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
Z
∗
∗
S[φ, φ , J, J ] = d4 x ∂µ φ∗ ∂ µ φ − m2 φ∗ φ + J ∗ φ + Jφ∗
Z
= d4 x −φ∗ ∂µ ∂ µ + m2 φ + J ∗ φ + Jφ∗
Z
= d4 x {−φ∗ OKG φ + J ∗ φ + Jφ∗ }
where the differential operator OKG is
OKG = ∂µ ∂ µ + m2 .
Introduce shifted fields:
0
φ(x) = φ (x) + i
Z
φ∗ (x) = φ0∗ (x) − i
0∗
= φ (x) + i
Z
d4 yDF (x − y)J(y)
Z
d4 yDF∗ (x − y)J ∗ (y)
d4 yDF (x − y)J ∗ (y)
where iDF (x − y) is the solution to
OKG iDF (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
( Z
Z
Z
4
S = d x − φ0∗ + i d4 yDF (x − y)J ∗ (y) OKG φ0 + i d4 yDF (x − y)J(y)
+J
=
Z
4
d x
(
∗
0
φ +i
0∗
0
− φ OKG φ − i
+i
+i
=−
Z
4
0∗
0
d xφ OKG φ + i
Z
)
Z
0∗
4
∗
d yDF (x − y)J(y) + J φ + i d yDF (x − y)J (y)
Z
4
Z
Z
Z
d4 yJ(x)DF (x − y)J ∗ (y)
d4 yJ ∗ (x)DF (x − y)J(y)
4
∗
d yJ(x)DF (x − y)J (y)
)
d4 xd4 yJ ∗ (x)DF (x − y)J(y).
The generating functional is then
Z
∗
∗
Z[J, J ∗ ] = DφDφ∗ eiS[φ,φ ,J,J ]
Z
4
4
∗
= Z0 exp − d xd yJ (x)DF (x − y)J(y) ,
where
Z0 =
Z
DφDφ∗ e−iφ
∗
OKG φ
,
and the two-point function is
E
D 1
∂
∂
∗
0 φ (x)φ(y) 0 =
−i
−i ∗
Z[J, J ∗ ] Z0
∂J(x)
∂J (y)
J=0
Z
∂
4
∗
= −i
i d xJ(x)DF (x − y)Z[J, J ] ∂J(x)
J=0
= DF (x − y).
Part 2: Photon propagator
The action for the photon field is
Z
1
S[Aµ , j µ ] = d4 x − Fµν (x)F µν (x) + jµ (x)Aµ (x)
4
Integrate by parts as described pedantically in Appendix 1:
Z
2
ν
1 µ
µ
4
A (x) ∂ gµν − ∂µ ∂ν A (x) + jµ (x)A (x)
= d x
2
(1)
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9
3
Go over to momentum space:
=
Z
d4 k
(2π)4
1 µ
A (k) −k 2 gµν + kµ kν Aν (−k) + jµ (−k)Aµ (k)
2
Go through mysterious Faddeev–Popov procedure to come up with magical
factor of 1/ξ to desingularize the kernel of the kinetic term, which is otherwise
singular as illustrated by brute force in Appendix 2:
1 µ
1
d4 k
ν
µ
2
A
(−k)
+
j
(−k)A
(k)
k
k
A
(k)
−k
g
+
1
−
µ
µ
ν
µν
(2π)4 2
ξ
Z
4
d k
1 µ
EM
ν
µ
A
(k)O
A
(−k)
+
j
(−k)A
(k)
≡
µ
µν
(2π)4 2
=
Z
(2)
EM
where the differential operator Oµν
is
1
EM
∂µ ∂ν
Oµν
= ∂ 2 gµν − 1 −
ξ
or in momentum space
1
EM
kµ kν
Oµν
= −k 2 gµν + 1 −
ξ
and has inverse
iD
µν
1
=− 2
k
kµ kν
µν
g − (1 − ξ) 2
k
(3)
defined such that the 4x4 matrix equation
O EM · iD = 1
is satisfied.
Now introduce shifted fields:
Aν (x) = A0ν (x) −
Z
d4 y iDνρ (x − y)jρ (y)
The action becomes
(
Z
Z
1 0µ
1
4
EM
0ν
0µ
d4 y jµ (x)iDµρ (x − y)jρ (y)
S= d x
A (x)Oµν A (x) − jµ (x)A (x) +
2
2
)
Z
0µ
4
µρ
+ jµ (x) A (x) − d y iD (x − y)jρ (y)
=
1
2
Z
EM
d4 xA0µ (x)Oµν
A0ν (x) −
1
2
Z
d4 xd4 y 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 integrate
the photon out of the QED lagrangian to give an effective electron-electron
interaction. If we wanted to calculate the expectation value of a product of
electron field operators, say for example the electron density at a point x, we
would be led to consider an expression of the form
Z
DψDψDA ψ(x)γ 0 ψ(x)eiL[ψ,ψ,A]
E
D
0
Z
ψ(x)γ ψ(x) =
(5)
DψDψDA eiL[ψ,ψ,A]
where the QED lagrangian is the spacetime integral of the QED lagrangian
density,
Z
Z
1
4
4
µ
µν
L[ψ, ψ, A] =
d x L[ψ, ψ, A] =
d x − Fµν F + ψ(i/
∂ − m)ψ − eψγµ ψA .
4
Then the photon part of the path integral is
Z
Z
1
DA exp i d4 x − Fµν F µν − ejµ Aµ
4
where jµ = eψγµ ψ is the electron current. The argument of the exponential
here is just the action (1). Transforming it into the form (4), the photon path
integral is
Z
Z
h
i
R 4 4
µν
ie2
EM
DA0 exp i d4 x A0µ (x)Oµν
A0ν (x)
× e− 2 d xd y jµ (x) iD (x−y)jν (y)
The first term is independent of the fermions, and since there are no photon
fields in the payload of the numerator in (6), this term cancels, leaving
Z
DψDψ ψ(x)γ 0 ψ(x)eiLeff [ψ,ψ]
E
D
0
Z
ψ(x)γ ψ(x) =
(6)
DψDψ eiLeff [ψ,ψ]
where the effective Lagrangian now contains a nonlocal term and hence cannot
be written as the space integral of a local density function:
Z
n
o
Leff [ψ, ψ] = d4 x ψ(x)(i/
∂ − m)ψ(x)
Z
Z
n
o
e2
−
d4 x d4 y ψ(x)γµ ψ(x) iDµν (x − y) ψ(y)γν ψ(y) .
2
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9
5
To look at the interaction term in k-space, we put
Z 4
d p1 −ip1 x
ψ(x) =
e
ψ(p1 )
(2π)4
Z 4
d p2 +ip2 x
ψ(x) =
ψ(p2 )
e
(2π)4
Z
qµ qν
d4 q e−iq(x−y)
µν
µν
iD (x − y) =
g − (1 − ξ) 2
(2π)4 q 2 + i
q
Z 4
d k1 −ik1 y
ψ(y) =
e
ψ(k1 ).
(2π)4
Z 4
d k2 +ik2 y
ψ(y) =
ψ(k2 )
e
(2π)4
The fermion-fermion interaction term becomes
Z
Z
n
o
e2
−
d4 x d4 y ψ(x)γµ ψ(x) iDµν (x − y) ψ(y)γν ψ(y)
2
i
h


qµ qν
µν
Z 4
d p1 d4 k1 d4 q  ψ(p1 − q)γµ ψ(p1 ) g − (1 − ξ) q2 ψ(k1 + q)γν ψ(k1 ) 
e2
.
=−


2
(2π)12
q 2 + i
The longitudinal part of iDµν actually doesn’t contribute here, because e.g.
i
h
h
i
ψ(k1 +q)/
q ψ(k1 ) = ψ(k1 +q) (/
k1 + /
q) − /
k 1 ψ(k1 ) = ψ(k1 +q) m − m ψ(k1 ) = 0
so the interaction term reduces to
Z 4
d p1 d4 k1 d4 q ψ(p1 − q)γ µ ψ(p1 )ψ(k1 + q)γµ ψ(k1 )
e2
−
.
2
(2π)12
q 2 + i
This looks exactly like the interaction term you get when you write down the
hamiltonian for the interacting homogenous electron gas, although in that case
there are no γµ matrices and the k integrals are only over 3-dimensional space.
Part b.
The lowest-order diagram is
p0
p
µ
p+p
= (−ie)2 (p − p0 )µ
0
ν
k
k0
−igµν
us (k)γ ν vs0 (k 0 )
(p − p0 )2
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9
6
The squared, averaged matrix element is (ignoring the electron mass)
1 4 (p − p0 )µ (p − p0 )ν e
Tr /
k γµ /
k 0 γν
4
(p − p0 )4
(p − p0 )µ (p − p0 )ν kµ kν0 + kν kµ0 − (k · k 0 )gµν
= e4
0
4
(p − p )
n h
ih
i
o
4
e
0
0
0
0
0 2
=
2
k
·
(p
−
p
)
k
·
(p
−
p
)
−
(k
·
k
)(p
−
p
)
(p − p0 )4
|M|2 =
We work in a frame such that
k = (E, 0, 0, E),
k 0 = (E, 0, 0, −E)
and
p = (E, |p| sin θ, 0, |p| cos θ),
p0 = (E, −|p| sin θ, 0, −|p| cos θ).
Then (7) reads
e4 E 2 − cos2 θ + 1
2
2|p|
e4 E 2
sin2 θ.
=
2|p|2
|M|2 =
Inserting the kinematic factors from P&S equation 4.84,
e4
dσ
=
sin2 θ
dΩ
256π 2 |p||E|
e4
√
=
sin2 θ.
2
256π E E 2 − m2
The total cross section is
Z
π
dσ
sin θdθ
dΩ
Z π
e4
√
sin3 θdθ
=
128πE E 2 − m2 0
e4
√
.
=
96πE E 2 − m2
σ = 2π
0
Part c.
The two first-order diagrams are
(7)
7
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9
k+q
µ
M1 =
ρ
σ
q
ν
q
k
= (−ie)2
Z
d4 k −igµρ
i
i
−igσν
ρ
σ
(2k
+
q)
(2k
+
q)
(2π)4
q2
(k + q)2 − m2
k 2 − m2
q2
k
ρ
µ
M2 =
σ
q
= (2ie2 )
Z
ν
q
ig ρσ
−igσν
d4 k −igµρ
(2π)4
q2
k 2 − m2
q2
The sum of the diagrams is
−igσν
−igµρ
ρσ
Π (q)
M1 + M2 =
q2
q2
where
ρσ
Π (q) = e
2
Z
d4 k (2k + q)ρ (2k + q)σ − 2g ρσ (k + q)2 − m2
(2π)4
[(k + q)2 − m2 ] [k 2 − m2 ]
(8)
Rewrite the denominator:
1
=
[(k + q)2 − m2 ] [k 2 − m2 ]
=
Z
1
0
Z
1
0
dx
{x [(k + q)2 − m2 ] + (1 − x) [k 2 − m2 ]}
2
dx
[k 2
2
+ 2xk · q + xq 2 − m2 ]
Shift variables to l = k + xq in (8):
d4 l 4lρ lσ + (1 − 2x)2 q ρ q σ − 2g ρσ l2 + (1 − x)2 q 2 − m2
,
(2π)4
[l2 − ∆]2
(9)
where we ignored numerator terms of odd order in l, and where
Πρσ (q) = e2
Z
dx
Z
∆ = −x(1 − x)q 2 + m2 .
8
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9
Using lρσ → d1 l2 g ρσ under the integral sign, we rewrite this as
Z
n
h
i
o
ρσ
2
dx g ρσ I1 (∆) + (1 − 2x)2 q ρ q σ − 2[(1 − x)2 q 2 − m2 ]g ρσ I2 (∆)
Π (q) = e
(10)
where
I1 (∆) =
=−
4
−2
d
i
(4π)2
Z
4
d
dd l
l2
d
2
(2π) (l − ∆)2
− 2 d2 Γ 1 − d2
d
∆1− 2
d
2i Γ 2 − 2
∆
(4π)2 ∆2− d2
(here we have used the fact that 1 − d2 Γ 1 − d2 = Γ 2 − d2 ) and
=−
Z
dd l
1
d
2
(2π) (l − ∆)2
i Γ 2 − d2
=
.
(4π)2 ∆2− d2
I2 (∆) =
Then (10) is
ie2
Π (q) =
(4π)2
ρσ
ie2
=
(4π)2
2
=
ie
(4π)2
Z
Z
Z
dx
dx
dx
Γ 2−
d
2
2− d
2
n
∆
Γ 2 − d2 n
o
−2∆ − 2(1 − x)2 q 2 + 2m2 g ρσ + (1 − 2x)2 q ρ q σ
d
∆2− 2
Γ 2−
d
2
d
∆2− 2
− 2(1 − x)(1 − 2x)q 2 g ρσ + (1 − 2x)2 q ρ q σ


o


− (1 − 2x)2 q 2 g ρσ − q ρ q σ − (1 − 2x)q 2 g ρσ

{z
}
|
→0
where the last term in the curly brackets integrates to 0, being odd under
x → (1 − x) while the denominator, which involves ∆, is even. Then the final
result is
iΠρσ (q) = (q 2 g µν − q ρσ )iΠ(q 2 )
iΠ(q 2 ) =
=
e2
(4π)2
e2
(4π)2
Z
1
0
Z
0
1
(1 − 2x)2
Γ(2 − d2 )
d
∆2− 2
dx
h
i
(1 − 2x)2 const. − log ∆ dx
where
const. =
2
− γ + log 4π
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
0
R1
(1 − 2x)2 log
m2
m2 −x(1−x)q 2
dx
9
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9
1
1
0
0
-1
-1
-2
-2
-2
-1
0
1
2
3
q 2 /m2
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
Z 1
e2
m2
2
iΠ(q 2 ) − iΠ(0) =
dx.
(11)
(1
−
2x)
log
(4π)2 0
m2 − x(1 − x)q 2
Values of this integral are plotted in Figure 1. The divergence at q 2 = 4m2 is
clear.
Appendix: Integration-by-parts of kinetic term in photon Lagrangian
Someday I hope to be able to perform in my head the kind of manipulations
that 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
4
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 9
10
density is
1
L = − Fµν F µν
4
1
= − (∂µ Aν − ∂ν Aµ )(∂ µ Aν − ∂ ν Aµ )
4
1
= − (∂0 A1 − ∂1 A0 )(∂ 0 A1 − ∂ 1 A0 )
2
1
− (∂0 A2 − ∂2 A0 )(∂ 0 A2 − ∂ 2 A0 )
2
1
− (∂0 A3 − ∂3 A0 )(∂ 0 A3 − ∂ 3 A0 )
2
1
− (∂1 A2 − ∂2 A1 )(∂ 1 A2 − ∂ 2 A1 )
2
1
− (∂1 A3 − ∂3 A1 )(∂ 1 A3 − ∂ 3 A1 )
2
1
− (∂2 A3 − ∂3 A2 )(∂ 2 A3 − ∂ 3 A2 ).
2
I get scared whenever I see something like Aµ or ∂ µ because I know there are a
bunch of hidden minus signs in there and they confuse me. To make sure that
all minus signs are explicit, let’s rewrite Fµν F µν with all minus signs displayed
explicitly. 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
1
L = − (−∂0 A1 − ∂1 A0 )(∂0 A1 + ∂1 A0 )
2
1
− (−∂0 A2 − ∂2 A0 )(∂0 A2 + ∂2 A0 )
2
1
− (−∂0 A3 − ∂3 A0 )(∂0 A3 + ∂3 A0 )
2
1
− (−∂1 A2 + ∂1 A2 )(−∂1 A2 + ∂1 A2 )
2
1
− (−∂1 A3 + ∂1 A3 )(−∂1 A3 + ∂1 A3 )
2
1
− (−∂2 A3 + ∂2 A3 )(−∂2 A3 + ∂2 A3 )
2
i
1h
= + (∂0 A1 )2 + (∂1 A0 )2 + 2(∂0 A1 )(∂1 A0 )
2
i
1h
+
(∂0 A2 )2 + (∂2 A0 )2 + 2(∂0 A2 )(∂2 A0 )
2
i
1h
+
(∂0 A3 )2 + (∂3 A0 )2 + 2(∂0 A3 )(∂3 A0 )
2
i
1h
(∂1 A2 )2 + (∂2 A1 )2 − 2(∂1 A2 )(∂2 A1 )
−
2
i
1h
−
(∂1 A3 )2 + (∂3 A1 )2 − 2(∂1 A3 )(∂3 A1 )
2
i
1h
−
(∂2 A3 )2 + (∂3 A2 )2 − 2(∂2 A3 )(∂3 A2 ) .
2
The next step is to integrate by parts, which entails making replacements like
(∂1 A3 )(∂3 A1 )
→
−A3 ∂1 ∂3 A1 .
Also, we break up the terms with prefactors of 2 into two separate terms, each of
which we integrate by parts in a different way. This gives a different Lagrangian
that integrates to the same thing as the old Lagrangian did, so we will just call
the new Lagrangian L as well:
i
1h
L = − A1 ∂02 A1 + A0 ∂12 A0 + A1 ∂1 ∂0 A0 + A0 ∂0 ∂1 A1
2
i
1h 2 2 2
A ∂0 A + A0 ∂22 A0 + A2 ∂2 ∂0 A0 + A0 ∂0 ∂2 A2
−
2
i
1h 3 2 3
A ∂0 A + A0 ∂32 A0 + A3 ∂3 ∂0 A0 + A0 ∂0 ∂3 A3
−
2
i
1h 2 2 2
A ∂1 A + A1 ∂22 A1 − A2 ∂2 ∂1 A1 − A1 ∂1 ∂2 A2
+
2
i
1h 3 2 3
A ∂1 A + A1 ∂32 A1 − A3 ∂3 ∂1 A1 − A1 ∂1 ∂3 A3
+
2
i
1h 3 2 3
A ∂2 A + A2 ∂32 A2 − A3 ∂3 ∂2 A2 − A2 ∂2 ∂3 A3 .
+
2
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:
0
(∂12 + ∂22 + ∂32 )
1 “ 0 1 2 3” B
∂1 ∂0
A A A A B
L=−
@
∂2 ∂0
2
∂3 ∂0
∂0 ∂1
(∂02 + ∂22 + ∂32 )
−∂2 ∂1
−∂3 ∂1
∂0 ∂2
−∂1 ∂2
2
(∂0 + ∂12 + ∂32 )
−∂3 ∂2
10
∂0 ∂3
A0
C B A1
−∂1 ∂3
CB
A @ A2
−∂2 ∂3
(∂02 + ∂22 + ∂12 )
A3
Appendix 2: Convincing the skeptics among us that −k 2 gµν + kµ kν is
singular, but −k 2 gµν + (1 − ξ1 )kµ kν 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 existence
of an eigenvector with eigenvalue 0, namely k µ 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-Popov
factor is nonsingular:
1
C
C
A
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