1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 − 1
A 1
A 2
B 1
B 2
= +
A 1
A 2
B 1
B 2
A 1
A 2
−B 1
−B 2
= +
A 1
A 2
B 1
B 2
B 1 =B 2 = 0
ψ 0 =
A 1
A 2
0
0
We see that thepositive energy solutions, for a free particle at rest, aredescribed by the
upper two component spinor. what we have calledψA. We are free to choose each component
of that spinor independently. For now, lets assume that thetwo components can be used to
designate the spin up and spin downstates according to some quantization axis.
For the “negative energy solutions” we have.
1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 − 1
A 1
A 2
B 1
B 2
=−
A 1
A 2
B 1
B 2
A 1 =A 2 = 0
ψ 0 =
0
0
B 1
B 2
Wecan describe two spin states for the “negative energy solutions”.
Recall that we have demonstrated that the first two componentsofψare large compared to the other
two for a non-relativistic electron solution and that the first two components,ψA, can be used as
the two component spinor in the Schr ̈odinger equation (with a normalization factor). Lets identify
the first component as spin up along the z axis and the second as spindown. (We do still have a
choice of quantization axis.) Define a 4 by 4 matrix which gives the z component of the
spin.
Σz =1
2 i(γ 1 γ 2 −γ 2 γ 1 )=
1
2 i[(
0 −iσx
iσx 0)(
0 −iσy
iσy 0)
−
(
0 −iσy
iσy 0)(
0 −iσx
iσx 0)]
=
1
2 i[(
σxσy 0
0 σxσy)
−
(
σyσx 0
0 σyσx)]
=
1
2 i(
[σx,σy] 0
0 [σx,σy])
=
(
σz 0
0 σz