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Solutions to Problems in Peskin and Schroeder,
An Introduction To Quantum Field Theory
Homer Reid
June 23, 2006
Chapter 9
Problem 9.1
Part a.
Part 1: Complex scalar propagator
The action for the scalars alone is
S[φ, φ∗, J, J∗] =
∫
d4x∂µφ
∗∂µφ−m2φ∗φ+ J∗φ+ Jφ∗
=
∫
d4x−φ∗
[∂µ∂
µ +m2]φ+ J∗φ+ Jφ∗
=
∫
d4x −φ∗OKGφ+ J∗φ+ Jφ∗
where the differential operator OKG is
OKG = ∂µ∂µ +m2.
Introduce shifted fields:
φ(x) = φ′(x) + i
∫
d4yDF (x− y)J(y)
φ∗(x) = φ′∗(x) − i
∫
d4yD∗
F (x− y)J∗(y)
= φ′∗(x) + i
∫
d4yDF (x− y)J∗(y)
where iDF (x − y) is the solution to
OKGiDF (x− y) = δ(4)(x− y).
1
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9 2
Then the action for the scalars becomes
S =
∫
d4x
−[
φ′∗ + i
∫
d4yDF (x − y)J∗(y)
]
OKG
[
φ′ + i
∫
d4yDF (x− y)J(y)
]
+ J∗
[
φ′ + i
∫
d4yDF (x − y)J(y)
]
+ J
[
φ′∗ + i
∫
d4yDF (x− y)J∗(y)
]
=
∫
d4x
− φ′∗OKGφ′ − i
∫
d4yJ(x)DF (x − y)J∗(y)
+ i
∫
d4yJ∗(x)DF (x− y)J(y)
+ i
∫
d4yJ(x)DF (x− y)J∗(y)
= −∫
d4xφ′∗OKGφ′ + i
∫
d4xd4yJ∗(x)DF (x− y)J(y).
The generating functional is then
Z[J, J∗] =
∫
DφDφ∗eiS[φ,φ∗,J,J∗]
= Z0 exp
−∫
d4xd4yJ∗(x)DF (x− y)J(y)
,
where
Z0 =
∫
DφDφ∗e−iφ∗OKGφ,
and the two-point function is
⟨
0∣∣∣φ∗(x)φ(y)
∣∣∣0
⟩
=1
Z0
[
−i ∂
∂J(x)
] [
−i ∂
∂J∗(y)
]
Z[J, J∗]∣∣∣J=0
=
[
−i ∂
∂J(x)
]
i
∫
d4xJ(x)DF (x− y)Z[J, J∗]
∣∣∣J=0
= DF (x− y).
Part 2: Photon propagator
The action for the photon field is
S[Aµ, jµ] =
∫
d4x
−1
4Fµν(x)F
µν(x) + jµ(x)Aµ(x)
(1)
Integrate by parts as described pedantically in Appendix 1:
=
∫
d4x
1
2Aµ(x)
[∂2gµν − ∂µ∂ν
]Aν(x) + jµ(x)A
µ(x)
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9 3
Go over to momentum space:
=
∫d4k
(2π)4
1
2Aµ(k)
[−k2gµν + kµkν
]Aν(−k) + jµ(−k)Aµ(k)
Go through mysterious Faddeev–Popov procedure to come up with magicalfactor of 1/ξ to desingularize the kernel of the kinetic term, which is otherwisesingular as illustrated by brute force in Appendix 2:
=
∫d4k
(2π)4
1
2Aµ(k)
[
−k2gµν +
(
1 − 1
ξ
)
kµkν
]
Aν(−k) + jµ(−k)Aµ(k)
≡∫
d4k
(2π)4
1
2Aµ(k)OEM
µνAν(−k) + jµ(−k)Aµ(k)
(2)
where the differential operator OEM
µν is
OEM
µν = ∂2gµν −(
1 − 1
ξ
)
∂µ∂ν
or in momentum space
OEM
µν = −k2gµν +
(
1 − 1
ξ
)
kµkν
and has inverse
iDµν = − 1
k2
(
gµν − (1 − ξ)kµkν
k2
)
(3)
defined such that the 4x4 matrix equation
OEM · iD = 1
is satisfied.Now introduce shifted fields:
Aν(x) = A′ν(x) −∫
d4y iDνρ(x− y)jρ(y)
The action becomes
S =
∫
d4x
1
2A′µ(x)OEM
µνA′ν(x) − jµ(x)A
′µ(x) +1
2
∫
d4y jµ(x)iDµρ(x − y)jρ(y)
+ jµ(x)
[
A′µ(x) −∫
d4y iDµρ(x− y)jρ(y)
]
=1
2
∫
d4xA′µ(x)OEM
µνA′ν(x) − 1
2
∫
d4xd4y jµ(x) iDµρ(x− y)jρ(y). (4)
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9 4
Aside. Effective fermion-fermion interaction in QED.
I think it is interesting to see how this procedure may be used to integratethe photon out of the QED lagrangian to give an effective electron-electroninteraction. If we wanted to calculate the expectation value of a product ofelectron field operators, say for example the electron density at a point x, wewould be led to consider an expression of the form
⟨
ψ(x)γ0ψ(x)⟩
=
∫
DψDψDA ψ(x)γ0ψ(x)eiL[ψ,ψ,A]
∫
DψDψDA eiL[ψ,ψ,A](5)
where the QED lagrangian is the spacetime integral of the QED lagrangiandensity,
L[ψ, ψ,A] =
∫
d4xL[ψ, ψ,A] =
∫
d4x
−1
4FµνF
µν + ψ(i/∂ −m)ψ − eψγµψAµ
.
Then the photon part of the path integral is
∫
DA exp
i
∫
d4x
[
− 1
4FµνF
µν − ejµAµ
]
where jµ = eψγµψ is the electron current. The argument of the exponentialhere is just the action (1). Transforming it into the form (4), the photon pathintegral is
[∫
DA′ exp
i
∫
d4x[
A′µ(x)OEM
µνA′ν(x)
]]
× e−ie2
2
R
d4xd4y jµ(x) iDµν (x−y)jν(y)
The first term is independent of the fermions, and since there are no photonfields in the payload of the numerator in (6), this term cancels, leaving
⟨
ψ(x)γ0ψ(x)⟩
=
∫
DψDψ ψ(x)γ0ψ(x)eiLeff[ψ,ψ]
∫
DψDψ eiLeff[ψ,ψ]
(6)
where the effective Lagrangian now contains a nonlocal term and hence cannotbe written as the space integral of a local density function:
Leff
[ψ, ψ] =
∫
d4x
ψ(x)(i/∂ −m)ψ(x)
− e2
2
∫
d4x
∫
d4y
ψ(x)γµψ(x) iDµν(x− y)ψ(y)γνψ(y)
.
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9 5
To look at the interaction term in k-space, we put
ψ(x) =
∫d4p1
(2π)4e−ip1xψ(p1)
ψ(x) =
∫d4p2
(2π)4e+ip2xψ(p2)
iDµν(x− y) =
∫d4q
(2π)4e−iq(x−y)
q2 + iε
(
gµν − (1 − ξ)qµqν
q2
)
ψ(y) =
∫d4k1
(2π)4e−ik1yψ(k1).
ψ(y) =
∫d4k2
(2π)4e+ik2yψ(k2)
The fermion-fermion interaction term becomes
− e2
2
∫
d4x
∫
d4y
ψ(x)γµψ(x) iDµν(x− y)ψ(y)γνψ(y)
= −e2
2
∫d4p1d
4k1d4q
(2π)12
ψ(p1 − q)γµψ(p1)[
gµν − (1 − ξ) qµqν
q2
]
ψ(k1 + q)γνψ(k1)
q2 + iε
.
The longitudinal part of iDµν actually doesn’t contribute here, because e.g.
ψ(k1+q)/qψ(k1) = ψ(k1+q)[
(/k1 + /q) − /k1
]
ψ(k1) = ψ(k1+q)[
m−m]
ψ(k1) = 0
so the interaction term reduces to
−e2
2
∫d4p1d
4k1d4q
(2π)12
ψ(p1 − q)γµψ(p1)ψ(k1 + q)γµψ(k1)
q2 + iε
.
This looks exactly like the interaction term you get when you write down thehamiltonian for the interacting homogenous electron gas, although in that casethere are no γµ matrices and the k integrals are only over 3-dimensional space.
Part b.
The lowest-order diagram is
p+ p′
k
p
k′
p′
ν
µ
= (−ie)2(p− p′)µ[ −igµν(p− p′)2
]
us(k)γνvs′(k
′)
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9 6
The squared, averaged matrix element is (ignoring the electron mass)
|M|2 =1
4e4
(p− p′)µ(p− p′)ν
(p− p′)4Tr
[/kγµ/k′γν
]
= e4(p− p′)µ(p− p′)ν
(p− p′)4[kµk
′
ν + kνk′
µ − (k · k′)gµν]
=e4
(p− p′)4
2[
k · (p− p′)] [
k′ · (p− p′)]
− (k · k′)(p− p′)2
(7)
We work in a frame such that
k = (E, 0, 0, E), k′ = (E, 0, 0,−E)
and
p = (E, |p| sin θ, 0, |p| cos θ), p′ = (E,−|p| sin θ, 0,−|p| cos θ).
Then (7) reads
|M|2 =e4E2
2|p|2[− cos2 θ + 1
]
=e4E2
2|p|2 sin2 θ.
Inserting the kinematic factors from P&S equation 4.84,
dσ
dΩ=
e4
256π2|p||E| sin2 θ
=e4
256π2E√E2 −m2
sin2 θ.
The total cross section is
σ = 2π
∫ π
0
dσ
dΩsin θdθ
=e4
128πE√E2 −m2
∫ π
0
sin3 θdθ
=e4
96πE√E2 −m2
.
Part c.
The two first-order diagrams are
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9 7
M1 =q
k + q
q
k
µ νρ σ
= (−ie)2∫
d4k
(2π)4
[−igµρq2
]
(2k + q)ρ[
i
(k + q)2 −m2
]
(2k + q)σ[
i
k2 −m2
] [−igσνq2
]
M2 =q q
k
µ νρ σ
= (2ie2)
∫d4k
(2π)4
[−igµρq2
] [igρσ
k2 −m2
] [−igσνq2
]
The sum of the diagrams is
M1 + M2 =
[−igµρq2
]
Πρσ(q)
[−igσνq2
]
where
Πρσ(q) = e2∫
d4k
(2π)4(2k + q)ρ(2k + q)σ − 2gρσ
[(k + q)2 −m2
]
[(k + q)2 −m2] [k2 −m2](8)
Rewrite the denominator:
1
[(k + q)2 −m2] [k2 −m2]=
∫ 1
0
dx
x [(k + q)2 −m2] + (1 − x) [k2 −m2]2
=
∫ 1
0
dx
[k2 + 2xk · q + xq2 −m2]2
Shift variables to l = k + xq in (8):
Πρσ(q) = e2∫
dx
∫d4l
(2π)44lρlσ + (1 − 2x)2qρqσ − 2gρσ
[l2 + (1 − x)2q2 −m2
]
[l2 − ∆]2,
(9)where we ignored numerator terms of odd order in l, and where
∆ = −x(1 − x)q2 +m2.
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9 8
Using lρσ → 1dl2gρσ under the integral sign, we rewrite this as
Πρσ(q) = e2∫
dx
gρσI1(∆) +[
(1 − 2x)2qρqσ − 2[(1 − x)2q2 −m2]gρσ]
I2(∆)
(10)where
I1(∆) =
(4
d− 2
)∫ddl
(2π)dl2
(l2 − ∆)2
= − i
(4π)2
(4d− 2
)d2Γ
(1 − d
2
)
∆1− d2
= − 2i
(4π)2Γ
(2 − d
2
)
∆2− d2
∆
(here we have used the fact that(1 − d
2
)Γ
(1 − d
2
)= Γ
(2 − d
2
)) and
I2(∆) =
∫ddl
(2π)d1
(l2 − ∆)2
=i
(4π)2Γ
(2 − d
2
)
∆2− d2
.
Then (10) is
Πρσ(q) =ie2
(4π)2
∫
dxΓ
(2 − d
2
)
∆2− d2
[−2∆ − 2(1 − x)2q2 + 2m2
]gρσ + (1 − 2x)2qρqσ
=ie2
(4π)2
∫
dxΓ
(2 − d
2
)
∆2− d2
− 2(1 − x)(1 − 2x)q2gρσ + (1 − 2x)2qρqσ
=ie2
(4π)2
∫
dxΓ
(2 − d
2
)
∆2− d2
− (1 − 2x)2
[q2gρσ − qρqσ
]− (1 − 2x)q2gρσ
︸ ︷︷ ︸
→0
where the last term in the curly brackets integrates to 0, being odd underx → (1 − x) while the denominator, which involves ∆, is even. Then the finalresult is
iΠρσ(q) = (q2gµν − qρσ)iΠ(q2)
iΠ(q2) =e2
(4π)2
∫ 1
0
(1 − 2x)2Γ(2 − d
2 )
∆2− d2
dx
=e2
(4π)2
∫ 1
0
(1 − 2x)2[
const. − log ∆]
dx
where
const. =2
ε− γ + log 4π
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9 9
-2
-1
0
1
2
3
4
5
6
7
8
9
-2 -1 0 1 2 3 4-2
-1
0
1
2
3
4
5
6
7
8
9
q2/m2
∫1 0(1
−2x
)2lo
g(
m2
m2−x(1−x)q
2
)
dx
Figure 1: Integral in (11) as evaluated by numerical quadrature.
is infinite but q-independent at d = 4. Isolating the q-dependence of Π, we have
iΠ(q2) − iΠ(0) =e2
(4π)2
∫ 1
0
(1 − 2x)2 log
(m2
m2 − x(1 − x)q2
)
dx. (11)
Values of this integral are plotted in Figure 1. The divergence at q2 = 4m2 isclear.
Appendix: Integration-by-parts of kinetic term in photon Lagrangian
Someday I hope to be able to perform in my head the kind of manipulationsthat are needed to go from the first to the second lines of P&S equation 9.51.However, at the moment I need to work it all out in detail. The Lagrangian
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9 10
density is
L = −1
4FµνF
µν
= −1
4(∂µAν − ∂νAµ)(∂
µAν − ∂νAµ)
= −1
2(∂0A1 − ∂1A0)(∂
0A1 − ∂1A0)
− 1
2(∂0A2 − ∂2A0)(∂
0A2 − ∂2A0)
− 1
2(∂0A3 − ∂3A0)(∂
0A3 − ∂3A0)
− 1
2(∂1A2 − ∂2A1)(∂
1A2 − ∂2A1)
− 1
2(∂1A3 − ∂3A1)(∂
1A3 − ∂3A1)
− 1
2(∂2A3 − ∂3A2)(∂
2A3 − ∂3A2).
I get scared whenever I see something like Aµ or ∂µ because I know there are abunch of hidden minus signs in there and they confuse me. To make sure thatall minus signs are explicit, let’s rewrite FµνF
µν with all minus signs displayedexplicitly. Basically we just go through and replace ∂0 = ∂0, ∂
i = −∂i, A0 = A0,
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9 11
Ai = −Ai. This gives
L = −1
2(−∂0A
1 − ∂1A0)(∂0A
1 + ∂1A0)
− 1
2(−∂0A
2 − ∂2A0)(∂0A
2 + ∂2A0)
− 1
2(−∂0A
3 − ∂3A0)(∂0A
3 + ∂3A0)
− 1
2(−∂1A
2 + ∂1A2)(−∂1A
2 + ∂1A2)
− 1
2(−∂1A
3 + ∂1A3)(−∂1A
3 + ∂1A3)
− 1
2(−∂2A
3 + ∂2A3)(−∂2A
3 + ∂2A3)
= +1
2
[
(∂0A1)2 + (∂1A
0)2 + 2(∂0A1)(∂1A
0)]
+1
2
[
(∂0A2)2 + (∂2A
0)2 + 2(∂0A2)(∂2A
0)]
+1
2
[
(∂0A3)2 + (∂3A
0)2 + 2(∂0A3)(∂3A
0)]
− 1
2
[
(∂1A2)2 + (∂2A
1)2 − 2(∂1A2)(∂2A
1)]
− 1
2
[
(∂1A3)2 + (∂3A
1)2 − 2(∂1A3)(∂3A
1)]
− 1
2
[
(∂2A3)2 + (∂3A
2)2 − 2(∂2A3)(∂3A
2)]
.
The next step is to integrate by parts, which entails making replacements like
(∂1A3)(∂3A
1) → −A3∂1∂3A1.
Also, we break up the terms with prefactors of 2 into two separate terms, each ofwhich we integrate by parts in a different way. This gives a different Lagrangianthat integrates to the same thing as the old Lagrangian did, so we will just callthe new Lagrangian L as well:
L = −1
2
[
A1∂20A
1 +A0∂21A0 +A1∂1∂0A
0 +A0∂0∂1A1]
− 1
2
[
A2∂20A
2 +A0∂22A
0 +A2∂2∂0A0 +A0∂0∂2A
2]
− 1
2
[
A3∂20A
3 +A0∂23A
0 +A3∂3∂0A0 +A0∂0∂3A
3]
+1
2
[
A2∂21A
2 +A1∂22A
1 −A2∂2∂1A1 −A1∂1∂2A
2]
+1
2
[
A3∂21A
3 +A1∂23A
1 −A3∂3∂1A1 −A1∂1∂3A
3]
+1
2
[
A3∂22A
3 +A2∂23A
2 −A3∂3∂2A2 −A2∂2∂3A
3]
.
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9 12
I will think of this as a kind of bilinear form between 4-dimensional vectors:
L = −1
2
“
A0A
1A
2A
3”
0
B
B
@
(∂2
1+ ∂2
2+ ∂2
3) ∂0∂1 ∂0∂2 ∂0∂3
∂1∂0 (∂2
0+ ∂2
2+ ∂2
3) −∂1∂2 −∂1∂3
∂2∂0 −∂2∂1 (∂2
0+ ∂2
1+ ∂2
3) −∂2∂3
∂3∂0 −∂3∂1 −∂3∂2 (∂2
0+ ∂2
2+ ∂2
1)
1
C
C
A
0
B
B
@
A0
A1
A2
A3
1
C
C
A
Appendix 2: Convincing the skeptics among us that −k2gµν + kµkν issingular, but −k2gµν + (1 − 1
ξ)kµkν is invertible with inverse given by
(3)
These should all work in both octave or matlab.
octave:1> g=[1 0 0 0; 0 -1 0 0; 0 0 -1 0; 0 0 0 -1];
octave:2> ku=rand(4,1) % k with raised index
ku =
0.880788
0.157326
0.075570
0.481218
octave:3> kl=g*ku; % k with lowered index
octave:4> k2=kl’ * ku
k2 = 0.51375
octave:5> kmat=-k2*g + kl * kl’
kmat =
0.262033 -0.138571 -0.066562 -0.423851
-0.138571 0.538505 0.011889 0.075708
-0.066562 0.011889 0.519465 0.036366
-0.423851 0.075708 0.036366 0.745324
octave:6> rank(kmat)
ans = 3
Sure enough, the matrix has less than full rank! The problem is the existenceof an eigenvector with eigenvalue 0, namely kµ itself.
octave:7> kmat*ku
ans =
5.5511e-17
-1.3878e-17
0.0000e+00
-5.5511e-17
On the other hand, the matrix modified to contain the magical Fadeev-Popovfactor is nonsingular:
octave:8> xi=rand
xi = 0.35839
octave:9> kmat2=-k2*g + (1-1/xi)*kl*kl’
kmat2 =
-1.902603 0.248076 0.119162 0.758798
0.248076 0.469442 -0.021285 -0.135536
0.119162 -0.021285 0.503530 -0.065104
0.758798 -0.135536 -0.065104 0.099185
octave:10> rank(kmat2)
ans = 4
Moreover, its inverse is just the matrix (3):
octave:11> iD=-(g - (1-xi)*ku*ku’/k2) / k2
iD =
-0.060627 0.336846 0.161802 1.030323
0.336846 2.006626 0.028901 0.184036
0.161802 0.028901 1.960341 0.088400
1.030323 0.184036 0.088400 2.509375
octave:12> kmat2*iD
ans =
1.00000 -0.00000 -0.00000 -0.00000
0.00000 1.00000 0.00000 0.00000
0.00000 0.00000 1.00000 0.00000
0.00000 0.00000 0.00000 1.00000
13
