The calculation of the first line, for example, proceeds as follows,
̄ur(k)us(k) =u†(k)γ^0 vs(k) =(
ξr†(√
k·σ)
−iξr†(√
k·σ ̄))t(
0 I
−I 0)(
(√
k·σ)ξs
i(√
k· ̄σ)ξs)= iξr†{
(√
k·σ)(√
k·σ ̄) + (√
k· ̄σ)(√
k·σ)}
ξs
= 2imξr†ξs= 2imδrs (23.20)These results may be used in turn to evaluate other spinor bilinears, such as the quantity ̄urγμus.
This is a 4-vector, which can only depend uponkμ, so that it must be of the form,
u ̄rγμus=kμMrs (23.21)The matrixMrs, which in principle could depend on the Lorentz-invariant quantitym, is actually
independent ofmon dimensional grounds. To evaluateMrs, we contract both sides withkμ, and
obtain,
m^2 Mrs= ̄urkμγμus=−imu ̄rus= 2m^2 δrs (23.22)where we have used the defining equation forusfor the second equality. A similar calculation may
be carried out forvs, and we find,
u ̄r(k)γμus(k) = 2kμδrs
v ̄r(k)γμvs(k) = 2kμδrs (23.23)Finally, we will also need
u†r(k)vs(−k) =v†r(k)us(−k) = 0 (23.24)23.3 Evaluation of the electric charge operator and Hamiltonian
Using the above spinor identities, it is now straightforward to further simplify the expression for
the electric charge, and we find,
Q=
∫ d (^3) k
(2π)^3
∑
s=1, 2
{
b†s(k)bs(k) +ds(k)d†s(k)
}
(23.25)
where we have made the following convenient choice for the normalization factor,
f(k^0 ) =
1
√
2 k^0(23.26)
To evaluate the Hamiltonian, we use the Dirac equation to recast its expression as follows,
H=∫
d^3 xψ†(
−γi∂i+m)
ψ=∫
d^3 xiψ†∂ 0 ψ (23.27)