Quantum Mechanics for Mathematicians

and an action of the groupG 0 of time-independent gauge transformations, with
Lie algebra the functionsφ(x). In this case the conditionμ= 0 will just be
Gauss’s law. To see this, note that the moment mapμ∈(LieG 0 )∗will be given
by

μ(A(x),−E(x))(φ(x)) =

∫

R^3

φ(x′)∇·Ed^3 x′

since

{μ,A(x)}={

∫

R^3

φ(x′)∇·Ed^3 x′,A(x)}

={

∫

R^3

(−∇φ(x′))E(x′)d^3 x′,A(x)}

=−∇φ(x)

(using integration by parts in the first step, the Poisson bracket relations in
the second). This agrees with the definition in section 15.3 of the moment
map, since∇φ(x) is the infinitesimal change inA(x) for an infinitesimal gauge
transformationφ(x). One can similarly show that, as required sinceEis gauge
invariant,μsatisfies
{μ,E(x)}= 0

46.4 Quantization in Coulomb gauge

A different method for dealing with the time-independent gauge transformations
is to impose an additional gauge condition. For any vector potential satisfying
the temporal gauge conditionA 0 = 0, a gauge transformation can be found such
that the transformed vector potential satisfies:

Definition(Coulomb gauge).A vector potentialAμis said to be in Coulomb
gauge if∇·A= 0.

To see that this is possible, note that under a gauge transformation one has

∇·A→∇·A+∇^2 φ

so such a gauge transformation will put a vector potential in Coulomb gauge if
we can find a solution to
∇^2 φ=−∇·A (46.13)

Using Green’s function methods like those of section 12.7, this equation forφ
can be solved, with the result

φ(x) =

1

4 π

∫

R^3

1

|x−x′|

(∇·A(x′))d^3 x′

Since in temporal gauge

∇·E=−

∂

∂t

∇·A