6Quantum electrodynamics
III Quantum Field Theory
6.4 Quantization of interactions
Let’s work out some quantum amplitudes for light interacting with electrons.
We again have the Lagrangian
L = −
1
4
F
µν
F
µν
+
¯
ψ(i
/
D − m)ψ,
where
D
µ
= ∂
µ
+ ieA
µ
,
For a change, we work in the Coulomb gauge
∇ · A
= 0. So the equation of
motion for A
0
is
∇
2
A
0
= e
¯
ψγ
0
ψ = ej
0
.
This equation has a solution
A
0
(x) = e
Z
d
3
x
0
j
0
(x
0
, t)
4π|x − x
0
|
.
In the Coulomb gauge, we can rewrite the Maxwell part (i.e.
F
µν
F
µν
part) of
the Lagrangian (not density!) as
L
A
=
Z
d
3
x
1
2
(E
2
− B
2
)
=
Z
d
3
x
1
2
(
˙
A + ∇A
0
)
2
−
1
2
B
2
=
Z
d
3
x
1
2
˙
A
2
+
1
2
(∇A
0
)
2
−
1
2
B
2
,
where the gauge condition means that the cross-term vanishes by integration by
parts.
Integrating by parts and substituting our A
0
, we get that
L
A
=
Z
d
3
x
1
2
˙
A
2
+
e
2
2
Z
d
3
x
0
j
0
(x)j
0
(x
0
)
4π|x − x
0
|
−
1
2
B
2
.
This is weird, because we now have a non-local term in the Lagrangian. This
term arises as an artifact of working in Coulomb gauge. This doesn’t appear in
Lorenz gauge.
Let’s now compute the Hamiltonian. We will use capital Pi (
Π
) for the
conjugate momentum of the electromagnetic potential
A
, and lower case pi (
π
)
for the conjugate momentum of the spinor field. We have
Π =
∂L
∂
˙
A
=
˙
A
π
ψ
=
∂L
∂
˙
ψ
= iψ
†
.
Putting all these into the Hamiltonian, we get
H =
Z
d
3
x
1
2
˙
A
2
+
1
2
B
2
+
¯
ψ(−iγ
i
∂
i
+ m)ψ − ej · A
+
e
2
2
Z
d
3
x
0
j
0
(x)j
0
(x
0
)
4π|x − x
0
|
,
where
j =
¯
ψγψ, j
0
=
¯
ψγ
0
ψ.
After doing the tedious computations, one finds that the Feynman rules are as
follows:
(i) The photons are denoted as squiggly lines:
Each line comes with an index
i
that tells us which component of
A
we
are talking about. Each internal line gives us a factor of
D
tr
ij
=
i
p
2
+ iε
δ
ij
−
p
i
p
j
|p|
2
,
while for each external line,
E
(i)
we simply write down a polarization vector
E
(i)
corresponding to the
polarization of the particle.
(ii) The ej · A term gives us an interaction of the form
This contributes a factor of
−ieγ
i
, where
i
is the index of the squiggly line.
(iii)
The non-local term of the Lagrangian gives us instantaneous non-local
interactions, denoted by dashed lines:
x
y
The contribution of this factor, in position space, is given by
i(eγ
0
)
2
δ(x
0
− y
0
)
4π|x − y|
.
Whoa. What is this last term saying? Recall that when we previously derived
our Feynman rules, we first obtained some terms from Wick’s theorem that
contained things like
e
ip·x
, and then integrated over
x
to get rid of these terms,
so that the resulting formula only included momentum things. What we’ve
shown here in the last rule is what we have before integrating over
x
and
y
. We
now want to try to rewrite things so that we can get something nicer.
We note that this piece comes from the
A
0
term. So one possible strategy
is to make it into a
D
00
piece of the photon propagator. So we treat this as
a photon indexed by
µ
= 0. We then rewrite the above rules and replace all
indices i with µ and let it range over 0, 1, 2, 3.
To do so, we note that
δ(x
0
− y
0
)
4π|x − y|
=
Z
d
4
p
(2π)
4
e
ip·(x−y)
|p|
2
.
We now define the γ-propagator
D
µν
(p) =
i
p
2
+iε
δ
µν
−
p
µ
p
ν
|p|
2
µ, ν 6= 0
i
|p|
2
µ = ν = 0
0 otherwise
We now have the following Feynman rules:
(i) The photons are denoted as squiggly lines:
Each internal line comes with an index
µ
ranging over 0
,
1
,
2
,
3 that tells
us which component of
A
we are talking about, and give us a factor of
D
µν
, while for each external line,
E
(µ)
we simply write down a polarization vector
E
(µ)
corresponding to the
polarization of the particle.
(ii) The ej · A term gives us an interaction of the form
This contributes a factor of
−ieγ
i
, where
i
is the index of the squiggly line.
Now the final thing to deal with is the annoying
D
µν
formula. We claim that
it can always be replaced by
D
µν
(p) = −i
η
µν
p
2
,
i.e. in all contractions involving
D
µν
we care about, contracting with
D
µν
gives
the same result as contracting with
−i
η
µν
p
2
. This can be proved in full generality
using momentum conservation, but we will only look at some particular cases.
Example. Consider the process
e
−
e
−
→ e
−
e
−
.
We look at one particular diagram
p, s
q, r
p
0
, s
0
q
0
, r
0
µ
ν
We have two vertices, which contribute to a term of
e
2
[¯u(p
0
)γ
µ
u(p)]D
µν
(k)[¯u(q
0
)γ
ν
u(q)],
where
k = p − p
0
= −q + q
0
.
We show that in this case, we can replace D
µν
(k) by
D
µν
(k) =
−iη
µν
k
2
.
The proof follows from current conservation. Recall that u(p) satisfies
(
/
p − m)u(p) = 0.
We define the spinor combinations
α
µ
= ¯u(p
0
)γ
µ
u(p)
β
µ
= ¯u(q
0
)γ
µ
u(q)
What we have is that
k
µ
α
µ
= ¯u(p
0
)(
/
p
0
−
/
p)u(p) = ¯u(p
0
)(m − m)u(p) = 0.
Similarly, k
µ
β
µ
is also zero. So our Feynman diagram is given by
α
µ
D
µν
β
ν
= i
α · β
k
2
−
(α · k)(β · k)
|k|
2
k
2
+
α
0
β
0
|k|
2
.
But we know that α
µ
k
µ
= β
µ
k
µ
= 0. So this is equal to
i
α · β
k
2
−
k
2
0
α
0
β
0
|k|
2
k
2
+
α
0
β
0
|k|
2
= i
α · β
k
2
−
1
|k|
2
k
2
(k
2
0
− k
2
)α
0
β
0
= i
α · β
k
2
−
|k|
2
|k|
2
k
2
α
0
β
0
= −i
α · β
k
2
= α
µ
−iη
µν
k
2
β
ν
.
What really is giving us this simple form is current conservation.
In general, in Lorenz gauge, we have
D
µν
= −
i
p
2
η
µν
+ (α − 1)
p
µ
p
ν
p
2
,
and the second term cancels in all physical processes, for similar reasons.
Charged scalars
We quickly go through the Feynman rules for charged complex scalar fields. We
will not use these for anything. The Lagrangian coming from minimal coupling
is
L = (D
µ
ψ)
†
D
µ
ψ −
1
4
F
µν
F
µν
We can expand the first term to get
(D
µ
ψ)
†
D
µ
ψ = ∂
µ
ψ
†
∂
µ
ψ − ieA
µ
(ψ
†
∂
µ
ψ − ψ∂
µ
ψ
†
) + e
2
A
µ
A
µ
ψ
†
ψ.
The Feynman rules for these are:
(i) The first interaction term gives us a possible vertex
p q
This contributes a factor of −ie(p + q)
µ
.
(ii) The A
µ
A
µ
ψ
†
ψ term gives diagrams of the form
p q
This contributes a factor of 2ie
2
η
µν
.