Gauge Theories in Particle Physics Third Edition Volume 1: From Relativistic Quantum ... · 2005-09-07 · Gauge Theories in Particle Physics Third Edition Volume 1: From Relativistic

Gauge Theories in Particle Physics

Third Edition

Volume 1: From Relativistic Quantum

Mechanics to QED

0750308648

Outline solutions for selected problems

I J R Aitchison and A J G Hey

Outline solutions for selected problems

Chapter 2

2.1 The quantum numbers of the ‘anything’ must be B = 1, Q = 0 in both cases, togetherwith S = −1 in the first case and S = +1 in the second; consider what are the lightestsingle-particle or two-particle states with these quantum numbers.

2.2 (i) 4-momentum conservation gives p+k = p′+k′ which may be written as p′ = (k−k′)+p;squaring both sides gives p′2 = q2 +2p · (k−k′)+M2c2, and the result follows by evaluat-ing the dot product in the frame such that p = (Mc,0). For highly relativistic electronsE ≈ c|k|, E′ ≈ c|k′|, whence q2c2 = (E − E′)2 − c2(k − k′)2 ≈ −2EE′ + 2EE′ cos θ =−4EE′ sin2 θ/2. For elastic scattering p′

2 = M2c2, and hence from the first resultq2 = −2M(E − E′). Combining this with the (approximate) formula for q2 in termsof θ gives the formula for E′.(ii) E′ = 4.522 GeV (negelecting the electron mass).(iii) The invariant mass of the produced hadronic state is (p′2/c2)1/2; using the first dis-played equation and the (approximate) expression for q2 in terms of θ gives 1238 MeV/c2

(approximately equal to the mass of the I = 3/2 ∆ resonance).(iv) The term ‘quasi-elastic peak’ in this context means the value of E′ corresponding toa collision between the incident electron and a single nucleon in the He nucleus, ignoringthe binding energy of the struck nucleon (i.e. it is treated like elastic scattering). Sincethe binding energy is of order 10 MeV, very much less than the electron energy, this isexpected to be a good first approximation to the kinematics. It gives E′ = 355.6 MeV.The struck nucleon is, however, moving by virtue of being bound inside the He nucleus(it has a bound state wavefunction and an associated momentum distribution). Ratherthan attempting a more realistic correction based on the nucleon wavefunction, we canget a fair idea of a typical nucleon momentum by using the ‘uncertainty relation’ estimatep ∼ ~/R where R is the nuclear radius. Here R ∼ 1.5fm, and p ∼ 130MeV/c, which givesa struck nucleon speed of order v/c ∼ 0.14. (Note that He is a tightly bound nucleus, andthe formula for the nuclear radius R = 1.1 × (A)1/3 is not really applicable - of course,we are making rough estimates anyway.) Considering configurations with the outgoingelectron moving parallel/antiparallel to the struck nucleon gives a typical shift in E′ oforder ± 50 MeV. Note that this is quite a bit bigger than the nucleon’s binding energy -the relativistic transformation has amplified the effect.

2.3 (i) 5.1 eV.(ii) (a) 1.29 (b) 0.75(iii) 〈σ1 · σ2〉 = +1 for S = 1, −3 for S = 0. Hyperfine splitting = 8.45 ×10−4 eV.(iv) 0.57.

2.4 One-gluon exchange; confinement.Ground state expected to be at the minimum of E(r) as a function of r, i.e. at r0 suchthat dE(r)/dr|(r=r0) = 0.Egr(cc) ≈ 3.23GeV.Threshold for production of ‘open charm’ (DD states) opens at about 3.73 GeV.

1

Outline solutions for selected problems 2

2.5 E± = 2a cos 2θ ± a; |+〉 = cos θ|1〉+ sin θ|2〉, |−〉 = − sin θ|1〉+ cos θ|2〉. System is in statecos θ|+〉 − sin θ|−〉 at time t = 0; evolves to

cos θexp[i(2a cos 2θ + a)t/~]|+〉 − sin θexp[i(2a cos 2θ − a)t/~]|−〉

at time t; amplitude of |2〉 in this state is

cos θ sin θexp[i(2a cos 2θ + a)t/~]− exp[i(2a cos 2θ − a)t/~];

modulus squared of this gives result.2.6 (iv) d = 2, R = 1mm: MP,3+d ≈ 1.54 TeV.

Chapter 3

3.1 From the inverse transformation and the chain rule

∂

∂t′=

∂t

∂t′∂

∂t+∂x1

∂t′∂

∂x1

we find∂

∂t′= γ

[∂

∂t− v

(− ∂

∂x1

)]and similarly for − ∂

∂x1′ .3.2 Six. F 21 = B3, F 32 = B1, F 13 = B2;F 10 = E1, F 20 = E2, F 30 = E3. Consider ν = 1

component of (3.17); LHS is

∂µFµ1 = ∂2F

21 + ∂3F31 + ∂0F

01

=∂B3

∂x2− ∂B2

∂x3− ∂E1

∂t= (∇×B)1 −

(∂E

∂t

)1

which verifies the first component of (3.8).3.3 It is advisable in such manipulations to let the whole expression act on an arbitrary

function F (x) say, which can be removed at the end. Thus

pe−iqf(x)F (x) = − i∂

∂xe−iqf(x)F (x)

= − i(−iq)∂f

∂xe−iqf(x)F + e−iqf(x) − i

∂F (x)∂x

= (−q)∂f∂x

e−iqf(x)F + e−iqf(x)pF (x);

multiplying this from the left by eiqf(x) gives the result after F is removed.

Chapter 4

4.1(b) ρ = i(φ∗φ− φ∗φ), j = i−1(φ∗∇φ− (∇φ∗)φ).For φ = Ne−ip·x, ρ = 2|N |2E, j = 2|N |2p, jµ = 2|N |2pµ.

4.2(ii) Consider for example βαi = −αiβ. Multiply from the left by β to obtain αi = −βαiβ.Now take the Trace and use Tr(βαiβ) = Tr(αiβ2) = Tr(αi).

4.4(b) (σ · a)(σ · b) = σiaiσjbj (sum on i and j) = σiσjaibj= (δij1 + iεijkσk)aibj = a · b1 + iσ · (a× b).

4.6(i) (σ · u)2 = 1. Hence σ · u 12 (1+σ · u)φ = 1

2 (σ · u+1)φ. Projection operator for σ · u = −1is 1

2 (1− σ · u).(ii) Take

φ+ =N

2(1+σ·u)

(10

)=N

2

(1 + cos θ sin θe−iφ

sin θeiφ 1− cos θ

)(10

)= N

(cos2 θ/2

sin θ/2 cos θ/2eiφ

)


and choose N = 1/(cos θ/2) for normalization. Possible φ− is(− sin θ/2e−iφ

cos θ/2

).

4.8 Consider j′y for example. We have

j′y = ω′†αyω′ = ω†e−iΣxα/2αyeiΣxα/2ω.

Now

eiΣxα/2 = 1 + iΣxα/2 +12(iΣxα/2)2 +

13!

(iΣxα/2)3 + . . .

= 1 + iΣxα/2−12(α/2)2 − 1

3!iΣx(α/2)3 + . . .

= cosα/2 + iΣx sinα/2

as in (4.80). Hence

j′y = ω†[(cosα/2− iΣx sinα/2)αy(cosα/2 + iΣx sinα/2)]ω.

But

Σxαy =(σx 00 σx

)(0 σyσy 0

)=(

0 σxσyσxσy 0

)=(

0 iσziσz 0

)= iαz;

similarly αyΣx = −iαz; and ΣxαyΣx = −αyΣ2x = −αy. Hence

j′y = ω†[cos2 α/2 αy + 2 sinα/2 cosα/2 αz − sin2 α/2 αy]ω

= ω†[cosα αy + sinα αz]ω= cosα jy + sinα jz.

4.10 Under (4.83):ψ′ψ′ = ψ′†βψ′ = ψ†e−iΣ·nθ/2βeiΣ·nθ/2ψ.

Now Σ =(

σ 00 σ

), β =

(1 00 −1

), and Σβ = βΣ, so that

e−iΣ·nθ/2β = βeiΣ·nθ/2

and the exponentials cancel.Under (4.90): use αxβ = −βαx.

4.11 (a) γµγν + γνγµ = 2gµν .(b) Taking the dagger of (iγµ∂µ−m)ψ = 0 gives −i(∂µψ†)γµ†−mψ† = 0, or ψ†(iγµ†←−∂µ +m) = 0 where the notation ←−∂µ means that the derivative acts on what is to the left of it,i.e. on ψ†. Multiply this equation from the right by γ0 = β and use γµ†β = βγµ (obviousfor µ = 0, check it for the space components); this gives ψ(i←−6∂ +m) = 0.(c) ∂µ(ψγµψ) = (ψ←−6∂ )ψ + ψ(6∂ψ) = −mψψ +mψψ = 0.

4.13 VKG = iq(∂µAµ +Aµ∂µ)− q2AµAµ.4.14 (ii) (σ ·∇)(σ ·A)ψ = σi∂iσjAjψ = σiσj∂i(Ajψ)

= (δij1 + iεijkσk)∂i(Ajψ) = ∂i(Aiψ) + iεijkσk∂i(Ajψ)= ∇ · (Aψ) + iσ · ∇× (Aψ).

4.15 (i) For a massive particle the direction of its momentum reverses on transforming to aframe moving parallel to, and faster than, it; this is not possible for a massless particlewhich moves at the speed of light. (ii) We give the solution in a form which adapts to thegeneral case. We have (E − σxpx)uR = 0. Mulitiplying from the left by (1− 1

2σxεx) andinserting a unit operator (to first order in εx) it follows that


(1− 12σxεx)(E − σxpx)(1−

12σxεx)(1 + 1

2σxεx)uR = 0.But (1 + 1

2σxεx)uR = u′R, and(1− 1

2σxεx)(E − σxpx)(1−12σxεx) = E − σxpx − Eσxεx + εxpx

(to first order in εx), which is just E′ − σxp′x, whence (E′ − σxp′x)u′R = 0 as required.(iv) See part (i). Since, for m 6= 0, we can reverse the helicity by a Lorentz transformation,we expect that in a Lorentz-invariant theory mass terms will mix states of opposite helicity.(v) let BR = (1 + 1

2σxεx), BL = (1− 12σxεx), so that u′R = BRuR, u

′L = BLuL. Equation

(4.98) is (E − σxpx)φ = mχ. So we haveBL(E − σxpx)BLB

−1L φ = mBLχ. But B−1

L = BR, and part (ii) showed that BL(E −σxpx)BL = E′ − σxp

′x. Hence it follows that (E′ − σxp

′x)φ

′ = mχ′ if φ′ = BRφ andχ′ = BLχ. Similarly for (4.99).

Chapter 5

5.1 Consider the term∫ L0

(∂φ∂x

)2

dx. We have

∂φ

∂x=

∞∑r=1

Ar(t)(rπL

)sin(rπxL

).

The quantity ‘(∂φ∂x

)2

’ is the square of this entire series - i.e. it involves not only terms ofthe form [

A1(t)(πL

)sin(πxL

)]2,

[A2(t)

(2πL

)sin(

2πxL

)]2,

etc., but also all the ‘cross’ terms such as

A1(t)(πL

)sin(πxL

)·A2(t)

(2πL

)sin(

2πxL

).

To make sure we are including all these cross terms, we use a different summation indexfor the two independent summations when writing the product:∫ L

0

(∂φ

∂x

)2

dx =∫ L

0

[ ∞∑r=1

Ar(t)(rπL

)sin(rπxL

)]

×

[ ∞∑s=1

As(t)(sπL

)sin(sπxL

)]dx

=∞∑

r,s=1

Ar(t)As(t)(rπL

)(sπL

)∫ L

0

sin(rπxL

)sin(sπxL

)dx. (∗)

We need to evaluate the integral:∫ L

0

sin(rπxL

)sin(sπxL

)dx

=∫ L

0

12

cos[(r − s)πx

L

]− cos

[(r + s)

πx

L

]dx

=12

L

π(r − s)sin[(r − s)πx

L

]− L

π(r + s)sin[(r + s)

πx

L

]L0

.

Here both r and s are positive integers, so the second term vanishes. The first does also,except when r = s, in which case it has the value L/2 as can be seen by evaluating the


integral directly for r = s. So we have∫ L

0

sin(rπxL

)sin(sπxL

)= L/2 for r = s

= 0 for r 6= s.

This means that in the sums over r and s in (∗), only the term in which r = s survives,after the integral over x has been done. So (∗) becomes

∞∑r=1

(Ar(t))2(rπL

)2

L,

and ∫ L

0

12ρc2(∂φ

∂x

)2

dx =(L

2

) ∞∑r=1

12ρω2

rA2r.

Similarly for the φ2 term.5.2 S =

∫ t00

[ 12mx2 +mgx]dt.

(a) x = at : S(a) =∫ t0

0

(12ma2 +mgat)dt =

12ma2t0 +

12mgat20.

Choose ‘a’ so that end point of this trajectory coincides with end point of the Newtoniantrajectory in (b), which is at x = 1

2gt20. So we set at0 = 1

2gt20 i.e. a = 1

2gt0. Then S(a) =38mg

2t30 = 0.375(mg2t30). (b): S(b) = 13mg

2t30 = 0.3333(mg2t30). (c): b = 12gt

−10 , S(c) =

720mg

2t30 = 0.35(mg2t30). So S(b) < S(c) < S(a).5.3 (a) We want to verify that ˙q = −i[q, H] is consistent with ˙q = p/m. We have

[q, H] = [q,1

2mp2 +

12mω2q2] =

12m

[q, p2]

=1

2mp [q, p] + [q, p] p using [A, BC] = B [A, C] + [A, B] C

=1

2m2ip =

imp

whence −i[q, H] = p/m.5.4 (a) From (5.61) and (5.62), q = (a + a†)/(2mω)

12 . So 1

2mω2q2 = 1

4ω(a + a†)2. Similarlyfor the 1

2m p2 term.

(b) [a, a†] = 1. [(a†a+ 12 )ω, a] = ω [a†a, a] = ω a† [a, a] + [a†, a] a = −ωa.

(c) We first prove that a(a†)n − (a†)na = n (a†)n−1, by induction: (i) it is true for n = 1;(ii) multiplying from the left by a† and using a†a = aa†−1 in the first term proves the re-sult for n+1 assuming it’s true for n. Letting this result act on |0〉 we deduce a(a†)n|0〉 =n(a†)n−1|0〉. Hence 〈0|(a)n(a†)n|0〉 = 〈0|(a)n−1a(a†)n|0〉 = n〈0|(a)n−1(a†)n−1|0〉, and us-ing induction again (or iterating this step another n−1 times) it follows that 〈0|(a)n(a†)n|0〉 =n!, provided |0〉 is normalized.

(d) n |n〉 =1√n!a†a(a†)n|0〉 =

1√n!a†n(a†)n−1 + (a†)na|0〉

= n1√n!

(a†)n|0〉 = n|n〉.

5.6 (a) From (5.116),

φ(x, t) =∫ ∞

−∞

dk2π√

2ω[a(k)eikx−iωt + a†(k)e−ikx+iωt]


and so π(y, t) is given by

π(y, t) =∫ ∞

−∞

dk′

2π√

2ω′[(−iω′)a(k′)eik′y−iω′t + (iω′)a†(k′)e−ik′y+iω′t].

Note the use of a different integration variable in π(x, t), to ensure that the operators insidethe integral for φ are treated independently of those inside the integral for π. In the com-mutator [φ(x, t), π(y, t)] there are therefore four types of term: [a(k), a(k′)], [a†(k), a†(k′)],which vanish by the second line of (5.117), and [a(k), a†(k′)], [a†(k), a(k′)]. The first ofthese latter two yields∫ ∞

−∞dk∫ ∞

−∞dk′

1(2π)2

1√4ωω′

eikx−iωt e−ik′y+iω′t(iω′)2πδ(k − k′)

=∫ ∞

−∞dk

12π

iω12ω

eik(x−y) =i2δ(x− y)

using (E.25). The [a†(k), a(k′)] term gives the same.(b)

[φ(x1, t1), φ(x2, t2)] =∫ ∞

−∞

d3k

(2π)3√

2E

∫ ∞

−∞

d3k′

(2π)3√

2E′

[(a(k)e−ik·x1 + a†(k)eik·x1), (a(k′)e−ik′·x2 + a†(k′)eik′·x2)].

The surviving terms are the ‘[a, a†]’ and ‘[a†, a]’ ones which give∫ ∞

−∞

dk

(2π)3√

2E

∫ ∞

−∞

dk′

(2π)3√

2E′[e−ik·x1 eik′·x2(2π)3δ3(k − k′)

−eik·x1 e−ik′·x2(2π)3δ3(k − k′)]

=∫ ∞

−∞

d3k

(2π)32E[e−ik·(x1−x2) − eik·(x1−x2)]

with k · (x1 − x2) = E(t1 − t2)− k · (x1 − x2). For t1 = t2 the integral is∫d3k

(2π)32E[eik·(x1−x2) − e−ik·(x1−x2)]

which vanishes since the integrand is an odd function of k. So D is Lorentz invariant (seeAppendix E) and vanishes when t1 = t2. It therefore also vanishes at all points whichcan be connected to ‘t1 = t2’ by a Lorentz transformation. Noting that for t1 = t2 theinvariant interval (x1−x2)2 is spacelike (i.e. is less than zero), we deduce that D vanishesfor all x1, x2 such that (x1 − x2)2 < 0.

5.7 We have

φ(x, t) =∫ ∞

−∞

dk2π√

2ω[a(k)eikx−iωt + a†(k)e−ikx+iωt]

∂φ

∂x=∫ ∞

−∞

dk2π√

2ω[a(k)(ik)eikx−iωt + a†(k)(−ik)e−ikx+iωt]

and so

12

∫ ∞

−∞dx(∂φ

∂x

)2

=12

∫ ∞

−∞

dk(2π)√

2ω[a(k)(ik)eikx−iωt + a†(k)(−ik)e−ikx+iωt].∫ ∞

−∞

dk′

(2π)√

2ω′[a(k′)(ik′)eik′x−iω′t + a†(k′)(−ik′)e−ik′x+iω′t].


There are four terms when the brackets are multiplied out. For example, the first is

12

∫dx∫

dk2π√

2ω

∫dk′

2π√

2ω′(ik)(ik′)a(k)a(k′)ei(k+k′)x−i(ω+ω′)t.

The integral over x yields (2π)δ(k + k′); the k′-integral can then be done leading to

12

∫dk

(2π)√

2ω12π

1√2ω

(ik)(−ik)a(k)a(−k)e−2iωt (∗∗)

using ω′ = |k′| = k = ω. The three other terms can be reduced similarly. Meanwhile,

π(x, t) = ˙φ(x, t) =

∫dx

2π√

2ω[a(k)(−iω)eikx−iωt + a†(k)(iω)e−ikx+iωt]

and so

12

∫π2dx =

12

∫dx∫

dk2π√

2ω[a(k)(−iω)eikx−iωt + a†(k)(iω)e−ikx+iωt].∫

dk′

2π√

2ω′[a(k′)(−iω′)eik′x−iω′t + a†(k′)(iω′)e−ik′x+iω′t].

For the ‘aa’ term the x-integral yields

12

∫dk

[(2π)√

2ω]2(−iω)2a(k)a(−k)e−2iωt

which cancels against (∗∗). Similarly, the ‘a†a†’ terms cancel, and the ‘aa†’ and ‘a†a’terms give (5.119).

5.8 (a) We have

˙φ(x, t) = − i[φ(x, t), H]

= − i[φ(x, t),12

∫π2(y, t) +

(∂φ

∂y

)2

dy].

φ commutes with (∂φ/∂y)2, and the non-vanishing term on the RHS is

− i12[φ(x, t),

∫π2(y, t)dy]

=−i2

∫π(y, t)[φ(x, t), π(y, t)] + [φ(x, t), π(y, t)]π(y, t)dy

=−i2

∫π(y, t)iδ(x− y) + iδ(x− y)π(y, t)dy = π(x, t)

as required.

Chapter 6

6.1 The RHS is

12

∫ ∞

−∞dt1

∫ t1

−∞dt2f(t1)f(t2) +

∫ ∞

t1

dt2f(t2)f(t1). (A)

The second term is12

∫ ∞

−∞dt1∫ ∞

t1

dt2 f(t2)f(t1).


With the usual convention for double integrals, in this expression the integral over t2 isunderstood to be performed first, holding t1 fixed, and then t1 is integrated over. Weproceed by changing the order of integration in this double integral. In order not to makea mistake, it is a good idea to draw a diagram of the t1 (horizontal) - t2 (vertical) plane,with the region of integration shaded over: it is the whole of the region lying above theline t1 = t2. When we write the integral in the other order (i.e. doing first the t1 integraland then the t2 one) we must be careful to arrange the limits so that the required regionis covered. The result is

12

∫ ∞

−∞dt2∫ t2

−∞dt1 f(t2)f(t1).

We now rename t1 as t2, and t2 as t1, showing that this term is in fact equal to the firstin (A).

6.2 In the rest frame of C,

mC = EA + EB =√m2

A + p2 +√m2

B + p2.

Taking the term√m2

A + p2 to the LHS, squaring, cancelling the p2, and squaring againwe arrive at

(m2A +m2

C −m2B)2 = 4m2

C(m2A + p2).

This gives |p| as required.6.3 (iii)

∂

∂t1[θ(t1 − t2)φ(x1, t1)φ(x2, t2) + θ(t2 − t1)φ(x2, t2)φ(x1, t1)]

= δ(t1 − t2)φ(x1, t1)φ(x2, t2) + θ(t1 − t2)(∂

∂t1φ(x1, t1))φ(x2, t2)

−δ(t2 − t1)φ(x2, t2)φ(x1, t1) + θ(t2 − t1)φ(x2, t2)(∂

∂t1φ(x1, t1))

The δ functions pick out only the point t1 = t2 in the terms multiplying them, and atthis (equal-time) point the φ fields commute according to (5.105), leading to the result

(( ˙φ(x1, t1) is short for ∂φ(x1, t1)/∂t1). (iv) Differentiating the result of (iii) with respect

to t1 we get

∂2

∂t21T (φ(x1, t1)φ(x2, t2) = δ(t1 − t2)π(x1, t1)φ(x2, t2) + θ(t1 − t2)¨φ(x1, t1)φ(x2, t2)

− δ(t2 − t1)φ(x2, t2)π(x1, t1) + θ(t2 − t1)φ(x2, t2)¨φ(x1, t1).

Again the δ functions force t1 to equal t2 in the terms multiplying them, which become

[φ(x1, t1), φ(x2, t2)] δ(t1 − t2) = −iδ(x1 − x2)δ(t1 − t2)

using (5.104) and (E.32). The remaining two terms are just T¨φ(x1, t1)φ(x2, t2). Sincethe operator −∂2/∂x2

1 +m2 doesn’t involve t1, it ‘passes through’ the θ-functions in theT -product, acting just on φ(x1, t1):

(− ∂2

∂x21

+m2)Tφ(x1, t1)φ(x2, t2) = T

((− ∂2

∂x21

+m2)φ(x1, t1)))φ(x2, t2)

.

Hence (∂2

∂t21− ∂2

∂x21

+m2

)Tφ(x1, t1)φ(x2, t2)


= −iδ(x1 − x2)δ(t1 − t2) + T

((∂2

∂t21− ∂2

∂x21

+m2)φ(x1, t1))φ(x2, t2)

.

The equation of motion for the free KG field φ implies that the second term vanishes.6.4

〈0|aA(p′A)φA(x1)|0〉 = 〈0|aA(p′A)∫

d3k

(2π)3√

2Ek[aA(k)e−ik·x1 + a†A(k)eik·x1 ]|0〉,

with Ek =√m2

A + k2. The aA(k) term gives zero on |0〉. To evaluate the ‘aAa†A’ term,

we use (6.46):aA(p′A)a†A(k) = (2π)3δ3(p′A − k) + a†A(k)aA(p′A)

leading to

〈0|aA(p′A)φA(x1)|0〉 = 〈0|∫

d3k

(2π)3√

2Ekeik·x1(2π)3δ3(p′A − k)|0〉

=1√2E′

A

eip′A·x1 ,

where E′A = (m2

A + p′A2)1/2.

6.5

〈0|TφC(x1)φC(x2)|0〉 = 〈0|θ(t1 − t2)φC(x1)φC(x2) + θ(t2 − t1)φC(x2)φC(x1)|0〉

= 〈0|θ(t1 − t2)

∫d3k

(2π)3√

2ωk[aC(k)e−ik·x1 + a†C(k)eik·x1 ]

×∫

d3k′

(2π)3√

2ωk′[aC(k′)e−ik′·x2 + a†C(k′)eik′·x2 ]|0〉+ similar term multiplying θ(t2 − t1)

where k0 = ωk =

√m2

C + k2 and k′0 = ω′k =√m2

C + k′2. The conditions 〈0|a†C(k) =

aC(k′)|0〉 = 0 reduce the θ(t1 − t2) term to

〈0|θ(t1 − t2)∫

d3k

(2π)3√

2ωk

∫d3k′

(2π)3√

2ωk′aC(k)a†C(k′)e−ik·x1 eik′·x2 |0〉.

UsingaC(k)a†C(k′) = (2π)3δ3(k − k′) + a†C(k′)aC(k)

this becomes

θ(t1 − t2)∫

d3k

(2π)3√

2ωk

∫d3k′

(2π)3√

2ωk′(2π)3δ3(k − k′)e−ik·x1 eik′·x2

=∫

d3k

(2π)32ωkθ(t1 − t2)e−iωk(t1−t2)+ik·(x1−x2).

Similarly for the θ(t2 − t1) term.6.6 We write (6.91) as

(−ig)2∫ ∫

d4x1d4x2ei(p′A−pB)·x1 ei(p′B−pA)·x2 f(x1 − x2)

where f(x1−x2) is given by the RHS of (6.98). Introducing x = x1−x2, X = (x1+x2)/2,the integral becomes∫ ∫

d4xd4X ei(p′A−pB)·(X+x/2) ei(p′B−pA)·(X−x/2) f(x)


since the Jacobian for the change of variables is unity (for instance, dt1dt2 → d(t1 −t2)d(t1 + t2)/2, etc.) The only place X appears is in the exponent, allowing the X-integralto be done using the 4-dimensional version of (E.26); we then get

(2π)4δ4(p′A − pB + p′B − pA)∫

d4x ei[(p′A−pB)−(p′B−pA)]·x/2 f(x).

The 4-momentum δ-function ensures that in the exponent p′A−pB is equal to −(p′B−pA),each being equal to q. Re-instating the factor (−ig)2 we obtain

(−ig)2(2π)4δ4(p′A + p′B − pA − pB)∫

d4x eiq·x f(x)

which is (6.99). Substituting now the RHS of (6.98) for f(x) and performing the x-integration gives a factor (2π)2δ4(q − k), and finally performing the k-integration gives(6.100).

6.7 The analogue of (6.91) for this contraction is

(−ig)2∫ ∫

d4x1d4x2 ei(p′A+p′B)·x1 e−i(pA+pB)·x2 〈0|T (φC(x1)φC(x2))|0〉.

Introducing x and X as in problem 6.6, this becomes

(−ig)2(2π)4δ4(p′A + p′B − pA − pB)∫

d4x ei(pA+pB)·x f(x).

Inserting (6.98) for f(x) and performing the x and then the k integrals leads to the sameexpression as (6.100) but with q replaced by pA + pB - that is, (6.101).

6.8 In the CM frame,

pA = (√m2

A + k2,k), p′B = (√m2

A + k′2,k′),

and |k| = |k′| by energy conservation. Hence

u = (pA − p′B)2 = (0, (k − k′))2 = −(k − k′)2 ≤ 0.

6.9 In the frame in which B is initially at rest, pB = (mB,0), pA = (√m2

A + p2,p) and

(pA · pB)2 −m2Am

2B = (mB

√m2

A + p2)2 −m2Am

2B

= m2Bp2.

Hence [(pA · pB)2 −m2Am

2B]1/2 = mB|p|. But in this frame mB = EB, and |p| = EA|v|.

Chapter 7

7.2 (a) The calculation is similar to that in problem 5.7.

N0φ = i

∫(φ† ˙

φ− φ ˙φ†) d3x = i

∫d3x

∫d3k

(2π)3√

2ω(a†(k)eik·x + b(k)e−ik·x)

×∫

d3k′

(2π)3√

2ω′(−iω′a(k′)e−ik′·x + iω′b†(k′)eik′·x)

−∫

d3k

(2π)3√

2ω(a(k)e−ik·x + b†(k)eik·x)

∫d3k′

(2π)3√

2ω′(iω′a†(k′)eik′·x − iω′b(k′)e−ik′·x)

.


After multiplying out the brackets, the terms a†b† and ab cancel using [a†(k), b†(k′)] =[a(k), b(k′)] = 0. The term a†a is∫

d3x

∫d3k

(2π)3√

2ω

∫d3k′

(2π)3√

2ω′ω′a†(k)a(k′)ei(ω−ω′)t−i(k−k′

)·x.

The x-integral gives (2π)3δ3(k−k′), which forces ω = ω′ and allows the k′-integral to bedone giving ∫

d3k

(2π)312a†(k)a(k).

The term aa† gives the same when normally ordered (throwing away a divergent piece);similarly the b†b terms.

7.3 The commutator is

[∫

d3k

(2π)3√

2ω(a(k)e−ik·x1 + b†(k)eik·x2),

∫d3k′

(2π)3√

2ω′(a†(k′)eik·x2 + b(k′)e−ik·x2)].

The non-vanishing terms involve [a(k), a†(k′)] and [b†(k), b(k′)]. The first of these gives∫d3k

(2π)3√

2ω

∫d3k′

(2π)3√

2ω′ e−ik·x1 eik′·x2(2π)3δ3(k − k′)

=∫

d3k

(2π)32ωe−ik·(x1−x2)

and the second gives

−∫

d3k

(2π)32ωeik·(x1−x2).

As long as (x1−x2) is spacelike ((x1−x2)2 < 0) we can transform x1−x2 into −(x1−x2)by a continuous Lorentz transformation, and so the second term cancels the first for(x1 − x2)2 < 0.Similar manipulations show that the two terms in the commutator can be re-written asfollows. The first is ∫

d3k

(2π)32ωe−ik·(x1−x2) = 〈0|φ(x1)φ†(x2)|0〉

which has the interpretation that a particle is created at x2 and destroyed at x1. Thesecond is

−∫

d3k

(2π)32ωeik·(x1−x2) = −〈0|φ†(x2)φ(x1)|0〉

which corresponds to the creation of an antiparticle at x1 and its destruction at x2. Com-pare the comments following (7.32). Thus for the commutator to vanish in the spacelikeregion, the contributions of these two processes must cancel, and this requires the compo-nents of ‘k’ to be the same in each case - that is, the masses of the particle and antiparticlemust be identical.

7.5 The LHS of the anticommutator (7.44) is∫d3k

(2π)3√

2ω

∑s=1,2

(cs(k)uα(k, s)e−iωt+ik·x + d†s(k)vα(k, s)eiωt−ik·x),

∫d3k′

(2π)3√

2ω′∑s′=1,2

(c†s′(k′)u†β(k

′, s′)eiω′t−ik′·y + ds′(k′)v†β(k

′, s′)e−iω′t+ik′·y)

.


From (7.41) the only non-vanishing terms involve cs(k), c†s′(k′) and d†s(k), ds′(k′),which give∫

d3k

(2π)3√

2ω

∫d3k′

(2π)3√

2ω′∑s=1,2

∑s′=1,2

[uα(k, s)u†β(k′, s′)e−i(ω−ω′)t eik·x e−ik′·y(2π)3δ3(k − k′)δss′ + similar term in vαv

†β ]

=∫

d3k

(2π)32ω[∑s=1,2

uα(k, s)u†β(k, s)eik·(x−y) +

∑s=1,2

vα(k, s)v†β(k, s)e−ik·(x−y)]. (C)

We need to evaluate∑s uα(k, s)u†β(k, s) and

∑s vα(k, s)v†β(k, s). Problem 7.8 below deals

with similar expressions and gives some hints. First, note that ‘uu†’ is a matrix (u†u issingle number), and uαu

†β is its (α, β) element. We can calculate the matrix u(k, s)u†(k, s)

straightforwardly using (4.105):

u(k, s)u†(k, s) = (ω +m)

(φs

σ·kω+mφ

s

)(φs† φs† σ·k

ω+m

)= (ω +m)

(φsφs† φsφs† σ·k

ω+mσ·kω+mφ

sφs† σ·kω+mφ

sφs† σ·kω+m

).

We now use∑s φ

sφs† = 1 (see problem 7.8) to obtain

∑s

u(k, s)u†(k, s) = (ω +m)

(1 σ·k

ω+mσ·kω+m

k2

(ω+m)2

).

Since k2 = ω2 −m2 this becomes(ω +m σ · kσ · k ω −m

)(u)

which is in fact the matrix βm + α · k + ω using (4.31). For the ‘∑vv†’ term, we first

change the integration variable k to −k so as to have the same exponent as the ‘∑uu†’

term. This means that in evaluating∑s v(k, s)v

†(k, s) we use (4.114) with p replaced by−k (and E by ω), which gives(

ω −m −σ · k−σ · k ω −m

). (v)

The sum of (u) and (v) is just 2ω times the unit 4 × 4 matrix, whose (α, β) element isδαβ . Inserting this into (C) gives (7.44).

7.6

Nψ =∫

d3x

∫d3k

(2π)3√

2ω

∑s

[c†s(k)u†(k, s)eik·x + ds(k)v†(k, s)e−ik·x]

×∫

d3k′

(2π)3√

2ω′∑s′

[cs′(k′)u†(k′, s′)e−ik′·x + d†s′(k′)v†(k′, s′)eik′·x].

Consider the term ‘c†scs′ ’. The integral over x gives (2π)3δ3(k − k′) and the k′-integralcan then be done giving the term∫

d3k

(2π)32ω

∑s,s′

c†s(k)cs′(k)u†(k, s)u(k, s).


Taking φ1, φ2 to be orthogonal (as is conventional), problem 4.12 implies that u†(k, s)u(k, s′) =2ωδss′ , so that this term is ∫

d3k

(2π)3∑s

c†s(k)cs(k).

The ‘dsd†s′ ’ term is handled similarly, using v†(k, s)v(k, s′) = 2ωδss′ . Now consider the

‘c†sd†s′ ’ term: the integral over x yields (2π)3δ3(k + k′) so that we need to evaluate

u†((ω,k), s)v((ω,−k), s′). It is easily verified from (4.105) and (4.114) that this vanishes;similarly for the ‘dscs′ ’ term.

HD =∫

d3xπ ˙ψ − LD =

∫d3xψ†(−iα ·∇ + βm)ψ

=∫

d3x

∫d3k

(2π)3√

2ω[∑s

c†s(k)u†(k, s)eik·x + ds(k)v†(k, s)e−ik·x]

× (−iα ·∇ + βm)∫

d3k′

(2π)3√

2ω′[∑s′

cs′(k′)u(k′, s′)e−ik′·x + d†s′(k′)v(k′, s′)eik′·x].

Now note that

(−iα ·∇ + βm)u(k′, s′)e−ik′·x = (α · k′ + βm)u(k′, s′)e−ik′·x

= ω′u(k′, s′)e−ik′·x.

In the ‘c†scs′ ’ term, the x-integral can be done, and then the k′-integral, leading to∫d3k

(2π)32ω

∑s,s′

c†s(k)cs′(k)u†(k, s)u(k, s′)ω.

Use of u†(k, s)u(k, s′) = 2ωδss′ then gives the first term in (7.50). For the ‘dsd†s′ ’ term

note that

(−iα ·∇ + βm)v(k′, s′)eik′·x = (−α · k′ + βm)v(k′, s′)eik′·x

= − ω′v(k′, s′)eik′·x.

Performing the x-integral and then the k′-integral leads to the second term of (7.50). The‘c†sd

†s’ and ‘dscs’ terms vanish as in the calculation of Nψ.

7.8 The solution to problem 7.5 showed that∑s

u(k, s)u†(k, s) = (βm+ α · k + ω).

Hence ∑s

u(k, s)u(k, s) = (β2m+ αβ · k + βω) = (m− γ · k + βω)

= (6k +m).

7.9〈0|T (ψα(x1)

¯ψ(x2))|0〉 =

〈0|θ(t1 − t2)∫

d3k

(2π)3√

2ω[∑s

cs(k)uα(k, s)e−ik·x1 + d†s(k)vα(k, s)eik·x1 ]

×∫

d3k′

(2π)3√

2ω′[∑s′

c†s′(k)uβ(k′, s′)eik′·x2 + ds′(k′)vβ(k′, s′)e−ik′·x2 ]|0〉


−〈0|θ(t2 − t1)∫

d3k

(2π)3√

2ω[∑s

c†s(k)uβ(k, s)eik·x2 + ds(k)vβ(k, s)e−ik·x2 ]

×∫

d3k′

(2π)3√

2ω′[∑s′

cs′(k)uα(k′, s′)e−ik′·x1 + d†s′(k′)vα(k′, s′)eik′·x1 ]|0〉

where in the exponents k0 = (k2 + m2)1/2 ≡ ω, and k0′ = (k′2 + m2)1/2 ≡ ω′. In theθ(t1 − t2) part, the terms in d†s and ds′ vanish, using d|0〉 = 0 = 〈0|d† from (7.36); and inthe θ(t2− t1) part, the terms in c†s and cs′ vanish similarly. In the θ(t1− t2) part we thenuse cs(k), c†s′(k′) = (2π)3δ3(k − k′)δss′ to reduce it to

θ(t1 − t2)∫

d3k

(2π)32ω

∑s

uα(k, s)uβ(k, s)e−ik·(x1−x2) =

θ(t1 − t2)∫

d3k

(2π)32ω(6k +m)αβe−ik·(x1−x2).

Similarly, the θ(t2 − t1) part reduces to

−θ(t2 − t1)∫

d3k

(2π)32ω

∑s

vα(k, s)vβ(k, s)e−ik·(x2−x1) =

−θ(t2 − t1)∫

d3k

(2π)32ω(6k −m)αβe−ik·(x2−x1).

So altogether we get

〈0|T (ψα(x1)¯ψ(x2))|0〉 =

∫d3k

(2π)32ω[θ(t1 − t2)(6k +m)αβe−iω(t1−t2)+ik·(x1−x2)

−[θ(t2 − t1)(6k −m)αβe−iω(t2−t1)+ik·(x2−x1)]

where 6k = ωβ − γ · k. We need to show that this is equal to∫d4k

(2π)4e−ik·(x1−x2)

i(6k +m)αβk2 −m2 + iε

. (T )

Consider the integral over k0 in (T ). There are poles at k0 = (k2 +m2 − iε)1/2 whichis in the lower half k0-plane, and which tends to the real value ω as ε → 0; and atk0 = −(k2 +m2 − iε)1/2 which is in the upper half k0-plane, and tends to −ω as ε → 0.On the other hand, the k0-dependence of the exponential is

e−ik0(t1−t2).

We must therefore close the k0-contour (see Appendix F) in the upper half k0-plane fort1 − t2 < 0, picking up the contribution from the pole at k0 = −ω; and in the lower halfk0-plane for t1 − t2 > 0, picking up the contribution of the pole at k0 = ω. This gives

θ(t1 − t2)∫

d3k

(2π)4

(−2πi2ω

)i(ωβ − γ · k +m)αβe−iω(t1−t2)+ik·(x1−x2)

+θ(t2 − t1)∫

d3k

(2π)4

(2πi−2ω

)i(−ωβ − γ · k +m)αβeiω(t1−t2)+ik·(x1−x2).

In the second integral, change k to −k to obtain

θ(t1 − t2)∫

d3k

(2π)32ω(ωβ − γ · k +m)αβe−iω(t1−t2)+ik·(x1−x2)

−θ(t2 − t1)∫

d3k

(2π)32ω(ωβ − γ · k −m)αβe−iω(t2−t1)+ik·(x2−x1)


as required.7.10 The Euler-Lagrange equations for Aν are (compare (5.134))

∂µ(

∂L∂(∂µAν)

)=

∂L∂Aν

,

where in this caseL = −1

2Fµν∂

µAν − jem νAν

with Fµν = ∂µAν − ∂νAµ. We use a ‘brute force’ approach. Consider the E-L equationswritten out in full for one component of Aν , say A1:

∂0

(∂L

∂(∂0A1)

)+ ∂1

(∂L

∂(∂1A1)

)+ ∂2

(∂L

∂(∂2A1)

)+ ∂3

(∂L

∂(∂3A1)

)=

∂L∂A1

.

We need to identify the terms in L which involve ∂0A1, ∂1A1, ∂2A1, ∂3A1. First, notethat ∂1A1 does not appear since Fµν = 0 when µ = ν. The relevant terms are containedin

−12(F01∂

0A1 + F10∂1A0 + F12∂

1A2 + F21∂2A1 + F13∂

1A3 + F31∂3A1)

= −12(∂0A1 − ∂1A0)∂0A1 + (∂1A0 − ∂0A1)∂1A0 + . . .

Now use ∂0A1 = −∂0A1, ∂1A2 = ∂1A2, etc, to pick out just the terms involving A1,which are

12(∂0A1)2 − 1

2(∂2A1)2 − 1

2(∂3A1)2

−(∂0A1)(∂1A0) + (∂2A1)(∂1A2) + (∂3A1)(∂1A3). (R)

We then find

∂L∂(∂0A1) = ∂0A1 − ∂1A0 = ∂1A0 − ∂0A1∂L

∂(∂2A1) = ∂1A2 − ∂2A1 = ∂1A2 − ∂2A1∂L

∂(∂3A1) = ∂1A3 − ∂3A1 = ∂1A3 − ∂3A1.

(D)

Substituting these expressions into the E-L equations for A1 we obtain

∂0(∂1A0 − ∂0A1) + ∂2(∂1A2 − ∂2A1) + ∂3(∂1A3 − ∂3A1) = −jem 1, (S)

or(∂0∂0 + ∂1∂1 + ∂2∂2 + ∂3∂3)A1 − ∂1(∂0A0 + ∂1A1 + ∂2A2 + ∂3A3) = jem 1

which is the ν = 1 component of (7.62) (written with the index ν lowered).Equations (D) can be written in covariant form as

∂L∂(∂µAν)

= ∂νAµ − ∂µAν .

Those skilled in manipulation of indices in tensor calculus can derive this result directlyfrom the covariant Lagrangian; (7.62) then follows trivially.

7.11 (b) The condition (7.90) is

(A(k2)gνµ +B(k2)kνkµ)(−k2gµσ + kµkσ) = gνσ.


Multiplying out the brackets,

−k2A(k2)gνσ +A(k2)kνkσ − k2B(k2)kνkσ +B(k2)kνk2kσ = gνσ,

giving (7.91).7.12 Referring to (R) of problem 7.10, the addition of the term − 1

2 (∂µAµ)2 to L means thatwe must add to (R) the additional terms

−12(∂1A

1)2 − (∂1A1)(∂0A

0 + ∂2A2 + ∂3A

3).

This produces a non-zero value for ∂L/∂(∂1A1), namely

∂L∂(∂1A1)

= −(∂1A1) + (∂0A0 − ∂2A2 − ∂3A3).

Including then the contribution of ∂1(∂L/∂(∂1A2)) in (S) of problem 7.10 (with jem 1 setto zero) leads to the required equation of motion for A1.

7.13 The εµ(k, λ)’s are given in (7.102)-(7.104), from which we note that, for each value ofλ, the only non-vanishing component of εµ(k, λ) is that in which µ = λ, which equals 1.Consider then the case µ = 0 in (7.115): the λ = 1, 2 and 3 terms all vanish, and the onlysurviving ν-component is ν = 0, for which the LHS is -1, verifying the result. Similarlyfor the other cases.

7.14 For (7.119): following the same steps as in (7.87)-(7.91), we require (M−1)µν where

Mµν = (−k2gµν + (1− 1/ξ)kµkν).

Let(M−1)νσ = A(k2)gνσ +B(k2)kνkσ.

The condition gµσ = Mµν(M−1)νσ gives

gµσ = (−k2gµν + (1− 1/ξ)kµkν)(A(k2)gνσ +B(k2)kνkσ)= − k2A(k2)gµσ − k2B(k2)kµkσ +A(k2)(1− 1/ξ)kµkσ +B(k2)(1− 1/ξ)kµkσk2.

To match coefficients of gµσ on both sides we require A = −1/k2, and then the vanishingof the kµkσ term gives B = (1− ξ)/(k2)2.

7.15 The Hamiltonian density isH = HS + H′S

where HS is the Hamiltonian density for the free complex scalar field,

HS = π†π + ∇φ† ·∇φ+m2φ†φ,

and H′S is the interaction part. But also the general definition of the Hamiltonian is

H = π˙φ+ π†

˙φ† − L.

Now

π =∂L

∂˙φ

= ˙φ† − iqφ†A0

from (7.134) - (7.136), and

π† = ˙φ+ iqφA0.


Substituting for ˙φ and ˙

φ† in favour of π and π†, we obtain

H = π(π† − iqφA0) + π†(π + iqφ†A0)

−(π + iqφ†A0)(π† − iqφA0)−∇φ† ·∇φ−m2φ†φ − Lint

= π†π + ∇φ† ·∇φ+m2φ†φ− q2φ†φ(A0)2 − Lint

which establishes (7.139).7.16 Consider first

∂2ν〈0|T (φ(x1)φ†(x2)|0〉 =

∂2νθ(t1 − t2)〈0|φ(x1)φ†(x2)|0〉+ θ(t2 − t1)〈0|φ†(x2)φ(x1)|0〉,

where ∂2ν = ∂/∂xν2 (similarly for ∂1µ). If ν = i, a spatial index, then the derivative passesthrough the θ-functions and we get simply 〈0|T (φ(x1)∂2iφ

†(x2))|0〉. If ν = 0, then thedifferentiation produces extra terms

−δ(t1 − t2)〈0|φ(x1)φ†(x2)|0〉+ δ(t2 − t1)〈0|φ†(x2)φ(x1)|0〉

which cancel using (7.32) (at equal times (x1 − x2)2 is spacelike). So

∂2ν〈0|T (φ(x1)φ†(x2))|0〉 = 〈0|T (φ(x1)∂2ν φ†(x2))|0〉.

Now consider

∂1µ∂2ν〈0|T (φ(x1)φ†(x2))|0〉 = ∂1µ〈0|T (φ(x1)∂2ν φ†(x2))|0〉 =

∂1µθ(t1 − t2)〈0|φ(x1)∂2ν φ†(x2)|0〉+ θ(t2 − t1)〈0|∂2ν φ

†(x2)φ(x1)|0〉.

As before, if µ = j the derivative passes through the θ-functions and we get 〈0|T (∂1j φ(x1)∂2ν φ†(x2)|0〉.

On the other hand, if µ = 0 we get

δ(t1 − t2)〈0|φ(x1)∂2ν φ†(x2)|0〉+ θ(t1 − t2)〈0|∂10φ(x1)∂2ν φ

†(x2)|0〉

−δ(t1 − t2)〈0|∂2ν φ†(x2)φ(x1)|0〉+ θ(t2 − t1)〈0|∂2ν φ

†(x2)∂10φ(x1)|0〉.

= 〈0|T (∂10φ(x1)∂2ν φ†(x2))|0〉+ 〈0|[φ(x1), ∂2ν φ

†(x2)]|0〉δ(t1 − t2).

If ν is a spatial index, the equal time commutator vanishes, as can be seen by differentiating(7.32) with respect to a spatial component of x2. But if ν = 0 then the equal timecommutator is

[φ(x1),˙φ†(x2)]δ(t1 − t2) = iδ3(x1 − x2)δ(t1 − t2).

These results establish (7.140).

Chapter 8

8.2 We use (8.25), and the mode expansion (7.16) in (8.23), to obtain

〈s+, p′|jµem,s(x)|s+, p〉 = ie√

4EE′〈0|a(p′)[φ†∂µφ− (∂µφ†)φ]a†(p)|0〉

= ie√

4EE′〈0|a(p′)∫

d3k′

(2π)3√

2ω′[a†(k′)eik′·x + b(k′)e−ik′·x]

×∫

d3k

(2π)3√

2ω[−ikµa(k)e−ik·x + ikµb†(k)eik·x]a†(p)|0〉

−(∂µφ†)φ term


where E =√M2 + p2, E′ =

√M2 + p′2, ω =

√M2 + k2, and ω′ =

√M2 + k′

2. Weevaluate the φ†∂µφ term. The a and b operators commute with each other, so we canmove the b(k) all the way to the right where it will give zero on |0〉; similarly the b†(k′)will give zero on 〈0|. This term therefore reduces to

ie√

4EE′〈0|a(p′)∫

d3k′

(2π)3√

2ω′a†(k′)eik′·x

∫d3k

(2π)3√

2ω− ikµa(k)e−ik·xa†(p)|0〉.

We then usea(k)a†(p) = a†(p)a(k) + (2π)3δ3(k − p),

and similarly for the product a(p′)a†(k′), together with

a(k)|0〉 = 〈0|a†(k′) = 0,

to get rid of all the operators. The δ-functions allow all the momentum integrals to beperformed, setting k = p and k′ = p′ so that ω = E, ω′ = E′, kµ = pµ, and k′

µ = p′µ.

This term becomes simplyepµe−i(p−p′)·x;

the (∂µφ†)φ term supplies the p′µ contribution to (8.27).8.3 We use (8.32) to obtain

〈s−, p′|jµem,s(x)|s−, p〉 = ie√

4EE′〈0|b(p′)same expression for

the current as in problem 8.2b†(p)|0〉.

This time, in the φ†∂µφ term of the current, we can move a(k′) to the right to annihilateon |0〉, and a†(k′) to the left to annihilate on 〈0|. This leaves an expression of the form

〈0|b(p′).....b(k′)....b†(k)....b†(p)|0〉.

But we must remember to normally order the operators in the current: this means thatthe above expression should in fact be

〈0|b(p′).....b†(k)....b(k′)....b†(p)|0〉.

We now use the commutation relations for b(p′) and b†(k) to get b†(k) anihilating on|0〉, and similarly for b(k′) and b†(p) to get b(k′) annihilating on |0〉. This leaves delta-functions δ(p−k′) and δ(k−p′), allowing the momentum integrals (in this first term) tobe performed, setting k = p′,k′ = p, and hence ω = E′, ω′ = E, k = p′ and k′ = p. Theresult is

−ep′µe−i(p−p′)·x

for this first term. The second supplies the −epµ part, and the whole matrix element isindeed the negative of the one in problem 8.2.

8.4 We consider a Lorentz transformation along the x1-axis. Then

j′x = ψ′†αxψ

′ = ψ†eαxϑ/2αxeαxϑ/2ψ

= ψ†αxeαxϑψ

= ψ†αx(coshϑ+ αx sinhϑ)ψ= coshϑ jx + sinhϑ ρ

as in (4.86), where we have used (4.90), and (4.91) with ϑ/2 replaced by ϑ.


8.5 (a) We are assuming E = E′, as in the required application. Then

u†(k′, s′ = 1)u(k, s = 1) = (E +m)(φ1† φ1† σ·k′

(E+m)

)( φ1

σ·k(E+m)φ

1

)

= (E +m)

1 + φ1†(

k · k′

(E +m)2+ i

σ · k′ × k

(E +m)2

)φ1

which reduces immediately to the given expression.

8.5 (b) We have

φ1 =(

10

), and φ2 =

(01

).

Also

σ ·A =(

Az Ax − iAyAx + iAy −Az

).

Then, by straightforward matrix multiplication we obtain, for instance,

φ1†σ ·Aφ2 = Ax − iAy.

Generally,φi†σ ·Aφj = (σ ·A)ij .

(c) The expression S of (8.46) is

S =12|u′†s′=1us=1|2 + |u′†s′=1us=2|2 + |u′†s′=2us=1|2 + |u′†s′=2us=2|2.

Part (a) evaluated u′†s′=1us=1, and u′

†s′=2us=2 is the same but with φ1 replaced by φ2.

Since φ1†φ2 = 0, we find

u′†s′=1us=2 = (E +m)

iφ1†σ · k′ × kφ2

(E +m)2

and similarly for u′†s′=2us=1. The result follows after noting that

|φ1†σ ·Aφ1|2 + |φ1†σ ·Aφ2|2 + |φ2†σ ·Aφ1|2 + |φ2†σ ·Aφ2|2 = 2A2.

(d)

S = (E = m)2[

1 +k2 cos θ

(E +m)2

]2+

(k2)2 sin2 θ

(E +m)4

= (E +m)2[

1 +(E −mE +m

)cos θ

]2+(E −mE +m

)2

sin2 θ

= [(E +m) + (E −m) cos θ]2 + (E −m)2 sin2 θ= (E +m)2 + 2k2 cos θ + (E −m)2

= 2(E2 +m2 + k2 cos θ)= 2[E2 +m2 + k2(1− 2 sin2 θ/2)]

= 4E2(1− k2

E2sin2 θ/2)

and the result follows using v = |k|/E.


8.6 The manipulations are similar to those in problem 8.2 except that anticommutation rela-tions are used.

8.7 γµ = (γ0, γ0α). Soγµ† = (γ0†,α†γ0†) = (γ0,αγ0).

Henceγ0γµ†γ0 = ((γ0)3, γ0α(γ0)2) = (γ0, γ0α) = γµ.

8.8

Tr[( 6k′ +m)γµ(6k +m)γν ] = Tr(6k′γµ 6kγν) +mTr(γµ 6kγν)+mTr(6k′γµγν) +m2Tr(γµγν).

The terms linear in m vanish by (8.73); (8.74) gives Tr(γµγν) = 4gµν , and (8.75) impliesthat

Tr[6k′γµ 6kγν ] = 4[k′µkν + k′νkµ − k′ · kgµν ].

These results establish (8.78).8.9

L00 = 2[2k′0k0 − (k′ · k)] + 2m2.

Here, k0 = (m2 + k2)1/2 = k′0 ≡ E; k′ · k = k0k′

0 − k · k′ = E2 − k2 cos θ. So

L00 = 2[E2 + k2 cos θ] + 2m2

= 2[E2 +m2 + k2 cos θ]

and the result follows as in problem 8.5(d).8.11 Using (8.27), (8.55), and the Fourier transform of (7.119) (compare (6.98) and (7.58)),

the term written explicitly in (8.93) becomes

(−i)2

2

∫ ∫d4x1 d4x2e(p+ p′)µe−i(p−p′)·x1

×∫

d4l

(2π)4e−il·(x1−x2) .

i[−gµν + (1− ξ)lµlν/l2]l2

.− eu(k′, s′)γνu(k, s)e−i(k−k′)·x2.

The integral over x1 gives (2π)4δ4(p+l−p′), and that over x2 gives (2π)4δ4(l+k′−k). Theintegral over l can now be done using one of these delta functions, which sets l = k−k′ = qeverywhere, including in the other delta function. This gives one half times (8.95) - theother half is supplied by the ‘(x1 ↔ x2)’ part of (8.93).

8.12 Using∂µe−i(p−p′)·x = −i(p− p′)µe−i(p−p′)·x,

we find from (8.27)

∂µ〈s+, p′|jµem,s(x)|s+, p〉 = −ie(p2 − p′2)e−i(p−p′)·x

which vanishes since p2 = p′2 = M2. Similarly,

∂µ〈e−, k′, s′|jµem,e(x)|e−, k, s〉 = ieu′(6k− 6k′)ue−i(k−k′)·x

which vanishes since 6 ku = mu and u′ 6 k′ = mu′. (The particles in the external statesare ‘on-shell’ - i.e. their wavefunctions satisfy the free-particle equations of motion, orequivalently their 4-momenta satisfy the energy-momentum condition for a free particle.]

8.13 We haveLµνTµν = 2[k′µkν + k′

νkµ + (q2/2)gµν ](p+ p′)µ(p+ p′)ν .


It is advantageous to write p′ = p + q, so that (p + p′)µ = (2p + q)µ, and similarly for(p+ p′)ν , and then drop the qµqν part using (8.188). Then

LµνTµν = 8[2p · k′ p · k + (q2/2)p2]= 8[2p · k′ p · k + (q2/2)M2].

8.14 We have (correcting the misprint in (8.134))

F (q2) =∫

e−iq·xρ(x)d3x =∫

e−iq·x e−|x|/a

8πa3d3x

=1

8πa3

∫e−i|q||x| cos θ e−|x|/a |x| d|x| 2πd(cos θ).

The integral over cos θ can be done with the result

1i|q||x|

(ei|q||x| − e−i|q||x|

)so that F (q2) becomes

F (q2) =1

4a3i|q|

∫ ∞

0

|x|e−|x|/a (ei|q||x| − e−i|q||x|) d|x|.

One way of doing the |x|-integral is to write it as

∂

∂(−1/a)

∫ ∞

0

e−|x|/a (ei|q||x| − e−i|q||x|) d|x|,

where now the simple exponential integrals can be done, and the differentiation withrespect to a performed, after using

∂

∂(−1/a)= a2 ∂

∂a.

This leads to (8.135).8.15 After replacing εµ by kµ, (8.161) becomes (after correcting the misprints in it)

−e2ε∗ν(k′, λ′)u(p′, s′)γν(6p+ 6k +m)

(p+ k)2 −m26ku(p, s)

−e2ε∗ν(k′, λ′)u(p′, s′)6k(6p− 6k′ +m)

(p− k′)2 +m2γνu(p, s).

In the first term, replace 6ku(p, s) by

(6k′ + 6p′ − 6p)u(p, s) = ( 6k′ + 6p′ −m)u(p, s) = ( 6p+ 6k −m)u(p, s)

using the Dirac equation and 4-momentum conservation. Also note that

(6p+ 6k +m)(6p+ 6k −m) = (p+ k)2 −m2.

The first term is then−e2ε∗ν(k′, λ′)uγνu.

In the second term, replace u(p′, s′)6k by

u(p′, s′)(6k′ + 6p′ − 6p) = u(p′, s′)(6k′ − 6p+m) = −u(p′, s′)(6p− 6k′ −m).


The second term becomese2ε∗ν(k

′, λ′)uγνu

which cancels the first. Similarly when ε∗ν(k′, λ′) is replaced by k′ν .

8.16 (a) We work in the massless limit, as in section 8.6.3. We have∑λ,λ′,s,s′

M(s)γe− M

(u)∗γe− =

e4

su

∑λ,λ′,s,s′

ε∗ν(k′, λ′)ερ(k′, λ′)εµ(k, λ)ε∗σ(k, λ)u(p′, s′)γν(6p+ 6k)γµu(p, s)u(p, s)γρ(6p− 6k′)γσu(p′, s′)

=e4

sugνρgµσTrγν(6p+ 6k)γµ 6pγρ(6p− 6k′)γσ 6p′

=e4

suTrγν(6p+ 6k)γµ 6pγν(6p− 6k′)γµ 6p′.

We now useγµ 6pγν(6p− 6k′)γµ = −2(6p− 6k′)γν 6p

from (J.5), and then

γν(6p+ 6k)(6p− 6k′)γν = 4(p+ k) · (p− k′)

from (J.4), and finallyTr(6p 6p′) = 4p · p′

from (J.30) to obtain∑λ,λ′,s,s′

M(s)γe− M

(u)∗γe− = −32

e4

su(p+ k) · (p− k′) p′ · p.

Now

(p+ k) · (p− k′) = p2 − p · k′ + k · p− k · k′

= − p′ · k + k · p− k · k′

since p · k′ = p′ · k in the massless limit for all external lines (as follows from squaringp− k′ = p′ − k). Hence

(p+ k) · (p− k′) = k · (p− p′ − k′) = −k2.

This therefore vanishes when the photon is on-shell (k2 = 0).8.17 The interference term calculated in problem 8.16 (a) had the value∑

λ,λ′,s,s′

M(s)γe− M

(u)∗γe− = −32

e4

su(p+ k) · (p− k′) p′ · p.

This time (with k2 = −Q2) we have

(p+ k) · (p− k′) = p2 − p · k′ + k · p− k · k′

= − p · k′ + k′ · p′ +Q2/2− k · k′

using k · p = k′ · p′ +Q2/2 (from squaring k + p = k′ + p′). So

(p+ k) · (p− k′) = Q2/2 + k′ · (p′ − p− k) = Q2/2− (k′)2 = Q2/2.


Hence using p · p′ = −t/2, we obtain

14

∑λ,λ′,s,s′

M(s)γe− M

(u)∗γe− = e4

2tQ2

su.

There are two such terms in the product(M(s)

γe− +M(u)γe−

)(M(s)

γe− +M(u)γe−

)∗,

which are equal.8.18 (a) We calculate LµνM

µνeff using (8.185) and (8.190).

LµνMµνeff = 4[k′µkν + k′νkµ + (q2/2)gµν ][2pµpν + (q2/2)gµν ]

= 4[4k′ · p k · p+ q2k · k′ + q2p2 + 4(q2/2)2]= 4[4k′ · p k · p+ q2(−q2/2) + q2M2 + (q2)2]= 8[2k′ · p k · p+ (q2)2/4 + q2M2/2].

In ‘laboratory’ frame with pµ = (M,0), neglecting the electron mass so that ω = |k| ≡k, ω′ = |k′| ≡ k′, and using (8.219), the preceding expression becomes

LµνMµνeff = 8[2kk′M2 + 4k2k′2 sin4 θ/2− 2kk′M2 sin2 θ/2]

= 8[2kk′M2(1− sin2 θ/2) + 4k2k′2 sin4 θ/2]

= [16kk′M2 cos2 θ/2][1 + 2kk′

M2sin2 θ/2 tan2 θ/2]

= [16kk′M2 cos2 θ/2][1− (q2/2M2) tan2 θ/2]

The first factor is exactly that in (K.24), leading to the ‘no-structure’ cross section.(b) Returning to line 3 of part (a):

LµνMµνeff = 4[4k′ · p k · p+ q2k · k′ + q2p2 + 4(q2/2)2].

With all particles massless, s = 2k ·p, t = −2k ·k′, and u = −2p ·k′. Hence our expressionbecomes

4[−us− t2/2 + t2] = 4[−us+ t2/2].

dσ/dt is then given bydσdt

=1

16π(4πα)2

t24[−u

s+

t2

2s2].

In the massless case we have s + t + u = 0 so that t2 = (u + s)2, leading to the requiredexpression for dσ/dt.From (K.41), (K.46) and (K.50) we find

dy =k′2

2πkM2π

dt2k′2

.

But in this frame s = (k + p)2 = 2kM . Hence dt = sdy and the result is established.8.19 (a) We shall label the 4-momentum and spin of the ingoing e− by k, s, of the ingoing e+

by k1, s1, of the outgoing µ− by p′, r′, and of the outgoing µ+ by p1, r1. The Mandelstamvariables are

Q2 = (k + k1)2 = (p′ + p1)2, t = (k1 − p1)2 = (k − p′)2, u = (k − p1)2 = (k1 − p′)2.

The amplitude is

iev(k1, s1)γµu(k, s)−igµνQ2

ieu(p′, r′)γνv(p1, r1).


Note that the vγµu factor depends only on the e− and e+ momenta, while the uγνv factordepends only on the momenta of the µ− and µ+.(b) The spin-averaged squared cross matrix element is then

|M|2 =14

(e2

Q2

)∑s,s1

v(k1, s1)γµu(k, s)u(k, s)γνv(k1, s1)

×∑r′,r1

u(p′, r′)γµv(p1, r1)v(p1, r1)γνu(p′, r′)

=(4πα)2

Q4L(e)µνL(µ)µν

whereL(e)µν =

12Tr[( 6k1 −m)γµ(6k +m)γν ]

andL(µ)µν =

12Tr[( 6p′ +M)γµ(6p1 −M)γν ],

using (7.61).(c) Using the trace theorems as in (8.78), the lepton tensors are

L(e)µν = 2[k1µkν + k1νkµ − (k1 · k)gµν ]

andL(µ)µν = 2[p′µpν1 + p′νpµ1 − (p′ · p1)gµν ]

where now (in the massless limit) Q2/2 = p′ · p1 = k1 · k. Hence

|M|2 =(e2

Q2

)4[2p′ · k1 p1 · k + 2p′ · k p1 · k1 −Q2p′ · p1 −Q2k1 · k + (Q2)2]

=(e2

Q2

)4[2

u2

4+ 2

t2

4−Q2Q2/2−Q2Q2/2 + (Q2)2]

=(e2

Q2

)2[t2 + u2].

In the massless limit, all 3-momenta have equal modulus which (slightly confusingly) wedenote by k. Then

t = −2k2(1− cos θ), u = −2k2(1 + cos θ)

so thatt2 + u2 = 8k4(1 + cos2 θ).

Since Q2 = 4k2 the result follows.Crossing symmetry implies that the amplitude for

e−(k, s) + e+(k1, s1)→ µ−(p′, r′) + µ+(p1, r1)

is equal to (minus) the amplitude for

e−(k, s) + µ−(−p1,−r1)→ µ−(p′, r′) + e−(−k1,−s1).

We can therefore obtain this amplitude from the one calculated in section 8.7 by makingthe replacements

p, r → −p1,−r1 and k′, s′ → −k1,−s1.


The spin labels disappear at the Trace stage, of course, and

q2(section 8.7) = (k − k′)2 → (k + k1)2 = Q2.

So (8.185) and (8.186) become

(8.185)→ 2[−k1µkν − k1νkµ + (Q2/2)gµν ]

and(8.186)→ 2[−p′µpν − p′νpµ + (Q2/2)gµν ]

leading to exactly the same result for |M|2.(d) Using the formula (6.129) for the differential cross section for elastic scattering in thecentre of mass, and replacing |Mfi|2 by |M|2, we obtain

σ =∫

dσdΩ

=∫

164π2Q2

16π2α2(1 + cos2 θ) 2π d cos θ

=α2

4Q22π(

cos θ +cos3 θ

3

)1

−1

= 4πα2/3Q2

as required.8.20 We have

u(p′)iσµν

2Mqνu(p) = iu(p′)

12i(γµγν − γνγµ

2M

)(p′ν − pν)u(p)

= − 14M

u(p′)(γµ 6p′ − γµ 6p− 6p′γµ + 6pγµ)u(p)

= − 14M

u(p′)[γµ 6p′ + 6pγµ]u(p) +12u(p′)γµu(p).

Now

u(p′)γµ 6p′u(p) = u(p′)γµγρp′ρu(p)= u(p′)(2gµρ − γργµ)p′ρu(p)= u(p′)2p′µu(p)− u(p′)6p′γµu(p)= u(p′)2p′µu(p)−Mu(p′)γµu(p),

and similarlyu(p′)6pγµu(p) = u(p′)2pµu(p)−Mu(p′)γµu(p).

These formulae establish the required result.

Chapter 9

9.1 Using (8.206) with F1 = 1 and κ = 0 for the current matrix elements, and cancelling afactor of e2, we obtain

Wµνel =

18πM

∑s,s′

u(p, s)γµu(p′, s′)u(p′, s′)γνu(p, s)(2π)4δ4(p+ q − p′) 1(2π)3

d3p′

2E′

=1

8πMTrγµ(6p′ +M)γν(6p+M)(2π)4δ4(p+ q − p′) 1

(2π)3d3p′

2E′


where the Trace is 2Mµν (see (8.186)). Equation (9.104) then becomes

dσ =(

4παq2

)2 14[(k · p)2 −m2M2]1/2

LµνMµν(2π)4δ4(p+ q − p′) d3p′

2E′(2π)3d3k′

2ω′(2π)3.

This is precisely the formula which which yields the cross section for elastic eµ scattering(see Appendix K for the flux and phase space factors).

9.2 (a) Omitting the terms involving qµ and qν from the expression (9.10) for Wµν , we needto evaluate

LµνWµν = 2[k′µkν + k′νkµ + (q2/2)gµν ][−gµνW1 + (pµpν/M2)W2]

= 2[(−2k′ · k − 2q2)W1 + (2p · k′ p · k + q2p2/2)W2/M2].

In the ‘laboratory’ system, and neglecting the electron mass (compare (8.217) and (8.218)),

p · k′ = ω′M, p · k = ωM, q2 = −2k · k′,

and so(2p · k′ p · k + q2p2/2)/M2 = 2ωω′ + q2/2 = 2ωω′ − k · k′.

Writing as usual k = ω = |k|, k′ = ω′ = |k′|, we have

k · k′ = kk′(1− cos θ) = 2kk′ sin2 θ/2

and2ωω′ − k · k′ = kk′(1 + cos θ) = 2kk′ cos2 θ/2.

HenceLµνW

µν = 4kk′[2W1 sin2 θ/2 +W2 cos2 θ/2]

and, from (9.104),

dσ =(

4παq2

)2 14[(k · p)2 −m2M2]1/2

4πM4kk′[2W1 sin2 θ/2 +W2 cos2 θ/2]d3k′

2ω′(2π)3.

Now (q2)2 = 16k2k′2 sin4 θ/2, and (neglecting the electron mass)[(k · p)2 −m2M2]1/2 = kM . Also,

d3k′ = k′2dk′dΩ.

Hence

dσ =α2

4k2 sin4 θ/2[2W1 sin2 θ/2 +W2 cos2 θ/2]dk′dΩ

as required.(b) We have

Q2 = 2kk′(1− cos θ) and ν = k − k′.

Hence

d cos θ dk′ =1∣∣∣∣∣ ∂Q2

∂ cos θ∂Q2

∂k′∂ν

∂ cos θ∂ν∂k′

∣∣∣∣∣dQ2 dν

=1∣∣∣∣ 2kk′ 2k(1− cos θ)

0 1

∣∣∣∣ dQ2 dν =1

2kk′dQ2 dν.


Using dΩ = 2πd cos θ and the result of part (a), this leads to (9.16) from (9.12).Similarly, since

x = Q2/2Mν and y = ν/k,

we have

dQ2dν =1∣∣∣∣∣ ∂x

∂Q2∂x∂ν

∂y∂Q2

∂y∂ν

∣∣∣∣∣dx dy

=1∣∣∣∣ 1

2Mν − Q2

2Mν2

0 1k

∣∣∣∣ = 2Mνk dx dy

= 2Mk2y dx dy

as in (9.19), leading straightforwardly to the formula for d2σ/dxdy.9.3 (a) The transverse polarization vectors are given in (9.38). These satisfy ε(λ = ±1) ·p = 0

in the laboratory frame, and also (9.40). Hence in the product εµε∗νWµν , with Wµν given

by (9.10), only the contraction with −gµνW1 survives, leading to

12

∑λ=±1

εµ(λ)ε∗ν(λ)Wµν =12[−εµ(λ = 1)ε∗µ(λ = 1)− εµ(λ = −1)ε∗µ(λ = −1)]W1

=12[1 + 1]W1 = W1

and (9.46) follows from (9.45).(b) From (9.47) the longitudinal/scalar virtual photon cross section is

σS = (4π2α/K)ε∗µ(λ = 0)εν(λ = 0)Wµν

where Wµν given by (9.10), and where εµ(λ = 0) is real and given by (9.41), and satisfiesq · ε = 0 (see (9.49)). Thus in the contractions with Wµν , terms involving qµ and qν canbe dropped. The W1 term in ‘ε · ε W ’ is then simply (−ε · ε)W1 = −W1 (note (9.42)),while the W2 term is

1M2

ε(λ = 0) · p ε(λ = 0) · p W2 =1

M2Q2(q3M)(q3M)W2

=(q3)2

Q2W2 =

Q2 + (q0)2

Q2W2 =

(1 +

ν2

Q2

)W2,

and (9.48) follows from these results.From (9.46) we obtain

W1 =K

4π2ασT,

and substituting this into (9.48) gives the required result for W2.9.4 In the limit m→ 0 the spinors φ and χ satisfy σ ·pφ = Eφ and σ ·pχ = −Eχ respectively,

where in both cases E = |p|. Hence φ satisfies

σ · p|p|

φ = φ

which shows it has positive helicity (compare (4.67)); similarly χ has negative helicity.(b) For example,

PRPL =(

1 + γ5

2

) (1− γ5

2

)=

14[1− γ2

5 ] = 0.


When m 6= 0, the operators PR and PL still project out the φ and χ components of the4-component spinor, but these 2-component objects are no longer (with m 6= 0) helicityeigenstates (since, for example, (σ · p/|p|)φ is no longer equal to φ or −φ).(c) We may write

uγµu = u†(PR + PL)γ0γµ(PR + PL)u.

We exploit the fundamental relation γµγ5 = −γ5γµ (see (J.11)). Consider one ‘cross’

term:

u†PRγ0γµPLu = u†γ0PLγ

µPLu

= u†γ0γµPRPLu

= 0.

Similarly for the term u†PLγ0γµPRu. The only surviving terms are

u†PRγ0γµPRu+ u†PLγ

0γµPLu

which is just uRγµu + uLγ

µu. Hence ‘R’ states connect only to ‘R’ states, and similarlyfor ‘L’ states, and so helicity (in the massless limit) is conserved.Note that ‘uR’ could perhaps more clearly be written as

uR

since we form it by taking the dagger of uR and then multiplying by γ0 - i.e. we take theDirac ‘bar’ of uR. uR is however the conventional notation.(d) In this case a typical cross term is

u†PRγ0γµγ5PLu = u†γ0PLγ

µγ5PLu

= u†γ0γµPRγ5PLu

= u†γ0γµγ5PRPLu

= 0,

and again helicity is conserved.(e) The Dirac mass term is

¯ψψ = ψ†(PR + PL)γ0(PR + PL)ψ.

Consider a ‘diagonal’ term:

ψ†PRγ0PRψ = ψ†γ0PLPRψ = 0,

and similarly for the other diagonal term. Only the ‘L-R’ and ‘R-L’ terms survive (thedaggers in the printed answer are a misprint).

9.5 Neglecting the positron and proton masses, their 4-momenta are pe+ = (k, 0, 0,−k), say,and pp = (p, 0, 0, p). Then

W 2CM = (pe+ + pp)2 = (k + p)2 − (k − p)2 = 4kp.

So WCM = 2√kp = 300.3 GeV.

A leptoquark of mass Mlq formed as a resonance state of the e+ and the struck quarkwould appear as a peak in the effective mass of the e+ and the quark, at an effective massequal to Mlq. In a simple parton model picture, this efective mass is

√(pe+ + xpp)2. So

we expect a peak whenp2e+ + 2x pe+ · pp + x2p2

p = M2lq,


or, neglecting the positron and proton masses, at x = M2lq/W

2CM.

9.6 (a) Using (9.92) for d2σ/dx1 dx2, we have

dσdq2

=∫

dx1 dx24πα2

9q2∑a

e2a[qa(x1)qa(x2) + qa(x1)qa(x2)]δ(q2 − sx1x2)

=4πα2

9q2

∫dx1 dx2

∑a

e2a[qa(x1)qa(x2) + qa(x1)qa(x2)]1sδ(q2/s− x1x2)

with the help of (E.29), and then writing 1/s = x1x2/q2 and τ = q2/s we obtain the

desired formula.(b)

dq2 dxF =

∣∣∣∣∣ ∂q2

∂x1

∂q2

∂x2∂xF∂x1

∂xF∂x2

∣∣∣∣∣ dx1 dx2

=∣∣∣∣ sx2 sx1

1 −1

∣∣∣∣ dx1 dx2

= − s(x1 + x2) dx1 dx2.

The minus sign can be absorbed by appropriate choice of limits in the q2−xF integration.In the variables (x1, x2), the integration is over the square 0 ≤ x1 ≤ 1, 0 ≤ x2 ≤ 1.Consider performing the integration holding x1 fixed and integrating over x2, and thenintegrating over x1. Take x1 = 1/2 as an example, with x2 running from x2 = 0 to x2 = 1.In the q2 − xF plane, this line maps into the line 2q2/s+ xF = 1/2, and it is traversed inthe sense of q2/s increasing (from 0 to 1/2) but xF decreasing (from 1/2 to -1/2). We canreverse the sense in which xF is covered by invoking the minus sign from the determinant.The variables x1 and x2 are given in terms of xF and τ by x1 − x2 = xF and x2 = τ/x1.So we have

x1 − τ/x1 = xF.

Solving for x1 (which is greater than 0) we find

x1 =12[xF + (x2

F + 4τ)1/2],

and hencex2 =

12[−xF + (x2

F + 4τ)1/2],

so that x1 + x2 = (x2F + τ)1/2. Hence

d2σ =4πα2

9q2∑a

e2a[qa(x1)qa(x2) + qa(x1)qa(x2)] dx1 dx2

=1

(x1 + x2)s. . . dq2 dxF

=1

(x2F + 4τ)1/2

4πα2τ

9q4∑a

e2a[qa(x1)qa(x2) + qa(x1)qa(x2)] dq2 dxF

which leads to the desired expression.9.7 Let the 4-momenta of the incoming q and q be k and k1 respectively, and let those of

the outgoing µ− and µ+ be p′ and p1. Then the q − q − γ vertex, for scalar quarks, isproportional to (k − k1)µ, while the γ − µ− − µ+ vertex is same as in problem 8.19(b).Thus in evaluating the unpolarized cross section we need the contraction

T = (k − k1)µ(k − k1)ν(p′µp1ν + p′νp1µ − (Q2/2)gµν)

= 2p′ · (k − k1) p1 · (k − k1)− (Q2/2)(k − k1)2.


Introduce the Mandelstam variables

s = (k + k1)2 = Q2, t = (k − p′)2 = (k1 − p1)2, and u = (k − p1)2 = (k1 − p′)2.

Then, negelecting lepton masses,

T = 2[− t2

+u

2][−u

2+t

2]− (Q2/2)(−Q2)

= − 12(t− u)2 +

12(Q2)2

= − 12(4k2 cos θ)2 +

12(4k2)2

∝ (1− cos2 θ)

where k is the CM momentum.

Chapter 10

10.1 We need to calculate the Jacobian for the change of variables

x1, x2, x3, x4 → x, y, z,X. (A)

Each of these variables is in fact a 4-vector, with (therefore) four components, as is explicitin the notation d4x1 etc. Thus there are a total of 16 variables altogether, and we certainlydon’t want to deal with a 16× 16 determinant. We can however write the 16-dimensionalintegration element as

[dx01 dx0

2 dx03 dx0

4] [dx11 dx1

2 dx13 dx1

4] [dx21 etc]

and imagine doing the transformation first for the ‘ 0 ’ components of the two sets in (A):

x01, x

02, x

03, x

04 → x0, y0, z0, X0,

then for the ‘ 1 ’ components, etc. The total Jacobian will then be the product of four4× 4 Jacobians, one for each of the components. But since they all have exactly the sameform, we shall suppress the explicit ‘component’ label, and write simply

J =

∣∣∣∣∣∣∣∣∣∂x∂x1

∂x∂x2

∂x∂x3

∂x∂x4

∂y∂x1

∂y∂x2

∂y∂x3

∂y∂x4

∂z∂x1

∂z∂x2

∂z∂x3

∂z∂x4

∂X∂x1

∂X∂x2

∂X∂x3

∂X∂x4

∣∣∣∣∣∣∣∣∣for one of the determinants. Evaluating the partial derivatives, we obtain

J =

∣∣∣∣∣∣∣∣1 0 −1 00 1 0 −10 0 1 −1

1/4 1/4 1/4 1/4

∣∣∣∣∣∣∣∣=

14

∣∣∣∣∣∣

1 0 −10 1 −11 1 1

∣∣∣∣∣∣−∣∣∣∣∣∣

0 1 −10 0 −11 1 1

∣∣∣∣∣∣

= 1.

Hence the full Jacobian is unity.


Inverting the relations giving x, y, z,X in terms of x1, x2, x3, x4, we find

x1 =14(4X + 3x− y + 2z)

x2 =14(4X − x+ 3y − 2z)

x3 =14(4X − x− y + 2z)

x4 =14(4X − x− y − 2z).

The exponentials in (10.3) then become

ei(p′A−pB

4 )·(4X+3x−y+2z) ei(p′B−pA

4 )·(4X−x+3y−2z)

which reduces to the exponentials given in (10.4). The propagator factors only depend onthe difference in the arguments of the two fields (see (6.98)).

10.2 The integral is ∫ 1

0

1[A+ x(B −A)]2

= − 1(B −A)

[1

A+ x(B −A)

]10

=1

A−B

[1B− 1A

]=

1AB

.

10.3 The [. . .] bracket is

[k2 − xk2 − (1− x)m2A + xq2 − 2x q · k + xk2 − xm2

B + iε]= [k2 − 2x q · k + xq2 − xm2

B − (1− x)m2A + iε]

= [(k − xq)2 − x2q2 + xq2 − xm2B − (1− x)m2

A + iε]= [k′2 − −x(1− x)q2 + xm2

B + (1− x)m2A+ iε]

which is just [k′2−∆+iε]. Note that we have replaced ‘(1−x)iε+ iε’ by ‘iε’ since all thatmatters is the sign of this infinitesimal imaginary part, and (1−x) can never be negative.

10.4 We have

Π[2]C (q2) =

−g2

8π2

∫ 1

0

∫ Λ

0

u2du(u2 + ∆)3/2

.

One way of evaluating the integral over u is to substitute u = ∆1/2 tan θ. Then

du = ∆1/2 sec2 θ dθ

and(u2 + ∆)3/2 = ∆3/2 sec3 θ.

The integral over u becomes∫tan2 θ

sec3 θsec2 θ dθ =

∫sin2 θ

cos θdθ

=∫ (

1cos θ

− cos θ)

dθ

= ln(sec θ + tan θ)− sin θ

= ln(

(u2 + ∆)1/2

∆1/2+

u

∆1/2

)− Λ

(u2 + ∆)1/2.


This expression vanishes at u = 0 and has the valueln(

Λ + (Λ2 + ∆)1/2

∆1/2

)− Λ

(Λ + ∆)1/2

at u = Λ (correcting the formula given in (10.51)).

10.5 Π[2]C (q2) is given by (10.42) and its dependence on q2 is contained in the quantity ∆ of

(10.43). Hence

dΠ[2]C (q2)dq2

= ig2

∫ 1

0

dx∫

d4k′

(2π)4−2

(k′2 −∆ + iε)3d∆dq2

= ig2

∫ 1

0

dx∫

d4k′

(2π)42

(k′2 −∆ + iε)3x(1− x).

The denominator of the k′ integral now behaves like (k′)6 at large values of k′, which willensure convergence. In more detail, we now have, in place of (10.44), a k′0 integral of theform ∫

dk′0

[(k′0)2 −A]3=

12∂2

∂A2

∫dk′0

[(k′0)2 −A]

which will result in a denominator (u2+∆)5/2 in (10.50), so that the u-integral is manifestlyconvergent, by counting powers of u at large values of u.

10.6 Consider the counter term

Lct =12δZC∂µφph,C∂

µφph,C

where normal-ordering is understood. Remembering that ‘L = T − V ’, the correspond-ing interaction potential is −Lct, and in the Dyson-Wick expansion it is ‘−i’ times theinteraction that appears. Hence the one-C matrix element required is

i2δZC〈C, k|∂µφph,C∂

µφph,C|C, k〉

=i22EδZC〈0|aC(k)

∫d3k1

(2π)3√

2E1

[aC(k1)(−ik1µ)e−ik1·x + a†C(k1)(ik1µ)eik1·x]

×∫

d3k2

(2π)3√

2E2

[aC(k2)(−ik2µ)e−ik2·x + a†C(k2)(ik2µ)eik2·x] a†C(k)|0〉

where normal-ordering is understood but not shown explicitly. The only terms which givea non-zero vev are those in which the number of a operators is equal to the number of a†

operators. One of these is

〈0|aC(k)(ik1µ)a†C(k1)(−ik2µ)aC(k2)a

†C(k)|0〉.

As usual, we write

aC(k)a†C(k1) = a†C(k1)aC(k) + (2π)3δ3(k − k1),

the a†C(k1) then annihilating on 〈0|, and similarly with aC(k2)a†C|0〉 which yields a factor

(2π)3δ3(k−k2). The k1 and k2 integrations can then be done, setting k1 = k and k2 = k,and hence E1 = E, and E2 = E, and kµ1 = kµ, kµ2 = kµ. This term therefore finally yields

i2δZCk

2.


The other surviving term is (after normal-ordering)

〈0|aC(k)(ik2µ)a†C(k2)(−ik1µ)aC(k1)a†C(k)|0〉.

This contributes exactly the same result, removing the factor 12 as required. The 1-C

matrix element of the counter term involving φ2ph,C is very similar, lacking the k2 factor

coming from the gradients.

Chapter 11

11.1 The Feynman rule for the counter term in the fermion propagator is given by (a) of (11.7).The fermion analogue of (10.63) is then

S =i

6p−m+ (Z2 − 1)6p− δm− Σ[2](p).

For 6p ≈ m we expand Σ[2](p) as

Σ[2](p) ≈ Σ[2](6p = m) + (6p−m)dΣ[2]

d6p

∣∣∣∣6p=m

,

so that

S ≈ i

6p−m+ (Z2 − 1)6p− δm− Σ[2](6p = m)− (6p−m)dΣ[2]

d6p

∣∣∣6p=m

=i

6p[Z2 − dΣ[2]

d6p

∣∣∣6p=m

] +m dΣ[2]

d6p

∣∣∣6p=m

− Σ[2](6p = m)− δm−m,

where Z2 and δm must be chosen so that this expression is equal to i/(6p − m). Thecoefficient of 6p therefore yields

Z2 = 1 +dΣ[2]

d6 p

∣∣∣∣6p=m

,

while the mass terms may be written as

mdΣ[2]

d6 p

∣∣∣∣6p=m

− Σ[2](6p = m)− δm−m = m(Z2 − 1)− Σ[2](6p = m)− (m0Z2 −m)−m

= −m+ (m−m0)Z2 − Σ[2](6p = m),

whence we requirem0 −m = −Z−1

2 Σ[2](6p = m).

When these values are substituted back into the first expression for S, we obtain (11.10).11.2 Consider QED for definiteness, with interaction Lagrangian

Lint(x) = −q ¯ψ(x)γµψ(x).

The amplitude for a typical closed fermion loop will involve (compare (10.3)) a sequenceof fermion propagators starting from a certain space-time point and ending at the samepoint:

〈0|T (ψ(x1)¯ψ(x2))|0〉〈0|T (ψ(x2)

¯ψ(x3))|0〉 . . . 〈0|T (ψ(xN )¯ψ(x1))|0〉. (A)


Note that each propagator involves fields from different interaction Lagrangians in a termof given order in the Dyson expansion analogous to (6.42), since each interaction involvestwo fields at the same point. In a given ‘Dyson’ term, however, the interactions may bewritten in any order within the overall time-ordering symbol T , for example (referring to(A)) as

〈0| . . . T. . . [ ¯ψ(x1)6A(x1)ψ(x1)][¯ψ(x2)6A(x2)ψ(x2)] . . . [

¯ψ(xN )6A(x1)ψ(xN )] . . . |0〉,

without introducing any sign changes. The product of contractions shown in (A) is nowobtained after permuting ¯

ψ(x1) through an odd number of fermion fields (namely thesingle field ψ(x1) and the appropriate number of pairs ¯

ψ(x2)ψ(x2), . . . ,¯ψ(xN )ψ(xN )),

which produces an overall minus sign.11.3 The amplitude arises from the second-order term in the Dyson expansion, which involves

the T -productT (¯ψα(x1)γ

µαβψβ(x1)

¯ψρ(x2)γνρσψσ(x2)),

where we have indicated the (summed over) Dirac matrix indices α, β, ρ, σ explicitly.Figure 11.2(b) corresponds to the contraction

. . . γµαβ〈0|T (ψβ(x1)¯ψρ(x2))|0〉γνρσ〈0|T (ψσ(x2)

¯ψα(x1))|0〉 . . .

which has the structure

γµαβSβρ(x1 − x2)γνρσSσα(x2 − x1)

=∑α

[γµS(x1 − x2)γνS(x2 − x1)]αα

which exhibits the required Trace.11.4 The result is (up to a factor of 2) the same as (8.78) with k′ = k + q.11.5 We have, from (11.32),

Π[2]γ (q2) = Π[2]

γ (q2)−Π[2]γ (0),

where from (11.23)

Π[2]γ (q2) = 8ie2

∫ 1

0

dx x(1− x)I(x, q2)

with

I(x, q2) =∫

d4k′

(2π)41

(k′2 −∆γ(x, q2) + iε)2,

where∆γ(x, q2) = −x(1− x)q2 +m2.

As noted in the text, the k′-integral in I(x, q2) has been evaluated in section 10.3.1 and10.3.2, with the result (correcting (10.51))

I(x, q2) =i

8π2

ln

(Λ + (Λ2 + ∆γ(x, q2))1/2

∆1/2γ (x, q2)

)− Λ

(Λ2 + ∆γ(x, q2))1/2

,

where Λ is the ultraviolet cut off. The quantity Π[2]γ (q2) involves the x-integral of the

subtracted quantityI(x, q2)− I(x, q2 = 0),

and we shall see that this has a well-defined finite limit as Λ→∞. We have

I(x, q2)− I(x, q2 = 0) =i

8π2

ln

(Λ + (Λ2 + ∆γ(x, q2))1/2

Λ + (Λ2 + ∆γ(x, q2 = 0))1/2∆1/2γ (x, q2 = 0)

∆1/2γ (x, q2)

)


− Λ(Λ2 + ∆γ(x, q2))1/2

+Λ

(Λ2 + ∆γ(x, q2 = 0))1/2

.

Now

limΛ→∞

Λ + (Λ2 + ∆γ(x, q2))1/2

Λ + (Λ2 + ∆γ(x, q2 = 0))1/2= 1

andlim

Λ→∞− Λ

(Λ2 + ∆γ(x, q2))1/2+

Λ(Λ2 + ∆γ(x, q2 = 0))1/2

= 0.

Noting also that ∆γ(x, q2 = 0) = m2, we obtain

Π[2]γ(q2) = − e2

2π2

∫ 1

0

dx x(1− x) ln[

m2

∆γ(x, q2)

]which is equivalent to (11.36).

11.7 For q2 = −q2 m2 we write

Π[2]γ (q2) =

2απ

∫ 1

0

dx x(1− x) ln[1 + q2x(1− x)/m2]

≈ 2απ

∫ 1

0

dx x(1− x) q2

m2x(1− x)

=2αq2

πm2

∫ 1

0

dx x2(1− x)2

which leads to (11.38).11.8 In this case we are interested in q2 = −|q2| m2. We write

Π[2]γ (|q2|) =

2απ

∫ 1

0

dx x(1− x) ln(|q2|m2

x(1− x) + 1)

=2απ

∫ 1

0

dx x(1− x) ln|q2|m2

x(1− x)[1 +m2

|q2|x(1− x)]

=2απ

∫ 1

0

dx x(1− x)

ln(|q2|m2

)+ ln[x(1− x)] +O

(m2

|q2|

).

Then using ∫ 1

0

x lnx dx =(x2

2lnx− x2

4

)1

0

= −14

and ∫ 1

0

x2 lnx dx =(x3

3lnx− x3

3

)1

0

= −19,

together with ∫ 1

0

x(1− x) ln(1− x) dx =∫ 1

0

x(1− x) lnxdx,

we arrive at (11.54).11.9 The e contributes 0.01639, the µ 0.00825 and the τ 0.00388 for a total of 0.0285.11.10 In the system ~ = c = 1, with one independent dimension which we take to be mass (see

Appendix B), the electromagnetic field strength tensor Fµν has dimension M2, so thatthe Maxwell action

−14

∫d4xFµνFµν


is dimensionless. Hence (FµνFµν)2 has dimension M8, and therefore the coupling constant‘E’ must have dimension M−4 so that

E

∫d4x (FµνFµν)2

is dimensionless.In terms of the fields E and B,

FµνFµν = 2(B2− E

2).

HenceE(FµνFµν)2 = 4E(B

2− E

2)2.

This fourth-order (in the fields) interaction will contribute to γ−γ scattering, which doesnot occur in the classical (Maxwell) theory. See for example The Quantum Theory ofFields, volume 1, by S. Weinberg (CUP), pages 533-4.

Documents

Gauge Theories in Particle Physics Third Edition Volume 1: From Relativistic Quantum ... · 2005-09-07 · Gauge Theories in Particle Physics Third Edition Volume 1: From Relativistic