constants.
ak,α|nk,α〉 =
√
nk,α|nk,α− 1 〉
a†k,α|nk,α〉 =
√
nk,α+ 1|nk,α+ 1〉
Thenk,αcan only take oninteger valuesas with the harmonic oscillator we know.
As with the 1D harmonic oscillator, we also can define thenumber operator.
H =
(
a†k,αak,α+
1
2
)
̄hω=
(
Nk,α+
1
2
)
̄hω
Nk,α = a†k,αak,α
The last step is tocompute the raising and lowering operators in terms of the original
coefficients.
ak,α =
1
√
2 ̄hω
(ωQk,α+iPk,α)
Qk,α =
1
c
(ck,α+c∗k,α)
Pk,α = −
iω
c
(ck,α−c∗k,α)
ak,α =
1
√
2 ̄hω
(ω
1
c
(ck,α+c∗k,α)−i
iω
c
(ck,α−c∗k,α))
=
1
√
2 ̄hω
ω
c
((ck,α+c∗k,α) + (ck,α−c∗k,α))
=
1
√
2 ̄hω
ω
c
(ck,α+c∗k,α+ck,α−c∗k,α)
=
√
ω
2 ̄hc^2
(2ck,α)
=
√
2 ω
̄hc^2
ck,α
ck,α=
√
̄hc^2
2 ω
ak,α
Similarly we can compute that
c∗k,α=
√
̄hc^2
2 ω
a†k,α
Since we now have the coefficients in our decomposition of the field equal to a constant times
the raising or lowering operator, it is clear thatthese coefficients have themselves become
operators.