A 1 (−E+mc^2 ) +pzc
A 2 (−E+mc^2 ) + (px+ipy)c
−A 1 pzc−A 2 (px−ipy)c+ (E+mc^2 )
−A 1 (px+ipy)c+A 2 pzc
= 0
A 1 =
−pzc
−E+mc^2A 2 =−(px+ipy)c
−E+mc^2
0
0
p^2 c^2
−E+mc^2 + (E+mc(^2) )
−−E−+pzmcc 2 (px+ipy)c+−−(pEx++mcipy 2 )cpzc
= 0
0
0
−E^2 +(mc^2 )^2 +p^2 c^2
−E+mc^2
0
= 0
E^2 =p^2 c^2 + (mc^2 )^2u~p=N
−pzc
−E+mc^2
−(px+ipy)c
−E+mc^2
1
0
u~p=0=N
0
0
1
0
This reduces to the spin up “negative energy” solution for a particleat rest as the momentum goes
to zero. The full solution is
ψ
(3)
~p =N
−pzc
−E+mc^2
−(px+ipy)c
−E+mc^2
1
0
ei(~p·~x−Et)/ ̄hwithEbeing a negative number. We will eventually understand the “negative energy” solutions in
terms of anti-electrons otherwise known as positrons.
Finally, the“negative energy”, spin down solutionfollows the same pattern.
−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
A 1
A 2
0
1
= 0
ψ~p(4)=N
−(px−ipy)c
−Ep+zmcc^2
−E+mc^2
0
1
ei(~p·~x−Et)/ ̄hwithEbeing a negative number.