6Quantum electrodynamics
III Quantum Field Theory
6.2 Quantization of the electromagnetic field
We now try to quantize the field, using the Lorenz gauge. The particles we
create in this theory would be photons, namely quanta of light. Things will go
wrong really soon.
We first try to compute the conjugate momentum
π
µ
of the vector field
A
µ
.
We have
π
0
=
∂L
∂
˙
A
0
= 0.
This is slightly worrying, because we would later want to impose the commutation
relation
[A
0
(x), π
0
(y)] = iδ
3
(x −y),
but this is clearly not possible if π
0
vanishes identically!
We need to try something else. Note that under the Lorenz gauge
∂
µ
A
µ
= 0,
the equations of motion tell us
∂
µ
∂
µ
A
ν
= 0.
The trick is to construct a Lagrangian where this is actually the equation of
motion, and then later impose ∂
µ
A
µ
= 0 after quantization.
This is not too hard. We can pick the Lagrangian as
L = −
1
4
F
µν
F
µν
−
1
2
(∂
µ
A
µ
)
2
.
We can work out the equations of motion of this, and we get
∂
µ
F
µν
+ ∂
ν
(∂
µ
A
µ
) = 0.
Writing out the definition of
F
µν
, one of the terms cancels with the second bit,
and we get
∂
µ
∂
µ
A
ν
= 0.
We are now going to work with this Lagrangian, and only later impose
∂
µ
A
µ
= 0
at the operator level.
More generally, we can use a Lagrangian
L = −
1
4
F
µν
F
µν
−
1
2α
(∂
µ
A
µ
)
2
,
where
α
is a fixed constant. Confusingly, the choice of
α
is also known as a
gauge. We picked
α
= 1, which is called the Feynman gauge. If we take
α →
0,
we obtain the Landau gauge.
This new theory has no gauge symmetry, and both
A
0
and
A
are dynamical
fields. We can find the conjugate momenta as
π
0
=
∂L
∂
˙
A
0
= −∂
µ
A
µ
π
i
=
∂L
∂
˙
A
i
= ∂
i
A
0
−
˙
A
i
.
We now apply the usual commutation relations
[A
µ
(x), A
ν
(y)] = [π
µ
(x), π
ν
(y)] = 0
[A
µ
(x), π
ν
(y)] = iδ
3
(x −y)δ
ν
µ
.
Equivalently, the last commutation relation is
[A
µ
(x), π
ν
(y)] = iδ
3
(x −y)η
µν
.
As before, in the Heisenberg picture, we get equal time commutation relations
[A
µ
(x, t),
˙
A
ν
(y, t)] = iη
µν
δ
3
(x −y).
As before, we can write our operators in terms of creation and annihilation
operators:
A
µ
(x) =
Z
d
3
p
(2π)
3
1
p
2|p|
3
X
λ=0
E
(λ)
µ
(p)[a
λ
p
e
ip·x
+ a
λ†
p
e
−ip·x
],
π
ν
(x) =
Z
d
3
p
(2π)
3
i
r
|p|
2
3
X
λ=0
(E
(λ)
(p))
ν
[a
λ
p
e
ip·x
− a
λ†
p
e
−ip·x
],
where for each
p
, the vectors
{E
(0)
(
p
)
, E
(1)
(
p
)
, E
(2)
(
p
)
, E
(3)
(
p
)
}
form a basis of
R
3,1
. The exact choice doesn’t really matter much, but we will suppose we pick
a basis with the following properties:
(i) E
(0)
(p) will be a timelike vector, while the others are spacelike.
(ii)
Viewing
p
as the direction of motion, we will pick
E
(3)
(
p
) to be a longi-
tudinal polarization, while
E
(1),(2)
(
p
) are transverse to the direction of
motion, i.e. we require
E
(1),(2)
µ
(p) ·p
µ
= 0.
(iii) The normalization of the vectors is given by
E
(λ)
µ
(E
(λ
0
)
)
µ
= η
λλ
0
,
We can explicitly write down a choice of such basis vectors. When
p ∝
(1, 0, 0, 1), then we choose
E
(0)
µ
(p) =
1
0
0
0
, E
(1)
µ
(p) =
0
1
0
0
, E
(2)
µ
(p) =
0
0
1
0
, E
(3)
µ
(p) =
0
0
0
1
.
Now for a general
p
, we pick a basis where
p ∝
(1
,
0
,
0
,
1), which is always possible,
and then we define
E
(
p
) as above in that basis. This gives us a Lorentz-invariant
choice of basis vectors satisfying the desired properties.
One can do the tedious computations required to find the commutation
relations for the creation and annihilation operators:
Theorem.
[a
λ
p
, a
λ
0
q
] = [a
λ†
p
, a
λ
0
†
q
] = 0
and
[a
λ
p
, a
λ
0
†
q
] = −η
λλ
0
(2π)
3
δ
3
(p −q).
Notice the strange negative sign!
We again define a vacuum |0i such that
a
λ
p
|0i = 0
for λ = 0, 1, 2, 3, and we can create one-particle states
|p, λi = a
λ†
p
|0i.
This makes sense for
λ
= 1
,
2
,
3, but for
λ
= 0, we have states with negative
norm:
hp, 0|q, 0i ≡ h0|a
0
p
a
0†
q
|0i
= −(2π)
3
δ
3
(p −q).
A Hilbert space with a negative norm means that we have negative probabilities.
This doesn’t make sense.
Here is where the Lorenz gauge comes in. By imposing
∂
µ
A
µ
= 0, we are
going to get rid of bad things. But how do we implement this constraint? We
can try implementing it in a number of ways. We will start off in the obvious
way, which turns out to be too strong, and we will keep on weakening it until it
works.
If we just asked for
∂
µ
A
µ
= 0, for
A
µ
the operator, then this doesn’t work,
because
π
0
= −∂
µ
A
µ
,
and if this vanishes, then the commutation conditions cannot possibly be obeyed.
Instead, we can try to impose this on the Hilbert space rather than on the
operators. After all, that’s where the trouble lies. Maybe we can try to split the
Hilbert space up into good states and bad states, and then just look at the good
states only?
How do we define the good, physical states? Maybe we can impose
∂
µ
A
µ
|ψi = 0
on all physical (“good”) states, but it turns out even this condition is a bit too
strong, as the vacuum will not be physical! To see this, we decompose
A
µ
(x) = A
+
µ
(x) + A
−
µ
(x),
where
A
+
µ
has the annihilation operators and
A
−
µ
has the creation operators.
Explicitly, we have
A
+
µ
(x) =
Z
d
3
p
(2π)
3
1
p
2|p|
E
(λ)
µ
a
λ
p
e
−ip·x
A
−
µ
(x) =
Z
d
3
p
(2π)
3
1
p
2|p|
E
(λ)
µ
a
λ†
p
e
ip·x
,
where summation over λ is implicit. Then we have
∂
µ
A
+
µ
|0i = 0,
but we have
∂
µ
A
−
µ
|0i 6= 0.
So not even the vacuum is physical! This is very bad.
Our final attempt at weakening this is to ask the physical states to satisfy
∂
µ
A
+
µ
(x) |ψi = 0.
This ensures that
hψ|∂
µ
A
µ
|ψi = 0,
as
∂
µ
A
+
µ
will kill the right hand side and
∂
µ
A
−
µ
will kill the left. So
∂
µ
A
µ
has
vanishing matrix elements between physical states.
This is known as the Gupta-Bleuler condition. The linearity of this condition
ensures that the physical states span a vector space H
phys
.
What does
H
phys
look like? To understand this better, we write out what
∂
µ
A
−
µ
is. We have
∂
µ
A
−
µ
=
Z
d
3
p
(2π)
3
1
p
2|p|
E
(λ)
µ
a
λ
p
(−ip
µ
)e
−ip·x
=
Z
d
3
p
(2π)
3
1
p
2|p|
i(a
3
p
− a
0
p
)e
ip·x
,
using the properties of our particular choice of the
E
µ
. Thus the condition is
equivalently
(a
3
p
− a
0
p
) |ψi = 0.
This means that it is okay to have a timelike or longitudinal photons, as long as
they come in pairs of the same momentum!
By restricting to these physical states, we have gotten rid of the negative
norm states. However, we still have the problem of certain non-zero states having
zero norm. Consider a state
|φi = a
0†
p
a
3†
p
|0i.
This is an allowed state, since it has exactly one timelike and one longitudinal
photon, each of momentum
p
. This has zero norm, as the norm contributions
from the
a
0
p
part cancel from those from the
a
3
p
part. However, this state is
non-zero! We wouldn’t want this to happen if we want a positive definite inner
product.
The solution is to declare that if two states differ only in the longitudinal
and timelike photons, then we consider them physically equivalent, or gauge
equivalent! In other words, we are quotienting the state space out by these
zero-norm states.
Of course, if we want to do this, we need to go through everything we do and
make sure that our physical observables do not change when we add or remove
these silly zero-norm states. Fortunately, this is indeed the case, and we are
happy.
Before we move on, we note the value of the Feynman propagator:
Theorem.
The Feynman propagator for the electromagnetic field, under a
general gauge α, is
h0|T A
µ
(x)A
ν
(y) |0i =
Z
d
4
p
(2π)
4
−i
p
2
+ iε
η
µν
+ (α −1)
p
µ
p
ν
p
2
e
−ip·(x−y)
.