6Quantum electrodynamics
III Quantum Field Theory
6.1 Classical electrodynamics
We begin by reviewing what we know about classical electrodynamics. Classically,
the electromagnetic field is specified by an electromagnetic potential
A
, from
which we derive the electromagnetic field strength tensor
F
µν
= ∂
µ
A
ν
− ∂
ν
A
µ
,
We can then find the dynamics of a free electromagnetic field by
L = −
1
4
F
µν
F
µν
.
The Euler-Lagrange equations give
∂
µ
F
µν
= 0.
It happens that F satisfies the mystical Bianchi identity
∂
λ
F
µν
+ ∂
µ
F
νλ
+ ∂
ν
F
λµ
= 0,
but we are not going to use this.
We can express these things in terms of the electromagnetic field. We write
A = (φ, A). Then we define
E = −∇φ −
˙
A, B = ∇ ∧A.
Then F
µν
can be written as
F
µν
=
0 E
x
E
y
E
z
−E
x
0 −B
z
B
y
−E
y
B
z
0 −B
x
−E
z
−B
y
B
x
0
Writing out our previous equations, we find that
∇ · E = 0
∇ · B = 0
˙
B = −∇ ∧ E
˙
E = ∇ ∧ B.
We notice something wrong. Photons are excitations of the electromagnetic
field. However, the photon only has two polarization states, i.e. two degrees of
freedom, while this vector field A
µ
has four. How can we resolve this?
There are two parts to the resolution. Firstly, note that the
A
0
field is not
dynamical, as it has no kinetic term in the Lagrangian (the
∂
0
A
0
terms cancel
out by symmetry). Thus, if we’re given
A
i
and
˙
A
i
at some initial time
t
0
, then
A
0
is fully determined by ∇ · E = 0, since it says
∇ ·
˙
A + ∇
2
A
0
= 0.
This has a solution
A
0
(x) =
Z
d
3
x
∇ ·
˙
A(x
0
)
4π(x − x
0
)
.
So
A
0
is not independent, and we’ve got down to three degrees of freedom.
What’s the remaining one?
The next thing to note is that the theory is invariant under the transformation
A
µ
(x) 7→ A
µ
(x) + ∂
µ
λ(x),
Indeed, the only “observable” thing is
F
µν
, and we can see that it transforms as
F
µν
7→ ∂
µ
(A
ν
+ ∂
ν
λ) − ∂
ν
(A
µ
+ ∂
µ
λ) = F
µν
.
It looks like that we now have an infinite number of symmetries.
This is a different kind of symmetry. Previously, we had symmetries acting
at all points in the universe in the same way, say
ψ 7→ e
iα
ψ
for some
α ∈ R
.
This gave rise to conservation laws by Noether’s theorem.
Now do we get infinitely many symmetries from Noether’s theorem? The
answer is no. The local symmetries we’re considering now have a different
interpretation. Rather than taking a physical state to another physical state,
they are really a redundancy in our description. These are known as local or
gauge symmetries. Seeing this, we might be tempted to try to formulate the
theory only in terms of gauge invariant objects like
E
and
B
, but it turns out
this doesn’t work. To describe nature, it appears that we have to introduce
quantities that we cannot measure.
Now to work with our theory, it is common to fix a gauge. The idea is that
we specify some additional rules we want
A
µ
to satisfy, and hopefully this will
make sure there is a unique
A
µ
we can pick for each equivalence class. Picking
the right gauge for the right problem can help us a lot. This is somewhat like
picking a coordinate system. For example, in a system with spherical symmetry,
using spherical polar coordinates will help us a lot.
There are two gauges we would be interested in.
Definition (Lorenz gauge). The Lorenz gauge is specified by
∂
µ
A
µ
= 0
Note that this is Lorenz, not Lorentz!
To make sure this is a valid gauge, we need to make sure that each electro-
magnetic potential has a representation satisfying this condition.
Suppose we start with
A
0
µ
such that
∂
µ
A
0µ
=
f
. If we introduce a gauge
transformation
A
µ
= A
0
µ
+ ∂
µ
λ,
then we have
∂
µ
A
µ
= ∂
µ
∂
µ
λ + f.
So we need to find a λ such that
∂
µ
∂
µ
λ = −f.
By general PDE theory, such a λ exists. So we are safe.
However, it turns out this requirement does not pick a unique representation
in the gauge orbit. We are free to make a further gauge transformation by with
∂
µ
∂
µ
λ = 0,
which has non-trivial solutions (e.g. λ(x) = x
0
).
Another important gauge is the Coulomb gauge:
Definition (Coulomb gauge). The Coulomb gauge requires
∇ · A = 0.
Of course, this is not a Lorentz-invariant condition.
Similar to the previous computations, we know that this is a good gauge.
Looking at the integral we’ve found for A
0
previously, namely
A
0
=
Z
d
3
x
0
∇ ·
˙
A(x
0
)
4π|x − x
0
|
,
we find that
A
0
= 0 all the time. Note that this happens only because we do
not have matter.
Here it is easy to see the physical degrees of freedom — the three components
in
A
satisfy a single constraint
∇ · A
= 0, leaving behind two physical degrees
of freedom, which gives the desired two polarization states.