"Z

MZ

10

20

30

40

86 88 90 92 94

Ecm [GeV]

AF

B(µ

)

AFB from fit

QED correctedaverage measurements

ALEPHDELPHIL3OPAL

MZ

AFB0

-0.4

-0.2

0

0.2

0.4

88 90 92 94

Fig. 1.12. Average over measurements of the hadronic cross-sections (left) and of the muon forward–backward asymmetry (right) by the fourexperiments, as a function of centre-of-mass energy. The full line represents the results of model-independent fits to the measurements, as outlinedin Section 1.5. Correcting for QED photonic effects yields the dashed curves, which define the Z parameters described in the text.

corrections only affect final states containing quarks. To first order in !S for massless quarks, the QCD corrections areflavour independent and the same for vector and axial-vector contributions:

RA,QCD = RV,QCD = RQCD = 1 + !S(m2Z)

"+ · · · . (1.38)

The hadronic partial width therefore depends strongly on !S. The final state QED correction is formally similar, butmuch smaller due to the smaller size of the electromagnetic coupling:

RA,QED = RV,QED = RQED = 1 + 34Q2

f!(m2

Z)

"+ · · · . (1.39)

The total cross-section arising from the cos#-symmetric Z production term can also be written in terms of the partialdecay widths of the initial and final states, $ee and $ff ,

%Zff

= %peakff

s$2Z

(s ! m2Z)2 + s2$2

Z/m2Z

, (1.40)

where

%peakff

= 1RQED

%0ff

(1.41)

and

%0ff

= 12"

m2Z

$ee$ff

$2Z

. (1.42)

The term 1/RQED removes the final state QED correction included in the definition of $ee.The overall hadronic cross-section is parametrised in terms of the hadronic width given by the sum over all quark

final states,

$had =!

q "=t

$qq. (1.43)

Figure 7.3: The hadronic cross-section at LEP e+e− colliders around the Z0 mass (see text for expla-nations) [32].

The ALEPH, DELPHI, L3, OPAL and SLD Collaborations / Physics Reports 427 (2006) 257 –454 277

0

10

20

30

86 88 90 92 94Ecm [GeV]

! had

[nb]

3"

2"

4"

average measurements,error bars increased by factor 10

ALEPHDELPHIL3OPAL

Fig. 1.13. Measurements of the hadron production cross-section around the Z resonance. The curves indicate the predicted cross-section for two,three and four neutrino species with SM couplings and negligible mass.

Assuming that the only invisible Z decays are to neutrinos coupling according to SM expectations, the number oflight neutrino generations, N!, can then be determined by comparing the measured R0

inv with the SM prediction for"!!/"!!:

R0inv = N!

!"!!

"!!

"

SM. (1.50)

The strong dependence of the hadronic peak cross-section on N! is illustrated in Fig. 1.13. The precision ultimatelyachieved in these measurements allows tight limits to be placed on the possible contribution of any invisible Z decaysoriginating from sources other than the three known light neutrino species.

1.5.3. Asymmetry and polarisationAdditional observables are introduced to describe the cos # dependent terms in Eq. (1.34) as well as effects related

to the helicities of the fermions in either the initial or final state. These observables quantify the parity violation ofthe neutral current, and therefore differentiate the vector- and axial-vector couplings of the Z. Their measurementdetermines sin2 #f

eff .Since the right- and left-handed couplings of the Z to fermions are unequal, Z bosons can be expected to exhibit a net

polarisation along the beam axis even when the colliding electrons and positrons which produce them are unpolarised.Similarly, when such a polarised Z decays, parity non-conservation implies not only that the resulting fermions willhave net helicity, but that their angular distribution will also be forward–backward asymmetric.

When measuring the properties of the Z boson, the energy-dependent interference between the Z and the purelyvector coupling of the photon must also be taken into account. This interference leads to an additional asymmetrycomponent which changes sign across the Z-pole.

Considering the Z exchange diagrams and real couplings only,2 to simplify the discussion, the differential cross-sections specific to each initial- and final-state fermion helicity are:

d$Ll

dcos#! g2

Leg2Lf(1 + cos#)2, (1.51)

d$Rr

dcos#! g2

Reg2Rf(1 + cos#)2, (1.52)

2 As in the previous section, the effects of radiative corrections, and mass effects, including the imaginary parts of couplings, are taken intoaccount in the analysis. They, as well as the small differences between helicity and chirality, are neglected here to allow a clearer view of the helicitystructure. It is likewise assumed that the magnitude of the beam polarisation is equal in the two helicity states.

Figure 7.4: The hadronic cross-section at LEP e+e− colliders constraining the number of light neutrinos[32].

decay width (i.e. when the Z0 decays in channels that cannot be detected). Of course, Γinv isdetermined by subtraction of measured quantities:

Γinv = Γtot − Γee − Γµµ − Γττ − Γhad

where Γhad is the decay width in hadronic channels. Assuming that only neutrinos contributeto Γinv, the number of neutrinos is then:

Nν =ΓinvΓνν

=ΓinvΓll

(ΓllΓνν

)

SM

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Tests of the theory 229

where ΓinvΓll

is taken from measurements assuming lepton universality and(

ΓllΓνν

)SM

is taken from

the prediction of the standard model.

In addition, the measurement of the differential cross-section as function of the scatteringangle of the out-going fermion with respect to the direction of the incoming e− gives access to thecouplings cV and cA from table 7.2 by measuring various asymmetries such as forward-backward:

AfFB =σF − σBσF + σB

where σF (σB) is the cross-section of forward (backward) fermions i.e. within 0 ≤ θ ≤ π2

(π2 ≤ θ ≤ π). When polarized electrons can be used such as at the SLC collider at SLAC, theleft-right asymmetry can be measured:

AfLR =σL − σRσL + σR

where σL (σR) is the cross-section of left(right)-handed polarized fermions. The differentialcross-section is given by [32]:

d σffd cos θ

=3

8σtotf f [(1− PeAe)(1 + cos2 θ) + 2(Ae − Pe)Af cos θ]

Pe representing the electron polarization and Af being the asymmetric parameter:

Af = 2

cfVcfA

1 +

(cfVcfA

)2

Hence, the measurements of the asymmetries give access to Af and thus tocfVcfA

. Now according

to the equality 7.40,

T3 = +12 ,

cfVcfA

= 1− 4 sin2 θwQ

T3 = −12 ,

cfVcfA

= 1 + 4 sin2 θwQ

Since for charged fermions, the up component of the isodoublet has always a positive electriccharge and the down component a negative charge, the 2 equalities above can be summarizedas:

cfV

cfA= 1− 4 sin2 θw|Q|

Therefore, the measurements of asymmetries in various channel (f = µ, τ, c, b) give an estimationof sin2 θw for those channels. We can then check the consistency of the values obtained withthe one measured with the ratio of the masses of the W and Z bosons (eq. 7.81). Due tohigher order corrections, the masses of heavy virtual particles enter into the game. Hence, thismeasurement gives an indirect sensitivity to the Higgs mass. This is basically what is shown inthe figure 7.5. The first 3 measurements on the top of the figure involve leptonic couplings onlywhile the last 3 involve couplings with leptons and quarks.

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230 The Standard Model382 The ALEPH, DELPHI, L3, OPAL and SLD Collaborations / Physics Reports 427 (2006) 257 –454

102

103

0.23 0.232 0.234

sin2!lepteff

mH

[GeV

]

"2/d.o.f.: 11.8 / 5

A0,lfb 0.23099 ± 0.00053

Al(P# ) 0.23159 ± 0.00041

Al(SLD) 0.23098 ± 0.00026

A0,bfb 0.23221 ± 0.00029

A0,cfb 0.23220 ± 0.00081

Qhadfb 0.2324 ± 0.0012

Average 0.23153 ± 0.00016

$% = 0.02758 ± 0.00035$%m = 178.0 ± 4.3 GeV

Fig. 7.6. Comparison of the effective electroweak mixing angle sin2 !lepteff derived from measurements depending on lepton couplings only (top) and

also quark couplings (bottom). Also shown is the SM prediction for sin2 !lepteff as a function of mH. The additional uncertainty of the SM prediction

is parametric and dominated by the uncertainties in !"(5)had(m2

Z) and mt , shown as the bands. The total width of the band is the linear sum of theseeffects.

7.3.5. DiscussionThe unexpectedly large shifts and differences observed in the various analyses for asymmetry parameters, effective

coupling constants, #f and sin2 !lepteff all show the consequences of the same effect. It is most clearly visible in the

effective couplings and sin2 !lepteff averages and stems from the measurements of A0

LR and A0,bFB .

The results as shown in Fig. 7.4 suggest that the effective couplings for b-quarks cause the main effect; both gVband gAb deviate from the SM expectation at the level of two standard deviations. In terms of the left- and right-handedcouplings gLb and gRb, which are much better aligned with the axes of the error ellipse, only gRb shows a noticeabledeviation from the expectation. The value of gLb, which is essentially equivalent to R0

b ! g2Rb +g2

Lb due to the smallnessof gRb, shows no discrepancy. The data therefore invite an economical explanation in terms of a possible deviation ofthe right-handed b-quark coupling alone, even at Born level (see Eq. (1.7)), from the SM prediction. This would affectAb and A0,b

FB , which both depend only on the ratio gRb/gLb, more strongly than R0b .

From the experimental point of view, no systematic effect potentially explaining such shifts in the measurementof A0,b

FB has been identified. While the QCD corrections are significant, their uncertainties are small compared tothe total errors and are taken into account, see Section 5.7.2. Within the SM, flavour specific electroweak radiativecorrections as listed above and their uncertainties are much too small to explain the difference in the extracted sin2 !lept

effvalues. All known uncertainties are investigated and are taken into account in the analyses. The same holds for theA0

LR measurement, where the most important source of systematic uncertainty, namely the determination of the beampolarisation, is small and well-controlled.

Thus the shift is either a sign for new physics which invalidates the simple relations between the effective parametersassumed in this chapter, or a fluctuation in one or more of the input measurements. In the following we assumethat measurement fluctuations are responsible. Furthermore, we largely continue to assume a Gaussian model forthe experimental errors, despite the fact that this results in a value for sin2 !lept

eff , with small errors, which is in pooragreement with both A0

LR and A0,bFB . As a direct consequence, the $2/dof in all analyses including these measurements

Figure 7.5: Top: determination of sin2 θw from various asymmetries measurements and bottom: predic-tion from the standard model taking into account higher order corrections (giving a sensitivity on mH).From [32].

7.4 Changes of concepts

As a conclusion, it is worth to make a break and examine the consequences of what we havelearned in this chapter. Let us put words behind the previous mathematical developments inorder to better understand how some concepts have to be revisited.

7.4.1 Mass in modern physics

Let us first recall how mass is understood in the theory of special relativity. In contrast toNewtonian mechanics, the mass of the system is not a measure of the amount of matter. QuotingEinstein [1], “the mass of a body is a measure of the energy contained in it”, meaning that themass is equivalent to the rest energy of a body (one just has to set p = 0 in formula 1.20).As recall in the first chapter, the mass is a 4-scalar and thus does not depend on the frame.In Newtonian mechanics, the mass of a body is also a measure of its inertia (and the sourceof gravitational force). It is no more the case in relativity. Indeed, inertia is the tendencyof an object to resist any change in its motion, and thus any change in its velocity (i.e. theacceleration). If the inertia only depended on a single number, the mass, we would still havein special relativity the Newtonian relationship between the force and the (usual) acceleration:~F = d~p

dt = m~a. However, in special relativity:

~F =d~p

dt=d(γm~v)

dt= m

(dγ

dt~v + γ~a

)= m

(γ3

c2(~v.~a)~v + γ~a

)

The first term in the parenthesis implies that the resistance of a body to the force acceleratingit, depends not only on the mass but also on the angles between the force and the velocity.

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Mass in modern physics 231

It should be noticed that both in Special relativity and Newtonian mechanics, the mass is anintrinsic property of the particle or the system.

Now, after this chapter, how the mass of an elementary particle is understood? We have seenthat the mass of elementary particles is due to the interaction of these particles with the Higgsfield. Is it so surprising that a field can create a mass for particles initially massless? Actually,it is even pretty common. Let us imagine a universe made of massless particles, but theseparticles would be tiny dipoles. There would be two kinds of particles in our imaginary world:the ones having their positive pole in the “upward” position and called here “plus-particles” andthe others having their poles oriented in the opposite direction called “minus-particules” (seefigure 7.6). Both are considered massless (in other words, the energy of those particle can be

+"#"

+"#"

+"#"

+"#"

+"#"

+"#"

E"

#V"

+V"

Figure 7.6: An imaginary universe made of dipole particles plunged into an electric field. See text.

very close to 0) and move freely. Now, for an unknown reason, there is a phase transition in ouruniverse: suddenly a static electric field appears (as created by infinitely spaced capacitor plates)oriented in the “upward” position. It could have been “downward” meaning that the theorydescribing our universe is symmetric in the two directions but in our example, this symmetryhas been broken in favor of the “upward” direction. Even in absence of particles, namely in thevacuum, there is now this electric field. Since the particles are electrically neutral, the field doesnot affect the motion of the particles: they still move freely, without friction17. What aboutthe energy of the particles? Well, clearly the situation of “plus-particles” is more stable thanthe one of “minus-particles”, and thus for a given motion, the energy of “minus-particles” islarger than the one of “plus-particles”. That means that if we still consider the “plus-particles”as massless, due to the field the “minus-particles” have necessarily a larger energy and thusa minimal energy (with respect to “plus-particles” +2|~d|| ~E| where ~d is the electric moment)that would be interpreted as a mass: the interaction with the field would give a mass to the“minus-particles”! Similarly, the interaction of massless particles (gauge bosons and fermions aswell) with the Higgs field generates the mass of the particles. It is important to realize that itmeans that the mass is no more an intrinsic property of the particle. If the field vanishes (as inthe early times of the universe), the particles would be massless. The intrinsic property is nowthe coupling to the Higgs field (at least for the moment since there is still no theory explainingthe values of those couplings in case of fermions whereas for gauge bosons, the couplings to theHiggs field are a direct consequence of the gauge symmetry).

17In many attempts to explain the Higgs mechanism for a non scientist audience, the field is presented as asource of friction explaining why particles cannot move at the speed of light and hence justifying their mass. Onelimitation of this analogy is that for particles at rest, there is no friction at all. The analogy given in the text ismore appropriate avoiding this source of confusion.

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232 The Standard Model

Finally, a last word about the mass of composite particles. The more common compositeparticles (nucleon, light mesons etc) are made of sub-particles being in first approximationmassless (u and d quarks have a negligible mass compared to the mass of these hadrons andgluons are massless). Consequently, the mass of composite particles is largely dominated by theQCD potential responsible for the cohesion of those particles, and hence largely dominated bythe interactions of the constituents with the gluon field. Again, mass results from an interactionwith a field.

7.4.2 The vacuum and the Higgs field

In Quantum Field Theory, the vacuum is a complicated notion. It is not empty at all, becauseof quantum fluctuations, source of virtual particles that emerge and quickly disappear. We havelearned that the Higgs field permeates the universe in such a way that the vacuum expectationvalue is different than zero (at anytime, anywhere, because Higgs field is static and infinitely long-lived), allowing elementary particles to acquire mass. The Higgs field carries non-zero quantumnumbers: Y = 1 and T3 = −1/2. Thus, the vacuum can be understood as an unlimited source ofthese quantum numbers which are constantly exchanged with the particles moving in, as shownin figure 7.7. The key here, is that only quantum numbers are exchanged, not momentumor energy, otherwise we would see free particles traveling with kinks in their trajectories. In

eR eL eR

pµ = 0pµ = 0

T3 = 12 , Y = 1

T3 = 0Y = 2

T3 = 12

Y = 1

1

eR eL eR

pµ = 0pµ = 0

T3 = 12 , Y = 1

T3 = 0Y = 2

T3 = 12

Y = 1

1

eR eL eR

pµ = 0pµ = 0

T3 = 12 , Y = 1

T3 = 0Y = 2

T3 = 12

Y = 1

1

eR eL eR

pµ = 0pµ = 0

T3 = 12 , Y = 1

T3 = 0Y = 2

T3 = 12

Y = 1

1

eR eL eR

pµ = 0pµ = 0

T3 = 12 , Y = 1

T3 = 0Y = 2

T3 = 12

Y = 1

1

Figure 7.7: Interaction of the Higgs field condensate (the vacuum expectation value) with a fermion.

the jargon of condensed matter physics, such vacuum with these special properties is calleda condensate : a coherent state described by a non-zero field corresponding to the minimalenergy. The universe is no longer symmetric since there are preferred weak charges (Y = 1 andT3 = −1/2). However, the electroweak symmetry is only hidden: there is still conservation ofthe weak hypercharges and isospin third components via the Higgs condensate even if for us,it appears as a mass term which seems to violate this symmetry. The effect of the condensateis to couple the two chiralities of the fermion (the probability to flip the chirality being givenby the Yukawa coupling), generating an effective mass even if the fermions had zero mass inthe original lagrangian. Recall that the hamiltonian of a free Dirac particle does not commutewith the chirality (see section 2.3.4.3). Hence, the mass eigenstate (i.e Hamiltonian eigenstate)is necessarily made with a superposition of left-handed and right-handed chiralities, preciselywhat is produced by the interaction with the Higgs condensate. Consequently, a real electron for

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The vacuum and the Higgs field 233

instance (by real, I mean the one that is observed with a finite mass), cannot have defined weakcharges. In figure 7.8, we recall how mass is generated from a massless particle whose helicity(or equivalently chirality) is flipped regularly. Since angular momentum is always conserved,

x"

t"

v"="c"

v"<"c"

R.H"

L.H"

Figure 7.8: Generation of a massive particle from massless left and right-handed states. See text forthe explanations.

the left handed particle must travel in the opposite direction of the right handed. The greenarrow represents the spin oriented in the +x-direction. The right-handed state is representedby the red arrow toward the +x direction while left-handed is toward −x-direction. The overallvelocity of the particle (red dashed line) is then smaller than c, signaling a massive particle.

I want to address another question which is frequently asked: can the Higgs field be con-sidered as an absolute reference frame (a kind of aether)? The answer is no. The Higgs fieldchanges some of the underlying properties of the vacuum, i.e. the empty space itself. There isno meaning to asking whether you are moving relative to empty space. The Higgs field does notfill the empty space with something, it is part of the empty space itself. Moreover, the vacuumexpectation value of the Higgs field is constant everywhere. All elementary particles interactwith it the same way no matter how you are moving in space and where you are moving inspace. The Higgs field is as unmeasurable as space itself. The only way to identify the presenceof the Higgs field is via its excitation, namely via the production of Higgs bosons.

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234 The Standard Model

7.5 Exercises

Exercise 7.1 Determination of the number of light neutrinos.At LEP (e+e− collider), the total decay width Γz of the Z0 boson has been measured accuratelyusing the Z−lineshape. In addition, using the reactions at

√s = MZ , e+e− → l+l− (l being a

charged leptons) and e+e− → jets, it was possible to estimate the branching ratio of Z0 → l+l−

and Z0 → hadrons denoted respectively by BR(l) and BR(h).

1. Assuming lepton universality (and neglected all fermion masses), explain why the invisibledecay width Γinv can be measured as Γinv = Γz[1 − BR(h) − 3BR(l)]. The measurementgave Γinv = 499.0± 1.5 MeV.

2. How the number of light neutrinos (masses smaller than MZ/2) and Γinv are related? Inorder to reduce the systematics errors, it is more accurate to use:

Nν =ΓinvΓl× Γthl

Γthν

where Γthl and Γthν are respectively the theoretical ( Standard Model) partial decay width ofZ into a charged leptons or into a neutrino and Γl = 83.984 ± 86 MeV is the measuredvalue at LEP.

3. We wish now to calculate Γthl and Γthν .

(a) determine the amplitude of Z0 → ff where f is any fermions. For the labeling, useεµ for the Z0 polarization and 4-momenta p1 for f and p2 for f .

(b) Show that the averaged squared amplitude |M|2 can be expressed as:

|M|2 = g2w

12 cos2 θw

(−gµν +

(p1+p2)µ(p1+p2)νm2Z

∑s1,s2

[u(p1)γµ(cV − cAγ5)v(p2)

] [u(p1)γν(cV − cAγ5)v(p2)

]∗

We remind the useful relation for averaging the polarization of massive spin 1 bosons:∑

λ

ε∗µ(λ)εν(λ) = −gµν +qµqνm2Z

(c) Neglecting the masses of fermions, show that:

(p1 + p2)µ[u(p1)γµ(cV − cAγ5)v(p2)

]= 0

(d) And conclude that:

|M|2 =g2w

12 cos2 θw(−gµν) (c2

V + c2A)Tr

[γµ /p1γ

ν/p2

]

(e) Finally, show that:

Γ(Z0 → ff) =g2w

48π cos2 θwMZ(c2

V + c2A)

4. Using the numerical value sin2 θw = 0.23, compute the number of neutrinos.

5. Additional question: calculate the lifetime of the Z0. Use MZ = 91.19 GeV and g2w =

0.426 GeV.

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Exercises 235

Exercise 7.2 Following a similar approach as in exercise 7.1, calculate the decay width of theW boson. Use the numerical values sin2 θw = 0.23, g2

w = 0.426 GeV and MW = 80.38 GeV.

Exercise 7.3 Partial decay width of the Higgs bosonWe are interested in a Higgs boson decaying into a pair of fermions: H → ff . For labeling, usep1 for f and p2 for f . The masse mf of the fermion is not neglected.

1. Draw the Feynman diagram and determine its amplitude.

2. Show that the squared average amplitude is |M|2 = 4√

2m2fGF (p1.p2 −m2

f ).

3. Conclude that:

Γ(H → ff) =GF

4π√

2MH m2

f

(1−

4m2f

M2H

) 32

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236 The Standard Model

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Chapter 8

Experimental aspects of the famousHiggs boson

In 2012, both ATLAS and CMS experiments announced the discovery of a new particle, a boson,with a mass about 125 GeV after having analyzed 2 years of data taking at the LHC. To ourcurrent knowledge, this particle has the properties of the famous Higgs boson. This topic isobviously a very hot topic continuously changing for 2 years, with impressive new measurementspublished for each international conference. Therefore, I have not (yet) produced written noteswhich would be obsolete too quickly. For the most up to date documentation, please see theslides distributed during the lecture.

237

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238 Experimental aspects of the famous Higgs boson

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Bibliography

[1] A. Einstein, Ann. Phys. 18, 639 (1905)

[2] J.L Basdevant et J. Dalibard, “Mecanique quantique”, Les editions de l’Ecole Polytechnique

[3] M. Le Bellac, “Physique quantique”, collection savoirs actuels, CNRS editions.

[4] Pierre Fayet, “Champs relativistes”, programme d’approfondissement de Physique, EcolePolytechnique

[5] Zhi-Qiang Shi, “The discussion about the spin states, the helicity states and the chiralitystates”, arXiv:1101.0481v1 [hep-ph] 3 Jan 2011.Palash B. Pal, “Dirac, Majorana and Weyl fermions”, arXiv:1006.1718 [hep-ph] 12 Oct2010.

[6] I.J.R. Aitchison & A.J.G. Hey “Gauge theories in particle physics”, Vol1, IOP 2003.

[7] I.J.R. Aitchison & A.J.G. Hey “Gauge theories in particle physics”, Vol2, IOP 2003.

[8] M. E. Peskin, D. V. Schroeder, “An Introduction To Quantum Field Theory”, WestviewPress Inc 1995.

[9] F. Halzen and D. Martin, “Quarks & Leptons”, Wiley, (1984)

[10] M.E. Levi et al., “Weak Neutral Currents in e+ e- Collisions at√s = 29 GeV,

Phys.Rev.Lett.51:1941,1983.

[11] B. Odom, D. Hanneke, B. D’Urso, and G. Gabrielse, Phys. Rev. Lett., 97, 030801 (2006).D. Hanneke, S. Fogwell, and G. Gabrielse, Phys. Rev. Lett., 100, 120801 (2008)D. Hanneke, S. Fogwell, and G. Gabrielse, arXiv:1009.4831, (2010)

[12] T. Aoyama, M. Hayakawa, T. Kinoshita, and M. Nio, “Tenth-Order QED Contribution tothe Lepton Anomalous Magnetic Moment – Sixth-Order Vertices Containing an InternalLight-by-Light-Scattering Subdiagram” , arXiv:1201.2461, (2012)T. Aoyama, M. Hayakawa, T. Kinoshita, and M. Nio, Phys. Rev. Lett. 99, 110406 (2007)

[13] http://en.wikipedia.org/wiki/Penning trap

[14] D.H. Perkins, “Introduction to High Energy Physics”, Addison-Wesley (1982)

[15] D. Griffiths, “Introduction to elementary particles”, Harper & Row, (1987)

[16] M. Breidenbach et al., “Observed Behavior of Highly Inelastic electron-Proton Scattering”,Phys.Rev.Lett.23:935-939, (1969)

239

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240 BIBLIOGRAPHY

[17] J. I. Friedman and H. W. Kendall, Ann. Rev. Nucl. Sci. 22, 203 (1972)

[18] W. Atwood et al., “Lectures on Lepton Nucleon Scattering and Quantum Chromodynam-ics”, Birkhauser (1982)

[19] Particle data group, “Review of Particle Physics”, Journal of Physics G Nuclear and ParticlePhysics, Vol 37 (2010)

[20] E. Lieder, C. Lorce, Physics Reports 541 (2014) 163-248.

[21] H. Burkhardt and B. Pietrzyk, Phys. Lett. B356, 398 (1995).

[22] Q.Ho-Kim, P.X. Yem, “Elementary Particles and their interactions”, Springer (1998).

[23] http://en.wikipedia.org/wiki/Colour confinement

[24] S. Bethke “Experimental Tests of Asymptotic Freedom”, Prog.Part.Nucl.Phys.58:351-386(2007)

[25] J.M Campbell, J.W. Huston, W.J. Stirling, “Hard Interactions of Quarks and Gluons: aPrimer for LHC Physics”, Rept.Prog.Phys.70:89 (2007), arXiv:hep-ph/0611148

[26] A. Lesor, arXiv:0911.0058v1 (2009)

[27] Carlo Giunti and Chung W. Kim“Fundamentals of Neutrino Physics and Astrophysics”,Oxford University press (2007).

[28] Daya Bay Collaboration “Observation of electron-antineutrino disappearance at Daya Bay”,Physical Review Letters 108 171803 (2012), arXiv:1203.1669

[29] S.M. Bilenky, C. Giunti, “Lepton Numbers in the framework of Neutrino Mixing”,Int.J.Mod.Phys. A16 (2001) 3931-3949, arXiv:hep-ph/0102320v1

[30] P Q Hung, and J J Sakurai, “The Structure of Neutral Currents”, Annual Review of Nuclearand Particle Science Vol. 31: 375-438 (1981)

[31] C. Campagnari, M. Franklin, “The discovery of the top quark”, Rev. Mod. Phys. 69,137–212 (1997)

[32] “Precision electroweak measurements on the Z resonance”, Physics Reports 427 (2006)257–454

[33] See for example

http://en.wikipedia.org/wiki/Euler_angles

[34] Janice B. Shafer, Joseph J. Murray, and Darrell O. Huwe, Phys. Rev. Lett. 10, 179 (1963)

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Appendix A

Fermi’s golden rule and timedependent perturbation theory

We are going to study collisions or decay processes in which the physical state of a system afterthe collision |f〉 differs from that before the collision |i〉. Problems of this kind is well describedby time dependent perturbation theory where the formalism is explained in quantum mechanicscourses such as in [2, p. 349]. A brief summary is given below.

A.1 Fermi’s golden rule

A.1.1 Quantum mechanics approach

Let us first consider a free particle which can be in a state satisfying the usual Schrodingerevolution equation:

H0Φn = EnΦn (A.1)

where the wave functions are normalized to one particle in a box of volume V: 〈m|n〉 =∫V φ∗mφnd

3x = δmn so that:

〈~x|n〉 = φn(~x) =1√Vei~pn.~x (A.2)

The particle is considered initially at a large distance from the region of nonzero potentialV (~x, t), and after passing through that region is again traveling as a free particle, but in adifferent state than before the interaction1. The question is to solve:

(H0 + V (~x, t))ψ = i∂ψ

∂t(A.3)

We develop ψ on the φn basis:

ψ =∑

n

an(t)φn(~x)e−iEnt (A.4)

Substituting equ. A.4 into A.3 and using A.1, we get:

i∑

n

dan(t)

dtφn(~x)e−iEnt =

n

V (~x, t)an(t)φn(~x)e−iEnt (A.5)

1The reader may notice that we implicitly assume a non-relativistic particle since the number of particlesremains constant before and after the interaction. However, the Fermi’s golden rule that will be establishedremains valid even in the case of relativistic particles as will be shown using quantum field theory.

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242 Fermi’s golden rule and time dependent perturbation theory

Now let us multiply by φ∗f and integrate over the volume, and we obtain the equations of thean coefficients:

dafdt

(t) = −i∑

n

an(t)

Vφ∗f (~x)V (~x, t)φn(~x) d3x ei(Ef−En)t (A.6)

where we have used the normalization condition. Before feeling the potential V during a timeT , the particle was in an eigenstate |i〉 until t = −T/2 and hence:

ai(t ≤ −T/2) = 1an6=i(t ≤ −T/2) = 0

so that:dafdt

(t ≤ −T/2) = −i∫

Vφ∗f (~x)V (~x, t)φn(~x) d3x ei(Ef−En)t (A.7)

Now, if we assume that the potential is small enough (justifying an approach with perturbationtheory), we make the assumption that the condition A.7 remains true even after −T/2 2 . Wecan then integrate A.7 to obtain:

af (t) = −i∫ t

−T/2dt′∫

Vφ∗f (~x)V (~x, t′)φi(~x) d3x ei(Ef−Ei)t

And in particular, when the interaction with the potential has ceased:

Sfi = af (T/2) = −i∫ T/2

−T/2dt

Vφ∗f (~x)V (~x, t)φi(~x) d3x ei(Ef−Ei)t (A.8)

This expression is valid as soon as Sfi 1. We can define the transition rate i.e. the transitionprobability from a state |i〉 to a state |f〉 per unit time as:

Γi→f = limT→+∞

|Sfi|2T

(A.9)

Now, let us consider that the potential does not dependent on time: V (~x, t) = V (~x). Thus,Γi→f simplifies to:

Γi→f = limT→+∞

1

T

∣∣∣∣∣−i∫

Vφ∗f (~x)V (~x)φi(~x) d3x

∫ T/2

−T/2dt ei(Ef−Ei)t

∣∣∣∣∣

2

= limT→+∞

|Mfi|2T

∣∣∣∣∣

∫ T/2

−T/2dt ei(Ef−Ei)t

∣∣∣∣∣

2

(A.10)where:

Mfi = 〈f |V |i〉 =

Vφ∗f (~x)V (~x)φi(~x) d3x (A.11)

The integral is easy to calculate:

∫ T/2

−T/2dt ei(Ef−Ei)t = 2

sin[(Ef − Ei)T2

]

Ef − Ei= T

sin[(Ef − Ei)T2

]

(Ef − Ei)T2And finally:

Γi→f = |Mfi|2 limT→+∞

Tsin2

[(Ef − Ei)T2

][(Ef − Ei)T2

]2 (A.12)

2More rigorously, what is done here is a development to first order.

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Quantum field approach 243

One can show that [3, p. 154]:

limT→+∞

Tsin2

[(Ef − Ei)T2

][(Ef − Ei)T2

]2 → limT→+∞

T2π

Tδ(Ef − Ei) = 2πδ(Ef − Ei) (A.13)

Thus, equ. A.12 simplifies to the famous Fermi’s golden rule:

Γi→f = 2π|Mfi|2δ(Ef − Ei) (A.14)

Γi→f is the transition rate, namely the transition probability per unit time from an initial stateto a final state. Mfi is the transition amplitude between the initial and final state and is givenat the first order of perturbation theory (Born approximation) by the matrix element of theperturbation between these 2 states: Mfi = 〈f |V |i〉. However Mfi can be computed at higherorder using the Feynman diagrams as we’ll see later on.

A.1.2 Quantum field approach

Consider the first order S[n=1] of the S-matrix 3.33 between a state |i〉 and |f〉:

〈f |S[1]|i〉 = −i∫ +∞

−∞dt 〈f |HI int|i〉

Replacing the bornes of integration by −T/2 and +T/2 and using the Schrodinger representationof the hamiltonian HI int = eitH0Hint e

−itH0 (equation 3.21), we have:

〈f |S[1]|i〉 = −i∫ +T/2

−T/2dt 〈f |eitH0Hint e

−itH0 |i〉 = −i∫ +T/2

−T/2dt eitEf 〈f |Hint|i〉 e−itEi

Supposing as in the previous section that the hamiltonian does not depend on time, it comes:

〈f |S[1]|i〉 = −i 〈f |Hint|i〉∫ +T/2

−T/2dt eit(Ef−Ei) = −i 〈f |Hint|i〉T

sin[(Ef − Ei)T2

][(Ef − Ei)T2

]

The transition rate is defined by the probability per time unit:

Γi→f = limT→+∞

| 〈f |S[1]|i〉 |2T

= | 〈f |Hint|i〉 |2 limT→+∞

Tsin2

[(Ef − Ei)T2

][(Ef − Ei)T2

]2

The limit converges to 2πδ(Ef − Ei) (equation A.13) and thus:

Γi→f = 2π| 〈f |Hint|i〉 |2δ(Ef − Ei)

which is the Fermi’s golden rule expression.

A.2 Transition rate

The Fermi’s golden rule given in equ. A.14 supposes that the final state |f〉 is precisely known. Inpractice, it is impossible and the final state is rather described by a set of final states belongingto a given phase space element. Let us consider an infinitesimal phase space element so that

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244 Fermi’s golden rule and time dependent perturbation theory

we can assume that |Mfi| remains constant for these dN states. The infinitesimal transitionprobability per unit time is then the sum of the probabilities of each individual states. Hence,A.14 becomes:

dΓi→f = 2π|Mfi|2δ(Ef − Ei)dNAs well known from quantum mechanics, in case of a single particle in a box of sides L, thecomponents of its momentum are quantised as: px,y,z = 2π

L nx,y,z, and therefore the number ofstates in d3~p = dpxdpydpz is dN = dnxdnydnz = V d3~p/(2π)3 leading to:

dΓi→f = 2π|Mfi|2δ(Ef − Ei) Vd3~p

(2π)3

with as before V = L3. The generalization to an initial state with ni particles and final statescontaining nf is straightforward:

dΓi→f = 2π|Mfi|2δ(E′1 + ...+ E′nf − E1 − ...− Eni)nf∏

k=1

V d3~p′k

(2π)3(A.15)

where variables related to the final state are denoted with a prime symbol (’). Now, we’re goingto factorize Mfi, so that the implicit dependency on the arbitrary volume V will be removed.To first order of perturbation, generalizing equ. A.11, Mfi is now:

Mfi = 〈f |V |i〉=∫V φ′∗1 (~x′1)...φ′∗nf (~x′nf ) V (~x′1, ..., ~x

′nf, ~x1, ..., ~xni) φ1(~x1)...φni(~xni) d

3x′1...d3x′nfd

3x1...d3xni

= V−ni+nf

2

∫V e

i~p1.~x1 ...ei~pni .~xni e−i~p′1.~x′1 ...e

−i~p′nf .~x′nf V (...) d3x′1...d

3x′nfd3x1...d

3xni

where we have inserted equ. A.2. Physics being invariant by translation, V (~x1, ..., ~xni , ~x′1, ..., ~x

′nf

)only depends on the differences of positions and not on the positions themselves, so we can makethe change of variables ~ri = ~xi − ~x1 with i > 1 and ~r′ = ~x′i − ~x1 without incidence on theperturbation V . Hence it comes:

Mfi = V−ni+nf

2

Vd3x1e

i(~p1+...+~pni−~p′1−...−~p′nf ).~x1 Vfi (A.16)

where:

Vfi =

Vd3r2...d

3rnid3r′1...d

3r′nf ei~p2.~r1 ...ei~pni .~rni e−i

~p′1.~r′1 ...e−i~p′nf .

~r′nf V (~r2, ..., ~rni , ~r′1, ..., ~r

′nf

)

As soon as L is large enough with respect to the range of the interaction described by V , Vdoesn’t depend on the volume V = L3. However, the integral in A.16 does depend on L. Let

us denote ~∆p = ~p1 + ...+ ~pni − ~p′1 − ...− ~p′nf . The integral∫V d

3x1ei ~∆p.~x1 appearing in A.16 is

easy to calculate:

∫V d

3x1ei ~∆p.~x1 =

∫ L/2−L/2 dx1e

i∆px.x1∫ L/2−L/2 dy1e

i∆py .y1∫ L/2−L/2 dz1e

i∆pz .z1

= L sin(∆pxL/2)∆pxL/2

Lsin(∆pyL/2)

∆pyL/2L sin(∆pzL/2)

∆pzL/2

We can then estimate |Mfi|2 inserting these last results:

|Mfi|2 = V−(ni+nf )L3Lsin2(∆pxL/2)

(∆pxL/2)2L

sin2(∆pyL/2)

(∆pyL/2)2L

sin2(∆pzL/2)

(∆pzL/2)2|Vfi|2

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Transition rate 245

We recognize the same functions as in equ. A.13 so that when L is large enough, |Mfi|2 convergesto:

|Mfi|2 = V1−(ni+nf )2πδ(∆px) 2πδ(∆py) 2πδ(∆pz)|Vfi|2 (A.17)

Coming back to A.15 and inserting A.17, the transition rate becomes:

dΓi→f = 2πV1−(ni+nf )2πδ(∆px) 2πδ(∆py) 2πδ(∆pz)|Vfi|2δ(E′1+...+E′nf−E1−...−Eni)nf∏

k=1

V d3~p′k

(2π)3

namely:

dΓi→f = V1−ni(2π)4δ(4)(p′1 + ...+ p′nf − p1 − ...− pni)|Vfi|2nf∏

k=1

d3~p′k(2π)3

(A.18)

where δ(4)(p′1 + ...) = δ(E′1 + ...)δ(p′1x + ...)δ(p′1z + ...)δ(p′1z + ...).

We want a formula of dΓi→f valid in relativistic regime. For non-relativistic quantum me-chanics (using the Schrodinger equation), the probability density is ρ = |φ|2, which leads withthe plane wave function A.2 to 1/V. The volume is not Lorentz invariant since it scales as 1/γand the probability density would be γ

V = EmV , namely proportional to E/V. When moving to

relativistic regime, one should use however the Klein-Gordon or Dirac equation and with thecorresponding plane wave solution3, the density probability gets an extra 2E factor (see chapter2):

ρ =2E

V (A.19)

The number of particle in the volume V is now 2E. This is precisely what was needed to complywith Lorentz transformation which required a density proportional to E/V. Each relativisticwave function now contributes to a multiplicative factor 2E in the matrix element squared.Let us define the new Lorentz invariant matrix element Mfi evaluated with the relativisticnormalization of wave functions. Hence, we have:

|Mfi|2 =

ni∏

k=1

2Ek

nf∏

k=1

2E′k |Vfi|2 ⇒ |Vfi|2 =

ni∏

k=1

1

2Ek

nf∏

k=1

1

2E′k|Mfi|2

so that the final transition rate (probability per unit time) is now:

dΓi→f = V1−ni(2π)4δ(4)(p′1 + ...+ p′nf − p1 − ...− pni)|Mfi|2ni∏

k=1

1

2Ek

nf∏

k=1

d3~p′k(2π)32E′k

(A.20)

In general, we square |Mfi| and sum or average over various degrees of freedom that are notobserved, such as spin or color. The |Mfi| appearing here implicitly includes these factors. Thisformula has been derived using the time dependent perturbation theory (especially for obtainingthe Fermi’s golden rule A.14). However, it is valid in general and the Fermi’s golden rule canbe obtained without using the perturbation theory.

3In case of Dirac’s solution, the plane wave is a more complicated object that the one used so far. It has 4components as shown in chapter 2.

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246 Fermi’s golden rule and time dependent perturbation theory

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Index

4-spinor, see Dirac-spinor4-vector

acceleration, 13current, 13definition, 11derivative, 12electromagnetic potential, 13momentum, 12polarization, 59position, 9velocity, 12

Abelian group, 67Action, 63–65Anomalous magnetic moment, 100Antiparticles, 25, 40, 42, 48Asymmetry

forward-backward, 229left-right, 229

Asymptotic freedom, 146

Baryon, 123, 126, 130number, 123

Bjorkenx-variable, 112, 115scaling, 112scaling violation, 120, 151

Boost, 11, see also Lorentz transformationBranching ratio, 28Breit-wigner distribution, 29

Cabibbo-Kobayashi-Maskawa (CKM) matrix, 171–

173, 223angle, 170

Callan-Gross formula, 114Charge conjugation, 40, 53, 181, 183, 220Chirality, 54, 82

projector, 55Clebsch-Gordan coefficients, 19

Clifford algebra, 44

Colour

confinement, 133

factor, 143–145, 149

Compton

scattering, 87

wavelength, 40

Condensate, 232

Continuity equation, 38

Coordinates

contravariant, 9

covariant, 10

cylindrical, 17

Covariante derivative, 68

CP violation, 182–184

Creation-annihilation operators, 41, 57, 60

Cross-sections

2→ 2, 33

Compton, 92

definition, 29

Mott, 108

Rosenbluth, 109

D’Alembertian operator, 12

Dalitz plot, 35

Decay

into 2 particles, 32

into 3 particles, 34

partial decay width, 28

rate formula, 26, 74

Deep inelastic scattering (DIS), 110

DGLAP evolution equation, 153

Dirac

adjoint equation, 45

adjoint spinor, 45

equation, 45

momentum space, 55

solution, 47–50, 52

hamiltonian, 50

247

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248 INDEX

lagrangian, 65spinors, 45, 56

Drell-Yan process, 154

Electromagnetic tensor, 13Euler angles, 21Euler-Lagrange equation, 64, 65

Fermiconstant, 160golden rule, 243theory, 159

Fine structure constant, 87Flux factor, 30Form factor, 108, 109

Gamma matrices, 44, 54Gauge

colour, 133Coulomb, 59Lorenz, 58non-abelian, see Yang-Mills theoriesprinciple, 68transformation

electroweak, 194, 200, 218QCD, 134, 137QED, 58, 67

unitary, 215Gell-Mann

-Nishijima formula, 125, 194matrices, 127, 134

Glashow-Iliopoulos-Maıani (GIM) mechanism,171

Gluondiscovery, 148field, 136properties, 138self-interaction, 142

Goldstone boson, 209, 212, 215Gordon decomposition, 97Gyromagnetic ratio, 95, 97, 100, 105

Hadronization, 147Helicity, 51, 82Hermitian conjugate, 45Higgs

boson, 213mass, 214

condensate, 232couplings, 216, 217, 221doublet, 210field, 212, 213mechanism, 217vacuum expectation value, 215

Hyperchargestrong, 125weak, 193

Initial State Radiation, 227Isospin

strong, 124weak, 190, 191

Kallen function, see Triangle functionKlein-Gordon equation, 39Klein-Nishina formula, 92

Lande g-factor, see Gyromagnetic ratioLeast action principle, 63–65Lepton number, 181Lifetime of particles, 29Lineshape (of Z0), 227Lorentz transformation, 10

Magnetic moment, 95, see also Anomalous mag-netic moment, 105

Mandelstam variables, 24, 25Mass

in relativity, 12, 230of fermions, 219, 220, 222, 223of gauge bosons, 214of Higgs boson, 214on-shell condition, 12

Maxwell’s equations, see Electromagnetic ten-sor

Meson, 123, 126, 130Minkowski metric, 10Momentum

4-momentum, 12operator, 38

Mott cross-section, 108

Neutral current, 173, 174Neutrino

CP violation, 179mass, 180

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INDEX 249

number of, 227oscillation, 177sterile, 194

Noether theorem, 66Nuclear magneton, 105Nucleon, 123

Parityand spherical harmonics, 17intrinsic, 53operator for spin 1/2, 53violation, 160, 161

PartonDistribution Function (PDF), 115–118model, 114

Pauliexclusion principle, 42matrices, 18, 124, 191spinors, 18

Penning-trap, 101Phase space, 26, 27, 31Polarization, 59

summation, 85Pontecorvo-Maki-Nakagawa-Sakata (PMNS) ma-

trix, 176, 223Probability current

Dirac, 46Klein-Gordon, 39Schrodinger, 38

Propagator, 79Proper time, 11Pseudorapidity, 15Pseudovector, 161

QCDcurrent, 136Feynman rules, 141gauge, 135lagrangian, 140potential, 144, 148

QEDcurrent, 67Feynman rules, 81gauge, 67, 68lagrangian, 68

Quadrilinear couplings, 202, 204, 205, 217Quark

colour, 129

from the sea, 116, see also Parton modelmodel, 126, 130radius, 119valence, 116, 117

Rapidity, 14Reduced rotation matrices, 21Renormalization, 93Rosenbluth cross-section, 109Running coupling constant, 94, 146

S-matrix, 72Sachs form factors, 109Scalar potential, 206, 210, 226Schrodinger equation, 38Screening, 95Space-time interval, 9Spherical harmonics, 17Spin, 17

addition, 19composite particles, 19operator, 18, 50spin-statistic theorem, 22summation (trace), 83

Spinorsdirac, see Dirac-spinorspauli, see Pauli-spinors

Splitting function, 152, 153Spontaneous symmetry breaking, 208, 209, 211,

215Strangeness quantum number, 125Structure constants, 135Structure functions, 111SU(2), 124, 189, 191, 193, 194SU(3)

colour, 133flavour, 126

T-product, 71Triangle function, 24Trilinear couplings, 202, 203, 216

V-A current, 163Vacuum

energy density, 226expectation value (v.e.v), 209, 215polarization, 95, 146

Virtual particles, 78

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250 INDEX

Weinberg angle, 195Wigner rotation functions, 21

Yang-Mills theories, 135, 200Yukawa coupling, 219

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INDEX 251

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