130_notes.dvi

so that
kρxρ=k·x=~k·~x−ωt.

We can then write the radiation field in a more covariant way.

A~(~x,t) =√^1

V

∑

k

∑^2

α=1

ˆǫ(α)

(

ck,α(0)eikρxρ+c∗k,α(0)e−ikρxρ

)

A convenient shorthand for calculations is possible by noticing that the second term is just the
complex conjugate of the first.

A~(~x,t) =√^1

V

∑

k

∑^2

α=1

ˆǫ(α)

(

ck,α(0)eikρxρ+c.c.

)

A~(~x,t) =√^1

V

∑

k

∑^2

α=1

ǫˆ(α)ck,α(0)eikρxρ+c.c.

Note again that we have made this a transverse field by construction. The unit vectors ˆǫ(α)are
transverse to the direction of propagation. Also note that we areworking in a gauge withA 4 = 0,
so this can also represent the 4-vector form of the potential. TheFourier decomposition of the
radiation fieldcan be written very simply.

Aμ=

1

√

V

∑

k

∑^2

α=1

ǫ(μα)ck,α(0)eikρxρ+c.c.

This choice of gauge makes switching between 4-vector and 3-vector expressions for the potential
trivial.

Let’sverify that this decomposition of the radiation field satisfies the Maxwell equation,
just for some practice. Its most convenient to use the covariantform of the equation and field.

✷Aμ= 0

✷

(

1

√

V

∑

k

∑^2

α=1

ǫ(μα)ck,α(0)eikρxρ+c.c.

)

=

1

√

V

∑

k

∑^2

α=1

ǫ(μα)ck,α(0)✷eikρxρ+c.c.

=

1

√

V

∑

k

∑^2

α=1

ǫ(μα)ck,α(0)(−kνkν)eikρxρ+c.c.= 0

The result is zero sincekνkν=k^2 −k^2 = 0.

Let’s alsoverify that∇·~ A~= 0.

∇·~

(

1

√

V

∑

k

∑^2

α=1

ˆǫ(α)ck,α(t)ei

~k·~x

+c.c.

)

=

1

√

V

∑

k

∑^2

α=1

ck,α(t)ˆǫ(α)·∇~ei

~k·~x

+c.c.

=

1

√

V

∑

k

∑^2

α=1

ck,α(t)ˆǫ(α)·~kei

~k·~x

+c.c.= 0

The result here is zero because ˆǫ(α)·~k= 0.