QuantumPhysics.dvi

Using the fact that creation operators always anti-commutewith one another, it is straightforward
to show that

b†σ 1 (~k 1 )···b†σi(~ki)···b†σj(~kj)···b†σn(~kn)|∅〉 (19.76)

=−b†σ 1 (~k 1 )···b†σj(~kj)···b†σi(~ki)···b†σn(~kn)|∅〉 (19.77)

so that the state obeys Fermi-Dirac statistics.

19.8 The photon spin and helicity

Just as we constructed the electro-magnetic momentum in terms of the Poynting vectors, so we
may also construct its total angular momentum,

J=

∫

d^3 x

(

x×(E×B)

)

(19.78)

Total angular momentum actually receives contributions from an orbital partLas well as from an
intrinsic photon spin partS, with the usual addition relation,J=L+S. We are concerned with
the spin part, and seek to separate it fromL. This separation cannot in fact be achieved in a gauge
invariant way. Thehelicity of the photon, i.e. the projection of spin onto the photon momentum
is the only part of the spin which can be isolated in a gauge invariant way.

We begin by producing a gauge dependent construction ofLandS, and we shall then extract
from these the gauge invariant helicity. UsingB=∇×Aand the double cross product formula,
we have

E×B =

∑

i

(

Ei(∇Ai)−(∇iA)Ei

)

=

∑

i

(

Ei(∇Ai)−∇i(AEi)

)

(19.79)

where we have used∇·E= 0 in going from the first to the second line. Next, we evaluate

x×

(

E×B

)

=

∑

i

(

Ei(x×∇)Ai−∇i(x×AEi) +E×A

)

(19.80)

The second term is clearly a surface term, and does not contribute to the integral when evaluating
J. The first term involves the orbital angular momentum operatorx×∇, and naturally is associated
with orbital angular momentum, though it is not gauge invariant. Thus, we define,

L =

∫

d^3 x

∑

i

Ei

(

x×∇

)

Ai

S =

∫

d^3 xE×A (19.81)

In transverse gauge,∇·A= 0, the orbital part is automatically transverse to momentum, so we
shall use this gauge to evaluate spin. ExpressingSin terms ofC(t,k),

S=

∫

d^3 k

(2π)^3

C(t,k)×∂tC(t,k) (19.82)