130_notes.dvi

H = (Fμ 4 )

∂Aμ

∂x 4

−L

=

1

2

(E^2 +B^2 )

if there are no source terms in the region.

Gauge symmetry may be used to put a condition on the vector potential.

∂Aν

∂xν

= 0.

This is called theLorentz condition. Even with this satisfied, there isstill substantial gauge
freedompossible. Gauge transformations can be made as shown below.

Aμ→Aμ+

∂Λ

∂xμ

✷Λ = 0

1.39 Quantization of the EM Field

The Hamiltonian for the Maxwell field may be used to quantize the field (See section 33) in much
the same way that one dimensional wave mechanics was quantized. The radiation field can be shown
to be the transverse part of the fieldA~⊥while static charges give rise toA‖andA 0.

Wedecompose the radiation field into its Fourier components

A~(~x,t) =√^1

V

∑

k

∑^2

α=1

ǫˆ(α)

(

ck,α(t)ei

~k·~x

+c∗k,α(t)e−i

~k·~x)

where ˆǫ(α) are real unit vectors, andck,αis the coefficient of the wave with wave vector~kand
polarization vector ˆǫ(α). Once the wave vector is chosen, the two polarization vectors must be
picked so that ˆǫ(1), ˆǫ(2), and~kform a right handed orthogonal system.

Plugging the Fourier decomposition into the formula for the Hamiltonian density and using the
transverse nature of the radiation field, we can compute the Hamiltonian (density integrated over
volume).

H =

∑

k,α

(ω

c

) (^2) [
ck,α(t)c∗k,α(t) +c∗k,α(t)ck,α(t)

]

This Hamiltonian will be used to quantize the EM field. In calculating the Hamiltonian, care has
been taken not to commute the Fourier coefficients and their conjugates.

The canonical coordinate and momenta may be found

Qk,α=

1

c

(ck,α+c∗k,α)