Solutions to Peskin & Schroeder
Chapter 9
Zhong-Zhi Xianyu∗
Institute of Modern Physics and Center for High Energy Physics,
Tsinghua University, Beijing, 100084
Draft version: November 14, 2012
1
Scalar QED
The Lagrangian for scalar QED reads
L = − 41 Fµν F µν + (Dµ ϕ)† (Dµ ϕ) − m2 ϕ† ϕ,
(1)
with
Fµν = ∂µ Aν − ∂ν Aµ ,
Dµ ϕ = (∂µ + ieAµ )ϕ.
(a) Expanding the covariant derivative, it’s easy to find the corresponding Feynman’s
rules:
= 2ie2 g µν
= −ie(p1 − p2 )µ
with all momenta pointing inwards.
The propagators are standard. We will work in the Feynman gauge and set ξ = 1,
then the propagator for photon is simply
−iηµν
,
p2 + iϵ
and the propagator for scalar is
i
.
p2 − m2 + iϵ
∗ E-mail:
xianyuzhongzhi@gmail.com
1
Notes by Zhong-Zhi Xianyu
Solution to P&S, Chapter 9 (draft version)
(b) Now we calculate the spin-averaged differential cross section for the process e+ e− →
ϕ∗ ϕ. The scattering amplitude is given by
iM = (−ie)2 v̄(k2 )γ µ u(k1 )
−i
(p1 − p2 )µ .
s
(2)
Then the spin-averaged and squared amplitude is
[
]
1 ∑
e4
|M|2 = 2 tr (p
k 1 (p
k2
/1 − p
/2 )/
/1 − p
/2 )/
4
4s
spins
]
e [
8(k1 · p1 − k1 · p2 )(k2 · p1 − k2 · p2 ) − 4(k1 · k2 )(p1 − p2 )2 .
4s2
4
=
(3)
We may parameterize the momenta as
with p =
√
k1 = (E, 0, 0, E),
p1 = (E, p sin θ, 0, p cos θ),
k2 = (E, 0, 0, −E),
p2 = (E, −p sin θ, 0, −p cos θ),
E 2 − m2 . Then we have
e4 p2
1 ∑
|M|2 =
sin2 θ.
4
2E 2
(4)
spins
Thus the differential cross section is:
( dσ )
( ∑
)
1
p
α2 (
m2 )3/2 2
1
=
|M|2 =
1− 2
sin θ.
4
2
2
dΩ CM
2(2E) 8(2π) E
8s
E
(5)
(c)
∫
∫
dd k
i
dd k
(p − 2k)µ (p − 2k)ν
2
−
(−ie)
d
2
2
(2π) k − m
(2π)d (k 2 − m2 )((p − k)2 − m2 )
(
)
∫
dd k 2ηµν (p − k)2 − m2 − (p − 2k)µ (p − 2k)ν
2
(
)
=−e
(2π)d
(k 2 − m2 ) (p − k)2 − m2
(
)
∫
∫ 1
2ηµν k ′2 + (1 − x)2 p2 − m2 + (1 − 2x)2 pµ pν + 4k ′µ k ′ν
dd k ′
= − e2
dx
(2π)d 0
(k ′2 − ∆)2
(
)
∫
∫
1
2ηµν k ′2 (1 − d2 ) + 2ηµν (1 − x)2 p2 − m2 − (1 − 2x)2 pµ pν
dd k ′
2
.=−e
dx
(2π)d 0
(k ′2 − ∆)2
d
d
[
2 ∫ 1
(1 − 2 )Γ(1 − 2 )2ηµν
−ie
=
dx
d/2
(4π)
∆2−d/2
0
(
)
)]
Γ(2 − d2 ) (
2 2
2
2
+
2η
(1
−
x)
p
−
m
−
(1
−
2x)
p
p
µν
µ
ν
∆2−d/2
2 ∫ 1
)
]
Γ(2 − d2 ) [ (
−ie
dx
2 (1 − x)2 − x(1 − x) p2 ηµν − (1 − 2x)2 pµ pν . (6)
=
d/2
2−d/2
(4π)
∆
0
(
)
We can symmetrize the integrand as (1 − x)2 → 12 (1 − x)2 + x2 , then we get
δΠµν = 2ie2 ηµν
δΠµν =
−ie2
(4π)d/2
∫
1
dx
0
)
Γ(2 − d2 )
(1 − 2x)2 (p2 ηµν − pµ pν .
2−d/2
∆
2
(7)
Notes by Zhong-Zhi Xianyu
2
Solution to P&S, Chapter 9 (draft version)
Quantum statistical mechanics
In this problem we study the path integral formulation in statistical mechanics. The
theory can be described by the partition function:
Z = tr e−βH ,
(8)
where H is the Hamiltonian of the system. It is a function of the generalized coordinates
q and the corresponding conjugate momentum p. In this problem, we simply assume
the Hamiltonian has the following form:
H=
p2
+ V (q).
2m
(9)
We assume the dimension of the configuration space is d, then both q and p have d components. Then we assume the eigenstates of both q and p form a complete orthonormal
basis of the Hilbert space:
∫
∫
dd p
1 = dd q |q⟩⟨q|;
1=
|p⟩⟨p|.
(10)
(2π)d
Then the partition function can be written as
∫
−βH
Z = tr e
= dd q ⟨q|e−βH |q⟩.
(11)
(a) Now we derive a path integral expression for the partition function. Following the
same way of deriving path integral in a quantum field theory, we separate the quantity
e−βH into N factors:
e−βH = e−ϵH · · · e−ϵH ,
(N factors),
then inserting a complete basis between each pair of adjacent factors, as
∫
−βH
e
= dd q1 · · · dd qN −1 ⟨q|e−ϵH |qN −1 ⟩⟨qN −1 |e−ϵH |qN −2 ⟩ · · · ⟨q1 |e−ϵH |q⟩.
Now we focus on one factors:
⟨qi+1 |e−ϵH |qi ⟩ = ⟨qi+1 |e−ϵ
and
)
1 2
2m p +V (q)
ϵ
|qi ⟩ = e−ϵV (qi ) ⟨qi+1 |e− 2m p |qi ⟩,
2
∫
2
dd pi+1 dd pi
⟨qi+1 |pi+1 ⟩⟨pi+1 |e−ϵp /2m |pi ⟩⟨pi |qi ⟩
(2π)d (2π)d
∫
dd p ip(qi+1 −qi ) −ϵp2 /2m [ m ]d/2 −m(qi+1 −qi )2 /2ϵ
=
e
e
= 2πϵ
e
.
(2π)d
ϵ
− 2m p2
⟨qi+1 |e
(
|qi ⟩ =
Inserting all this into the partition function, we get:
N ∫
[ m ]N d/2 ∏
]
[
m(qi+1 − qi )2
Z=
− ϵV (qi ) ,
dd qi exp −
2πϵ
2ϵ
i=0
with qN +1 = q0 .
Now let N → ∞, then we have
∫
I
[
]
Z = Dq exp − β dτ LE (τ ) ,
3
(12)
(13)
Notes by Zhong-Zhi Xianyu
Solution to P&S, Chapter 9 (draft version)
where the integral measure is defined by
Dq = lim
N →∞
[
m ]N d/2 ∏ d
d qi ,
2πϵ(N )
i=0
N
(14)
and LE (τ ) is a Lagrangian in Euclidean form:
LE (τ ) =
m ( dq )2
+ V (q(τ )).
2 dτ
(15)
Note that the periodic integral on τ comes from the trace in the partition function.
(b) Now we study an explicit example, a simple harmonic oscillator, which can by
defined by the Lagrangian
LE = 12 q̇ 2 + 21 ω 2 q 2 .
(16)
Our task is to complete the path integral to find a expression for the partition function
of harmonic oscillator. This can be easily done by a Fourier transformation of the
coordinates q(τ ) with respect to τ . Since the “time” direction is periodic, the Fourier
spectrum of q is discrete. That is,
∑
q(τ ) = β −1/2
e2πinτ /β qn ,
(17)
n
Then we have:
∫
∫
]
1 ∑ [( 2πi )2
dτ LE (τ ) = dτ
mn + ω 2 qm qn e2πi(m+n)τ /β
2β m,n
β
]
]
1 ∑ [( 2πi )2
1 ∑ [( 2π )2 2
=
mn + ω 2 qm qn δm,−n =
n + ω 2 qn q−n
2 m,n
β
2
β
n∈Z
]
1 ∑ [( 2π )2 2
n + ω 2 |qn |2
(18)
=
2
β
n∈Z
Then the path integral can be written as
∫
∫ ∏
[
)
]
β ( 4π 2 n2
−βω 2 q02
2
2
Z = C dq0 e
dReqn dImqn exp −
+
ω
|q
|
n
2
β2
n>0
]−1
C ∏ [ 4π 2 n2
C ∏[
1 ( βω )2 ]−1
2
=
+
ω
=
1
+
ω n>0
β2
ω n>0
(πn)2 2
∑
]
[
= C sinh−1 (βω/2) = C
exp − βω(n + 12 ) .
(19)
n≥0
(c) From now on we will consider the statistics of fields. We study the statistical
properties of boson system, fermion system, and photon system.
For a scalar field, the Lagrangian is given by
∫
]
(
)2
1[ 2
(20)
ϕ̇ (τ, x) + ∇ϕ(τ, x) + m2 ϕ2 (τ, x) .
LE (τ ) = d3 x
2
Following the method we used to deal with the simple harmonic oscillator, here we
decompose the scalar field ϕ(τ, x) into eigenmodes in momentum space:
∫
∑
d3 k ik·x
2πinτ /β
−1/2
e
ϕ(τ, x) = β
e ϕn,k .
(21)
(2π)3
n
4
Notes by Zhong-Zhi Xianyu
Solution to P&S, Chapter 9 (draft version)
Then the Lagrangian can also be rewritten in terms of modes, as
]
∫
∫
∑ ∫ d3 kd3 k ′ 1 [( 2πi )2
′
′
2
dτ LE (τ ) = dτ d3 x
n
n
−
k
·
k
+
m
(2π)6 2β
β
′
n,n
′
′
× ϕn,k ϕn′ ,k′ ei2π(n +n)τ /β+i(k+k )·x
[
]
∫
d3 k ( 2π )2 2
1 ∑
2
2
n
+
k
+
m
|ϕn,k |2
=
2 n
(2π)3
β
[
)
]
∫
∑ (( 2π )2
d3 k 1 2
2
2
2
2
ω |ϕ0,k | +
=
n + ωk |ϕn,k | ,
(2π)3 2 k
β
n>0
(22)
where ωk2 = k2 + m2 . Then the partition function, as a path integral over the field
configurations can be represented by
[
((
)
]
∫ ∏
2π )2 2
Z=C
Reϕn,k Imϕn,k exp − β
n + ωk2 |ϕn,k |2
(23)
β
n>0,k
By the calculation similar to that in (b), we get
)]−1
[
∏ [ ∏ ( 4π 2 n2
∏
(
2
Z=C
ωk
+
ω
=
C
exp
− βωk n +
k
2
β
n>0
k
)]
.
(24)
k
This product gives the meaning to the formal expression
proper regularization.
(d)
1
2
[
det(−∂ 2 + m2 )
Then consider the fermionic oscillator. The action is given by
∫
∫
(
)
S = dτ LE (τ ) = dτ ψ̄(τ )ψ̇(τ ) + ω ψ̄(τ )ψ(τ ) .
]−1/2
with
(25)
The antiperiodic boundary condition ψ(τ + β) = −ψ(τ ) is crucial to expanding the
fermion into modes:
∑
e2πiτ /β ψn .
(26)
ψ(τ ) = β −1/2
n∈Z+1/2
Then the partition function can be evaluated to be
[
(
) ]
∫ ∏
∑
2πin
Z=
dψ̄n dψn − β
ψ̄n
+ ω ψn
β
n
n∈Z+1/2
∞ (
)
∏
4π 2 (n +
+ ω = C(β)
β
β2
n=0
n∈Z+1/2
)
(
(
)
= C(β) cosh 12 βω = C(β) eβω/2 + e−βω/2 ,
= C(β)
∏
( 2πin
1 2
2)
+ ω2
)
(27)
with the form of a two-level system, as expected.
(e) Finally we consider the system of photons. The partition function is given by
[∫
]
∫
( 1
)
3
2 µ
2
Z = DAµ DbDc exp
dτ d x − 2 Aµ ∂ A − b∂ c
[
]4·(−1/2)
= C(β) det(−∂)
· det(−∂ 2 ),
5
(28)
Notes by Zhong-Zhi Xianyu
Solution to P&S, Chapter 9 (draft version)
where the first determinant comes from the integral over the vector field Aµ while the
second one comes from the integral over the ghost fields. Therefore
[
]2·(−1/2)
Z = C(β) det(−∂ 2 )
,
(29)
which shows the contributions from the two physical polarizations of a photon. Here we
see the effect of the ghost fields of eliminating the additional two unphysical polarizations
of a vector field.
6