With this matrix defining the spin, thethird component is the one with spin upalong the z
direction for the “negative energy solutions”. We could also define 4by 4 matrices for the x and y
components of spin by using cyclic permutations of the above.
So the four normalized solutions for a Dirac particle at rest are.
ψ(1)=ψE=+mc (^2) ,+ ̄h/ 2 =
1
√
V
1
0
0
0
e−imc(^2) t/ ̄h
ψ(2)=ψE=+mc (^2) ,−h/ ̄ 2 =
1
√
V
0
1
0
0
e−imc(^2) t/ ̄h
ψ(3)=ψE=−mc (^2) ,+ ̄h/ 2 =
1
√
V
0
0
1
0
e+imc(^2) t/ ̄h
ψ(4)=ψE=−mc (^2) ,− ̄h/ 2 =
1
√
V
0
0
0
1
e+imc(^2) t/ ̄h
Thefirst and third have spin upwhile the second and fourth have spin down. The first and second
are positive energy solutions while thethird and fourth are “negative energy solutions”, which
we still need to understand.
36.6.2 Dirac Plane Wave Solution
We now have simple solutions for spin up and spin down for both positiveenergy and “negative
energy” particles at rest. Thesolutions for nonzero momentum are not as simple.
(
γμ
∂
∂xμ+
mc
̄h)
ψ= 0ψ~p(x) =u~pei(pρxρ)/ ̄h
(
γμipμ
̄h+
mc
̄h)
ψ= 0γi=(
0 −iσi
iσi 0)
γ 4 =(
1 0
0 − 1
)
[(
0 ~p·~σ
−~p·~σ 0)
+
(
−Ec 0
0 Ec)
+mc]
u~pei(pρxρ)/ ̄h= 0
(
−E+mc^2 c~p·~σ
−c~p·~σ E+mc^2)
u~p= 0σx=(
0 1
1 0
)
σy=(
0 −i
i 0)
σz=(
1 0
0 − 1
)
−E+mc^20 pzc (px−ipy)c
0 −E+mc^2 (px+ipy)c −pzc
−pzc −(px−ipy)c E+mc^20
−(px+ipy)c pzc 0 E+mc^2
u~p= 0