To evaluateHin terms of the vector potential, in transverse gauge, we usethe expressionsE=
−∂tAandB=∇×A, which themselves may be expressed in terms of the oscillator variables
aα(k) as follows,
E(t,x) = −∫ d (^3) k
(2π)^3
∂tC(t,k)eik·x
B(t,x) = i
∫
d^3 k
(2π)^3
k×C(t,k)eik·x (19.28)
The electric contribution to the Hamiltonian is calculatedas follows,
∫
d^3 xE^2 =
∫
d^3 x
∫ d (^3) k
(2π)^3
∂tC(t,k)eik·x
∫ d (^3) k′
(2π)^3
∂tC(t,k′)eik
′·x
(19.29)
The use of the Fourrier equation
∫
d^3 xei(k+k
′)·x
= (2π)^3 δ(3)(k+k′) (19.30)
implies that thek′-integral can be carried out explicitly, and results in the following expression for
the electric part of the Hamiltonian,
1
2
∫
d^3 xE^2 =
1
2
∫ d (^3) k
(2π)^3
∂tC(t,k)·∂tC(t,−k)
=
1
2
∫ d (^3) k
(2π)^3
∂tC(t,k)·∂tC(t,k)∗ (19.31)
The magnetic part is computed analogously, and we find,
1
2
∫
d^3 xB^2 =
1
2
∫ d (^3) k
(2π)^3
k^2 C(t,k)·C(t,k)∗ (19.32)
As a result, the full Hamiltonian is given by
H=
1
2
∫
d^3 k
(2π)^3(
∂tC(t,k)·∂tC(t,k)∗+k^2 C(t,k)·C(t,k)∗)
(19.33)Next, we proceed to expressHin terms of the oscillator degrees of freedomaα(k), with the help
of (19.24). First, evaluate
∂tC(t,k)·∂tC(t,k)∗+k^2 C(t,k)·C(t,k)∗=|k|∑α,β(
εα(k)·εβ(k)∗aα(k)αβ(k)∗+ε∗α·εβaα(−k)∗αβ(−k))
(19.34)Using orthogonalityεα·ε∗β=δαβ, we find,
∂tC(t,k)·∂tC(t,k)∗+k^2 C(t,k)·C(t,k)∗
=|k|∑
α(
aα(k)αα(k)∗+aα(−k)∗αα(−k))
(19.35)