Peskin and Schroeder: Intro to QFT (Solutions)

Daniel Davies

UC Santa Cruz

Abstract

This document represents my attempt to learn basic quantum field theory in the summer of 2017,before having taken the class with Michael Dine in the fall. The tentative goal is to work throughthe first 7 chapters of Peskin and Schroeder, as this is what Michael attempts to cover in the firstquarter. Ultimately - though it may take well over a year - I would like this document to includemy solution to every problem in the book and be available online for future students to learn from.I consistently find points of nuance when solving these problems, as well as things that I believe theauthors should have said more explicitly.

1 Chapter 1

There are no problems in chapter 1

2 Chapter 2: Quantization of Scalar Fields

2.1 Classical Electromagnetism

The action of electrodynamics free of sources is given by

S[A] = −1

4

∫d4xFµνF

µν

Where Fµν = ∂µAν − ∂νAµ.

(a)The Euler Lagrange Equations (ELE) read:

0 = ∂µδS

δ(∂µAν)

0 =1

4

∫d4x∂µF

µν

which implies 0 = ∂µFµν

With the identification Ei = −F 0i and εijkBk = −F ij we see that the ELE is equivalent to:

∇ · E = 0

∇×B = 0

The remaining Maxwell equations are obtained via the Bianchi identity:∂[µFνσ] = 0, which is most easily

understood in the language of differential forms (since F = dA and d2 = 0).

1

(b) An energy momentum tensor that satisfies the relations (2.17) in P&S would be:

Tµν = −Fµσ∂νAσ +1

4ηµνFσρF

σρ

which, as we expected, is not a symmetric tensor. If we add the term ∂λ(FµλAν) the result is clearlysymmetric. Note the term:

Fµλ∂λAν − Fµσ∂νAσ

By renaming indices in the second portion (σ → λ) we note that this term is equivalent to:

FµλFλν =1

2εµλαβFαβFλν

which is now clearly symmetric under the exchange of µ and ν. The purely temporal component of thistensor is therefore nonzero:

T 00 = F 0λFλ0 +

1

4η0

0FσρFσρ

= E2 +1

42(B2 − E2)

=1

2(E2 +B2)

and similarly:

T i0 = F iλFλ0 +1

4ηi0FσρF

σρ

= −εiλσBσEλ= (E ×B)i

2.2 The Complex Scalar Field

The action of a free complex scalar field is given by

S =

∫d4x(∂µφ

∗∂µφ−m2φ∗φ)

(a) The conjugate momenta are given by

π = ∂0φ∗

π∗ = ∂0φ

and thus the Hamiltonian is

H =

∫d3x(π∗π +∇φ∗ · ∇φ+m2φ∗φ)

To compute the equations of motion for φ(x) and φ∗(x) we need to first compute the canonical commu-tation relations. This can be done by recognizing that the fields can be written

2

φ = φ1 + iφ2

φ∗ = φ1 − iφ2

for real fields φ1 and φ2. If ap, a†p and bp, b†p form the operator basis for the respective real fields,then it is clear the complex fields (and their conjugate momenta) have the form

φ(x) =1

(2π)3

∫d3p

1√2ωp

(αp + β†−p

)eip·x

φ∗(x) =1

(2π)3

∫d3p

1√2ωp

(βp + α†−p

)eip·x

π(x) =−i

(2π)3

∫d3p

√ωp2

(βp − α†−p

)eip·x

π∗(x) =−i

(2π)3

∫d3p

√ωp2

(αp − β†−p

)eip·x

for αp = ap + ibp and βp = ap − ibp. Since by construction, commutators between the two real fieldsvanish, we immediately find (up to normalization)

[φ(x), π(y)] = iδ(3)(x− y)

[φ∗(x), π∗(y)] = iδ(3)(x− y)

Now, using the Heisenberg equations of motion:

i∂0φ(x) = [φ(x), H]

=

∫d3y[φ(x), π∗(y)π(y) +∇φ∗(y) · ∇φ(y) +m2φ∗(y)φ(y)]

=

∫d3y(iπ∗(y)δ(3)(x− y))

= iπ∗(x)

i∂0π∗(x) = [π∗(x), H]

=

∫d3y[π∗(x), π∗(y)π(y) +∇φ∗(y) · ∇φ(y) +m2φ∗(y)φ(y)]

= −i∫d3y(∇(δ(3)(x− y)) · ∇φ(y) +m2φ(y)δ(3)(x− y))

= i(∇2φ(x)−m2φ(x))

where the step from the second-to-last line to the last line was done by integrating the differentialoperators by parts and evaluating the delta function. Hence as seen in the text, this implies the Klein-Gordon equation:

(∂µ∂µ +m2)φ(x) = 0

(b) Inserting the field definitions into the Hamiltonian yields

H =1

2(2π)6

∫d3x

∫d3p

∫d3q

(−√ωpωq(αp − β†−p)(βq − α

†−q) +

m2 − p · q√ωpωq

(βq + α†−q)(αp + β†−p)

)ei(p+q)·x

3

Integrating over position space yields a factor of (2π)3δ(3)(p + q), and following by integrating overq-space will send q −→ −p:

H =1

2(2π)3

∫d3p

(−ωp(αp − β†−p)(β−p − α†p) +

m2 + p · pωp

(β−p + α†p)(αp + β†−p)

)=

1

(2π)3

∫d3p(α†pαp + β†pβp)ωp

Hence the theory contains two sets of particles, both of mass m. This is evident when the vacuumexpectation value (vev) of the Hamiltonian is computed. The orthogonality conditions of the creationand annihilation operators will evaluate the integrand at zero momentum: ω0 = m.

(c) The conserved charge generated by a global U(1) symmetry of the theory is given by:

Q =i

2

∫d3x(φ∗π∗ − πφ)

Doing the same as for part (b), we find

Q =1

4(2π)6

∫d3x

∫d3p

∫d3q

√ωqωp

((βp + α†−p)(αq − β

†−q)− (βq − α†−q)(αp + β†−p)

)ei(p+q)·x

=1

4(2π)3

∫d3p

((βp + α†−p)(α−p − β†p)− (β−p − α†p)(αp + β†−p)

)=

1

2(2π)3

∫d3p

(α†pαp − β†pβp

)Hence the two particles have charge of equal magnitude but opposite sign. The eigenstates generatedby the β operators are the antiparticles of those states generated by the α operators. In analogy of thesingle particle harmonic oscillator, we recognize the integrand of Q as the operator which counts thestate; in our case of a quantized scalar field, this operator counts the number of quanta. In the contextof Noether’s theorem, we might conclude that a global U(1) symmetry implies conservation of particlenumber. Thus if Q is to be conserved, the creation of a quanta of this field must always come with thecreation of its antiparticle.

(d) We recall that a symmetry is any transformation of the field and its derivatives which leaves theextrema of the action (or equivalently the Euler-Lagrange equations) invariant. Simply stated, theLagrangian must transform at most by a total divergence:

L −→ L+ α∂µJ µ

This implies the existence of a conserved current vector

jµ =∑n

δLδ(∂µφn)

∆φn − J µ

Now consider the case of a free scalar field theory containing n identical complex fields, which we willlabel with the vector Φ:

L = ∂µΦ†∂µΦ−m2Φ†Φ

4

The set of fields Φ transforms under the defining representation of SU(n):

Φ −→ eiθaTa

Φ

Φ† −→ Φ†e−iθaTa

where the T a’s are the traceless, Hermitian n × n matrices that form a basis of the Lie algebra su(n).The θa’s are finite real numbers which parametrize the rotation in the n-dimensional space. For thepurpose of this problem we will take them to be infinitesimal. Since they do not depend on spacetimecoordinates, the Lagrangian is invariant under the action of this group. Hence J µ is zero. We then have

jµa = ∂µΦ†∆Φa + ∆Φ†a∂µΦ

Given infinitesimal θa

∆Φa = iT aΦ

∆Φ†a = −iΦ†T a

thusjµa = i

(∂µΦ†T aΦ−Φ†T a∂µΦ

)or in component form:

jµa = i (∂µφ∗i (Ta)ijφj − φ∗i (T a)ij∂

µφj)

and Qa = i

∫d3x

(πi(T

a)ijφj − φ∗i (T a)ijπ∗j

)Which, up to sign convention and the factor of 1/2 is the expression we were looking for. Note that for nscalar fields, there are n2−1 conserved charges associated with the global SU(n) symmetry; one for eachgenerator of the Lie algebra. For a pair of fields, the T a’s are replaced with the three Pauli sigma matrices.

The generalization of part (c) becomes:

Q =i

2

∫d3x

(Φ† ·Π† −Π ·Φ

)which we associate with a transformation of the form eiα ⊗ Idn×n

Exercise: Consider the case of finite θa

2.3 Two-Point Correlation Function

The amplitude for the quanta of a real scalar field to propagate from y to x is given by

D(x− y) = 〈0|φ(x)φ(y) |0〉 =

∫d3p

(2π)3

1

2Epe−ip·(x−y)

For this problem, we consider points that are space-like separated so that (x − y)2 = −r2. First weconvert to spherical coordinates and define (x− y) = ∆x:

5

D(x− y) =1

(2π)3

∫ 2π

0

dφp

∫ 1

−1

d cos(θ)

∫ ∞0

dpp2

2Epe−ip∆x cos(θ)

=1

8π2

∫ 1

−1

d cos(θ)

∫ ∞0

dpp2√

p2 +m2e−ip∆x cos(θ)

= − 1

4π2

∫ ∞0

dpp√

p2 +m2

sin (p∆x)

∆x

Shamelessly citing the Wikipedia page for Bessel functions, we note that the modified Bessel functionsKn(x) satisfy:

Kn(x) =π

2

inJ−n(ix)− i−nJn(ix)

sin(nπ)for some half-integer n

K0(ω) =

∫ ∞0

dtcos(tω)√t2 + 1

we recognize that by using t = p/m and ω = m∆x

D(x− y) = − m

4π2∆x

d

dωK0(ω)

∣∣∣ω=m∆x

So finally we have

D(x− y) =m

4π2|x− y|K1(m|x− y|)

3 Chapter 3: Quantization of Spinor Fields

3.1 The Lorentz Group

The commutation relations for the Lie algebra of the Lorentz Group SO(1,3) are the following:

[Jµν , Jρσ] = i(gνρJµσ − gµρJνσ − gνσJµρ + gµσJνρ)

(a) For generators defined by Li = 12εijkJ

jk and Ki = J0i infinitesimal Lorentz transformations can bewritten:

Φ −→ (1− iθiLi − iβiKi)Φ

Explicitly:

[Li, Lj ] = iεijkLk

[Ki,Kj ] = −iεijkLk[Li,Kj ] = iεijkKk

It is trivial to show that the linear combinations A± = 1√2(L± iK) satisfy:

[A±i , A±j ] = iεijkA

±k

[A±i , A∓j ] = 0

6

Hence so(1, 3)C ∼= sl(2,C) ⊕ sl(2,C). Many texts incorrectly state that the Lorentz algebra is isomor-phic to the direct sum of the su(2) Lie algebras. What they forget is that we are considering complexlinear combinations of the basis vectors, and hence are dealing with the complexified algebras. By theUnitarian trick developed by Hurwitz and Weyl, we may get away with this by considering holomorphicrepresentations of these algebras and the compact subgroups which form the image of the exponentialmap. For example, the complexified Lorentz group contains SU(2) × SU(2) as a compact subgroup,and so it is justified to construct representations of SO(1,3) from tensor product representations of realSU(2) representations.

(b) We may define a representation of the lie algebra so(1, 3)C by the following:

π(n,m)(Li) = J(n)i ⊗ Idm + Idn ⊗ J (m)

i

π(n,m)(Ki) = i(J(n)i ⊗ Idm − Idn ⊗ J (m)

i )

Where the J (s) are the corresponding generators of the (2s+1 dimensional) spin-s representation of su(2).One can check that the π(n,m)’s obey the previous commutation relations. Additionally, the Casimir op-erators of the representation correspond to a field with angular momentum n+m.

The ( 12 , 0) representation is the following:

Li 7→σi2

Ki 7→ iσi2

so a typical Lorentz transform becomes (by exponentiating the π(1/2,0)):

Λ(θ, β) = exp

(iθ − β

2· σ)

For the (0, 12 ) representation, the K’s change sign and thus the negative sign next to the β turns positive.

In the infinitesimal case, these transformations are exactly the ones described in the text up to a signconvention in equation (3.37).

(c) From equation (3.37) we know the infinitesimal Lorentz transformations on the Weyl spinors are

ψR −→(

1− iθ · σ2

+ β · σ2

)ψR

ψL −→(

1− iθ · σ2− β · σ

2

)ψL

If we define ψ′ = ψTLσ2 and the identity σ2σσ2 = −σT , we arrive at the transformation rule

ψ′ −→ ψ′(

1 + iθ · σ2

+ β · σ2

)Now let’s define the matrix

Vµ = V0 × Id2×2 + V · σ = ψR ⊗ ψ′

7

then Vµ transforms like

Vµ −→(

1− iθ · σ2

+ β · σ2

)Vµ

(1 + iθ · σ

2+ β · σ

2

)= Vµ + i

θ

2· [Vµ,σ] +

β

2· Vµ,σ

= Vµ + σ · (V × θ) + V0(β · σ) + V · β

From this we identify V ′0 = V0 + V · β and V ′ = V + (V × θ) + V0β , which is clearly the effect of aninfinitesimal boost and rotation on a four-vector. The statement we make here is that the representation(

1

2, 0

)⊗(

0,1

2

)=

(1

2,

1

2

)is the defining representation of the Lorentz algebra.

3.2 The Gordon Identity

The function u(p) is a solution to the Dirac equation:

(iγµ∂µ −m)u(p)eip·x = 0

so that(γµpµ −m)u(p) = 0

andu(p′)(γµp′µ −m) = 0

To derive the Gordon identity, consider the following product:

u(p′)iσµν(p′ν − pν)u(p)

We use the fact that −iσµν is the antisymmetric part of the tensor product of two gamma matrices(whereas the symmetric part is the metric ηµν to write

u(p′)iσµν(p′ν − pν)u(p) = u(p′)[(ηµν − γµγν)(p′ν − pν)

]u(p)

= u(p′)[(p′ − p)µ + γµ/p− γνγµp′ν

]u(p)

= u(p′)[(p′ − p)µ + γµ/p+ /p

′γµ − 2p′µ]u(p)

= u(p′)[γµ/p+ /p

′γµ − (p+ p′)µ]u(p)

= u(p′)[γµ2m− (p+ p′)µ

]u(p)

It was important that we do the manipulation to get the slashed operators on the correct side of the γµ,otherwise there would be no way to write things in terms of the mass. This result implies, with a littlerearrangement:

u(p′)γµu(p) = u(p′)

((p+ p′)µ

2m+iσµν(p′ − p)ν

2m

)u(p)

Which is Gordon’s Identity, that we were aiming to prove.

8

3.3 Spinor Products

(a) The value of /k0uR0 can be explicitly computed:

/k0uR0 = /k0 /k1uL0

= (k0k1)µνγµγνuL0

= (k0k1)µν(ηµν − iσµν)uL0

=(k0 · k1 − i(k0k1)µνσ

µν)uL0

= 0

The first term vanishes by definition, and the second term vanishes because we are contracting anantisymmetric tensor σµν with a symmetric one (k0k1)µν .

Similarly, from the definition of uL/R(p):

uL/R(p) =1√

2p · k0/puR/L(0)

we see that

/puL/R(p) =1√

2p · k0/p/puR/L(0)

=1√

2p · k0

pµpνγµγνuR/L(0)

=1√

2p · k0

pµpνηµνuR/L(0)

=p · p√2p · k0

uR/L(0)

which vanishes identically for lightlike momentum (p2 = 0).

(b) From equation (3.59) of the text we have

uL(0) =

(√k0 · σξ√k0 · σξ

)=√E

(√1 + σ3ξ√1− σ3ξ

)

=√

2E

ξ000ξ1

To make this a true left-handed spinor, we take ξ0 = 1 and ξ1 = 0 so that

uL(0) =√

2E

1000

And from uR(0) = /k1uL(0) = −γ1uL(0) we have

9

uL(0) =√

2E

0001

From here, it is just tedious matrix multiplication to find

uL(p) =1√

p0 + p3

−p1 + ip2

p0 + p3

00

and uR(p) =1√

p0 + p3

00

p0 + p3

p1 + ip2

3.4 Majorana Fermions

(a) We wish to show that the equation

iσµ∂µχ− imσ2χ∗ = 0

is Lorentz invariant and implies the Klein-Gordon equation. We start by recalling the transformationrule for the left-handed spinor χ(x):

χ(x) −→ Λ( 12 ,0)χ(Λ−1x)

So we have

iσµ∂µχ(x) −→iσµ∂µΛ( 12 ,0)χ(Λ−1x)

= iσµΛ( 12 ,0)(Λ

−1)νµ(∂νχ)(Λ−1x)

= iΛ(0, 12 )(Λ)µρ σρ(Λ−1)νµ(∂νχ)(Λ−1x)

= iΛ(0, 12 )σν(∂νχ)(Λ−1x)

where from the second to third line we have used the transformation property (3.29) for vectors (orequivalently, problem 3.1c):

Λ−1(0, 12 )

V µΛ( 12 ,0) = ΛµνV

ν

One must be careful though when reading the text, as in equation (3.29), Peskin and Schroeder do notdistinguish the spinor transformations on either side. They omit the zeros, adding confusion to whichtransformation is which.

Now, we see how the conjugate field transforms:

χ∗(x) −→ Λ∗( 12 ,0)χ

∗(Λ−1x)

Furthermore, we make use of the identity σ2σσ2 = −σ∗ to show that

σ2Λ∗( 12 ,0) = Λ(0, 12 )σ2

10

(This can be seen by simply examining the taylor series of Λ∗( 12 ,0)

). Hence

−imσ2χ∗(x) −→ −imΛ(0, 12 )σ2χ

∗(Λ−1x)

and we see, when combined with the previous result, that the field equation is Lorentz invariant.

To show that it implies the Klein-Gordon equation, conjugate:

σµ∂µχ = mσ2χ∗ and σ∗ν∂νχ

∗ = −mσ2χ

By taking the second derivative of the left equation and multiplying by σ∗ν then rearranging, we find

σ2σ∗νσ2σ

µ∂µ∂νχ+m2χ = 0

Noting that σν = (Id,−σ) and σν = (Id,σ) we see immediately that σ2σ∗νσ2 = σν

By explicit calculation, one can show that σ(ν σµ) = ηνµ, and thus the components of the spinor obeythe Klein-Gordon equation:

(∂µ∂µ +m2)χ = 0

(b) The action of the field is given by

S =

∫d4x[χ†iσµ∂µχ+

im

2

(χTσ2χ− χ†σ2χ

∗) ]To see this is real, first consider the term

∫d4xχ†iσµ∂µχ

Under conjugation, this becomes

∫d4x∂µχ

†(−iσµ)χ

And after integrating by parts (the field vanishes at infinity):

= −∫d4xχ†(−iσµ)∂µχ

which is identical to the term we started with. The Lagrangian density may not be real, but the actionis. Next we consider the adjoint of the second term:[ im

2

(χTσ2χ− χ†σ2χ

∗) ]† = − im2

(χ†σ†2(χT )† − (χ∗)†σ†2(χ†)†

)= − im

2

(χ†σ2χ

∗ − χTσ2χ)

=im

2

(χTσ2χ− χ†σ2χ

∗)Hence the action is real. If we now vary the action with respect to χ† we find

δS

δχ†= 0⇒ iσµ∂µχ− imσ2χ

∗ = 0

11

Which is the field equation we have been using. (Note: the term χTσ2χ can be equivalently written(−χ†σ2χ

∗)∗ and when differentiated, is identical to the other mass term, ridding us of the factor of 2).

(c) Defining the Dirac spinor by

ψ =

(χ1

iσ2χ∗2

)We have the Lagrangian:

L =(χ†1 −iχT2 σ2

)(0 11 0

)(−m iσµ∂µiσµ∂µ −m

)(χ1

iσ2χ∗2

)=(χ†1 −iχT2 σ2

)(iσµ∂µ −m−m iσµ∂µ

)(χ1

iσ2χ∗2

)=(χ†1 −iχT2 σ2

)(iσµ∂µχ1 − imσ2χ∗2

−mχ1 − σµσ2∂µχ∗2

)= iχ†1σ

µ∂µχ1 + iχ†2σµ∂µχ2 + im(χT2 σ2χ1 − χ†1σ2χ

∗2)

which includes the same kinetic terms as before (but now for each field) and a mass term that mixes thefields.

(d) The Dirac Lagrangian exhibits a global U(1) symmetry with Noether current jµ = ψγµψ whichsplits into

jµ = χ†1σµχ1 − χ†2σµχ2

Now computing the divergence of this current:

∂µjµ = χ†1σ

µ∂µχ1 + (∂µχ†1)σµχ1 − χ†2σµ∂µχ2 − (∂µχ

†2)σµχ2

= χ†1σµ∂µχ1 + (χ†1σ

µ∂µχ1)† − χ†2σµ∂µχ2 − (χ†2σµ∂µχ2)†

= m(χ†1σ2χ

∗2 + (χ†1σ2χ

∗2)† − χ†2σ2χ

∗1 − (χ†2σ2χ

∗1)†)

= m(χ†1σ2χ

∗2 + χ1σ2χ

T2 − χ

†2σ2χ

∗1 − χ2σ2χ

T1

)= 0

where we have used the identities σµ∂µχ1 = mσ2χ∗2 and σµ∂µχ2 = mσ2χ

∗1 from the Dirac equation. The

final step is immediate when we recognize that the components are all real.

In the case of the single spinor field with current jµ = χ†σµχ, we calculate the divergence to be

∂µjµ = m

(χTσ2χ+ χ†σ2χ

∗)Each term is real. We recognize the second term to be the negative conjugate of the other, thus theycancel and the current is conserved. The conserved charge is

Q =

∫d3xχ†χ

which when quantized just counts the number of quanta. As in chapter 2, global symmetries implyparticle conservation laws in free field theories.

12

If we wanted to construct a theory of N of these non-interacting spinor fields, we could just define thecomposite spinor:

Ξ =

χ1

...χN

and then the Lagrangian

L = iΞ†σµ∂µΞ− imΞTσ2Ξ∗

is O(N) invariant and yields the correct field equations for each χ. (Note: by σ in this equation, what isreally meant is IdN×N ⊗ σ, and so forth).

(e) To properly quantize the Majorana field, first notice that if χ2 = χ1 = χ, the Dirac Lagrangianreduces (up to a factor of 2) to the Majorana Lagrangian. We may then take inspiration from thequantization of the Dirac field for the present case. Consider that quantization:

ψ(x) =1

(2π)3

∫d3p

1√2Ep

∑s

(us(p)asp + vs(−p)bs†−p

)e−ip·x

where the Fourier components are given by:

us(p) =

(√σµpµξ

s

√σµpµξ

s

)vs(p) =

( √σµpµξ

s

−√σµpµξs)

and ξs is a 2-component spinor that represents the basis of the helicity operator.

This looks suspiciously like the composite spinor we made in part c. Suppose in quantizing the Majoranafield we replaced us(p) with

√σµpµξ

s and vs(−p) with iσ2(√σµpµξ

s)∗. We quickly find

iσ2(√σµpµξ

s)∗ = −√σµpµiσ2(ξs)∗

and we now have an ansatz for our quantized field:

χ(x) =1

(2π)3

∫d3p

√σµpµ√2Ep

∑s

(ξsaspe

−ip·x − iσ2(ξs)∗as†p eip·x)

We have replaced the operator bs†p with as†p because the fields χ and χ∗ obey the same field equations:the Majorana is it’s own anti-particle. Clearly:

χ†(x) =1

(2π)3

∫d3p

1√2Ep

∑s

((ξs)†as†p e

ip·x + iσ2(ξs)Taspe−ip·x)√σµpµ†

So the anticommutator is

χ(x), χ†(y) =1

(2π)6

∫d3pd3q

1

2√EpEq

√σµpµ

∑s,r

(ξsξr†asp, ar†q e−ip·x+iq·y + (ξs)∗(ξr)T as†p , arqeip·x−iq·y

)√σνqν

†

Now using the fact asp, ar†q = (2π)3δr,sδ(3)(p− q) we evaluate:

13

χ(x), χ†(y) =1

(2π)3

∫d3p

1

2Ep

√σµpµ

∑s

(ξsξs†e−ip·(x−y) + (ξs)∗(ξs)T eip·(x−y)

)√σνpν

†

=1

(2π)3

∫d3p

1

2Ep

√σµpµ

(Id2×2e

−ip·(x−y) + Id2×2eip·(x−y)

)√σνpν

†

=1

(2π)3

∫d3p

1

2Ep

(e−ip·(x−y) + eip·(x−y)

)× (Id2×2Ep − σ · p)

=1

(2π)3

∫d3pe−ip·(x−y) × Id2×2

= δ(3)(x− y)× Id2×2

where we have used the spin sum relations on page 49 of Peskin and Schroeder, and taken advantage ofthe invariant measure to flip the sign of the momentum of half the integrand, canceling the σ · p term.Evidently, this is the correct quantization of the Majorana field.

If we wanted to compute the Majorana Hamiltonian by hand, we would now have all the machinerynecessary to do so. This is overkill though, since we have quantized the Majorana field using theparticular spinor construction of part (c). We found the Majorana Lagrangian was twice the DiracLagrangian in that construction, so the same can automatically be said of the Hamiltonian.

Thus (from equation (3.104)) we have

HM =1

2HD =

1

2(2π)3

∫d3pEp

∑s

as†p asp + bs†p b

sp

But the Majorana is its own anti particle, so b=a and we arrive at the diagonalized Hamiltonian:

HM =1

(2π)3

∫d3pEp

∑s

as†p asp

3.5 Supersymmetry

I will attempt this problem when I know what is going on.

3.6 Fierz Transformations

(a) We begin by establishing a basis for an arbitrary 4×4 matrix Γ as Id, γµ, σµν , γµγ5, γ5

Then imposing the normalization condition Tr(ΓAΓB) = 4δAB we evaluate (noting that the identityalready satisfies this condition):

Tr(γ0γ0) = 4

Tr(γiγi) = −4

Tr(σ0iσ0i) = −4

Tr(σijσij) = 4

Tr(γ0γ5γ0γ5) = −4

Tr(γiγ5γiγ5) = 4

Tr(γ5γ5) = 4

14

all of this implies that the normalized basis is

Id, γ0, iγi, iσ0i, σij , iγ5γ0, γ5γi, γ5

(b) For this part, we use the notation uiΓuj = (Γ)ij and start with the product expansion:

(ΓA)ij(ΓB)kl =∑C,D

CCDAB (ΓC)il(ΓD)kj

Multiplying on the left by (ΓF )jk(ΓE)li and we find

Tr(ΓFΓBΓEΓA) =∑C,D

CCDAB Tr(ΓFΓD)Tr(ΓEΓC)

Impose the normalization conditions on the right hand side and simplify for the constant:

CEFAB =1

16Tr(ΓFΓBΓEΓA)

or equivalently:

(ΓA)ij(ΓB)kl =1

16

∑C,D

Tr(ΓCΓAΓDΓB)(ΓC)il(ΓD)kj

(c) First, we evaluate (u1u2)(u3u4):

(u1u2)(u3u4) =1

16

∑C,D

Tr(ΓCΓD)(u1ΓCu4)(u3ΓDu2)

=1

4

∑C

(u1ΓCu4)(u3ΓCu2)

=1

4

[(u1u4)(u3u2) + (u1γ

µu4)(u3γµu2) +1

2(u1σ

µνu4)(u3σµνu2)

− (u1γ5γµu4)(u3γ

5γµu2) + (u1γ5u4)(u3γ

5u2)]

The second calculation is quite similar. Consider the trace:

Tr(ΓCγµΓDγµ)

in general: ΓCγµ =

∑E cEγ

µΓE but clearly when taking the trace, only the term proportional to ΓDwill survive. (here we use the cyclic property of the trace and the fact γµγµ = 2 × Id. Hence the onlypart of the sum that matters is when C = D, and we have

(u1γµu2)(u3γµu4) =

1

16

∑C

Tr(ΓCγµΓCγµ)(u1ΓCu4)(u3ΓCu2)

The explicit form is not particularly enlightening at this point in time, but can be calculated quite easily,as before.

15

3.7 CPT

In this problem, the following transformations are useful (where |η|2 = 1):

Pψ(t, x)P = ηγ0ψ(t,−x) and Pψ(t, x)P = η∗ψ(t,−x)γ0

Tψ(t, x)T = −γ1γ3ψ(−t, x) and T ψ(t, x)T = ψ(−t, x)γ1γ3

CψC = −γ2ψ∗ and CψC = (−γ2ψ)T γ0

(a) Using this, we compute the C, P, T and CPT transformation rules of the bilinear ψσµνψ:

PψσµνψP = ψγ0σµνγ0ψ(t,−x) =

−(ψσ0iψ)(t,−x)

(ψσijψ)(t,−x)

so the tensor object picks up a negative sign for each of the spatial indices it carries. On the contrary,the Time reversal operator yields a negative sign only when there is no temporal index (we saw this withthe vector term as well):

T ψσµνψT = −ψγ1γ3σµνγ1γ3ψ(−t, x) =

(ψσ0iψ)(−t, x)

−(ψσijψ)(−t, x)

Finally,

CψσµνψC = (−γ2ψ)T γ0σµν(−γ2ψ∗)

= (ψ)∗γ0γ2γ0σµνγ2ψ∗

= −(ψ)∗γ2σµνγ2ψ∗

= −(ψγ2σ∗µνγ2ψ)∗

= −ψσµνψ

where we have used the fact that ψσµνψ is real, and the identity

γ2σ∗µνγ2 = σµν

(b) We remind ourselves again of the quantization of a complex scalar field:

φ(x) =1

(2π)3

∫d3p

1√2Ep

(ape−ip·x + b†pe

ip·x)If we demand Pφ(x)P = φ(−x) we have:

∫d3p

1√2Ep

(PapPe

−ip·x + Pb†pPeip·x) =

∫d3p

1√2Ep

(ape

ip·x + b†pe−ip·x)

by simply redefining the integration variable in the right hand side so that the exponentials match, wediscover the transformation rules:

PapP = a−p and Pb†pP = b†−p

16

Following a similar procedure, we quickly deduce (from the condition Cφ(x)C = φ∗(x)) the chargeconjugation rules:

CapC = ba and CbpC = ap

To evaluate the Time reversal operator rules, first write the time-dependent field:

φ(t, x) =1

(2π)3

∫d3p

1√2Ep

(ape−i(Ept+p·x) + b†pe

i(Ept+p·x))

Then if Tφ(t, x)T = φ(−t, x):

∫d3p

1√2Ep

(TapTe

i(Ept+p·x) + Tb†pTe−i(Ept+p·x)

)=

∫d3p

1√2Ep

(ape−i(p·x−Ept) + b†pe

i(p·x−Ept))

(Note: as per equation (3.133), passing the right T operator through the exponential changes it’s overallsign) So finally,

TapT = a−p and TbpT = b−p

Now we can evaluate the transformation rules of jµ = i(φ∗∂µφ−φ∂µφ∗). The results are easily obtained:

Pjµ(x, t)P =

j0(t,−x)

−j(t,−x)

Tjµ(x, t)T = −iT (φ∗∂µφ− φ∂µφ∗)T =

j0(−t, x)

−j(−t, x)

CjµC = i(φ∂µφ∗ − φ∗∂µφ) = −jµ

(c)

3.8 Bound States

I will attempt the problem in the near future.

4 Chapter 4: Interacting Fields

4.1 Classical Source Field

The Hamiltonian of this problem is

H = HKG −∫d3xj(x, t)φ(x)

(a) We wish to evaluate the probability of the source creating zero particles, or more precisely, theoverlap of the new ground state with itself after a long time has passed:

〈Ω|Ω〉 = 〈0| T ei∫d4xj(x,t)φ(x) |0〉

17

This follows directly from the discussion of section 4.2

(b) Since j(x) is not a quantum field, it cannot be contracted with any other states in any meaningfulway. Therefore, only the even powers of the exponential Taylor series will contribute to this amplitude.So to order j2 we have

〈Ω|Ω〉 = 〈0|0〉 − 〈0| T∫∫

d4xd4yj(x)j(y)φ(x)φ(y) |0〉

= 1−∫∫

d4xd4yj(x)j(y)DF (x− y)

= 1− 1

(2π)3

∫d3p

∫∫d4xd4yj(x)j(y)

e−ip(x−y)

2Ep

= 1− 1

(2π)3

∫d3p

1

2Epj(p)j(−p)

= 1− 1

(2π)3

∫d3p

1

2Ep|j(p)|2

where we have separately evaluated the integrals over x and y space to express j(x) in terms of its Fouriermodes. This is precisely the expression we expected to find.

(c) The Feynman diagram for part (b) would simply be a single propagator that begins at a point x andterminates at a point y. We cannot make sense of the source field in this context though. Consider thej2n term in this series (guaranteed to be non-vanishing):

(i)2n

(2n)!

∫...

∫Π2ni=1d

4xij(xi) · · · j(x2n)× 〈0|

all contractions of 2n fields|0〉

Since the contractions will be of the form of n Feynman propagators, all of which are identical, everyterm is exactly the same. The symmetry factor is the number of ways you can pair off 2n fields, whichis (2n− 1)(2n− 3)(2n− 5) · · · (3)(1) = (2n− 1)!!. A little bit of work will yield

(2n− 1)!!

(2n)!=

1

2nn!

So the term in the series simplifies to

(−1)n

2nn!

(∫∫d4x1d

4x2j(x1)j(x2)DF (x1 − x2)

)n=

1

n!

(−λ2

)nSo clearly, the series exponentiates and we have P (0) = |e−λ/2| = e−λ

(d) To find the probability that the source creates a particle with momentum ~k we evaluate the followingk-space probability density:

p(k) = | 〈k| T ei∫d4xj(x,t)φ(x) |0〉 |2

Because one of the fields in the series must be contacted with the outgoing state, we see that only oddterms contribute. Consider the j2n+1 term:

18

(i)2n+1

(2n+ 1)!

∫· · ·∫

Π2n+1i=1 d4xij(xi) · · · j(x2n+1) 〈k|

all contractions of 2n+1 fields

|0〉

=iS

(2n+ 1)!(−λ)n

∫d4xj(x)eik·x

=iS

(2n+ 1)!(−λ)nj(k)

The symmetry factor S is simple. There are 2n+1 ways of contracting fields with the outgoing state, andtherefore (2n-1)!! ways of contracting the remaining fields. Hence S = (2n+ 1)!!. It follows from before:

(2n+ 1)!!

(2n+ 1)!=

1

2nn!

Therefore the term in the series simplifies to

1

n!

(−λ2

)nj(k)

Exponentiating and taking the modulus squared we find:

p(k) = |j(k)e−λ/2|2 = |j(k)|2e−λ

And integrating over the possible momenta:

P (1) =1

(2π)3

∫d3k|j(k)|2

2Eke−λ = λe−λ

(e) The amplitude of producing n particles in particular momentum states k1, k2, ...kn is

〈k1 · · · kn| T ei∫d4xj(x,t)φ(x) |0〉

Only terms in this series of order jn+2p for non-negative integer p will contribute to this sum. Let’sevaluate the n+2p th term:

(i)n+2p

(n+ 2p)!

∫...

∫Πn+2pi=1 d4xij(xi) · · · j(xn+2p)× 〈0|

all contractions of n+2p fields

|0〉

As before, this will simplify to give us p copies of λ and n plane waves of the form exp iki · xi. All thatremains is the symmetry factor:

S(−λ)p

(n+ 2p)!inΠn

i=1

∫d4xij(xi)e

iki·xi

There are (n+2p)!n!(2p)! ways to contract the outgoing states with the n+2p fields, and then (2p− 1)!! ways of

contracting the remaining fields. Hence

S =(n+ 2p)!

n!(2p)!(2p− 1)!! =

(n+ 2p)!

2pn!p!

So our series evaluates to:

19

in

n!Πni=1j(ki)×

∞∑p=0

1

p!

(−λ2

)pthus by taking the modulus we have

p(k1, k2, ...kn) =1

(n!)2Πni=1|j(ki)|2e−λ

and integrating over the 3n-dimensional momentum space we have:

P (n) =1

(n!)2Πni=1

(1

(2π)3

∫d3ki|j(ki)|2

2Eki

)e−λ

=λne−λ

(n!)2× number of ways of arranging the n final states

=λne−λ

n!

which is a Poisson distribution.

(f) It is fairly easy to show this is normalized. Summing over n we have

∑n

P (n) = e−λ∞∑n=0

λn

n!

= e−λeλ

= 1

The mean value is

〈N〉 = e−λ∞∑n=0

nλn

n!

= e−λλd

dλ

∞∑n=0

λn

n!

= e−λλd

dλeλ

= λ

And the variance is obtained by solving for 〈N2〉:

〈N2〉 = e−λλd

dλ

(λd

dλ

∞∑n=0

λn

n!

)

= e−λλd

dλ

(λd

dλeλ)

= λ2 + λ

so the variance is 〈N2〉 − 〈N〉2 = λ. This is a well known fact about the Poisson distribution.

20

4.2 Decay of a Scalar Particle

The interaction Hamiltonian is

HInt = −µ∫d3x(Φφφ)(x)

So to first order in µ, the transition amplitude that describes the decay of the massive Φ is

〈φp1φp2 |Φ〉 = iµ

∫d4x 〈φp1φp2 | contractions of the term(Φφφ)(x) |Φ〉

= 2iµ(2π)4δ(4)(kµ − (p1 + p2)µ)

From this we conclude M = −2µ and from equation (4.86) we have

Γ =1

2M

∫∫d3p1d

3p2

(2π)6

4µ2

4Ep1Ep2(2π)4δ(4)(kµ − (p1 + p2)µ)

which when evaluated in the rest frame of the decaying particle sets E1 = E2 = M/2 and p1 = p2 =√M2 − 4m2

2. Thus

Γ =µ2

8pi2M

∫d3p

δ(M − 2Ep)

E2p

=µ2

8pi2M4π

∫dpp2

E2p

δ(M − 2Ep)

=µ2

2πM

(M2 − 4m2

M2

)

The lifetime is given by

τ = Γ−1 =2πM

µ2

(1− 4m2

M2

)−1

If M = 2m, this lifetime diverges, indicating that the decay product cannot carry momentum away fromthe rest frame origin, and the outgoing state is not well-defined.

4.3 The Linear Sigma Model

(a) Since the contraction of two fields Φi(x) and Φj(y) is δijDF (x− y) we see that in the single-vertex(O(λ)) case, only those terms which contract an incoming or outgoing state with the field operator willcontribute to the T matrix. Other terms that involve contractions like above correspond to disconnecteddiagrams with extra loops on them. These add only to the trivial component of the S matrix.

Hence to O(λ), the non-trivial part of the S matrix is given by

〈Φm(p3)Φn(p4)|T |Φl(p1)Φk(pk)〉 = −iλ4

∫d4x 〈Φm(p3)Φn(p4)| contractions with 4 fields |Φl(p1)Φk(pk)〉

= −i2λ(δmnδlk + δmlδnk + δmkδln)(2π)4δ(4)(p1 + p2 − p3 − p4)

21

Where we have used the fact that if an internal field Φi is contracted with (for example) an incomingstate |Φl〉 , the outgoing state 〈Φm| contracted with the other internal field Φi forces m=l. The symmetryfactor for each term is 8 because the number of ways we can equivalently contract indices is 4, and theexchange factor for the particles is 2.

Then with the following constraints, the value of M is:

M =

−2λ l = m = 1; k = n = 2

−2λ l = k = 1;m = n = 2

−6λ l = k = m = n = 1

And from equation (4.85) we have the center of mass differential cross section:

(dσ

dΩ

)CM

=

λ2/16π2E2

CM first 2 cases

9λ2/16π2E2CM final case

(b) By spontaneously breaking the symmetry as described in the problem, we expect to create Goldstonemodes and one massive scalar. Consider the transformation:

Φi(x) −→ πi(x)

ΦN (x) −→ v + σ(x)

Where the vev is found by minimizing the new potential: v =√µ2/λ. So Φi ·Φi = πi ·πi+v2 +2vσ+σ2

and

V (πi, σ) = −1

2µ2π2 − v2

2− µ2vσ − 1

2µ2σ2

+λ

4

(π4 + 4σ(v3 + π2v) + σ2(6v2 + 2π2) + σ3(4v) + σ4 + 2π2v2

)=λ

4π4 + µ2σ2 + µ

√λσ3 +

λ

4σ4 + µ

√λσπ2 +

λ

2σ2π2 + f(v)

So the scalar field σ picks up a mass of mσ =√

2µ and has a modified potential. The mass term for thepions has vanished completely. All the pion interactions scale with λ and so will vanish as λ −→ 0.

The propagator for the pion fields are simple:

Dπ(k) =iδij

k2 + iεfor i,j denoting the fields being contracted

For the massive scalar, nothing changes:

Dσ(k) =i

k2 − 2µ2 + iε

The vertex factors can be easily identified (at least to the first-order) by looking at the constants in thepotential and counting symmetry factors. They are:

A vertex involving 4 pions is the same as before: −2iλ(δmnδlk + δmlδnk + δmkδln)Using this, we immediately see that a vertex with 4 σ’s carries a term: −6iλA vertex with 2 pions and two σ’s is: −2iλδij

22

A vertex with 2 pions and 1 σ is: −2iµ√λδij

And finally, a vertex involving 3 σ’s is: 6iµ√λ

(c) If we want to calculate the pion-pion elastic scattering to forst order in λ we need only consider thepossible diagrams with that factor. The 4 vertex diagram will contribute, as well as the exchange of avirtual σ between the pions, as the two vertices together each contribute a factor

√λ. These are the

diagrams given in the book on page 129. Let’s start with the 4 vertex diagram:

The amplitude (A(4)) for this process is easily evaluated. It is equal to

A(4) = −2iλ(δijδlk + δilδjk + δikδlj)(2π)4δ(4)(p1 + p2 − p3 − p4)

Let’s now evaluate the amplitude A(1) for the process πiπj −→ σ −→ πkπl:

A(1) = (−2iµ√λ)2δijδkl

∫∫d4x1d

4x2

∫d4k

(2π)4

i

k2 − 2µ2 − iεe−i(p1+p2)·x1ei(p3+p4)·x2

= −4µ2λδijδkl(2π)8δ(4)(p1 + p2 − p3 − p4)

∫d4k

(2π)4

i

k2 − 2µ2 − iεδ(4)(k − p1 − p2)

=−4iµ2λδijδkl

(p1 + p2)2 − 2µ2(2π)4δ(4)(p1 + p2 − p3 − p4)

From similar calculations with the other two diagrams, we find the scattering matrix M is

M = −2λ[ 2µ2δijδkl

(p1 + p2)2 − 2µ2+

2µ2δikδjl

(p3 − p1)2 − 2µ2+

2µ2δilδkj

(p3 − p2)2 − 2µ2+ (δijδlk + δilδjk + δikδlj)

]If we set the initial momenta to be zero then the first term in the sum simplifies to −δijδkl. Fromconservation of momentum, the next two terms simplify in similar ways and clearly, they all cancel withthe 4-vertex term in the series. Thus the amplitude vanishes identically.

In the case that N=2 and there is only 1 species of pion (i=j=l=k=1), the matrix simplifies to

M = −2λ[ 2µ2

(p1 + p2)2 − 2µ2+

2µ2

(p3 − p1)2 − 2µ2+

2µ2

(p3 − p2)2 − 2µ2+ 3]

If we preform a Taylor expansion in the momentum, we find

M = 2λ

((p1 + p2)2 + (p3 − p2)2 + (p3 − p1)2

2µ2+O(p4)

)So we see that since the pion mass is zero, M∼ O(p4).

(d) If we add the small symmetry breaking term to our potential, the new potential becomes:

V (Φ) = −µ2

2Φ2 +

λ

4Φ4 − aΦN

And so the condition for the ground state of ΦN (which we will label f(a)) is

a = (−µ2 + λf2(a))f(a)

23

Since a is very small, we preform a Taylor expansion of the ground state and keep only up to the linearterm:

a = (−µ2 + λ(f(0) + f ′(0)a)2)(f(0) + f ′(0)a)

Simplifying and grouping terms by a, we find this is equivalent to:

[λf3(0)− µ2f(0)

]+ a[1 + µ2f ′(0)− 3λf2(0)f ′(0)

]= 0

so we see that

f(0) =

õ2

λand f ′(0) =

1

2µ2

Therefore the new ground state is

f(a) =

õ2

λ+

a

2µ2

which agrees in the limit a −→ 0 with our previous ground state. We now simply replace v in our firstexpression on page 21 with f(a):

V (πi, σ) = −1

2µ2π2 − 1

2f2(a)− µ2f(a)σ − 1

2µ2σ2

+λ

4

(π4 + 4σ(f3(a) + π2f(a)) + σ2(6f2(a) + 2π2) + σ3(4f(a)) + σ4 + 2π2f2(a)

)

To first order in a we have:

f2(a) =µ2

λ+

a

µ√λ

f3(a) =µ3

λ3/2+

3a

2λ

After some tedious arithmetic we find:

V (πi, σ) =1

2

(a√λ

µ

)π2 +

λ

4π4 + aσ +

1

2

(2µ2 +

3a√λ

µ

)σ2

+

(µ√λ+

aλ

2µ2

)σ3 +

λ

4σ4 +

(µ√λ+

aλ

2µ2

)σπ2 +

λ

2σ2π2

So the pions acquire a mass m2π =

a√λ

µand the σ mass is adjusted accordingly. The vertex involving 2

pions and a σ now carries a factor −2i(µ√λ+ aλ/2µ2)δij

We can modify the vertex weights fairly easily:

2 pion, 1 σ:

24

−2iµ√λδij −→ −2i

(µ√λ+

aλ

2µ2

)δij

From this and the new masses, we have the new amplitude

A(1) =−4i(µ

√λ+ aλ/2µ2)2δijδkl

(E1 + E2)2 − (~p1 + ~p2)2 − (2µ2 + 3a√λ/µ)

(2π)4δ(4)(p1 + p2 − p3 − p4)

Before, we relied on the fact that at threshold ~pIn = 0, this simplifies to the exact contribution (up tothe negative sign) from the 4-vertex term, and cancels with it. This is no longer the case. Instead, nowthis term may be expanded in a and the linear term does not vanish. In fact:

iM = ia3λ3/2

µ(δijδlk + δilδjk + δikδlj)

4.4 Rutherford Scattering

The interaction Hamiltonian in this problem is

HInt = e

∫d3xψγµψAµ(x)

for a classical vector field Aµ

(a) To lowest order in e, the T matrix element for an electron scattering off a localized potential is

〈p′| iT |p〉 = −ie∫d4xAµ(x) 〈p′| contractions of ψγµψ |p〉

the only contractions that exist are ψ with the incoming state and ψ with the outgoing state. Recallingthe Feynman rules, we find:

〈p′| iT |p〉 = −ie∫d4xu(p′)γµu(p)e−i(p−p

′)·xAµ(x)

= −ieu(p′)γµu(p)Aµ(p′ − p)

where the Fourier transform is defined with no negative sign next to the four-momentum variable in theexponential.

(b) We can define an incoming electron state of ”size” (impact parameter) b in the following way:

|ψ〉in =1

(2π)3

∫d3p

1√2Ep

ψ(p) |p〉 e−i~b·~p

And so the differential cross section is

dσ =

∫d2b

d3pf(2π)3

1

2Ef

∣∣∣ 〈pf |ψ(b)〉In∣∣∣2

Let’s first evaluate the term∣∣∣ 〈pf |ψ(b)〉In

∣∣∣2:

∣∣∣ 〈pf |ψ(b)〉In∣∣∣2 =

1

(2π)6

∫∫d3pd3k

1√2Ep2Ek

ψ(p)ψ∗(k) 〈pf |p〉 〈pf |k〉∗ e−i(~p−~k)·~b

25

To first order, we have 〈pf |p〉 = iM(2π)δ(Ep − Ef ) so we simplify:

∣∣∣ 〈pf |ψ(b)〉In∣∣∣2 =

1

(2π)4

∫∫d3pd3k

1√2Ep2Ek

ψ(p)ψ∗(k)M∗(p)M(k)e−i(~p−~k)·~bδ(Ep − Ef )δ(Ek − Ef )

Now integrating over the impact parameter space:

∫d2b∣∣∣ 〈pf |ψ(b)〉In

∣∣∣2 =1

(2π)2

∫∫d3pd3k

1√2Ep2Ek

ψ(p)ψ∗(k)M∗(p)M(k)δ(2)δ(Ep−Ef )δ(Ek−Ef )(~k⊥−~p⊥)

Following the treatment of euation (4.77) we can integrate over the parallel momentum space and gaina factor of the initial velocity in the denominator. Thus we have:

∫d2b∣∣∣ 〈pf |ψ(b)〉In

∣∣∣2 =1

vi(2π)2

∫d3p

1

2Epψ(p)ψ∗(p)|M|2δ(Ep − Ef )

Following the text further, we pull everything out of the integral besides the ψ:

∫d2b∣∣∣ 〈pf |ψ(b)〉In

∣∣∣2 =1

vi2Ep(2π)2|M|2δ(Ep − Ef )

∫d3pψ(p)ψ∗(p)

=2π

vi2Ep|M|2δ(Ep − Ef )

So finally,

dσ =1

vi

1

2Ei

d3pf(2π)2

1

2Ef|M|2δ(Ef − Ei)

By integrating over the magnitude of outgoing momentum: |pf | we isolate the angular distribution:

dσ =dΩ

(2π)2

1

2Ei

∫ ∞0

dpfvi|M|2

p2f

2Efδ(Ef − Ei)

=dΩ

(2π)2

1

2Ei

∫ ∞0

dpfvi|M|2

p2f

2Ef

Eikδ(p− k)

=dΩ

(2π)2

1

2Ei

∫ ∞0

dpfv2i

|M|2p2f

2Efδ(p− k)

So finally,dσ

dΩ=

1

(4π)2|M(k, θ)|2

(c) In the non-relativistic case, we have E ≈ m and k ≈ mvi. Given the definition of M:

iM = −ieus(p)γµAµ(p− k)ur(k)

and the localized potential

A0(p− k) =Ze

|p− k|2

26

we find that the polarized differential cross section is

dσ

dΩ

(r,s)

=1

(4π)2

Z2e4

|p− k|4us(p)γ0ur(k)ur(k)γ0us(p)

Averaging over the incoming spin r and summing over the outgoing spin s:

dσ

dΩ=

1

32π2

Z2e4

|p− k|4Tr(γ0(/p+m)γ0(/k +m))

One can check that the trace simplifies to 8(m2 + k2 sin2(θ/2)) so that

dσ

dΩ=Z2α2

4k4

(m2 + k2 sin2(θ/2))

sin4(θ/2)

In the non-relativistic limit, this simplifies quite nicely:

dσ

dΩ=

Z2α2

4m2v4

(1 + v2 sin2(θ/2))

sin4(θ/2)

For forward scattering, where θ is sufficiently small, we can discard the second term in the parenthesis,finally recovering the Rutherford formula:

dσ

dΩ=

Z2α2

4m2v4 sin4(θ/2)

5 Chapter 5: Elementary Processes in QED

5.1 Coulomb Scattering

The first part of this problem was done in problem 4.4c

To compute the Mott formula in an alternative way, consider the (spin-averaged) matrix element forthe electron-muon scattering that we have already seen. As in section 5.4 (but now for the case of themassive electron):

1

4

∑Spin

|M|2 =8e4

q4

[(p1 · p′2)(p2 · p′1) + (p1 · p2)(p′1 · p′2)−m2

e(p2 · p′2)−m2µ(p1 · p′1) + 2m2

em2µ

]In the limit of large mµ, we can choose a particular coordinate system in which:

p1 = (E, 0, 0, k)

p2 = (mµ, 0, 0,−k)

p′1 = (E, k sin θ, 0, k cos θ)

p′2 = (mµ,−k sin θ, 0,−k cos θ)

One finds that in these coordinates, the matrix element is given by:

27

1

4

∑Spin

|M|2 =e4

2k4 sin4(θ/2)

[2m2

µ(m2e−k2 sin2(θ/2))+mµ(4Ek2 sin2(θ/2))+(2k4 sin2(θ/2)+m2

ek2 cos(θ))

]

Using another result from that section:

(dσ

dΩ

)CM

=1

4E2CM(4π)2

1

4

∑Spin

|M|2

We see that only terms quadratic in mµ will survive in the numerator, since the center of mass energyis well approximated by the muon mass. Hence

(dσ

dΩ

)CM

=α2

4k4 sin4(θ/2)

(m2e − k2 sin2(θ/2)

)which since ~k = me

~β, is equivalent to the expression in the book.

5.2 Bhabha Scattering

Two particle elastic scattering in QED is (at the tree level) represented by contractions of the matrixelement:

〈kk′| ψ /Aψψ /Aψ |pp′〉

There are only two unique ways to contract these operators so that the incoming and outgoing statesare the pair e−e+, and these produce a matrix element:

iM = (−ie)2

(v(p′)γµv(k′)

−itu(k)γµu(p)− u(k)γµv(k′)

−isv(p′)γµu(p)

)Where we are using the Mandelstam variables:

t = (k − p)2

s = (p+ p′)2

u = (k′ − p) = (k − p′)2

and the spin indices have been dropped for the sake of legibility. Averaging over the incoming spinsand summing over the outgoing spins of the squared matrix element yields (when the electron mass isnegligible compared to it’s energy):

1

4

∑Spin

|M|2 = 2e4[( t

s

)2

+(st

)2

+ u2

(1

t2+

1

s2

)]− e4

4st

[Tr(γµ/k

′γν/kγµ/pγν/p

′) + c.c.]

The reader should check that, using the trace theorems of page 134-5 of the textbook, the trace termsimplifies to −8u2. Simplifying, we find:

1

4

∑Spin

|M|2 = 2e4[( t

s

)2

+(st

)2

+ u2

(1

t+

1

s

)2 ]

28

Identifying s = E2CM we have immediately

(dσ

dΩ

)CM

=2α2

4s

[( ts

)2

+(st

)2

+ u2

(1

t+

1

s

)2 ]and thus after integrating over the azimuthal angle:

(dσ

d cos θ

)CM

=πα2

s

[( ts

)2

+(st

)2

+ u2

(1

t+

1

s

)2 ]In the center of mass frame, s = 2p · p′ = 2p2(1− cos θ), so the differential cross section diverges roughlyas θ−6 for θ −→ 0. This is due to the production of an on-shell photon in the annihilation diagram.

5.3 Spinor Products

(a) Using the definitions provided in the problem statement, we have

|s(p1, p2)|2 = uR(p1)uL(p2)uL(p2)uR(p1)

=1

4(p1 · k0)(p2 · k0)uL(0) /p1 /p2uR(0)uR(0) /p2 /p1uL(0)

=1

16(p1 · k0)(p2 · k0)Tr(

(1− γ5) /k0 /p1 /p2(1 + γ5) /k0 /p2 /p1

)Let’s now evaluate the product

/p /k0/p = pµpνkσγµγσγν

= pµpνkσγµ(2gσν − γνγσ)

= 2(p · k0)/p− /p/p /k0

= 2(p · k0)/p− (p · p) /k0

= 2(p · k0)/p

since all these momenta are lightlike. Thus we have

Tr( /k0 /p1 /p2 /k0 /p2 /p1) = 2(p2 · k0)Tr( /k0 /p1 /p2 /p1)

= 4(p2 · k0)(p1 · p2)Tr( /k0 /p1)

= 16(p2 · k0)(p1 · p2)(k0 · p1)

From here it is easy to see that the traces with a single γ5 will vanish, and the last term (quadratic inγ5) is identical to the first. The γ5’s are separated by an odd number of gammas, and there is an overallnegative sign. Thus

|s(p1, p2)|2 =1

16(p1 · k0)(p2 · k0)32(p2 · k0)(p1 · p2)(k0 · p1)

= 2(p1 · p2)

29

(b) Consider the ordered product of n gamma matrices with a γ5 at the ith location (if none is present,let i = 0):

O(n, i) = γµ1γµ2 · · · γ5γµi+1 · · · γµn

And the reverse-ordered product:

O(n, i) = γµn · · · γµi+1γ5 · · · γµ2γµ1

Here we will use the property that for µ = 0, 1, 2, 3 the transpose is given by

(γµ)T

= (−1)µγµ

and (γ5)T

= γ5

Consider then the transpose invariance of the trace:

Tr(O(n, i)) = Tr(OT (n, i))

= (−1)µ1+···+µnTr(O(n, i))

where the index sum excludes the location of the γ5 matrix.

The constant out front cam be simplified to the following:

(−1)µ1+···+µn = (−1)s

where s is the number of odd indices not equal to 5. It stands to reason that if n is odd, the tracevanishes entirely. If n is even, we have an equality:

Tr(γµ1γµ2 · · · γµn) = Tr(γµn · · · γµ2γµ1)

But does the case of an odd total of 1’s and 3’s actually vanish? The total number of gammas, n, mustbe even for the trace to not vanish identically. This is guaranteed by the fact that it can always bewritten in terms of (n-2) matrices, and the fact that an individual gamma is traceless. If s is odd, thenwe have either an even number of γ1’s with an odd number of γ3’s or vice versa. By permuting, we pickup some overall sign change, but by arranging the gammas together by index and squaring them awaywe are capable of writing the trace in the following fashion (given that γr is to be interpreted as someproduct of r matrices of the set γ5, γ0, γ2 for r odd):

±Tr(γ3γr)

where the sign is a factor of the exact permutation we began with and how many of each pair of gammaswe squared away. It is unimportant. By cyclicity of the trace, we see

Tr(γ3γr) = Tr(γrγ3) = (−1)rTr(γ3γr)

so clearly it vanishes. Our reasoning is valid, and the identity holds.

To prove the equality stated in the problem:

30

uL(p1)γµuL(p2) =1

2√

(p1 · k0)(p2 · k0)uR(0) /p1γ

µ/p2uR(0)

=1

4√

(p1 · k0)(p2 · k0)Tr( /p1γ

µ/p2(1 + γ5) /k0)

=(p1)α(p2)β(k0)δ

4√

(p1 · k0)(p2 · k0)

[Tr(γαγµγβγδ) + Tr(γαγµγβγ5γδ)

]=

(p1)α(p2)β(k0)δ

4√

(p1 · k0)(p2 · k0)

[Tr(γδγβγµγα) + Tr(γδγ5γβγµγα)

]=

(p1)α(p2)β(k0)δ

4√

(p1 · k0)(p2 · k0)

[Tr(γβγµγαγδ) + Tr(γ5γβγµγαγδ)

]=

(p1)α(p2)β(k0)δ

4√

(p1 · k0)(p2 · k0)

[Tr(γβγµγαγδ)− Tr(γβγµγαγ5γδ)

]=

1

4√

(p1 · k0)(p2 · k0)Tr( /p2γ

µ/p1(1− γ5) /k0)

=1

2√

(p1 · k0)(p2 · k0)uL(0) /p2γ

µ/p1uL(0)

= uR(p2)γµuR(p1)

(c) To prove this particular Fierz identity, let’s examine the right hand side:

2[uL(p2)uL(p1) + uR(p1)uR(p2)]ab =1

2√

(p1 · k0)(p2 · k0)[ /p2(1 + γ5) /k0 /p1 + /p1(1− γ5) /k0 /p2]ab

= Cαβδ[γβ(1 + γ5)γδγα + γα(1− γ5)γδγβ ]ab

= Cαβδ[γ(β|δ|α) + γ[β|δ|α]γ5]ab

Where, for example, γ[β|δ|α]γ5 is a rank-3 tensor (part of the Clifford algebra) that has been anti-symmetrized in its 1st and 3rd indices. From this we recognize that the matrix which represents thisproduct is a linear combination of vector and pseudo-vector quantities. If we call this matrix M as thetext does and expand in a basis:

M =

(1− γ5

2

)γµV

µ +

(1 + γ5

2

)γµW

µ

Then as described in the problem,

V µ =1

4Tr(γµ(1− γ5)M)

=1

4Tr(γµ4uL(p2)uL(p1))

= uL(p1)γµuL(p2)

Since (1± γ5)/2 act as projection operators onto the right/left states. Likewise:

Wµ = Tr(γµuR(p1)uR(p2)) = uR(p2)γµuR(p1) = V µ

Then from the definition of M:

[M ]ab = uL(p1)γµuL(p2)[γµ]ab

31

and the identity is proven.

(d)

(e)

5.4 Positronium Lifetimes

32