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Topics in Standard Model

Alexey BoyarskyAutumn 2013

Course material

Books used in this course:

Bjorken-Drell: James D. Bjorken, Sidney David Drell Relativisticquantum mechanics, Vol. 1, McGraw-Hill, 1964

Aitchison-Hey: I.J.R. Aitchison, A.J.G. Hey, Gauge Theories inParticle Physics, Vol. 1: From Relativistic Quantum Mechanics toQED, 3rd edition

Cheng-Li: T.-P. Cheng and L.-F. Li, Gauge theory of elementaryparticle physics. Clarendon Press, Oxford, 1982.

Landau-Lifshitz-vol2: L.D. Landau, E.M. Lifshitz (1975). TheClassical Theory of Fields. Vol. 2 (4th ed.). Butterworth-Heinemann.

Landau-Lifshitz-vol3: L.D. Landau, E.M. Lifshitz (1977). Quantum

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 1

Course material

Mechanics: Non-Relativistic Theory. Vol. 3 (3rd ed.). PergamonPress.

Landau-Lifshitz-vol4: V.B. Berestetskii, E.M. Lifshitz, L.P. Pitaevskii(1982). Quantum Electrodynamics. Vol. 4 (2nd ed.). Butterworth-Heinemann

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Maxwell’s equations

¥ In 1860s Maxwell wrote a set of field equations describingelectromagnetism

div ~E = ρ (1)

div ~B = 0 (2)

curl ~E = −1c

∂ ~B

∂t(3)

curl ~B =1c

∂ ~E

∂t+ ~ (4)

we will always work in vacuum, so that ε = µ = 1 and as a result ~D = ~E and ~B = ~H! The

CGS system of units is used, so that ε0 = µ0 = 1 and dimensions of ~E and ~B are the same!

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Maxwell’s equations

¥ Check that as a consequence of Maxwell’s equations (1)–(4)

1. One can introduce functions A0 and ~A such that

~B = curl ~A

~E =− 1

c

∂ ~A

∂t− ~∇A0

(5)

and Eqs. (2)–(3) are satisfied identically2. Eqs. (5) define A0 and ~A up to an arbitrary scalar function λ(t, ~x) such that

changing~A→ ~A+ ~∇λ

A0 → A0 −1

c

∂λ

∂t

(6)

does not change the l.h.s. of Eq. (5)3. Derive equations on ~A and ~A0 based on Eqs. (1) and (4)4. Show that current should be conserved ∂ρ

∂t + div~ = 0

5. Show that in a space without sources ~E and ~B obey wave equation:

1

c2∂2 ~E

∂t2−∆~E = 0 (7)

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Maxwell’s equations

¥ Maxwell has predicted that in the empty space electric andmagnetic fields can exist – electromagnetic waves.

¥ Equation (7) means that the electromagnetic wave propagatesthrough the space with the speed of light c.

¥ Hertz had discovered radio-waves and confirmed this prediction

¥ Maxwell’s discovery meant that there is no “action-at-a-distance”but rather two electric (magnetic) bodies interact via emitting andreceiving electromagnetic waves

¥ Maxwell’s equations possess Lorentz symmetry .1 To see this,define 4-vector Aµ = (A0, ~A) (based on (5)) and define the fieldstrength

Fµν ≡ ∂Aµ∂xν

− ∂Aν∂xµ

(8)1To remind yourself about 4-vectors, Lorentz transformations and Lorentz covariant formulation of

electrodynamics, see Landau & Lifshitz, vol. 2, §§4–7, §§16–17, §23

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Maxwell’s equations

where xµ = (ct, ~x).

¥ Find the relation between Fµν and ~E, ~B so that (8) is equivalent to (5)

¥ The Maxwell’s equations take the form

∂Fµν

∂xµ= jν (9)

where jµ = (ρ,~) – 4-vector of electromagnetic current and Fµν =ηµαηνβFαβ where ηαβ is the Minkowski metric

¥ Lorentz symmetry is a rotation in the 4-dimensional space (givenby the 4×4 antisymmetric matrix Λµν) so that any vector translatesas

Xµ = ΛµνXν (10)

and field strength Fµν translates as

Fµν = ΛµαΛνβF

αβ (11)

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 6

Maxwell’s equations

¥ Show that (9) is equivalent to (1) and (4)

¥ Show that Eq. (9) with jµ = 0 can be rewritten as 4 wave equations for 4components of Aµ, supplemented by the Lorentz gauge condition ∂µAµ = 0

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 7

Photons

¥ To explain some experiments (black body radiation; photoelectriceffect) one was led to assume that the electromagnetic radiation isquantized and that a “quantum of light” – photon – exists (Einstein

1905)

¥ This has been confirmed by Compton in 1923 (Comptonscattering)

`

λ′ − λ =h

mec(1− cos θ)

– X-rays, scattering of atomicelectrons change wave length

– Naturally explained by theelastic scattering of two particles(electron and photon)

– Wave would not change itsfrequency (Thomson scattering)

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Quantum Mechanics

¥ The atomic physics led to the creation of Quantum mechanics

¥ The quantum system is described by Schrodinger equation:

i~∂ψ(x, t)∂t

= Hψ(x, t) (12)

where the operator H is called Hamiltonian

¥ The operator Hamiltonian is built based on correspondenceprinciple: if a classical system is described by the HamiltonianH(x, p) then the quantum system is described by H(x,−i~∂x). Forexample for a particle of mass m in external field we get

Classical H(x, p) =p2

2m+ V (x) ⇒ Quantum H = − ~

2

2m∂2

∂x2+ V (x)

(13)

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Solution of Schodinger equation

¥ Recall some important facts about the Schrodinger equation:2

– Consider

0 =

Zd

3x

»ψ∗“i~∂ψ

∂t−Hψ

”− ψ

“−i~∂ψ

∗

∂t−Hψ

∗”–

⇔ ∂

∂t

Zd

3x|ψ(x)|2 = 0

(14)

– |ψ(x)|2 is interpreted as probability density and∫d3x|ψ(x)|2 =

const as the full probability– ~J = i~

2m(ψ∗∇ψ − ψ∇ψ∗) is interpreted as a probability densitycurrent so that

∂|ψ|2∂t

+ div ~J = 0 (15)

Check this relation– Uncertainty principle ∆x∆p ∼ ~

2To remind yourself about this material, see e.g. Landau & Lifshitz, Vol. 3, §19

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 10

Solution of Schodinger equation

– Solution of the Schodinger equation for the hydrogen atom waspredicting the energy spectrum

En = −mee4

2~2n2, n = 1, 2, . . . (16)

for each n the degeneracy (the number of states) of the energylevel is

∑n−1l=0 (2l + 1) = n2.

This latter fact contradicted to the observations as the lowest energy level ofHydrogen contained two electrons

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Pauli Hamiltonian

¥ In order to explain the emission spectra of elements Pauliintroduced a spin (“a two-valued quantum degree of freedom” ).Its interpretation was not known. Pauli postulated Pauli (or Pauli-Schrodinger) equation

HPauli =1

2m

[~σ · (−i~~∇− e ~A

)]2

+ eA01 (17)

where Pauli matrices ~σ = (σx, σy, σz) are defined as

σx =(

0 11 0

), σy =

(0 −ii 0

), σz =

(1 00 −1

)(18)

and 1 is 2 × 2 identity matrix. Pauli Hamiltonian acts on the 2-component wave-function

i~∂

∂t

(ψ↑ψ↓

)= HPauli

(ψ↑ψ↓

)(19)

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Pauli matrices

Show that

– Pauli matrices are Hermitian– For each Pauli matrix σ2

i = 1– For any two Pauli matrices

{σi, σj} ≡ σiσj + σjσi = 2δij1 i, j = 1, 2, 3 (20)

– σxσy = iσz and in general

[σi, σj] = 2iεijkσk (21)

– As a consequence of (20) and (21) one finds

(~σ · ~X)(~σ · ~Y ) = ~X · ~Y 1 + i~σ · ( ~X × ~Y ) (22)

– Use the identity (22) to rewrite the Pauli Hamiltonian as

HPauli =1

2m(−i~~∇− e ~A)

2 − ~e2m

~σ · ~B + eA01 (23)

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Pauli matrices

– Compute the matrix U = ei2ασz (as the Taylor expansion of the exponent) and

show that matrix U is unitary. Try to repeat this for arbitrary ei2~α·~σ

– Show that any 2× 2 unitary matrix U with detU = 1 can be represented as

U = exp“ i2(n0 + ~n · σ)

”; n

20 + ~n

2= 1 (24)

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Quantum mechanics and relativity

Attempts to blend quantum mechanics and special relativity led to theproblem of infinities.3

¥ Electromagnetic energy of uniformly charged ball of radius reshould be smaller than the total mass of electron (mec

2)?

mec2 >

e2

re=⇒ re >

e2

mc2= 3× 10−13 cm (25)

– larger than the size of atomic nucleus (Hydrogen rH ∼ 1.75 ×10−13 m)

¥ Pauli’s idea of spin meant that electron possesses magneticmoment: µe = e~

2mec. The energy of magnetic sphere with radius

re would be µ2er3e

which would exceed mec2 for r ∼ 1

me

(e~c4

)2/3(even

larger distances than (25!)3See an interesting exposition of the historical perspectives at

http://people.bu.edu/gorelik/cGh_Bronstein_UFN-200510_Engl.htm, Sec. 4

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 15

Relativistic Schodinger equation 4

¥ Free Schrodinger equation is not Lorentz invariant (indeed, it isobtained from E = p2

2m dispersion relation)

¥ Replace r.h.s. by relativistic dispersion relation E =√p2c2 +m2c4

Relativistic Schrodinger equation ? i~∂ψ

∂t=

√−~2~∇2 +m2c4 ψ (26)

¥ How to make sense of square root? Try to take square of Eq. (26)?

− ~2∂2ψ

∂t2=

(−~2c2∇2 +m2c4

)ψ ⇔

(¤ +

(mc~

)2)ψ = 0 (27)

This is Klein-Gordon equation. Here ¤ = 1c2∂2

∂t2− ~∇2 is the

D’Alembert operator. Show that ¤ is Lorentz invariant

3The presentation of this topic follows Bjorken & Drell, Chap. 1, Sec. 1.1–1.3

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 16

Problems with Klein-Gordon equation

¥ By taking square of (26) we have included negative energy states(with E = −

√p2c2 +m2c4)

¥ Probabilistic interpretation is gone? Indeed, let us checkprobability density |ψ|2:

∂

∂t

∫d3x |ψ(x)|2 6= 0 (28)

¥ Repeating trick similar to Eq. (14) we find that the following currentis conserved as a consequence of Klein-Gordon equation:

Jµ = ψ∗∂ψ

∂xµ− ψ

∂ψ∗

∂xµ(29)

that is1c

∂J0

∂t+ div ~J = 0 (30)

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Problems with Klein-Gordon equation

¥ Can we use J0 as the probability density? No!

J0 = ψ∗∂ψ

∂t− ψ

∂ψ∗

∂t= |ψ|2∂(argψ)

∂t< 0 (31)

for any negative energy solution

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 18

Dirac equation 5

¥ Dirac (1928) had suggested that one needs linear in time evolution.Indeed, if i~∂tψ = Hψ than the probability density of the form (28)is conserved for any Hermitian Hamiltonian (c.f. (14)).

¥ Dirac proposed the form

i~∂ψ

∂t=

[−i~c

(αx

∂

∂x+ αy

∂

∂y+ αz

∂

∂z

)+ βmc2

]

︸ ︷︷ ︸Dirac Hamiltonian

ψ (32)

¥ Take square of Dirac Hamiltonian. We should arrive to the Klein-Gordon equation. This means that for i, j = 1, 2, 3

αiαj + αjαi = 2δij (33)

αiβ + βαi = 0 (34)4The presentation of this topic similar to that of Bjorken & Drell, Chap. 1, Sec. 1.1–1.3

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 19

Dirac equation 7

and additionally β2 = 1

¥ ⇒Both αi nor β are not ordinary numbers (they do not commute).

¥ Eq. (33) looks very much like Eq. (20). But guess that αi = σi iswrong there are only 3 Pauli matrices and we need 4

¥ Another guess:6

αi =(

0 σiσi 0

); β =

(1 00 −1

)(35)

Show that properties (33)–(34) are satisfied. Find different representation ofαi, β

6Show that αi, β are even-dimensional matrices

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 20

Properties of Dirac gamma-matrices 8

¥ Define 4 γ-matrices (or Dirac gamma-matrices):

γ0 = β; γi = βαi (36)

¥ Define 4-vector γµ = (γ0, γ1, γ2, γ3)

¥ Dirac equation can be rewritten as

i~γµ∂ψ

∂xµ= 0 (37)

– explicitly Lorentz-covariant

¥ Properties of γ-matrices:

γµγν + γνγµ = 2ηµν1 (38)7See Bjorken & Drell, Chap. 2, Sec. 2.1–2.2

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 21

Properties of Dirac gamma-matrices 9

¥ Defineσµν =

i

2[γµ, γν] (39)

¥ Show that [σµν, σλρ] commute as operators of Lorentz rotation in3 + 1

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Standard Model: great success of particle physics

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 23

Standard Model

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Standard Model

¥ The Standard Model describes all the confirmed data obtainedusing particle accelerators and has enabled many successfultheoretical predictions.

¥ It has been tested with amazing accuracy, and its calculablequantum corrections play an essential role. Its only missing featureis a particle, called the Higgs boson, whose coupling to the otherparticles is believed to generate their masses.

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Standard Model and particle astrophysics

The SM has also been successful in explaining phenomena in theEarly Universe and certain astrophysical systems.

¥ It plays an essential role in the epoch of Big Bang nucleosynthesis– observed abundances of elements are sensitive to details of theStandard Model.

¥ Energy production of main-sequence stars is controlled by nuclearphysics and weak β decay.

¥ Weak neutral-current processes are important for neutrino emissionfrom the collapsing core and ejection of outer layers of thesupernova progenitor.

¥ Propagation of cosmic rays through the Universe.

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 26

End of Lecture 1

End of Lecture I

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