Physics 217. Quantum Field Theory. Professor Dine
Fall, 2011. Homework Set 1.1
. SOLUTIONS
1. Fill in the details of the derivation of the effective Hamiltonian for an electron in a background
electromagnetic field. Check that you get the correct g value for the magnetic moment, and
that you find the correct spin-orbit coupling in the field of an atomic nucleus.
Solution: It is remarkable how two facts about the electron emerge so easily from the Dirac
equation. Below I repeat some of the material from the handout, with annotations relevant
to this problem:
~ =0
((p0 − eA0 − m)φ − ~σ · (~
p − eA)χ
(1)
and
~ = 0.
((p0 − eA0 + m)χ − ~σ · (~
p − eA)φ
(2)
for the upper and lower components of the Dirac spinor. We can solve for χ in terms of φ.
We will first work to first order in fields. However, for the hydrogen atom problem, powers
~ = 0 and work systematically order by
of A0 are of order powers of p~2 , so there we will set A
2
0
order both in p and A . In the present approximation we write:
χ=
=
p0
p0
1
~
~σ · (~
p − eA)φ
− eA0 + m
(3)
1
1
~ +
~σ · (~
p − eA)φ
eA0~σ · p~φ
0
+m
(p + m)2
Now substitute back in the equation for φ:
(p0 − eA0 − m)φ +
0
1
~ σ · (~
~ + ~σ · p~ eA
~
σ
·
(~
p
−
e
A)~
p
−
e
A)φ
~σ · p~φ+
p0 + m
(p0 + m)2
(4)
Using the identity σi σj = δij + iijk σk , we can rewrite this expression as:
1
1
~+A
~ · p~) + iijk (pi Aj + Ai pj )σ k φ −
p~2 + e(~
p·A
~σ · p~eA0~σ · p~φ .
0
0
p +m
(p + m)2
(5)
i
j
(It will be convenient to leave the last term in this form). Now the term iijk (p A +
Ai pj ), would vanish, except that pi and Aj don’t commute, and we obtain, from this term,
~ · ~σ . The term involving A0 can be rewritten as:
ijk ∂i Aj σ k = B
(p0 +eA0 −m)φ+
1
~σ · p~eA0~σ · p~
(p0 + m)2
= −~σ · p~~σ · p~
= −~
p2
(p0
eA0
ih̄~σ · p~∂j A0 σ j
−
2
+ m)
(p0 + m)2
3A0
eh̄
~ + ~σ · (E
~ × p~))
+
(i~
p·E
(p0 + m)2 (p0 + m)2
(6)
2. Peskin 2.1.
Solution:
a. It is convenient to rewrite:
1
L = − F µν ∂µ Aν
2
µν
where we have used the antisymmetry of F . So differentiating ∂µ Aν gives
(7)
∂µ F µν = 0.
(8)
(Note the symmetries among indices; differentiating the F µν term gives the same result; in
particular, if the lagrangian includes Aµ jµ , we obtain ∂µ F µν = j ν .)
b. The “naive” stress tensor is:
Tνµ =
∂L
∂ν Aρ − Lδνµ
∂(∂µ Aρ )
(9)
1 2 µ
δν .
= −F µρ ∂ν Aρ + Fµν
4
In the second line, we have used the Lagrangian in the form of eqn. 1, and noted that
differentiating the F term gives the same result as the ∂A term (thus accounting for a factor
of 2).
Now if we add ∂λ K λµν , we obtain an object which is conserved, if we differentiate ∂µ , since
∂µ ∂λ K λµν = 0.
(10)
In particular, we can take K λµν = F µλ Aν . Then the term we add to Tνµ is
∂λ (F µλ Aν ) = f µλ ∂λ Aν + ∂λ F µλ Aν .
(11)
The second term vanishes by the equations of motion. Rewriting the first term (just relabeling
the indices) as F µρ ∂ρ Aν , we have
2 µ
Tνµ = F µρ Fρν + 1 over4Fρσ
δν .
(12)
For example,
1 ~2 ~2
T00 = F 0i Fi0 − (E
−B )
2
1 ~2 ~2
= (E
+ B ).
2
(13)
3. Peskin 2.2.
Solution:
a) Conjugate momenta:
Π = φ̇∗
Π∗ = φ̇.
Hamiltonian is constructed as usual. Heisenberg equation of motion for φ follows almost by
inspection from [φ(~x, t), Π(~x0 , t)] = iδ(~x − ~x0 )l:
iφ̇(~x, t) = [H, φ(~x, t)] = iΠ∗ .
Z
iΠ̇(~x, t) = [H, Π(~x, t)] =
(14)
~ 2 φ∗ (~x0 , t)φ(~x0 , t) + m2 φ2 φ(~x0 t) ]
d3 x0 [Π(~x, t), −∇
(15)
where in the gradient term we have integrated by parts to get this into a convenient form.
Evaluating the commutator:
Z
~ 2 φ∗ (~x0 , t) + m2 φ∗ .
d3 x0 iδ(~x − ~x0 ) −∇
iΠ̇(~x, t) = [H, Π(~x, t)] =
(16)
With Π = iφ̇∗ , this is the Klein-Gordan equation for φ∗ .
b) Because φ is complex, we have to treat the field and its complex conjugate separately. As
we have encountered for the Dirac field, we have
Z
φ(~x, t) =
φ∗ (~x, t) =
1
d3 p
p
a(~
p)e−ip·x + b† (p)ei·x
3
(2π)
2E(p)
(17)
d3 p
1
†
ip·x
−i·x
p
a
(~
p
)e
+
b(p)e
(2π)3 2E(p)
(18)
Z
The Hamiltonian is:
Z
H=
d3 p †
†
a
(~
p
)a(~
p
)
+
b
(~
p
)b(~
p
)
ω(~
p)
(2π)3
(19)
up to a normal ordering constant.
c) We first need to understand the conserved current. The lagrangian is symmetric under
φ → eiα φ. Taking α infinitesimal, and a function of x, we can implement the usual Noether
procedure:
δL = i∂µ α(∂ µ φ∗ φ − φ∗ ∂ µ φ)
(20)
so that the conserved current is the coefficient of ∂ µ α:
j µ = i(∂ µ φ∗ φ − φ∗ ∂ µ φ).
(21)
The charge is the integral of j 0 . This is
Z
Q=
d3 xi(Πφ − φ∗ Π)
(22)
This differs by a sign and a factor of two from Peskin and Schroeder, but this can be absorbed
into the definition of the current (the current is still conserved); it corresponds to making the
replacement α → −α/2. Plugging in our mode expansion above, gives, not surprisingly:
1
Q=
2
Z
d3 p †
†
a
(~
p
)a(~
p
)
−
b
(~
p
)b(~
p
)
(2π)3
corresponding to particles of charge 1/2, anti-particles of charge −1/2.
d) This is a simple exercise with the commutation relations.
(23)