Forμ= 3 or 4,aμν=δμνand the requirement is fairly obviously satisfied. Checking the requirement
forμ= 1, we get.
γ 1 cos^2
θ
2+γ 1 γ 1 γ 2 cosθ
2sinθ
2−γ 1 γ 2 γ 1 cosθ
2sinθ
2−γ 1 γ 2 γ 1 γ 1 γ 2 sin^2θ
2= a 1 νγνaμν =
cosθ sinθ 0 0
−sinθ cosθ 0 0
0 0 1 0
0 0 0 1
γ 1 cos^2
θ
2+ 2γ 2 cos
θ
2sin
θ
2−γ 1 sin^2
θ
2= cosθγ 1 + sinθγ 2
γ 1 cosθ+γ 2 sinθ = cosθγ 1 + sinθγ 2This also proves to be the right transformation ofψ so that theDirac equation is invariant
under rotationsabout thekaxis if we transform the Dirac spinor according toψ′=Srotψwith
the matrix
Srot= cosθ
2+γiγjsinθ
2andijkis a cyclic permutation.
Despite the fact that we are using a vector of constant matrices,γμ, the Dirac equation is covariant
if we choose the right transformation of the spinors. This allows us to move from one coordinate
system to another.
As an example, we might try our solution for a free electron with spin up along the z axis at rest.
ψ(1)=ψE=+mc (^2) ,+ ̄h/ 2 =
1
√
V
1
0
0
0
e−imc(^2) t/ ̄h
1
√
V
1
0
0
0
eipρxρ/ ̄hThe solution we found for a free particle with momentum~pwas.
ψ~p(1)=√
E+mc^2
2 EV
1
0
pzc
E+mc^2
(px+ipy)c
E+mc^2
eipρxρ/ ̄hImagine we boost the coordinate system along the−xdirection withvc=β. We can transform the
momentum of the electron to the new frame.
p′ν=a(μνboost)pν=
γ 0 0 −iβγ
0 1 0 0
0 0 1 0
iβγ 0 0 γ
0
0
0
imc
=
βγmc
0
0
iγmc
The momentum along the x direction ispx=βγmc=mcsinhχ. We now have two ways to get the
free particle state with momentum in the x direction. We can use our free particle state
ψ(1) =√
E+mc^2
2 EV′
1
0
0
pxc
E+mc^2
ei(~p·~x−Et)/ ̄h