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1

COLLIDER PHYSICSLECTURE NOTES

Lecture notes based on a course given by Maxim Perelsteinat Cornell University on Collider physics. The course

goes over the basics of colliders and provides anintroduction to Madgraph. The notes focus on the collider

aspect of the course. If you find any errors pleaselet me know at [email protected]

Presented by: Maxim Perelstein

LATEX Notes by: JEFF ASAF DROR

Lectures given in 2009Notes take in 2013

CORNELL UNIVERSITY

mailto:[email protected]

Contents

0.1 Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1 Introduction 4

1.1 Master Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2 Example 1:e+e− → µ+µ− . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 QED 15

2.1 Initial State Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2 Final State Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.3 e+e− → µ+µ−γ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.A Calculating the Splitting Probability . . . . . . . . . . . . . . . . . . . . 30

2.B Phase Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.B.1 Two Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.B.2 Three Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3 QCD 38

3.1 e+e− → Hadrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.2 Hadronization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.3 e+e− → Jets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.4 Jet Algorithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.4.1 Sterman-Weinberg Jets . . . . . . . . . . . . . . . . . . . . . . . . 45

3.4.2 JADE Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.5 Event Shape Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.6 Parton Branching and Showering . . . . . . . . . . . . . . . . . . . . . . 49

4 Deep Inelastic Scattering 56

A Homework 60

A.1 Assignment 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

A.1.1 Strange Observations . . . . . . . . . . . . . . . . . . . . . . . . . 61

A.1.2 x1 and x2 Distributions . . . . . . . . . . . . . . . . . . . . . . . . 62

2

0.1. PREFACE 3

0.1 Preface

This set of notes was written based on lectures given in 2009 by Maxim Perelstein. Thecourse is intended for graduate students which have taken an introductory course toQuantum Field Theory. The aim of the course is to teach about collider physics throughanalytic and numerical techniques with emphasis on the use of Madgraph[1]. These notesfocus primarily on the analytic side do not provide an introduction to Madgraph. Thetext is composed of material primarily presented in class as well as some small additions.Any long extra derivations done by the author are left to the appendices.

Chapter 1

Introduction

The two main parameters in a collider are

• Type: e+e−, pp, pp, e±p

• Center of mass energy - Tells you the scale you are probing in. Another way to lookat it is because of E = mc2 the particles you are capable of producing have massescorresponding to this energy.

Consider an electron-positron collision,

e− e+

we have

p− = Eb (1, 0, 0, 1) (1.1)

p+ = Eb (1, 0, 0− 1) (1.2)

(Eb is the beam energy) which gives

s = (p+ + p−)2 = 2p+ · p− = 4E2b (1.3)

so we have

ECM = 2Eb (1.4)

The relevant quantities are event rate, luminosity, and cross sectionWe have

ρA

`A

A ρB

`B

4

5

σ =# of events

ρAρB`A`BA(1.5)

Normally we don’t quote the cross sections like this but instead we define the luminosity,

L = ρAρB`A`BA ·# of beam crossings

second(1.6)

The event rate is given by

R = Lintσ (1.7)

• Lint is given by accelerator physicists

• σ is given by theorists

• The rates are taken from the experiment

where

Lint =

∫dtL (1.8)

The units of cross section are in barns, or more commonly in pb.

There is a way to calculate the cross section of collision very heuristically without usingQFT. The quantum mechanical aspects of a beam are given by its Compton wavelength,

λc ∼1

Eb(1.9)

The area of the beam or the “cross section” is thus

σ ∼ 4πλ2c =

4π

E2b

(1.10)

This is indeed roughly a maximum value for the cross section. For example at LEP:

4π

(100GeV)2≈ 500nb (1.11)

The true value for example

σ(e+e− → Z,Eb ≈MZ

)≈ 40nb (1.12)

The reason for the suppression is the fact that the size of the couplings isn’t taken intoaccount.

The “rough” expression turns out to have the right scaling with the energy.

6 CHAPTER 1. INTRODUCTION

1.1 Master Formula

We have all seen the master formula for 2→ N cross section

pA pB

...

dσ =1

8EAEB

phase space︷︸︸︷N∏i=1

d3pi(2π)3

1

2Ei(2π)4δ4

(pA + pB −

N∑i=1

pi

)· |M (pApB → pi)|2︸︷︷︸TeV Matrix Element

(1.13)

Number of variables is 3N − 4.As a special case we can consider 2→ 2 process,

pA pB

The number of variables is 3(2) − 4 = 2. However, in practice nothing depends on theazimuthal angle since we can place the collision on a plane.

The task is to find

dσ

d cos θ=

1

8E2A

2∏i=1

d3pi(2π)3

(2π)4δ (pA + pB − p1 − p2) |M|2 (1.14)

=1

8E2A

1

2E2

p21dp1dφ

(2π)32E1

δ(Ecm − E1 − E2) |M|2 (1.15)

The φ integral is trivial and we have

dσ

d cos θ=

1

8E2A

1

2E2

p21dp1dφ

(2π)32E1

δ(Ecm − E1 − E2) |M|2 (1.16)

=

1

16πp1

s3/2|M|2 ,

√s < m1 +m2

0, otherwise(1.17)

If m1 = m2 then

|p1| =√s

2

velocity of p1︷︸︸︷√1− 4m2

2

s(1.18)

anddσ

d cos θ=

1

32πs

√1− 4m2

1

s|M|2 (1.19)

1.2. EXAMPLE 1:e+e− → µ+µ− 7

The amplitude is a Lorentz-invariant quantity which depends on the 4-momenta.There are very few Lorentz invariant quantities that you can form out of the momenta.We have

M = f(pi · pj ,m2k

) (1.20)

for N ≥ 3 we can also have εαβγδpαi p

βj p

γkp

δ` , but this is equal to zero for N = 2 since

εαβδδpα1p

β2p

γ3p

δ4 = εαβγδp

α1p

β2p

γ3 (p1 + p2 − p3)δ = 0 (1.21)

We already know that there is only one variable in the problem, so these must all berelated. While the dot-products are fine Lorentz invariant quantities, the more standardthing to do is to define the Mandelstam variables,

s = (pA + pB)2 = E2cm = (p3 + p4)2 (1.22)

t = (pA − p1)2 = (pB − p2)2 (1.23)

u = (pA − p2)2 = (pB − p1)2 (1.24)

1.2 Example 1:e+e− → µ+µ−

e−

e+

γ

µ+

µ−

We will take√s MZ so the weak interaction does not come into play and we can

consider the single diagram.This calculation can be done in two different ways. It can be done the standard way

using traces and also the more instructive way using helicity. Helicity is an operatorwhich can be defined for any particle with any spin and is given by

h = S · p (1.25)

where p ≡ p|p| . For a massless fermion, helicity is roughly equal to chirality.

Recall that the Lorentz group can be broken up as SO(3, 1) ≈ SU(2)× SU(2). TheWeyl spinors are in (2, 1) and (1, 2). The four components are natural in QED, howeverelsewhere it makes more sense to use Weyl spinors since they are the fundamental objectsof the Lorentz group. Chirality is what labels our particles. An electron can have left orright handed chirality. s The Lagrangian is given by

Lint =

charge︷︸︸︷g eγµeAµ = g

(e†Lσ

µeLAµ + e†RσµeRAµ

)(1.26)


since QED conserves parity the left and right components have the same coupling. Thesecond equation is really the way you should think about the physics. If we take |eL〉 (astate produced by a eL field) and act with it using the helicity operator,

h |eL〉 = −1

2|eL〉 , h |eR〉 = +

1

2|eR〉 (1.27)

If you work out the action of the creation and annihilation operators you find the confusingpart,

eL 3(e−L , e

+R

), eR 3

(e−R, e

+L

)(1.28)

You may wonder why one needs to worry about these details as we end up summing overspins. However, especially in the weak interactions it is often useful to avoid summingover the spins and calculating matrix elements individually.

Calculating these gives,

M(e−Le+R → µ−Rµ

+L) =M(e+

Le−R → µ+

Rµ−L) = −g2(1− cos θ) (1.29)

M(e−Le+R → µ−Lµ

+R) =M(e−Re

+L → µ−Rµ

+L) = −g2 (1 + cos θ) (1.30)

These are found in Peskin and Schroeder on pages 143-146.Its interesting to note that not all the possible combinations of collisions are possible.

e.g. the amplitude for two left handed particles colliding or being emitted is zero (inreality this is only true for mµ = me = 0). The final state polarization isn’t observablein current detectors. For an unpolarized beam,

1

4

∑helicity

|M|2 = g4(1 + cos4 θ

)(1.31)

and

dσ

d cos θ=

g4

32πs

(1 + cos2 θ

)(1.32)

=πα2

2s

(1 + cos2 θ

)(1.33)

(1.34)

Integration of cos θ gives the final cross section,

σe+e−→µ+µ− =4πα2

3s(1.35)

We will refer to this cross section several times throughout the lectures.We now move on to calculate the same expression but for scalar muons. We have

L = gAµ

(e†Lσ

µeL + e†RσµeR

)(1.36)

1.2. EXAMPLE 1:e+e− → µ+µ− 9

The notation is a little bit unfortunate. For the negatively charged particle, e†L createsa left handed electron and e†R creates a right handed electron, while for the positivelycharged particle the opposite is true.

The final result is given by

dσ

d cos θ∗=

g4

32πssin2 θ (1.37)

This can be understood solely on angular momentum arguments. We have

eL eRθ

If the outgoing particles are coming out at very small θ then they have L = 0 (sinceorbital angular momentum is r × p). In the massless limit we have that chirality is thesame as helicity and hence the initial state has Jz = −1. Since the muons are scalar andhave no spin, we must get a vanishing amplitude at θ = 0.

Thus far we have completely ignored the Z boson. We now restore it,

e

e

Z

µ

µ

This is a pretty simple addition however there are a couple of issues that come up withthese additions. With the photons they were always off-shell since the photon “mass”,p2 was the center of mass energy. With the Z it can be on-shell and the particle can bethought of as a 2 to 1 particle creation process followed by a particle decay. The naivepropagator of the vector particle is

−igµνp2 −M2

Z + iε(1.38)

However, what happens if√s = MZ (our particle is on-shell)? We have a singularity that

we need to make sense of. This can be done using higher order corrections. We have

= + + ...


This infinite series can be resummed. If you define

iΠ(p2) = (1.39)

you have

=−igµν

p2 −M2Z − Π(p2)

(1.40)

The physical mass is at

p2 −M2Z − Π(p2)

∣∣∣∣p2=M2

Z

= 0 (1.41)

In QED that’s all there is to it and is done when renormalizing the theory. However,when discussing the Z boson, Π(p2) has both a real and imaginary part. This can bebest understood using the optical theorem.

Consider the following,

Im

( )∼∑

f

∫dΠ2dΠ2

∣∣∣∣ ∣∣∣∣2Now the difference between the photon and the Z is that for the photon, it can’t decay, sothe diagram that is begin “squared” on the RHS of the optical theorem doesn’t contributeand the imaginary part of Π is zero. However, since the Z boson can decay this is nolonger zero. Then the equation for the Z mass is strange since equation 1.41 can’t besolved for real p2. So what people do to define the Z mass to be when

p2 −M2Z − ReΠ(p2)

∣∣∣∣p2=M2

Z

= 0 (1.42)

Now the propagator no longer vanishes at p2 = M2phys.

We can go a bit further and expand the real part ofΠ(p2) around the physical mass(the pole), MZ . We define the original boson mass above as MZ,0 and the physical massas MZ . The denominator is then given by

p2−M2Z︷︸︸︷

p2 −M2Z,0 −

(ReΠ(p2)

∣∣∣∣p2=M2

Z

+d

dp2ReΠ(p2)

∣∣∣∣p2=M2

Z

· (p2 −M2Z) + ...+ iImΠ(p2)

)=(1 +

d

dp2ReΠ

)︸︷︷︸

Z−1

(p2 −M2

Z

)+ iImΠ(p2) (1.43)

we have the normal expansion of Π and in addition we have the extra imaginary part.Further we denote Z as the wave function renormalization. So to first order (ignoring

1.2. EXAMPLE 1:e+e− → µ+µ− 11

the “dots” above) we have,

=−gµνZ

p2 −M2Z + iZImΠ(p2)

(1.44)

So to finish we need to calculate the imaginary part of Π. Roughly speaking looking atthe optical theorem we see that it is given by the width. By the definition of the particlewidth we have

ImΠ(p2) =Z−1

2

∫dΠ1dΠ2 |M(Z → p1p2)|2 (1.45)

= Z−1MZΓZ (1.46)

[Q 1: where did the Z−1 factor come from? Was it because we have one external Zboson?] so if you are doing this experiment near the Z pole then you need to correctyour propagator by writing

=−gµν

p2 −M2Z + iMZΓZ

(1.47)

This is the famous Breit-Wigner shape.Now if s ≈M2

Z then

γ∼ 1

s∼ 1

M2Z

(1.48)

Z ∼ 1

MZΓZ(1.49)

So if the width of the Z boson is small and you are colliding particles around the Z mass,to first approximation only the Z boson contributes. This is the case for the real Z boson,where ΓZ ≈ 2GeV.

We haveL = gLe

†Lσ

µeLZµ + gRe†Rσ

µeRZµ (1.50)

where

gL =e

swcw

(−1

2+ s2

w

), gR =

eswcw

(1.51)

Recall that for e+e− → γµ → µ+µ− we have

e− e+ µ− µ+ M/e2

L R L R 1 + cos θL R R L 1− cos θR L L R 1− cos θR L R L 1 + cos θ


Everything is the same for the Z boson. The only change in the different couplings andthe different Breit Wigner shape. We get,

e− e+ µ− µ+ ML R L R g2

L(1 + cos θ)(sfBW )L R R L gLgR(1− cos θ)(sfBW )R L L R gLgR(1− cos θ)(sfBW )R L R L g2

R(1 + cos θ)(sfBW )

where

fBW =1

s−M2Z + iMZΓZ

(1.52)

To find the cross sections we square the amplitudes and we have

dσ

d cos θ=

1

32πs|sfBW |2

(g4L + g4

R)(1 + cos θ)2 + 2g2Lg

2R(1− cos θ)2

(1.53)

Unlike in QED, we now get a term linear in cos θ given by

2(g2L − g2

R)2 cos θ (1.54)

From this one can define what is called a forward-backward asymmetry,

Afb =σ(cos θ > 0)− σ(cos θ < 0)

σ(cos θ > 0) + σ(cos θ < 0)(1.55)

Now we continue with studying the structure of the diagram in the Z channel. Nowthat the Z is on-shell, we can think of the process as producing a Z which subsequentlydecays. This can be done with as many particles you produce. We will now address thequestion of when you are allowed to split up a diagram. Consider the matrix element forthe process above,

M = [vBΓµuA]−gµν + pµpν

M2Z

p2 −M2Z + iMZpZ

[u1Γνu2] (1.56)

where Γµ denotes some vertex factors. Now recall the spin sum formula,

− gµν +pµpνM2

Z

→1∑

a=−1

ε∗aµ (p)εaν(p) (1.57)

This requires the particle being studied to be on-shell. If this is the case then we have

M =∑a

[vBΓµuA]ε∗aµ (p)εaν(p)

p2 −M2Z + iMZΓZ

[u1Γνu2] (1.58)

=∑a

M(e+e− → Za)fBWM(Za → µ−µ+) (1.59)

1.2. EXAMPLE 1:e+e− → µ+µ− 13

This process is known as “factorization of the matrix element”. This gets us half of theway to factorizing the cross section, we still need to factorize the phase space factor. Thecross section is

dσ =1

8EAEBdΠ1dΠ2(2π)4δ4(pA + pB − p1 − p2) |M|2 (1.60)

To factorize this we need to insert a 1,

1 =

∫dΠZ(2π)3δ3(pA + pB − pZ)2MZ (1.61)

Its straightforward to check that this is equal to one. Inserting in this relation,

dσ =1

2s

∑ab

∫dΠZ(2π)3δ3(pA + pB − pZ)M(e+e− → Za)M∗(e−e+ → Zb)

2Mz |fBW (s)|2 dΠ1dΠ2(2π)4δ4(pZ − p1 − p2)M∗(Za → µ+µ−)M(Zb → µ+µ−)(1.62)

Now this is almost factorized. However there are two “problems”. There is a missingdelta function for the energy of the Z boson and there is also the propagator factor. Itturns out that in the limit of narrow width these two problems cancel each other. This iscalled the “narrow-width approximation”. It consists of taking the Breit Wigner shapeand approximating it as,

MZ |fw(s)|2 =MZ

(s−MZ)2 + Γ2M2Z

≈ π

ΓZMZ

δ(EA + EB −MZ) (1.63)

The coefficient in front can be derived by integrating the Breit Wigner shape from −∞→∞.

Applying the narrow with approximation gives

dσ =∑ab

(1

8EAEB

∫dΠz(2π)4δ4(pA + pB − pZ)M(e+e− → Za)M†(e+e− → Zb)

1

ΓZ

1

2MZ

dΠ1dΠ2(2π)4δ4(pZ − p1 − p2)M∗(Za → µ+µ−)M(Zb → µ+µ−)

)(1.64)

≡ 1

ΓZtr[P abdDab

](1.65)

here we defined P as the production or “Spin Density Matrix” and D is the decay matrixwhich are given by

P ab ≡ 1

8EAEB

∫dΠz(2π)4δ4(pA + pB − pz)M(e+e− → Za)M†(e+e− → Zb) (1.66)

dDab ≡ 1

2MZ

dΠ1dΠ2(2π)4δ4(pZ − p1 − p2)M†(Za → µ+µ−)M(Zb → µ+µ−) (1.67)


Note that these are “almost” the expressions for the cross section for production anddecay distributions. If you have unpolarized production (This is NOT the case for theZ!) then

dσ = σ(e+e− → Z)1

ΓZdΓ(Z → µ+µ−) (1.68)

σ = σ(e+e− → Z)Br(Z → µ+µ−) (1.69)

This is okay for total rates, but not distributions.We now work an example. Recall that we have

L = gLe†Lσ

µeLZµ + gRe†Rσ

µeRZµ (1.70)

We have two possible initial states, e+Le−R and e+

Re−L . Lets look at angular momentum,

e+L e−R e+

R e−R

This are the only two possibilities. Because of this the spin-density matrix is actuallypretty simple. The off-diagonal matrix elements are zero. We only have P−− ∝ g2

L andP++ ∝ g2

R. In the electron positron collisions if you are given unpolarized beams thenyou will be dominated with outgoing left-handed particles since in the SM gL gR. Wedid one part of the calculation, now we need to calculate the decay distributions,

dD−−

d cos θ= g2

L(1 + cos θ)2 + g2R(1− cos θ)2 (1.71)

dD++

d cos θ= g2

R(1 + cos θ)2 + g2L(1− cos θ)2 (1.72)

and we have

σ−dD−−

d cos θ+ σ+

dD++

d cos θ∝ (g4

L + g4R)(‘1 + cos θ)2 + 2g2

Lg2R(1− cos θ)2 (1.73)

as expected.Now imagine what would happen if we forgot about the polarization of the beam. For

an unpolarized beam you would have

dΓ

d cos θ=

1

2

(dD++

d cos θ+dD−−

d cos θ

)(1.74)

=(g2L + g2

R

)(1 + cos2 θ) (1.75)

and you would conclude that Afb = 0 which is clearly wrong.In general you don’t need to factorize your amplitudes. Alternatively you just use

Breit Wigner shapes for your intermediate particles. However, this greatly simplifies thecalculations. For example, if you are working at the Tevatron and calculating the crosssection for qq → g → tt→ bb`¯νν. Then without factorization you will be working witha 6 body final state!

Chapter 2

QED

2.1 Initial State Radiation

So far we have been considering the 2 → 2 process. What can we do different? We canconsider the 2→ 3 process,

e

e

γ/Z

µ

µ

+

e

e

γ/Z

µ

µ

In general we must include the diagrams for the photon emitted by the muons in ourdiscussion. However, when dealing with Feynman diagrams you can get physically in-teresting results if you are dealing with a subset of the diagrams IF the subset is gaugeinvariant. This doesn’t give the correct results since there are interference terms, howeverthe result can hold in some approximation. It just so happens that each of these diagramsis gauge invariant on its own (we will check this later on).

We begin with a naive guess of the cross section.

σ2→3 ∼α

2πσ2→2 ∼

1

300σ2→2 (2.1)

This is very small. So the naive guess about such processes is that they have littleimportance except for precision measurements. However, it turns out that this guessisn’t quite correct. The reason for this is because of the propagator of the electron (oranti-electron) between the emission of the photon and the annihilation can blow up.

Consider the first diagram. Take the incoming electron momentum to be pA and thephoton momentum to be pγ. Then

M∝ 1

(pA − pγ)2 −m2e

=1

−2pA · pγ(2.2)

15

16 CHAPTER 2. QED

We take the four momentum to be (for now we ignore me though we will put it back inlater),

pA = (E, 0, 0, E) (2.3)

pγ = (zE,p⊥,√z2E2 − p2

⊥) (2.4)

where we have defined z as the fraction of the electron energy taken by the photon. Thedot product is given by

pA · pγ = zE2(1−√

1− p2⊥

z2E2) (2.5)

This goes to zero in two different limits.

1. When z → 0. This limit is a bit subtle. We have 0 < p⊥ < zE. So we can writep⊥ = xzE where x is the fraction of energy in the transverse direction (this takescare of the fact that p⊥ isn’t really independent of z. It is then obvious that in thez → 0 limit the dot product vanishes. This corresponds to a photon that carriesvery little energy and is known as a “soft singularity”

2. When p⊥ → 0. This means the photon is emitted parallel to the electron motion.This is called a “collinear singularity”

So this is bad. If you try to calculate the cross section for these tree-level Feynmandiagram you integrate over all phase space configurations and you get results that don’tmake sense. The resolution comes at two different levels.

1. Soft photons are drowned out in detector noise.

2. In practice you measure muons using the muon detectors and photons using theEM Calorimeter. The EM calorimeter cannot cover the entire solid angle of theinteraction point. There needs to be holes where the beams get in,

e+e−θmin

If γ are emitted with θ < θmin then they won’t be observed. This is nice since mostof these photons are unobservable and we don’t need to worry about them. Thisdeals with collinear photons.

2.1. INITIAL STATE RADIATION 17

Where do we draw the line between observable and unobservable collinear photons?To decide what kind of photons are “observed” we define a value Q such that if

pγT > Q (2.6)

we call it a 2→ 3 process. But there is still a problem. What will happen for the eventsthat p⊥ is small? These events will not disappear, they will just look like a 2→ 2 processbut with one difference, the energy of the electron will be smaller. This is also the casefor the collinear photons. In short p⊥ < Q reactions modify the rate of 2→ 2 processes.We need to know what this shift is.

The key observation here is that when p⊥ is small, pγ ≈ zpA. In this case

pA · pγ = zp2A ≈ 0 (2.7)

and hence the virtual electron is almost on-shell. This will be useful below.The amplitude takes the form,

M2→3 = vBγµ/pA − /pγ −me

−2pA · pγ(eγα)uA × (...) (2.8)

We want take the collinear limit, so

pA · pγ = zE2(p2⊥/2z

2E2)

= p2⊥/2z (2.9)

and hence we can replace the denominator with the above. Furthermore, the electron isroughly on shell so we can make the spin sum replacement,

M2→3 ≈ vBγµ

∑λ u

λ(pA(1− z))uλ(pA(1− z))

−2(p2⊥/2z)

(eγα)uA × (...) (2.10)

where the momenta of the virtual electron is just given by

pA − pγ = (1− z)pA (2.11)

We now just rewrite the term above as,

M2→3 = −e zp2⊥

∑λ

e[uλ(pA(1− z))γαuA

]︸︷︷︸Mλ

e−→e−γ

[vBγµuλ(pA(1− z))]× (...)︸︷︷︸Mλ

2→2(pA(1−z),pB→pi)

(2.12)

The first factor, known as the splitting amplitude doesn’t know anything about the restof the diagram. This result is very general.

We pause for a moment to prove gauge invarience. This element is multiplied by apolarization vector εα. Switching εα → pαγ gives,[

uλ/pγuA

]=[uλ(/pA − /peλ

)uA

](2.13)

=[uλ (me −me)uA

](2.14)

= 0 (2.15)

18 CHAPTER 2. QED

This justifies considering this single diagram in our calculation.This is what the matrix element look like, the next step is to consider the cross section.

We have the usual master formula,

dσ2→3 =1

8EAEBdΠγ

From M2→3︷︸︸︷(z

p2⊥

)2

×1

2

∑λ,λ′,h,a

Me−h→e−λ γa

(z)M∗e−h→eλ′γa

(z) · dΠ1,2

Mλ2→2(s(1− z))Mλ′,∗

2→2(s(1− z))εa(pγ)δ4(pA + pB − pγ − p1 − p2)(2π)4 (2.16)

where λ and λ′ are the spins from the matrix element, h is the spin of the electron, and a isthe spin of the photon. Furthermore, s is the Mandelstam variable. Notice the matchingof the spin indices, λ, λ′, between the different contributions. This is the same result wesaw in our first example of factorization above. The 2 → 2 amplitude will depend onlyon s as we are working in the s channel. However the effective “s” is modified since

seff = (pB + (1− z)pA)2 (2.17)

= 2pA · pB (1− z) (2.18)

= s (1− z) (2.19)

Note thatM(e−L → e−Rγ) =M(e−R → e−Lγ) = 0 (2.20)

This is easy to see from the structure of the e− → e−γ amplitude since,[uλ,†PLγ

0γαPRuA]

= 0 (2.21)

Note that if the initial state radiation was through a parity violating vertex this wouldnot have been true.

This means we only have non-zero amplitudes if

λ = h (2.22)

λ′ = h (2.23)

This is convenient since that means we can just sum the spins in the usual way,

dσ2→3 =1

2sdΠγ

(z

p2⊥

)21

2

∑h,a

|Mhha(z)|2 dΠ1,2

∑h′

∣∣∣Mh′

2→2(s(1− z))∣∣∣2 × δ4(...)(2π)4

(2.24)Doing the explicit calculation gives the same result for h = +1/2 and h = −1/2. This

calculation is done in the appendix (2.A) and gives,

1

2

∑h,a

|Mhha(z)|2 =2e2p2

⊥z(1− z)

[1 + (1− z)2

z

](2.25)


Note that a similar calculation is given in Peskin and Schroeder, 576-578.Inserting in this result gives,

dσ2→3 =1

2sdΠγ

(z

p2⊥

)22e2p2

⊥z(1− z)

[1 + (1− z)2

z

]dΠ1,2

∑h

∣∣Mh2→2(s(1− z))

∣∣2 (2.26)

= dΠγ

(z

p2⊥

)22e2p2

⊥z

[1 + (1− z)2

z

]× 2× dσ2→2(s(1− z)) (2.27)

where we have used

σ2→2 (s(1− z)) =1

2s(1− z)dΠ1,2

1

2

∑h

|M2→2|2 (2.28)

Now we have

dΠγ =d3pγ(2π)3

1

2Eγ=d2p⊥d(zEA)

(2π)3

1

2zEA=p⊥dp⊥

8π2

dz

z(2.29)

So we have

dσ2→3 =

∫ 1

0

dz

z

∫ Q

0→me

p⊥dp⊥8π2

(z

p2⊥

)24e2p2

⊥z

[1 + (1− z)2

z

](2.30)

=

∫ 1

0

dzα

π

[1 + (1− z)2

z

]log

Q

me

(2.31)

Notice that above we changed the limit of integration from 0 → me. This is because itturns out that a massive electron cannot omit perfectly collinear photons (this can bechecked using momentum conservation in a few lines). [Q 2: How do we know the limitchanges exactly to me, or is this an approximation?] We define

fe(1− z,Q) ≡ α

π

[1 + (1− z)2

z

]log

Q

me

(2.32)

and so

dσ2→3(p⊥ < Q) =

∫ 1

0

dzfe(1− z,Q)× dσ2→2(s(1− z)) (2.33)

This formula is very simple and has a beautiful interpretation. fe(x,Q) is the prob-ability to find an electron of energy xE in a beam supposedly consisting of electrons ofenergy E! This is analogous to the parton distribution functions of QCD. So inside amoving electron there is in fact a distribution of electrons with lower energy due to initialstate radiation. However, this assumes that the electrons emitted a photon. This is ofcourse not always true. To include the correction for when we have a true 2→ 2 processwe write,

fe(x,Q) = δ(1− x) +α

π

1 + x2

1− xlog

Q

me

(2.34)

20 CHAPTER 2. QED

Note that this equation is obviously not normalized at this point. The delta functionis normalized and there is an extra correction which will not integrate to zero. We willremedy this soon.

However, there is an issue with this formula. It is not well defined when z → 0 (orequivalently, when 1− x→ 0). fe is actually a divergent integral, and so is ill-defined wewill return to this in a moment.

Before we go on, consider the following point. In our example, e− → e−γ was followedby e−e+ → X but the collinear photon itself can also enter the subsequent reactions, e.g.

e

e

e

Z

e

µ

The cross section for this would be very similar,

dσ2→3(pe⊥ < Q) =

∫dzfγ(z,Q)dσ2→2(zs) (2.35)

where to this order in perturbation theory we have (by energy conservation)

fγ(z,Q) = fe(1− z,Q) (2.36)

The form of the z dependence arises from the fact that we still want to define z as theenergy fraction taken by the photon. Before we wrote dσ2→2(zs), where the term inparenthesis was the energy the center of mass energy of the 2 → 2 collision. In the new2→ 2 collision the energy of the 2→ 2 collision is given by

√(1− z)s.

Much more complicated processes are also possible. e.g.

As you go to higher and higher energies it becomes more and more important to includethe effect that an “electron” beam isn’t just electrons.

Thus far we have only dealt with the collinear divergence and we found that it is nota true divergence and it goes like ∼ log Q

me.

We now confront the z → 0 divergence. When z → 0,

Eγ → zEA → 0 (2.37)


and we have “soft photons”. Recall from QFT that the soft-photon divergence cancelsbetween diagrams with real soft photon emission and virtual photons with the virtualmomentum, `→ 0,

∫0dΠγ +

2

+

2

= finite

In order to solve for the effect of soft photon emission you need to calculate a com-plicated loop diagram. Instead, we can use a much easier way, and that’s by usingnormalization, ∫ 1

0

fe(x,Q) = 1 (2.38)

The expression we wrote above for fe is clearly not normalized. By matching ordersof α it has a chance at being normalized if we write the more appropriate form,

fe(x,Q) = (1 + αA+ ...)δ(1− x) +α

2π

1 + x2

1− xlog

Q

me

+O(α2) (2.39)

The difficulty is that ∫ 1

0

dx

1− x(2.40)

is not well defined. Mathematically you can define

1

(1− x)+

=1

1− x,∀x 6= 1 (2.41)

and ∫ 1

0

f(x)

(1− x)+

dx =

∫ 1

0

f(x)− f(1)

1− xdx (2.42)

In other words we are subtracting away the singularity. Now we have∫ 1

0

dx(1 + x2)

(1− x)+

=

∫ 1

0

x2 − 1

1− x= −

∫ 1

0

(x+ 1)dx = −3

2(2.43)

This implies that

A =3

2

α

2πlog

Q

me

(2.44)

and

fe(x,Q) = δ(1− x) +α

2π

[1 + x2

(1− x)+

+3

2δ(1− x)

]log

Q

me

(2.45)

This expression is well defined as x → 1 but it is still highly singular. As x → 1 thephotons get softer and softer and the probability of emitting another photon is gettinglarger and larger. Perturbation theory in this sense is breaking down. The computation

22 CHAPTER 2. QED

is pretty good for most values of x but as you get within a few percent of 1 it breaks down.When that happens you need to try to understand the systematics of the calculation andmake some approximation to get the leading behaviour of the process. If you can thensum up all the multiple photons contributions then you can achieve this. We will not dothe rigorous calculation, instead we use a trick to make the calculation straightforward.In QED this is just a beautiful trick. In QCD it is necessary.

The idea to find equations for how fe changes as a function of Q. These will not bedone perturbatively. When we talk about these probability distribution functions we talkabout their dependence on their energy, but they also have a second important depen-dence on Q. At first this dependence may seem strange since why would the compositionof an electron beam dependence on this value that we choose Q. However, thinking morecritically Q really tells us when we have or don’t have a photon. So depending on whatenergies we still count as photons we have a difference beam composition. i.e. this valuetells you when you call the photon part of the beam and when you call it a true emission.It’s value is up to us. When we want to study a 2→ 2 process, we can decide what Q is.Imagine that we know fe for some value of Q and then we increase Q → Q + ∆Q. Forsimplicity we focus on the photons. Consider

e−(x′EA)

γ(xEA, Q < p⊥ < Q+ ∆Q)

e−

where the splitting fraction is then z = xx′< 1.

fγ(x,Q+ ∆Q) = fγ(x,Q) +

∫ 1

0

dx′fe(x′, Q)

∫ 1

0

dzδ(x− zx′)απ

1 + (1− z)2

zlog

Q+ ∆Q

Q(2.46)

The first term is the probability of finding a photon with p⊥ < Q and the second term isjust the probability of finding a photon with Q < p⊥ < Q + ∆Q. The second term canbe understood as follows.

dzδ(x− zx′)απ

1 + (1− z)2

zlog

Q+ ∆Q

Q= fγ(1− z,Q+ ∆Q)− fγ(1− z,Q) (2.47)

is the probability of finding an electron beam with energy fraction x′ having a photonwith energy fraction x between Q and Q+ ∆Q, dx′fe(x

′, Q) is the probability of findingan electron beam with energy fraction x′.

We have,

fγ(x,Q+ ∆Q) = fγ (x,Q) +α

π

∫ 1

x

dz

zfe

(xz,Q) 1 + (1− z)2

zlog

Q+ ∆Q

Q(2.48)

2.2. FINAL STATE RADIATION 23

where we have used the fact that in order for the delta function to not have been zero werequire, x

x′< z < 1. Furthemore,

dfγd logQ

=α

π

∫ 1

x

dz

zfe

(xz,Q) 1 + (1− z)2

z(2.49)

where in the last step we took the infinitesimal limit for ∆Q. We define a “splittingfunction” Pe−→γ(z) and we can write the final result as

dfγd logQ

=α

π

∫ 1

x

dz

zPe−→γ(z)fe

(xz,Q)

(2.50)

In the treatment above we didn’t include all the possibilities. You could also have photonscoming from other photons and photons coming from positrons. We worked out Pe−→γ,the complete set is in Peskin and Schroeder, p.587. Including all the possibilities you get,

dfγ(x,Q)

d logQ=α

π

∫ 1

x

dz

z

(Pe−→γ(z)fe−(

x

z,Q) + Pγ→γ(z)fγ(

x

z,Q)

+ Pe+→γ(z)fe+(x

z,Q)

)(2.51)

These are known as the Gribov-Lipatov equations. In principle you can now calculate allthe fi’s at a given energy and solve the differential equation to the get the distributionsto all orders in Q.

At Q = me you must just have

fe−(x,me) = δ(1− x) (2.52)

fγ(x,me) = fe+(x,me) = 0 (2.53)

2.2 Final State Radiation

Recall that we have

dσ2→3 =α

π

dz

z(1 + (1− z))2dp⊥

p⊥dσ2→2(s(1− z)) (2.54)

Consider a collision occurring in space-time:

e− e+

τ `

24 CHAPTER 2. QED

The factorization can happen if you have physics at two separate length scales. Thetypical length of decay and the energy of the collision. Where the photon will be emittedis dependent on the energy of the photon emitted. We have

τ ∼ 1√−(pA − pγ)2

(2.55)

` ∼ 1√s

(2.56)

where the value of τ was found from taking the Fourier Transform of roughly on-shellpropagator of the virtual electron. If τ ` then we have factorization. Essentially allfactorization examples on a fundamental level involve separation of length scales.

The same considerations apply to final state radiation (FSR). As an example consider

e+e− → µ+µ−γ (2.57)

or diagrammatically,

µ+

γ

µ−

One physical difference between IRS is that a collinear γ can be detected in this case!

EM cal

A soft γ is still undetectable though. Set Eminγ .

We won’t go through the detailed calculation, but the ideas are the same here as theywere in the ISR case. We use equation (2.54) as a template. The primary contributionwill be at small x since that is where we now have our soft divergence.

Here we take z → x[Q 3: I’m not sure what is the reason to make this substitution].

Furthermore, we work in the limit of small x so 1 + (1− x)x1−−→ 2.

dσ2→3

dx

∣∣∣∣x1

=2α

π

1

x

∫ xEµ

mµ

dp⊥p⊥

dσ2→2 (2.58)

As before we let the limit of the p⊥ integral have a lower bound of the particle emittingthe particle. Furthermore, the maximum p⊥ is just the energy of the photon.

2.2. FINAL STATE RADIATION 25

We have

σ2→3 ≈2α

π

∫ O(1)

xmin

dx

x

∫ xEµ

mµ

dp⊥p⊥

dσ2→2 (2.59)

where xmin = Eminγ /Eµ. We don’t really know the upper-bound since we are assuming

that x is small. So where the cut-off for “small”? But we are still okay since we have alogarithmic integral over x. A log(O(1)) ≈ 0 so we ignore the extra term and we have

dσ2→3 =2α

π

(log

EµEminγ

)(log

Eµmµ

) ≡σ0︷︸︸︷σLO2→2 +

(terms with 1log or no log

)(2.60)

=α

2π

(log

s

(Eminγ )2

)(log

s

m2µ

)σ0 +

(terms with 1log or no log

)(2.61)

The two log form is called the “Sudakov double Logarithm”. The photons with energieslower then Emin

γ still get counted in the detector but they are detected as 2→ 2 collisions.We now want to quantify the effect of these photons.

The rate of γ’s with Eγ < Eminγ (which we call “real γ” to distinguish it from diagrams

with virtual photons) is IR divergent as x→ 0 (as the energy of the photon goes to zero).This comes up since we are only working in leading order. The solution is to introducea “regulator” µ that represents the photon mass, a trick often used in a QFT class. Wenow take the upper-bound to be Emin

γ and the lower bound to be µ. We have the samecalculation as before.

∆σreal γ

2→2 =α

2πlog

(Eminγ )2

µ2log

s

m2µ

σ0 + ... (2.62)

On top of this contribution there are also contributions from

2

+

Calculating these diagrams to leading order and next-to-leading order result

∆σLO+NLO2→2 = σ0

(1− α

2πlog

s

µ2log

s

m2µ

+ ...

)(2.63)

To get the full 2 → 2 rate we need to add this result with the calculation above forthe soft photons that we miss from the 2→ 3 calculation. This gives,

σ2→2 = σ0

(1− α

2πlog

s

(Eminγ )2

logs

m2µ

)(2.64)

The total result is well defined (independent of regulator)!There are several interesting facts that we can glean from this calculation.

26 CHAPTER 2. QED

1. Radiative correction to σ2→2(no γ’s of E > Eminγ ) is proportional to σ

πlog2. For

example, for√s = 100GeV,Emin

γ = 1GeV, we have ∆σNLOσLO

≈ 8% instead of thenaive guess of α

π≈ 0.3%. So the radiative corrections are not small due to the

logarithmic factors.

Any kind of precision result requires resummation: this is achieved by techniquessimilar to GL equation for ISR in the previous lecture. We will not do this herebut the answer is

σ2→2 = σ0 exp

(− α

2πlog

s

(Eminγ )2

logs

m2µ

)+ (no log2) (2.65)

The lesson here is when comparing different orders in perturbation theory you needto be careful due to the potential for large logs.

2. You may think that we are in deep trouble, since you need to go to higher andhigher orders in perturbation theory to get precision results and this was in QED!The situation is even more bleak in QCD. Fortunately, leading log corrections cancelin “inclusive” cross section,

σ(µ+µ− or µ+µ−γ) = σ2→2(no γ in Eγ < Eminγ ) + σ2→3(Eγ > Emin

γ ) (2.66)

= σ0 + (no log2 ) (2.67)

Although we have not yet shown in it turns out that there are no single log terms either.

This result is quite general: inclusive observables behave better in perturbation theorythen exclusive. It has a simple physical reason, which is easy to see in the space-timepicture.

L ``

σ0 describes the production process (shown as a gray circle). If the produced muonhappens to be a little bit off-shell then it splits.

We have

σ2→3 ∼ Prob(µ− → µ−γ

)σ0 (2.68)

σ2→2 ∼ Prob (no splitting)σ0 (2.69)

2.3. e+e− → µ+µ−γ 27

so we have

σtot ∼1︷︸︸︷

Prob(µ− → µ−γ) + Prob(no split)σ0 (2.70)

∼ σ0 ! (2.71)

The only caveat is that this again requires separation of scales. The production scale mustbe much smaller then the decay length. The factorization and conservation of probabilityis behind this cancellation. In practical terms this is very nice. If you are worried aboutlarge logs, you could either do the resummation which takes a lot of work or you couldtake the cheap way out and just work with inclusive variables.

2.3 e+e− → µ+µ−γ

Up to now we have avoided discussion of any interference terms. While this was wrong wegot sensible results since we considered a gauge invariant subset of diagrams. We will nowoutline the necessary steps to do the exact calculations without making approximations.We consider e+e− → γ → µ−µ+γ however, we will cheat a little. In general we shouldalso have interference terms between initial and final state radiation. We omit the initialstate radiation diagrams. The justification for this is that we do the calculations as amodel for e+e− → γ → qqg. These diagrams do not have any initial state gluon radiation.We will just go over the basic ideas and we will fill them in in the HW. We have twodiagrams

+

and as we just learned there are also these virtual diagrams.The available phase space is

3 · 3− 4− 1 = 4 (2.72)

since the naive phase space is 3 ·3, we have 1 energy conserving delta function which takesaway 4 degrees of freedom. Furthermore, we will work in the center of mass frame. Thisconstrains the decay products to be in a plane as their momenta need to add to zero. Thisplane can be in any orientation without a change in the cross section. This invarienceleads to a variable that we can integrate over leaving one less degree of freedom.

We have

q ≡ pA + pB =(√

s,0)→ xi ≡

2pi · qq2

=2Ei√s

(2.73)

28 CHAPTER 2. QED

Energy conservation constrains the possible energy fractions of the final states,∑i

Ei =√s ⇒ x1 + x2 + x3 = 2 (2.74)

where 0 ≤ xi ≤ 1. We can use the last condition on xi to restrict the free parameters. This“restriction” is different from the phase space degrees of freedom calculation above. Therewe argued that the cross section will be invariant under a rotation. The condition onthe xi’s is built into the momentum conserving delta function. Knowing these conditionshelps us choose appropriate parameters to use in our cross section. We choose x1, x2 andthe direction of e.g. γ with respect to e− as our free parameters. The three body phasespace can be split up as follows,∫ 3∏

i=1

dΠi(2π)4δ(4)(q −∑i

pi) =s

128π3

∫dx1dx2dΩ (2.75)

where dΩ is the overall rotation of the final state particles. This expression is derived inappendix 2.B.2.

Now,1

4

∑spins

|M|2 = LµνMµν (2.76)

where Lµν(Mµν) involves the electron (muon) trace.There is a useful trick we can use. By Lorentz invarience the structure of Mµν is

very simple. After integration over all angles the only vector left in the problem is themomenta going into the right vertex, qµ, so we can write,∫

dΩMµν = (Agµν +Bqµqν) · f(x1, x2) (2.77)

(obviously Lµν , the electron contribution, will not depend on Ω)One can show that

qµMµν = qνM

µν = 0 (2.78)

This implies that A = −B (these are secretly just Ward Identities).So we have

1

4

∑spins

|M|2 = (gµνLµν − qµqνLµν)Af(x1, x2) (2.79)

where

f(x1, x2) =1

3

∫dΩgµνMµν (2.80)

So in fact you don’t need to calculate Mµν but just its trace.Doing the calculation gives

dσ

dx1dx2

=1

2sgµνL

µν · f(x1, x2) =4πα2

3s︸︷︷︸σ0

α

2π

x21 + x2

2

(1− x1)(1− x2)(2.81)

There are two singularities

2.3. e+e− → µ+µ−γ 29

1. x1 = 2p1·qq2 → 1. This occurs when 2p1 · q = q2, or equivalently,

(p2 + p3)2 = (q − p1)2 = q2 − 2q · p1x1→1−−−→ 0 (2.82)

Hence, when the invariant mass of 1 and 2 is zero. If two massless momenta, k1

and k2 are collinear then,

m2inv ≡ (k1 + k2)2 = (E1 + E2)2 − (E1 + E2)2 = 0 (2.83)

So the first singularity corresponds to particle two being roughly collinear with thephoton.

23

1

2. Likewise for the second singularity we have x2 → 1 and particle 1 is roughly collinearwith the photon.

2

31

These are our old friends, the collinear singularities.We need to regulate these singularities. There are two ways,

• We can do what we just did for the singularity calculation and use small photonmass, µ (changes xmin 6= 0) and reintroduce mµ.

• Alternatively you can solve the problem by dimensional regularization: go to d =4 − 2ε. At this point the calculation gets complicated enough that you probablywant to use dim-reg.

Repeating the calculation in dim-reg brings about a few changes. The result is,

dσ

dx1dx2

= σ0α

2π

3(1− ε)2

(3− 2ε)Γ(2− 2ε)︸︷︷︸H(ε)

·x21 + x2

2 − ε(2− x1 − x2)

(1− x1)1+ε(1− x2)1+ε(2.84)

30 CHAPTER 2. QED

Integration gives

σ(ε) = σ0α

2πH(ε) ·

[2

ε2+

3

ε+

19

2+O(ε)

](2.85)

The 2ε2

term corresponds to the log sµ2 log s

µ2 divergence while the 3ε

corresponds to thesingle logs.

Also in dim-reg one can show that the contribution to the 2 → 2 cross section fromNLO calculations is,

σNLO2→2 (ε) = σ0α

2πH(ε) ·

[− 2

ε2− 3

ε− 8 +O(ε)

](2.86)

which implies that

σinclusive = σ2→2 + σreal γ2→3 = σ0

(1 +

α

2π

(19

2− 8

)+O(α2)

)(2.87)

= σ0

(1 +

3

4

α

π+O(α2)

)(2.88)

This is of order 0.3% as promised.

2.A Calculating the Splitting Probability

[Q 4: This section is under construction]We now calculate the averaged splitting probability. The derivation for this result was

omitted in Maxim’s lectures but done in Peskin and Schroeder. We want to be consistentin our calculation and approximate the final result to order p2

⊥. We are considering theprocess,

p

q

k

θ

We want to consider a real electron which splits off into an on-shell photon and a finalelectron to be slightly off-shell. We use a small angle approximation,

θ ≈ p⊥zp

(2.89)

This gives

pµ = p(1, 0, 0, 1) (2.90)

qµ = zp

(1,p⊥zp, 0, 1− p2

⊥2z2p2

)(2.91)

kµ =

((1− z)p,−p⊥, 0, (1− z)p+

p2⊥

2zp

)(2.92)

2.A. CALCULATING THE SPLITTING PROBABILITY 31

The difficulty in this calculation is that since the electron is off-shell we can’t use the usualspin-sum formula. Instead we construct the spinors explicitly. For the initial electrontravelling in the z direction we have,

us(p) =√

2p

(

0 00 1

)ξs(

1 00 0

)ξs

(2.93)

=

√2p

0010

,√

2p

0100

(2.94)

where ξs is the spinor and is (1, 0)T for a right handed particle and (0, 1)T for a left-handed one. To calculate the final electron spinor we take the spinor above and rotate itin the xz plane. This rotation is accomplished using,

e−θS1,3

=

(cos θ

2− iσ2 sin θ

20

0 cos θ2− iσ2 sin θ

2

)(2.95)

We now make a small angle approximation:

sinθ

2≈ tan

θ

2≈ − p⊥

2p(1− z), cos θ ≈ 1 (2.96)

Acting on the spinor above with energy 2p(1− z) gives,

us =

√2p(1− z)

pL

2p(1−z)100

,√

2p(1− z)

001p⊥

2p(1−z)

(2.97)

QED interactions conserve chirality and a left-handed electron only interacts with a left-handed electron. So we only have either a left-left interaction or right-right. We begin bystudying the left-left interaction.The last ingredient we need is the polarization vector.The polarization vector must be perpendicular to the direction of motion of the photon.We must have

ε · q = 0 (2.98)

(and ε0 = 0). Our condition implies

p⊥ε1 +2z2p2 − p2

⊥2zp

ε3 = 0 (2.99)

ε3 =p2⊥ − 2z2p2

2zpp⊥ε1 (2.100)

32 CHAPTER 2. QED

and hence

ε =

(1, a,

p2⊥ − 2z2p2

2zpp⊥

)(2.101)

=

(1, a,− zp

p⊥

)+O(

p2⊥

z2p2) (2.102)

where a is arbitrary. A convenient choice turns out to be a = i. We have also yet tonormalize the polarization vector such that ε2 = 1. This gives the final result

ε∗L =1√2

(1, i,−p⊥

zp

)+O(

p2⊥

z2p2) , ε∗R =

1√2

(1,−i,−p⊥

zp

)+O(

p2⊥

z2p2) (2.103)

where we chose our convention of εL having a −i to conform with normal conventionsof circular polarization. Furthermore the right handed partner was found as the vectorperpendicular to the left handed polarization vector.

We can now finally calculate the amplitude,

iMLLL = [u(k)(−ieγµ)u(p)] ε∗µ (2.104)

MLLL = −ig√

2p(1− z)(

0 0 p⊥2p(1−z) 1

)( 0 σi

−σi 0

)0100

√2p1√2

(1,+i,−p⊥

zp

)i(2.105)

= i√

2ep⊥√

1− z(

1

1− z+

1

z

)(2.106)

= iep⊥

√2(1− z)

z(1− z)=︸︷︷︸

by Parity

iMRRR (2.107)

A similar calculation yields,

iMLLR = iep⊥2(1− z)

z

by Parity︷︸︸︷= iMRRL (2.108)

Furthermore, we have,

iMRLL =MRLR =MLRL =MLRR = 0 (2.109)

Summing over all spins:

|M|2 =1

2

∑pol

|M| (2.110)

= 2e2p2⊥

(1 + (1− z)2

z2(1− z)

)(2.111)

2.B. PHASE SPACE 33

2.B Phase Space

In this section we calculate the three body phase space. The calculation is based on alovely note by Hitoshi Murayama[2].

2.B.1 Two Body

Here we quickly summarize some properties of two body phase space. In the center ofmass frame, two body phase space can always be written in the form,∫

dΦ2 =

∫d cos θ

2

dφ

2π

β

8π(2.112)

where

β ≡√

1− 2(m21 +m2)

s+

(m21 −m2

2)2

s2(2.113)

The energies of the outgoing particles are related to their masses and incoming energiesby,

E1 ≡√s

2

(1 +

m21

s− m2

2

s

)(2.114)

E2 ≡√s

2

(1 +

m22

s− m2

1

s

)(2.115)

Furthermore, their momenta are equal and given by

p1 = p2 =

√s

2β (2.116)

2.B.2 Three Body

We can now move onto three body phase space. We begin by introducing some convenientnotation. We have,

q 23→

1

3

2

we depict the second particle as a vector boson, however the results are more general andapply to any 3 body phase space. We take the initial center of mass energy to be

√s.

Further we denote the virtual particle which goes on to produce particles 2 and 3 by q23

34 CHAPTER 2. QED

and its mass as s23 ≡ (p2 + p3)2. Recall that we can split up any three body phase spaceinto a product of 2 body phase spaces,∫

dΦ3 =

∫ds23

2πdΦ2(p1, q23)dΦ2(p2, p3) (2.117)

The phase spaces are Lorentz invariant quantities. Since the phase space takes a simpleform in the zero momentum frame it is convenient to consider dΦ2(q1, q23) in the centerof mass frame and dΦ2(p2, p3) in the zero momentum frame for q23.∫

dΦ3 =

∫ds23

2π

d cos θ1

2

dφ1

2π

β1

8π

d cos θ23

2

dφ23

2π

β23

8π(2.118)

where we have

β1 ≡√

1− 2(m21 + s23)

s+

(m21 − s23)2

s2(2.119)

β23 ≡

√1− 2(m2

2 +m23)

s23

+(m2

2 −m23)2

s223

(2.120)

Note that we don’t need to worry about the momenta of the virtual particles explicitlyyet since we took dΦ2(p2, p3) to be the zero momentum frame of particle q23.

We define the angle θ23 to be the polar angle from −p1 and θ1 to be the polar angleof 1 frame some arbitrary axis. Then rotations about p1 leave the cross section invariant.Hence we can integrate over this angle,∫

dΦ3 =

∫ds23

2π

d cos θ1

2

dφ1

2π

β1

8π

d cos θ23

2

β23

8π(2.121)

Now the angles of θ1 and φ1 move the angle of p1 around and with it the entire systemby the dependence of θ23. Everything is usually set by θ23 and invariant under rotationsof the θ1 and φ1. This holds true if every axis picked out by θ1 is equally valid. This is nolonger the case of the particle spin is being measured. We will not make any assumptionsabout the isotropy until the very end of the calculation.

Our goal now is to try to rewrite the phase space in terms of convenient variables.One such set of variables are the energy fractions of the particles. The energy fractionsobey the constraint,

x1 + x2 + x3 = 2 (2.122)

so they are not independent. We want to rewrite two of the variables above (which wechoose to be s23 and cos θ23) in terms of x1 and x2. We need to find their Jacobian.

From the energies of two particles splittings we know that,

x1 = 1 +m2

1

s− s23

s(2.123)

2.B. PHASE SPACE 35

Now we we just need to find x2 in terms of s23 and cos θ23. To do this we need torelate the quantities in the center of mass frame to the rest frame of q23. Since the finalmomenta are on a plane we set the y component to be zero, and q23 to be along the zaxis,

q23 =

(E23, 0, 0,

√s

2β1

)(2.124)

p1 =

(E1, 0, 0,−

√s

2β1

)(2.125)

where

E23 =

√s

2

(1 +

s23

s− m2

1

s

)(2.126)

We have,

γ =E23√s23

(2.127)

γβ =p23√s23

=1

2

√s

s23

β1 (2.128)

and

p2 =(E2, p2 sin θ23, 0, p2 cos θ23

)(2.129)

(2.130)

where

E2 ≡√s23

2

(1 +

m22

s23

− m23

s23

)(2.131)

p2 ≡√s23

2β23 (2.132)

From here we can finally extract the energy fraction,

x2 =2√s

(γE2 + γβp2 cos θ23

)(2.133)

where γ, β, E2, p2 each contain s23 dependence. Explicitly we have,

x2 =2√s

1

2

√s

s23

(1 +

s23

s− m2

1

s

) √s23

2

(1 +

m22

s23

− m23

s23

)+ (2.134)

1

2

√s

s23

√1− 2

(m2

1 + s23

s

)+

(m2

1 − s223

s

)2

× (2.135)

√s23

2

√1− 2

(m2

2 +m23

s23

)+

(m2

2 −m23

s23

)2

cos θ23

(2.136)

36 CHAPTER 2. QED

We are almost done. We just need to find the Jacobian that relates x1, x2 to cos θ23

and s23 as well as the ranges of x1 and x2. In fact we don’t need to find the entireJacobian. From equation 2.123 we already know the Jacobian has the form

J =

(∂s23

∂x1

∂s23

x2

∂ cos θ23

∂x1

∂ cos θ23

∂x2

)=

(−s 0

∂ cos θ23

∂x1

∂ cos θ23

∂x2

)(2.137)

Thus to find the determinant we only need ∂ cos θ23

∂x2.

Unfortunately, the s23 dependence of x2 is a bit nasty. The calculation gets signifi-cantly simpler in the massles limit. There we have,

x2 =2√s

(√s

4

(1 +

s23

s

)+

√s

4

√1− 2s23

s+s2

23

s2cos θ23

)(2.138)

with s23

s= 1− x1. Inserting in this relation,

x2 = 1− x1

2+x1

2cos θ23 (2.139)

Hence we have,∂ cos θ23

∂x2

=2

x1

(2.140)

and the determinant of the Jacobian is,

|detJ | = 2s

x1

(2.141)

Lastly, we have the range of integration.

1− x1 < x2 < 1 (2.142)

Finally we have,∫dΦ3 =

∫ 1

0

dx1

∫ 1

1−x1

dx2x12s

x1

1

2π

1

2

1

(8π)2

∫d cos θ1

2

dφ1

2π(2.143)

=s

128π3

∫ 1

0

dx1

∫ 1

1−xdx2

∫dΩ

4π(2.144)

We are left with an integral over a solid angle. We usually do not measure the spinsof the final state particles. In this case the overall rotation of the final particles can bedropped. Finally we have, ∫

dΦ3 =s

128π3

∫ 1

0

dx1

∫ 1

1−xdx2 (2.145)

So while naively, you may think that any x2 within the range of 0 to 1 is valid phasespace, this turns out not to be the case. However, there is no preference to certain regionsof phase space since there is no dependence on x1 or x2 . The phase space is shown below,

2.B. PHASE SPACE 37

x1

x2

0 10

1

Chapter 3

QCD

3.1 e+e− → Hadrons

e

e

γ/Z

q → jet

q → jet

The probability of hadronization is 1. If you assume that the scale of hadronization ismuch smaller then the scale of the parton process thenx,

σLO(e+e− → hadrons) = σLO(e+e− → qq) (3.1)

Hadronization doesn’t “interfere” with the short-distance physics. This is actually an-other instance of factorization.

If√s MZ(just convenient to only study the photon channel but trivial to include

the Z) then

σLO(e+e− → qq) =4πα2

3s︸︷︷︸σ0≡σLO(e+e−→µ+µ−)

∑q

3Q2q (3.2)

where the charge factor arises since the quark vertex contributes a fraction of either 2/3or 1/3, there are three colors, and we sum over all the possible quarks that can form atthe energy threshold,

√s.

You can now define

R =σ(e+e− → had)

σ0

=∑q

3Q2q (3.3)

38

3.1. e+e− → Hadrons 39

Keep in mind that this formula is only valid away from threshold, since close to thresholdthere are corrections associated with many gluon emission effects. As an example in thelimit that Mb

√sMZ we have

R = 3

((1

3

)2

· 3 +

(2

3

)2

· 2

)=

11

3= 3.67 (3.4)

However at√s ≈MZ we have

RLOZ =

3∑

q

(gq 2L + gq 2

R

)gµ 2L + gµ 2

R

=Γ (Z → hadrons)

Γ (Z → µ+µ−)(3.5)

The leading order theory is RZ = 20.09 and the experimental result is 20.79±0.04. Theseare slightly inconsistent since we haven’t included the next to leading order correction.

As before the cross section can be calculated. You need to include the NLO corre-lations with a real and virtual gluon. The results are the same as for µ−µ+, except forcolor factors (see equation 2.88),

σ(e+e− → hadrons) = σLO(e+e− → qq) + σNLO(e+e− → qqg) +O(α2s) (3.6)

= σLO(e+e− → qq)[1 +

αsπ

+O(α2s)]

(3.7)

where we lost the 3/4 factor due to color.In perturbation theory,

σhad =

O(1)+O(αs)︷︸︸︷σqq + σqqg︸︷︷︸

O(αs)+..

+

O(α2s)+...︷︸︸︷

σqqgg + σqqqq ... (3.8)

Now of course αs is not a constant but instead it runs, which we can calculate throughthe RG equation. e.g. calculating

Such a calculation gives,

dαs(µ2)

d log µ2= −bα2

s (1 + b1αs + ...) (3.9)

where

b =33− 2nf

12π(3.10)

40 CHAPTER 3. QCD

where nf is the number of flavors. Note that αs depends on the renormalization scale soit can’t be a direct observable. For any observable O we must have,

dO

d log µ2= 0 (3.11)

This must be true for any possible choice of µ2. Since αs is dependent on µ2, the equationshould hold order-by-order in αs. In general we get

R(s) = R0(s) ·

[1 +

αs(µ2)

π+∞∑n=2

Cn ·(αs(µ

2)

π

)n](3.12)

The dependence of Cn can be deduced as follows. We have two scales in our process, sand our renormalization scale. We must have Cn to be dimensionless. This can only be

true if Cn = Cn

(µ2

s

). We have

dR

d log µ2= R0

[1

π

dαsd log µ2

+dC2(µ2/s)

d log µ2

(αsπ

)2]

(3.13)

By looking at equation 3.9 we see that the lowest order term of dαs/d log µ2 is of order 2and given by −bα2

s.So we can write

− bπα2s +

dC2

d log µ2

(αsπ

)2

= 0 (3.14)

⇒ dC2

d log µ2= πb (3.15)

This doesn’t tell us what C2 is but just says how it depends on µ2.An explicit calculation gives,

C2(µ2/s = 1) = 1.986− 0.115nf (3.16)

where nf is the number of flavors in your theory.Using the running equation (3.15) we find,

C2(µ2/s) = C2(1)− πb logs

µ2(3.17)

So the running of αs in fact gives you the running of C2 as well. The lesson is thateven though observables do not depend on your choice of scale, since we do cutoff ourperturbation theory at some order we do have some residual error. We have,

dR∣∣n−loop

d log µ2= O(αn+1

s ) (3.18)

3.2. HADRONIZATION 41

The best choice of scale is unknown. One can test examples where two or more termsin perturbation theory are known. One can compore for example NLO answer at variousµ2 to the known NNLO answer which is ∼ µ2 independent. Typically choose

µ =√s (3.19)

and use1

2

√s . µ . 2

√s (3.20)

for systematic error.

3.2 Hadronization

It would be nice to ask some questions beyond just the inclusive cross section. Forexample,

• What type of hadrons are there?

• What are their energies?

It turns out that the first question is inherently non-perturbative. We can make modelsbut there is no real field theory computation that will give you this. However, the secondone can be solved.

Since we cannot know exactly what happens in hadronization, we consider some mod-els. The model we use is known as the “color flux tube model”. This model is more thenjust a handwaving explaination and is actually the engine behind Pythia.

Consider two quarks moving away from each other. They constantly exchange gluonsboth pertubatively and non-pertabatively. This is important because if they never talkedto each other then each of the quarks would stay in their initial colored states of thefundamental and antifundamental representations. Instead, a flux tube builds,

q(1)q(2)

Flux Tube

The flux tube has some tension,

K ∼ GeV

fm(3.21)

and the tube kind of jiggles around a little bit. You can think of a regular tube that youare stretching. This jiggling gives a transverse momenta given by,

p⊥ ∼ ΛQCD (3.22)

42 CHAPTER 3. QCD

An important quantity for this model is the invariant mass of the quarks,

Minv(q, q) = (p1 + p2)2 = s (3.23)

This tube isn’t perfectly strong. At one point there will be a breakdown, where thetube splits,

qq qq

The two new quarks are roughly at rest. The invariant mass of the new meson is

M2inv(q1q) = (p1 + pq)

2 = 2√sp⊥ ∼

√sΛQCD = s

(ΛQCD√

s

)(3.24)

More breakdown gives Minv ≈ 1 GeV. The probability of which type of meson forms areinput parameters.

The general feature is

phad⊥ ∼ ΛQCD √s ∼ phad‖ (3.25)

So what you end with, regardless of the details, is a collimated beam of hadrons.This collimated set of hadrons is known as jets. This hierarchy between the parallel andperpendicular momenta is model independent. We have,

qhadrons

with θ ∼ ΛQCD/√s.

At high√s, can discuss energy/angular distributions of jets but not their composition.

The basic rules of jets are as follows.

1. Jets from u, u, d, d, s, s are essentially indistinguishable1.

2. Gluon jets look similar to q, but can be distinguished to some extent.

3. Jets from b/b and c/c can be “tagged”.

There are two ways to tag jets. The first way is using the vertex detector.

b

Si Det.

1This isn’t completely correct since we can somewhat measure the initial charge.

3.3. e+e− → Jets 43

The decay lengths (forgetting time dialation) for the b and c charms are,

cτb ≈ 500µm (3.26)

cτc ≈ 300µm (3.27)

The second way to tag jets is to use semileptonic decays. The b quark decays withbranching ratio of about 20% by

b→ c`ν (3.28)

The c quark forms a jet and if the lepton is a muon it can be detected using the muondetector.

A sample process is shown below,

µ Det.

H Cal.×

b

It’s important to note that the muon will be inline with the jet since they both come fromthe same particle. This signiture is quite unique. For any other process, the hadronicjet and the muon are not very unlikley to be inline with each other. This allows anotherway to tag the b quark. A similar proceduer can be done to tag the c quark. In realitythis tagging technique can only tell you that either a c or a b quark was there, it cannotdifferentiate between the two.

For the LHC the total tagging efficiency is ∼ 50%.

3.3 e+e− → Jets

We now begin to talk about jets order by order in perturbation theory. Consider a 2→ 2process,

e− e+

j2

j1

θ

This process is given by

dσ(e+e− → 2j)

d cos θ∝ 1 + cos2 θ + (Z-exchange) (3.29)

At NLO, things get interesting. There are basically two kinds of events,

44 CHAPTER 3. QCD

1. Non-singular regions: Every particle hadronizes independently and we have 3 jets.

j1

j2

j3

2. Singular regions: Some particles are collimated (θ → 0)

j1

j3

j2

θ

In this case two jets can merge into one and you have to do a 2 jet correction.

We now attempt to deal with the singular events. Say pg⊥ > Q to call an event a“3-jet” event. Then,

σ3 =αsπ

(C1 log2 s

Q2+ C2 log

s

Q2+ C3

)+O

(α2s,

ΛQCD

Q

)(3.30)

σ2 = σLO2 + σLO2

αsπ

(−C1 log2 s

Q2− C2 log

s

Q2+ C4

)(3.31)

where the form of σ2 can be inferred from σ3 by the condition since we already know theinclusive rate is (we mentioned this earlier),

σ2 + σ3 = σLO2

(1 +

αsπ

)(3.32)

Individually, σ2 and σ3 depend sensitivity on the exact criteria used to separate 2-jet and 3-jet events. One should be careful, there are “theory jets” defined in terms ofpartons and “experimental-jets” defined in terms of hadrons. We’re now talking about“theoretical jet algorithms”.

3.4 Jet Algorithms

Clealy defining jets is not a trivial task. However, to make the most out of the dataat colliders such as the LHC, we need to find algorithms to cluster partilces into jets.Several such algorithsm exist.

3.4. JET ALGORITHMS 45

3.4.1 Sterman-Weinberg Jets

At the time this algorithm was defined people only conceived of 2-3 jets so this methodis not really useful for modern colliders but it is nevertheless instructive.

j1

j2

An event is classified as a 2 jet event if at least a fraction of 1 − ε of the event’s totalenergy is contained in the two cones of opening half-angle δ. Algebraically,

E(within both cones) ≥ (1− ε)E(total) (3.33)

The idea is if you have a real three jet event, then a lot of energy will go into the thirdjet. So the energy within the first two cones will be larger then the fraction of the totalenergy.

The parameters here are δ and ε and can be anything but typically one should avoidusing the extreme values of δ too close to 0 or ε too close to 0 or 1.

We can make a prediction for how many two jet events we will have. Consider anevent with two quarks and a gluon. We denote the fraction of the momentum carriedby the quark and antiquark by x1 and x2 respectively and the momentum carried by thegluon by x3. Momentum conservation says that

x3 = 2− x1 − x2 < 1 (3.34)

so

x1 + x2 > 1 (3.35)

We have the top right triangle phase space we saw for three body phase space in x1 andx2. If x1 or x2 is close to 1 we have a singularity and a much larger probability to producea collinear gluon. In this case the Sterman Weinberg algorithm will see a two jet eventsince the gluon jet will be part of the quark jet. We have, [Q 5: insert simulated phasespace data]

46 CHAPTER 3. QCD

0 1

1

x1

x2 2 jet events

We cannot have events in the bottom left due to the result above. Recall that the crosssection is singular when x1, x2 → 1. In these regions we would interpret the events to betwo jet events.

The filled region is where the 2 jet events will lie. One can calculate the fraction ofevents that will have 2 jets. We must integrate over the jet energies that obey equation3.33. It is given by

f2 ≡σ2−jet

σtotal= 1− 8CF

αs2π

log

1

8

(log

(1

2ε− 1

)− 3

4+ 3ε

)+ (3.36)

π2

12− 7

16− ε+

3

2ε2 +O(δ2 log ε)

(3.37)

The particular result is not especially important but we recognize that at leading orderall events are 2 jets.

3.4.2 JADE Algorithm

The JADE algorithm defines a 3-jet event as

min (pi + pj)2 ≈ min 2EiEj (1− cos θij) > y · s (3.38)

with i, j = q, q, g. y is some dimensionless number quantifying how small this invariantmass should be before you call it a 2 jet event(recall that the invariant mass is zero if twoparticles are collinear). If none of the particles are collinear then the minimum invariantmass will be large, satisfying the JADE algorithm. The smaller the value of y the largeryour 3-jet rate is. As before one can calculate the fraction of 3-jet events,

f3 = CFαs2π

[2 log2 y

1− y+ (3− 6y) log

y

1− 2y+

5

2− 6y − 9

2y2 + 4Li2

y

1− y− π2

3

](3.39)

The key point is what you call a 2-jet event vs 3-jet event is a function of the algorithmas well as the parameter that describes your algorithm, in this case y. In principle youcan make y very small, but for very small y, f3 ≥ f2 and perturbation theory breaksdown (αs log2 y ∼ O(1)). These are not QCD innate non-perturbative divergences butcan be resummed and eliminated.

3.4. JET ALGORITHMS 47

1

y0.05 0.1

f2

f3

For multi-jets: Start with n partons; if

min(pi + pj)2 < ys (3.40)

cominbe i and j into a jet with pij = pi + pj and iterate. The interesting part about the4-jet rate is that it is the first order in which the 3 gluon vertex arises. Sample “righthalf” of 4-jet diagrams are,

This was the real direct evidence of the triplet-gluon vertex. This algorithm still has afew small issues since there are some combinations of partons which are designated asjets when they really shouldn’t be.

An important variation of the JADE algorithm is known as the kT algorithm. Hereyou compute (c.f. 3.38),

2min(E2i , E

2j

)(1 + cos θij) > ys (3.41)

For θij 1 we have 1− cos θij ≈ 12θ2ij we have

min(E2i , E

2j

)θ2ij = min(p2

i,⊥, p2j,⊥) (3.42)

where min(p2i,⊥, p

2j,⊥) is the transverse momentum of i relative to j (if i is the softer

particle). So what the algorithm does is group partons as different jets if they have havelarge p⊥ from one another.

There is something interesting you do to measure αs from f3. To second order in αs

f3 = f(y)f(αs(µ

2)

π) +

(αs(µ

2)

π

)2

·G(s

µ2, y) (3.43)

where G is some unknown second order correction to f3.

48 CHAPTER 3. QCD

But observables must be independent of renormalization scale,

df3

d log µ2= 0 (3.44)

This lets you find how G changes with renormalization scale.If µ2 ↑ then of course αs ↓. This means that G(µ

2

s) ↑ which implies that G is a

monotonically decreasing function of s. Looking at expression 3.43 we see that the firstorder correction has no dependence on the energy. Thus we also know that the fractionof 3 jet events will also go down as a function of s,

3.5 Event Shape Variables

We have some collection of hadron and measure their momenta pi. We don’t care howmany at this point. We want to find some function that we can measure experimentallyand calculate theoretically. Define

X ≡ f(pi) (3.45)

We want “IR safe” observables. These satisfy the following condition. Suppose we havepi → pj + pk,

pipj

pk

with pk = Apj (collinear) or |pk| |pj| (soft) then an IR safe observable obeys,

X → X +O(|pk|2

)(3.46)

In other words an IR safe observable isn’t strongly effected by collinear or soft particles.Its not hard to construct IR safe observables. This can be done by having linear

combinations of momenta. In that case small momenta do not effect the distribution andcollinear splittings drop out. One such variable is called the thrust,

T ≡ maxn

∑i |pi · n|∑i |pi|

(3.47)

where n is some unit vector. The maximum ranges over all possible unit vectors.To see that thrust is infrared safe, consider a particle emits a soft particle. This soft

particle would not show up in the computation of thrust since its momenta is small. Nowsuppose that a particle momenta splits such that p→ (1− λ)p, λp. In this case

|p · n| → (1− λ) |p · n|+ λ |p · n| = |p · n| (3.48)

and the same occurs in the denomenator. Thus emitted a collinear or soft particle hasno effect on the thrust making it IR safe.

Suppose you have two hadrons going back to back. Then you have,

3.6. PARTON BRANCHING AND SHOWERING 49

p2p1

which gives T = 1 (and the maximizing unit vector is along the direction of motion ofone of the particles).

The opposite case is a spherical distribution,

in this case any direction is equally as good at maximizing the sum and we have∑i |pi · n|∑i |pi|

=

∫d3p |p cos θ|∫d3p |p|

∑i |cos θi|∑

i 1(3.49)

=

∫ π0d cos θ

∫ 2π

0dφ |cos θ|

4π(3.50)

=

∫ π/20

d cos θ∫ 2π

0dφ cos θ

2π(3.51)

=1

2(3.52)

Hence the thurst tells you how spread out your distribution is with

1

2≥ T ≥ 1 (3.53)

Theory predicts to leading order (where we only have 2 jets) that T = 1 since thehadrons will just come out back to back. To next to leading order we have,

dσ

dT=

∫ ∫dx1dx2

dσqqgdx1dx2

δ (T − f xi) (3.54)

where f xi is some function of the energies for such a distribution. The reasoning forthis form is as follows. The dσqqg

dx1dx2δ (T − f xi) term is the contribution to the cross-

section with a given thrust, and we integrate over all possible xi’s. One can show thatfor 3 jets the thrust is, [Q 6: check]

f xi = max xi (3.55)

3.6 Parton Branching and Showering

We already saw that 3− jet/2− jet ratio is enhanced compared to the naive expectionO(1) · αs by soft/collinear logs. For example in the JADE algorithm,

f3

f2

∼ αsπ

log2 y (3.56)

50 CHAPTER 3. QCD

for small y. One approach is to just keep y large enough so that this is not a problemand we consistently have significantly more 2-jet events then 3-jet events. But this meansthrowing away a lot of information that’s available in the data and potentially useful.Another approach is to try to predict leading-log terms to all orders in perturbationtheory. This is possible due to factorization.

Calculating the full cross-section for different events can be very difficult for manyjets. This is possible for three jets but not going to be possible for more compilcated cases.However, we can still look at the more singular regions and get approximate answers. Welook at “Branching”:

θqs,a

We focus on collinear factorization, soft factorization can also be done but a bit morecomplicated. Thus we work in the θ → 0 limit.

M2→3 =∑s,a

Ms,a2→2

1

tMqs,a→qg (3.57)

where the invariant mass of the quark prior to branching is

t ≡ (pq + pg)2 = 2EqEg(1− cos θ) ≈ EqEgθ

2 (3.58)

We have t the initial invariant mass of the collision.

s is the spin of the virtual quark and a is its color. We have our energy fractiondefinitions,

xq ≡2Eq√t

, xg = 1− xq (3.59)

Now we want to square the ampltidue,

|M2→3|2 =1

t2

∑s1,s2,j,k

(∑s,a

Ms,a2→2M

(qs,a → qs1,jgs2,k

))×(∑

s′,b

Ms′,b2→2M

(qs′,b → qs1,jgs2,k

))∗(3.60)

In QED we argued that when a soft photon is emitted the polarization index does notchange, s1 = s = s′. This is still true here since QCD preserved parity. Now how do westudy the color structure? The color structure is just assocated with the color contributionat the splitting amplitude vertex, which contributes a

(T k)aj

and the complex conjugate


which contributes a(T k)bj,∑k

(T k)aj

(T k)bj

=δajδbj −

1

Nδajδ

bj (3.61)

=

(1− 1

N

)δajδ

bj (3.62)

∑j−−→(

1− 1

N

)δab (3.63)

[Q 7: How did we get this relation? (T k)aj(Tk) bj = C2(r)δab and I believe C2(r) = 3.‘]

Thus we have

|M2→3|2 =1

t2

∑s,a

|Ms,a2→2|

2∑s,s2,j,k

∣∣M(qs,i → qs,jgs2,k)∣∣2 (3.64)

The splitting contribution is∑s,s2,j,k

∣∣M(qs,i → qs,jgs2,k∣∣2 = t · 4g2 · C · F qg

s (xq) (3.65)

which gives

dσ3 ≈∑s

dσs2dt

tdxq

dφ

2π

αs2πCF qg

s (xq) (3.66)

Here we have three integration variables. The energy of the incoming particle, t, theenergy taken by the outgoing particle, and the angular distribution.

Then you can define the splitting function as

P sqq(xq) ≡

∫dφ

2πCF qg

s (xq) (3.67)

For unpolarized production ∑s

dσs2Psqq(xq) = dσ2Pqq(xq) (3.68)

where

Pqq ≡1

Npol

∑s

P sqq (3.69)

One can show that

Pqq(x) =4

3

1 + x2

1− x(3.70)

Up to a color factor, this the same splitting function for an electron to split into a photon.We have a probability that the quark will emit a gluon at some particle fraction of theenergy. The three particle cross-section is then given by,

dσ3 =dt

t· dxq ·

αs2πdσ2Pqq(xq) (3.71)

52 CHAPTER 3. QCD

The important point is that we have 2 singularities, one for small t and another forx → 1. So when you try to integrate these you get logs which can be large. So we needto include multiple splittings. However, this is easy to sum since you can just apply thisprocess again and again.

t0x0=1

t1x1

t2x2

t3x3

where remember here ti denotes the invariant mass going into splitting i and xi is thequarks energy fraction after splitting. xi and ti determine the kinematics completely. Theprocess in a collider is as follows. First we produce particles, then the particles shower.Once the energy of the new particles is below a certain threshold we have hadronization.

This factorization is possible because the time scale to emit an extra gluon is muchsmaller then the time of the process itself [Q 8: show].

In general you can have many gluon emissions. We don’t want to sum the seriesexactly, we just want to calculation the largest term. By energy conservation you musthave,

x0 > x1 > x2 > ... > xN (3.72)

We focus on the logN enhanced region. Remember that our whole calculation here as-sumed that the split quark was assumed to be a little bit off-shell. If you take this a littlemore off-shell and give it a small mass, then the more it increases the invariant mass ofthe initial quark, t. If you take it too much off-shell then you no longer have a singularregion. [Q 9: Understand this argument better and sharpen it.] So we have,

t0 t1 ... tN ≈ 1GeV (3.73)

We have some sort of “parton evolution” process. We can represent a “path” in t − xspace. While t is not time, it is a variable that’s monotonically changing. We can drawa plot of the evolution,

th ≈ 1GeVt0 ≈√s

0

1

← increasing t

xq


where th is the hadronization invariant mass. If you know this path then you know thatenergies of all your gluons and quarks emitted from the shower. Monte Carlo generatorscompute the Probability of each particles path,

P (ti, xi) (3.74)

Now there is a subtely with the value the initial invariant mass. The factorizationmethod discussed above only works if t

√s. However, in practice what people do is

fix the initial t such that t0 =√s. However, it turns out that this doesn’t matter since

the formula that we will see in a minute will gaurentee that we will see no splitting untilwe get to lower t values. The probability of splitting is essentially zero until t

√s. We

will see this effect soon.We now attempt to estimate this. Suppose you start at t0. We define the probability

of no branching until t1 by

p(t1) (3.75)

Suppose now we move from t→ t+ ∆t with ∆t < 0. Then

dp(t1) = p(t1)∆t

t

∫dxαs2πPqq(xq) (3.76)

This gives a very simple differential equation for p(t1). The solution to this equation is,

∆(t1) ≡ exp

[+

∫ t1

th

dt

t

∫dxαs2πPqq(x)

](3.77)

with dt < 0 and P (t1) = ∆(t0)/∆(t1). This is known as the Sudakov form factor.There are some complication associated with this showering algorithm.

1. There are other breakings. We only included q → qg but we can also have,

which adds Pqg, Pqg, and Pgq.

Monte Carlo must include relative possibilities of various splittings,

∆i(t1) = exp

[+

∫ t1

t0

dt

t

∫dxαs2π

∑j

Pji(x)

](3.78)

2. Soft singularities since ∫ 1

dxP (x) (3.79)

54 CHAPTER 3. QCD

is actually divergent! In the context of QCD is it actually clear what is happening.We have,

t = x(1− x)1− cos θ

2(3.80)

so even if you keep θ fixed, taking x → 1 takes t → 1. Emitting a very soft gluonthe quark that emits this gluon has to be off-shell by a very small amount. A verysoft gluon can only be a emitted by a very slightly offshell quark. But once a gluongets very soft it is unobservable due to hadronization. We have a natural cut-offfor our integral,

t > tmin (3.81)

or

x(1− x) >tmint

(3.82)

This is still not the end of the story since in our discussion we assumed that theemitted gluon is just collinear. However, there is also a soft singularity. These softsingularities need to be handeled seperately. We will not discuss this complicationfurther.

3. Coherence effects. Consider first QED. We can have,

γ

e+

e−

γ

θ

θee

E0

with an invariant mass of the emitting photon and electron of,

t ≈ EγEpEγθ2 ≈ zE2

0θ2 (3.83)

in the limit that z → 0 and θ → 0.

It turns out when θ < θee this cross section is heaviliy suppressed. This is calledthe Chudakov effect. We now show this in space-times terms. First how long doesit take for the electron to emit a photon? We can find this using the propagator,

1

t≈ 1

zE20θ

2≈ 1

p20

=1

E20 − p2

0

≈ 1

2E0 (E0 − |p0|)(3.84)

where we used the fact that E0 ≈ |p0|. Fourier transforming this propagator givesthe time between emission,

∆t ∼ 1

E0 − |p0|≈ 1

zE0θ2(3.85)


We can estimate the physical separation of e+e− at the moment of the photonemission,

∆b = ∆t · θee =θee

zE0θ2=

θeeEγθ2

(3.86)

If

pγ⊥ <1

∆b(3.87)

then the photon is radiated off e+ and e− add coherently. The probability ofemission is proportional to net charge which is zero. Suppressed at,

Eγθ <Eγθ

2

θee(3.88)

which implies that θ > θee.

In QCD we have many splittings. In every splitting we tend to have subsequentsplittings with small angles. We force “angular ordering” and the emission angle ineach subsequent splitting must be less than in the previous splitting.

Chapter 4

Deep Inelastic Scattering

p

P →X

γ,Z,W

e or νk → k′ →

p→ p′ →

e′ or ν ′

There are two possible types of momenta transfers: neutral and charged. For a neutralcurrent interaction we have one possible interaction: e−p → e−X. The charged currenthas two possibilities: e−p→ νX , νp→ e−X, where X can be “anything hadronic”.

The momentum transfer is q ≡ k − k′. Note that

q2 = −2k · k′ = −2EkEk′(1− cos θ) < 0 (4.1)

Since q2 is negative it is convenient to define Q2 = −q2 > 0. We work in the regime that

Q2 ΛQCD (4.2)

this is called the “parton picture” and the interaction occurs with one parton in theproton and the rest are spectators. In the cm frame

√s =

√(k + P )2 Mproton (4.3)

this implies that we can write,P ≈ Ep(1, 0, 0, 1) (4.4)

the partons havepz ≈ Ep (4.5)

and p⊥ ≈ ΛQCD. We ignore this value. The momenta of the quark is then given by

p = ξP (4.6)

56

57

where ξ ∈ (0, 1).The probability to find parton of flavor f and momentum fraction ξ is given by

ff (ξ)dξ (4.7)

ff are called the parton distribution functions (pdf).In this naive parton model we have,

σ(e−(k) + p(P )→ e−(k′) +X

)=

∫ 1

0

dξ∑f

ff (ξ)σ(e−(k) + qf (ξP )→ e−(k′) +X

)(4.8)

If we consider the parton level process,

q q

γ

ek → k′ →

p→ p′ →

e

Recall that1,

|M|2 = 3q2fe

2(1 + cos2 θ

)= 3q2

fe2 2(t2 + u2)

s2(4.9)

where we have used t = −12s(1 − cos θ), u = −1

2s (1 + cos θ). We can use crossing sym-

metry to find the s channel diagram:

|Ms−channel|2 =1

33q2fe

2 2(s2 + u2)

t2(4.10)

where, we had to be a little careful with the color factors. We can now use dt = 12sd cos θ

to rewrite these expressions:dσ

dt=

2πα2Q2f

s2

s2 + u2

t2(4.11)

where we define Qf ≡ qf/e. The Mandelstam variables are just

t = q2 = −Q2 (4.12)

s = (k + ξP )2 = ξ2k · P = ξs (4.13)

u = −s− t = Q2 − ξs (4.14)

since t = −12s(1− cos θ) it implies that

Q2 =1

2ξs (1− cos θ) ∈ [0, ξs] (4.15)

1We denote the parton level Mandelstam variables with a hat and the proton level variables withouta hat.

58 CHAPTER 4. DEEP INELASTIC SCATTERING

Since 1− cos θ ≤ 2 it implies

ξ ≥ Q2

s(4.16)

The physics of this is that if Q is very small then you impart very little energy to theelectron and there won’t be a large momentum fraction [Q 10: Check this!].

The cross-section is given by

dσ

dQ2=

2πα2s

Q4

∫ 1

Q2/s

dξ∑f

Q2fff (ξ)

[1 +

(1− Q2

ξs

)2]

(4.17)

Another useful observable is

x =Q2

2P · q(4.18)

We have,

p+ k = p′ + k′ ⇒ p+ q = p′ ⇒ (p+ q)2 = 0 (4.19)

This implies

2p · q = −q2 = Q2 (4.20)

so

x =2p · q2P · q

= ξ (4.21)

This “new” quantity isn’t really new at all but just the momentum fraction.

There is also one more quantity that is used in the literature denoted by

y ≡ 2P · qs

=Q2

xs∈ [0, 1] (4.22)

We have,

d2σ

dxdQ2=

2πα2s

Q4

∑f

Q2fff (x)

[1 +

(1− Q2

xs

)2]

(4.23)

d2σ

dxdy=

2πα2s

x2y2s

(∑f

Q2fff (x)

)(1 + (1− y)2

)(4.24)

The dependence of 1 + (1 − y)2 is called the Callan-Gross relation and is due to thefermionic nature of quarks. This type of experiment showed that quarks are fermions.

For the other processes one can show that,

59

d, u u, d

W+

ν e

dσ

dxdy=G2F s

π

[xfd + xfu(1− y)2

]

u, d d, u

W−

ν e+

dσ

dxdy=G2F s

π

[xfu(1− y)2 + xfd

]

We have several normalization “sum rules”:∫ 1

0

dxx∑i

fi(x) = 1 (4.25)

and ∫ 1

0

dx [fu(x)− fu(x)] = 2,

∫ 1

0

dx [fd(x)− fd(x)] = 1 (4.26)

These are the electric charge sum rules.This method has its limitations. In particular we can’t go down to arbitrary small x,

xmin =Q2

s(4.27)

Since Q2 ΛQCD is required to be able to factor short and long distance physics we needhigh energies to go down and measure pdf’s at small momenta fractions. At HERA forexample s = 320GeV and

xmin ∼ 10−5 − 10−4 (4.28)

Appendix A

Homework

A.1 Assignment 4

In this assignment we simulate e+e− → Jets with and without Pythia. The simulationis easy enough to run. We work at

√s = MZ . We begin by computing the number the

number of events with each number of jets. We find:

0 1 2 3 41

10

100

1000

104

Numberof Jets

Num

bero

fEve

nts

We then go on to measure the thrust for each event. To find the thrust we take the 3momentum of each event and take its dot product with a radial unit vector,

r = cos θ sinφ, sin θ sinφ, cosφ (A.1)

we use Mathematica to maximize

T =

∑i |pi · r|∑i |pi|

(A.2)

60

A.1. ASSIGNMENT 4 61

with respect to φ, θ. This is a computationally expensive process. Instead of doing thisfor all events we only do calculate for the first 3000 for Pythia and first 600 for withoutPythia. With Pythia we find:

0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

Thrust

Prob

abili

ty

and without:

0.6 0.7 0.8 0.9 1.00.0

0.2

0.4

0.6

0.8

Thrust

Prob

abili

ty

A.1.1 Strange Observations

• Both cases have a similar “lowest thrust” values of about 0.4 (below the allowedminimum for thrust). Possibly this has to do with particles not being detected asthey are emitted along the beam line, but I don’t quite get how.

62 APPENDIX A. HOMEWORK

A.1.2 x1 and x2 Distributions

The x1 and x2 distributions in the Pythia 3 jet events is

0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00.00

0.05

0.10

0.15

x Value

Prob

abili

ty

x1

x2

Bibliography

[1] J. Alwall et al. Madgraph 5: Going beyond. JHEP, 1106, 2011.

[2] H. Murayama. Notes on phase space. hitoshi.berkeley.edu/233B/phasespace.pdf, 2007.

63

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