It will be convenient to parametrize the corresponding solutions in a basis where the chirality matrix
is diagonal. In this basis, we decompose the 4-component spinors into blocks of two-component
left- and right-spinors,
us(k) =(
uLs(k)
uRs(k))
vs(k) =(
vLs(k)
vRs(k))
(23.12)The equations (23.11) then reduce to
(
−im k·σ
k·σ ̄ −im
)(
uLs(k)
uRs(k))
= 0
(
+im k·σ
k·σ ̄ +im)(
vLs(k)
vRs(k))
= 0 (23.13)or in components,
(k·σ)uRs(k) = +imuLs(k)
(k· ̄σ)uLs(k) = +imuRs(k)
(k·σ)vRs(k) = −imuLs(k)
(k· ̄σ)vLs(k) = −imuRs(k) (23.14)Since we have
(k·σ)(k·σ ̄) =m^2 I 2 (23.15)the general solutions to these equations may be parametrized by two pairs of independent 2-
component spinorsξsandηs, by,
us(k) =(
(√
k·σ)ξs
i(√
k· ̄σ)ξs)
vs(k) =(
(√
k·σ)ηs
−i(√
k·σ ̄)ηs)
(23.16)There is no problem in defining the square roots of the matricesk·σandk· ̄σ, since they are
Hermitian, and may be diagonalized. The spinorsξsandηsare arbitrary basis vectors, and may
be conveniently chosen to obey the following normalizationconditions,
ξs†′ξs=η†s′ηs = δs′,s
ξs†′ηs=ηs†′ξs = 0 (23.17)One may then choose the following explicit basis,
ξ 1 =η 1 =(
1
0)
ξ 2 =η 2 =(
0
1)
(23.18)With the help of these normalizations, it is now straightforward to evaluate
u ̄r(k)us(k) = +2mδrs u ̄r(k)vs(k) = 0
v ̄r(k)vs(k) =− 2 mδrs ̄vr(k)us(k) = 0 (23.19)