Peskin & Schroesder-An Introduction to Quantum Field Theory 习题答案 Ch02

(
)(
)]
+ m2 a†peip·x + bpe−ip·x aqe−iq·x + b†qeiq·x
∫∫ = d3x
d√3p
d√3q
[
(2π)3 2Ep (2π)3 2Eq (
)
× (EpEq + p · q + m2) a†paqei(p−q)·x + bpb†qe−i(p−q)·x
(
)]
− (EpEq + p · q − m2) bqaqe−i(p+q)·x + a†pb†qei(p+q)·x
Solutions to Peskin & Schroeder Chapter 2
Zhong-Zhi Xianyu∗
Institute of Modern Physics and Center for High Energy Physics, Tsinghua University, Beijing, 100084
d√3q (2π)3 2Eq
)(
[( bpe−ip·x
+ a†p )]
e−ip·x + a†peip·x · aqe−iq·x + b†qeiq·x
) eip·x
∂ ∂t
( aq
e−iq·x
+
) b†qeiq·x
∫ =
d3x
∫ (
d√3p (2π)3 2Ep
d√3q (2π)3 2Eq
)(
[( Eq bpe−ip·x + )]
+ (Eq + Ep) a†paqei(p−q)·x − bpb†qe−i(p−q)·x
∫ =
d√3p
d√3q
(2π)3 [
2Ep (2π)3 (
2Eq
)
× (Eq − Ep) bpaqe−i(Ep+Eq)t − a†pb†qei(Ep+Eqt) (2π)3δ(3)(p + q)
(
)
]
+ (Eq + Ep) a†paqei(Ep−Eq)t − bpb†qe−i(Ep−Eq)t (2π)3δ(3)(p − q)
The Hamiltonian:
∫ H=
d3x
(πϕ˙
+
π∗ ϕ˙ ∗
−
) L
=
∫
d3x
(π∗π
+
∇ϕ∗
·
∇ϕ
+
m2
ϕ∗
) ϕ.
(12)
(b) Now we Fourier transform the field ϕ as:
∫ ϕ(x) =
d3p (2π)3
√1 2Ep
( ape−ip·x
+
) b†peip·x ,
∫ =
∫ =
d3p (2π)32Ep
·
2Ep(a†pap
−
d3p (2π)3
(a†pap
−
b†pbp
) ,
bpb†p)
(18)
where the last equal sign holds up to an infinitely large constant term, as we did when calculating the Hamiltonian in (b). Then the commutators follow straightforwardly:
)( a†peip·x aqe−iq·x
−
) b†qeiq·x
− Ep bpe−ip·x − a†peip·x aqe−iq·x + b†qeiq·x
∫ =
∫ d3x
d√3p (2π)3 2Ep
(
d√3q (2π)3 2Eq
[ (Eq
−
( Ep) bpaqe−i(p+q)·x
)]
−
) a†pb†qei(p+q)·x
∫ H=
d3x
(ϕ˙ ∗ ϕ˙
+
∇ϕ∗
·
∇ϕ
+
m2
ϕ∗
) ϕ
∫∫ = d3x
d√3p
d√3q
[
(2π)3 2Ep (2π)3 2Eq
(
)(
)
× EpEq a†peip·x − bpe−ip·x aqe−iq·x − b†qeiq·x
(
)(
)
+ p · q a†peip·x − bpe−ip·x aqe−iq·x − b†qeiq·x
(b) The energy-momentum tensor The energy-momentum tensor can be defined to be the N¨other current of the space-time translational symmetry. Under space-time translation the vector Aµ transforms as,
∫ =
∫ =
d3x
Ep2 + p2 + m2 2Ep
(a†pap + bpb†p)
d3x
Ep(a†pap
+
b†pbp
+
[bp,
b†p
) ].
(15)
Note that the last term contributes an infinite constant. It is normally explained as the
− δνλδµκ,
∂Fµν = 0. ∂Aλ
Then from the first equality we get:
∂ ∂(∂λAκ)
( Fµν
F
µν
)
=
4F
λκ.
Now substitute this intoቤተ መጻሕፍቲ ባይዱEuler-Lagrange equation, we have
( ∂L ) 0 = ∂µ ∂(∂µAν ) −
T 00 = 1 (EiEi + BiBi), T i0 = T 0i = ϵijkEj Bk, etc.
(8)
2
2 The complex scalar field
The Lagrangian is given by:
L = ∂µϕ∗∂µϕ − m2ϕ∗ϕ.
(9)
(a) The conjugate momenta of ϕ and ϕ∗:
Solution to P&S, Chapter 2 (draft version)
Rewrite this in terms of the creation and annihilation operators:
∫ Q=i
d3x (ϕ∗ϕ˙ − ϕ˙∗ϕ)
=
i
∫ −
∫ d3x
∂( ∂t bp
d√3p (2π)3 2Ep
a matrix in fundamental (self) representation of U (2) group. The U (2) group, locally
∫ =
d√3p
d√3q
(2π)3 [
2Ep (2π)3
2Eq (
)
× (EpEq + p · q + m2) a†paqei(Ep−Eq)t + bpb†qe−i(Ep−Eq)t (2π)3δ(3)(p − q)
(
)
]
− (EpEq + p · q − m2) bqaqe−i(Ep+Eq)t + a†pb†qei(Ep+Eq)t (2π)3δ(3)(p + q)
∂iEi = 0,
ϵijk∂j Bk − ∂0Ei = 0,
∗E-mail: xianyuzhongzhi@gmail.com
ϵijk∂j Ek = 0,
∂iBi = 0. (4)
1
Notes by Zhong-Zhi Xianyu
Solution to P&S, Chapter 2 (draft version)
(c) The theory is invariant under the global transformation: ϕ → eiθϕ, ϕ∗ → e−iθϕ∗.
The corresponding conserved charge is:
∫ Q=i
d3x
(ϕ∗
ϕ˙
−
ϕ˙ ∗
) ϕ.
(17)
3
Notes by Zhong-Zhi Xianyu
π=
∂L ∂ϕ˙
= ϕ˙∗,
π˜
=
∂L ∂ ϕ˙ ∗
=
ϕ˙
=
π∗.
(10)
The canonical commutation relations:
[ϕ(x), π(y)] = [ϕ∗(x), π∗(y)] = iδ(x − y),
(11)
The rest of commutators are all zero.
δµAν = ∂µAν .
(5)
Thus
T˜µν
=
∂L ∂(∂µAλ)
∂ν
Aλ
−
ηµν
L
=
−F
µλ∂ν Aλ
+
1 4
ηµν FλκF
λκ.
(6)
Obviously, this tensor is not symmetric. However, we can add an additional term ∂λKλµν to T˜µν with Kλµν being antisymmetric to its first two indices. It’s easy to see that this term does not affect the conservation of T˜µν. Thus if we choose Kλµν = F µλAν , then:
∫
1 S=−
4
d4x Fµν F µν ,
with Fµν = ∂µAν − ∂ν Aµ.
(1)
(a) Maxwell’s equations We now derive the equations of motion from the action. Note that
∂Fµν ∂(∂λAκ)
=
δµλδνκ
[Q, a†] = a†, [Q, b†] = −b†.
(19)
We see that the particle a carries one unit of positive charge, and b carries one unit of negative charge.
(d) Now we consider the case with two complex scalars of same mass. In this case the Lagrangian is given by
∂L ∂Aν
= −∂µF µν
(2)
This is sometimes called the “second pair” Maxwell’s equations. The so-called “first pair” comes directly from the definition of Fµν = ∂µAν − ∂νAµ, and reads
Draft version: November 8, 2012
1 Classical electromagnetism
In this problem we do some simple calculation on classical electrodynamics. The
action without source term is given by:
(13)
thus:
∫ ϕ∗(x) =
d3p (2π)3
√1 2Ep
( bpe−ip·x
+
) a†peip·x .
(14)
2
Notes by Zhong-Zhi Xianyu
Solution to P&S, Chapter 2 (draft version)
Feed all these into the Hamiltonian:
L = ∂µΦ†i ∂µΦi − m2Φ†i Φi,
(20)
where Φi with i = 1, 2 is a two-component complex scalar. Then it is straightforward to
see that the Lagrangian is invariant under the U (2) transformation Φi → UijΦj with Uij
T µν
= T˜µν + ∂λKλµν
= F µλFλ ν +
1 4
ηµν FλκF
λκ.
(7)
Now this tensor is symmetric. It is called the Belinfante tensor in literature. We can also rewrite it in terms of Ei and Bi:
vacuum energy. We simply drop it:
∫ H=
d3x
Ep(a†pap
+
b†p
bp
) .
(16)
√ Where we have used the mass-shell condition: Ep = m2 + p2. Hence we at once find two sets of particles with the same mass m.
∂λFµν + ∂µFνλ + ∂ν Fµλ = 0.
(3)
The familiar electric and magnetic field strengths can be written as Ei = −F 0i and ϵijkBk = −F ij, respectively. From this we deduce the Maxwell’s equations in terms of Ei and Bi: