ExpressingCin terms of the oscillator modes, we obtaint-dependent terms andt-independent
terms. The former vanish upon integration, as they are odd ink. Thet-independent terms combine
as follows,
S=i∫ d (^3) k
(2π)^3
∑
α,β
(
εα(k)×ε∗β(k)
)
aα(k)a†β(k) (19.83)
Note that the combinationεα(k)×ε∗β(k) is automatically anti-symmetric inαandβ, and thus
parallel tok.
It is instructive to work out the Hilbert space of one-photonstates at fixed value of their
momentum. To do so, we choose the momentum along thez-axis, so that
k= (0, 0 ,k) ε 1 (k) = (1, 0 ,0)
ε 2 (k) = (0, 1 ,0) (19.84)
The polarization states may be denoted by
|k, 1 〉 = a† 1 (k)|∅〉
|k, 2 〉 = a† 2 (k)|∅〉 (19.85)
Measuring thez-component of spin (which is equivalent to helicity, sincekpoints in thez-direction),
gives the following,
Sz|k,γ〉 = i
∑
β
(
εγ(k)×ε∗β(k)
)
z
a†β(k)|∅〉
= i
∑
β
(
εγ(k)×ε∗β(k)
)
z
|k,β〉 (19.86)
Working this out separately for the casesγ= 1,2, we obtain,
Sz|k, 1 〉 = i
(
ε 1 (k)×ε∗ 2 (k)
)
z
|k, 2 〉= +i|k, 2 〉
Sz|k, 2 〉 = i
(
ε 2 (k)×ε∗ 1 (k)
)
z
|k, 1 〉=−i|k, 2 〉 (19.87)
Forming the combinations with circular polarization givesthe eigenstates ofSz, as follows,
Sz|k,±〉=±|k,±〉 |k,±〉=
1
√
2
(
|k, 1 〉±i|k, 2 〉)
(19.88)As a result, the photon states contains just two spin states,with eigenvalues± ̄h, and we conclude
that the photon is spin 1.