Quantum Mechanics for Mathematicians

the Coulomb gauge condition automatically implies that Gauss’s law (46.4) will
hold. We thus can take as phase space the solutions to the Maxwell equations
satisfying the two conditionsA 0 = 0,∇·A= 0, with phase space coordinates
the pairs (A(x),−E(x)) satisfying the constraints∇·A= 0,∇·E= 0.
In Coulomb gauge the one Maxwell equation (46.3) that is not automatically
satisfied is, in terms of the vector potential

−

∂^2 A

∂t^2

=∇×(∇×A)

Using the vector calculus identity

∇×(∇×A) =∇(∇·A)−∇^2 A

and the Coulomb gauge condition, this becomes the wave equation
(
∂^2
∂t^2

−∇^2

)

A= 0 (46.14)

This is just three copies of the real Klein-Gordon equation for massm= 0,
although it needs to be supplemented by the Coulomb gauge condition.
One can proceed exactly as for the Klein-Gordon case, using the Fourier
transform to identify solutions with functions on momentum space, and quan-
tizing with annihilation and creation operators. The momentum space solutions
are given by the Fourier transformsA ̃j(p) and a classical solution can be written
in terms of them by a simple generalization of equation 43.8 for the scalar field
case

Aj(t,x) =

1

(2π)^3 /^2

∫

R^3

(αj(p)e−iωpteip·x+αj(p)eiωpte−ip·x)

d^3 p

√

2 ωp

(46.15)

where

αj(p) =

A ̃j,+(p)

√

2 ωp

, αj(p) =

A ̃j,−(−p)

√

2 ωp

Hereωp=|p|, andA ̃j,+,A ̃j,−are the Fourier transforms of positive and negative
energy solutions of 46.14.
The three componentsαj(p) make up a vector-valued functionα(p). Solu-
tions must satisfy the Coulomb gauge condition∇·A= 0, which in momentum
space is
p·α(p) = 0 (46.16)

The space of solutions of this will be two dimensional for each value ofp, and
we can choose some orthonormal basis

 1 (p), 2 (p)

of such solutions (there is a topological obstruction to doing this continuously,
but a continuous choice is not necessary). Here theσ(p)∈R^3 forσ= 1,2 are
called “polarization vectors”, and satisfy

p· 1 (p) =p· 2 (p) = 0,  1 (p)· 2 (p) = 0, | 1 (p)|^2 =| 2 (p)|^2 = 1