Take thecomplex conjugatecarefully remembering thatx 4 andA 4 will change signs.
(
∂
∂xi
+
ie
̄hcAi)
γi∗S∗Cψ+(
−
∂
∂x 4−
ie
̄hcA 4
)
γ 4 ∗S∗Cψ+
mc
̄hSC∗ψ= 0Multiply from the leftbySC∗−^1.
(
∂
∂xi
+
ie
̄hcAi)
SC∗−^1 γ∗iSC∗ψ+(
−
∂
∂x 4−
ie
̄hcA 4
)
SC∗−^1 γ 4 ∗S∗Cψ+
mc
̄hψ= 0Compare this to theoriginal Dirac equation,
(
∂
∂xμ
+
ie
̄hcAμ)
γμψ+
mc
̄hψ= 0
(
∂
∂xi+
ie
̄hcAi)
γiψ+(
∂
∂x 4+
ie
̄hcA 4
)
γ 4 ψ+
mc
̄hψ= 0The two equations will be thesame if the matrixSCsatisfiesthe conditions.
S∗−C^1 γi∗S∗C=γi
SC∗−^1 γ∗ 4 SC∗=−γ 4.Recalling theγmatrices in our representation,
γ 1 =
0 0 0 −i
0 0 −i 0
0 i 0 0
i 0 0 0
γ 2 =
0 0 0 − 1
0 0 1 0
0 1 0 0
−1 0 0 0
γ 3 =
0 0 −i 0
0 0 0 i
i 0 0 0
0 −i 0 0
γ 4 =
1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 − 1
note thatγ 1 andγ 3 are completely imaginary and will change sign upon complex conjugation, while
γ 2 andγ 4 are completely real and will not. The solution in our representation (only) is
SC∗=SC∗−^1 =SC=SC−^1 =γ 2.It anti-commutes withγ 1 andγ 3 producing a minus sign to cancel the one from complex conjugation.
It commutes withγ 2 giving the right + sign. It anti-commutes withγ 4 giving the right - sign.
Thecharge conjugate of the Dirac spinoris given by.
ψ′=γ 2 ψ∗Of course a second charge conjugation operation takes the state back to the originalψ.
Applying this to the plane wave solutions gives.
ψ(1)~p =√
mc^2
|E|Vu(1)~p ei(~p·~x−Et)/ ̄h → −√
mc^2
|E|Vu(4)−~pei(−~p·~x+Et)/ ̄h≡√
mc^2
|E|Vv(1)~p ei(−~p·~x+Et)/ ̄hψ(2)~p =√
mc^2
|E|Vu(2)~p ei(~p·~x−Et)/ ̄h →√
mc^2
|E|Vu(3)−~pei(−~p·~x+Et)/h ̄≡√
mc^2
|E|Vv(2)~p ei(−~p·~x+Et)/h ̄