We again see that for a non-relativistic electron, the last two components are small compared to
the first. This solution is that for apositive energy electron. The fact that thelast two
components are non-zero does not mean it contains “negativeenergy” solutions.
If we make the upper two components those of aspin down electron, we get the next solution
following the same procedure.
−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
0
1
B 1
B 2
= 0
B 1 pzc+B 2 (px−ipy)c
−E+mc^2 +B 1 (px+ipy)c−B 2 pzc
−(px−ipy)c+B 1 (E+mc^2 )
pzc+B 2 (E+mc^2 )
= 0
B 1 =
(px−ipy)c
E+mc^2
B 2 =−pzc
E+mc^2
pz(px−ipy)c^2 −pz(px−ipy)c^2
E+mc^2
−E+mc^2 + p(^2) c 2
E+mc^2
0
0
= 0
0
−E^2 +(mc^2 )^2 +p^2 c^2
E+mc^2
0
0
= 0
E^2 =p^2 c^2 + (mc^2 )^2u~p=N
0
1
(px−ipy)c
E+mc^2
−pzc
E+mc^2
u~p=0=N
0
1
0
0
This reduces to the spin down positive energy solution for a particle at rest as the momentum goes
to zero. The full solution is.
ψ(2)~p =N
0
1
(px−ipy)c
E+mc^2
−pzc
E+mc^2
ei(~p·~x−Et)/ ̄hNow we take a look at the“negative energy” spin up solutionin the same way.
−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