We should find four solutions. Lets start with one that gives aspin up electronin the first two
components and plug it into the Dirac equation to see what the third and fourth components can
be for a good solution.
−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
1
0
B 1
B 2
= 0
−E+mc^2 +B 1 pzc+B 2 (px−ipy)c
B 1 (px+ipy)c−B 2 pzc
−pzc+B 1 (E+mc^2 )
−(px+ipy)c+B 2 (E+mc^2 )
= 0
Use the third and fourth components to solve for the coefficients and plug them in for a check of the
result.
B 1 =pzc
E+mc^2
B 2 =
(px+ipy)c
E+mc^2
−E+mc^2 + p(^2) c 2
E+mc^2
pz(px+ipy)c^2 −pz(px+ipy)c^2
E+mc^2
0
0
= 0
−E^2 +(mc^2 )^2 +p^2 c^2
E+mc^2
0
0
0
= 0
This will be a solution as long asE^2 =p^2 c^2 + (mc^2 )^2 , not a surprising condition. Adding a
normalization factor, the solution is.
u~p=N
1
0
pzc
E+mc^2
(px+ipy)c
E+mc^2
u~p=0=N
1
0
0
0
This reduces to the spin up positive energy solution for a particle at rest as the momentum goes to
zero. We can therefore identify this as that same solution boostedto have momentum~p. The full
solution is.
ψ(1)~p =N
1
0
pzc
E+mc^2
(px+ipy)c
E+mc^2
ei(~p·~x−Et)/ ̄h