Quantum Mechanics for Mathematicians

Taking

Ĥ=^1

2

∫

R^3

:(|Ê|^2 +|B̂|^2 ):d^3 x=

1

2

∫

R^3

:

(

|

∂Â

∂t

|^2 +|∇×Â|^2

)

:d^3 x

=

∫

R^3

ωp(a† 1 (p)a 1 (p) +a† 2 (p)a 2 (p))d^3 p

one can show, using equations 46.17, 46.18 and properties of the polarization
vectorsj(p), that one has as required

∂Â

∂t

= [Â,−iĤ]

∂

∂t

(aσ(p)e−iωpt)|t=0=−iωpaσ(p) = [aσ(p),−iĤ]

The first of these uses the position space expression in terms of fields, the second
the momentum space expression in terms of annihilation and creation operators.

46.5.2 Spatial translations

For spatial translations, we have

U(a, 1 ) =e−ia·

P̂

whereP̂is the momentum operator. It has the momentum space expression

P̂=

∫

R^3

p(a† 1 (p)a 1 (p) +a† 2 (p)a 2 (p))d^3 p

which satisfies

∇(aσ(p)eip·x) =ipaσ(p)eip·x= [−iP̂,aσ(p)eip·x]

In terms of position space fields, one has

P̂=

∫

R^3

:Ê×B̂:d^3 x

satisfying
∇Âj= [−iP̂,Âj]

One way to derive this is to use the fact that, for the classical theory,

PEM=

∫

R^3

E×Bd^3 x

is the momentum of the electromagnetic field, since one can use the Poisson
bracket relations 46.7 to show that

{PEM,Aj(x)}=∇Aj(x)