ψ~p(1) =
coshχ 2 0 0 sinhχ 2
0 coshχ 2 sinhχ 2 0
0 sinhχ 2 coshχ 2 0
sinhχ 2 0 0 coshχ 2
1
√
V
1
0
0
0
eipρxρ/ ̄h=
1
√
V
coshχ 2
0
0
sinhχ 2
eipρxρ/ ̄h=
1
√
V
coshχ
2
1
0
0
tanhχ 2
eipρxρ/ ̄h1 + coshχ
2=
1 +eχ+e−χ
2
2=
eχ+ 2 +e−χ
4=
(
eχ(^2) +e
−χ
2
2
) 2
= cosh^2χ
2cosh
χ
2=
√
1 + coshχ
2=
√
1 +γ
2=
√
E+mc^2
2 mc^2ψ~p(1) =1
√
V
√
E+mc^2
2 mc^2
1
0
0
tanhχ 2
eipρxρ/ ̄h=
1
√
γV′√
E+mc^2
2 mc^2
1
0
0
tanhχ 2
eipρxρ/h ̄=
√
E+mc^2
2 EV′
1
0
0
tanhχ 2
eipρxρ/ ̄hIn the last step the simple Lorentz contraction was used to setV′ = Vγ. Thisboosted state
matches the plane wave solution including the normalization.
36.10Parity
It is useful to understand theeffect of a parity inversion on a Dirac spinor. Again work with
the Dirac equation and its parity inverted form in whichxj→−xjandx 4 remains unchanged (the
same for the vector potential).
γμ∂
∂xμψ(x) +mc
̄hψ(x) = 0γμ∂
∂x′μψ′(x′) +mc
̄hψ′(x′) = 0ψ′(x′) = SPψ(x)
∂
∂x′j= −
∂
∂xj