Note that the eigenvalues for the upper and lower components have the same possible values, but are
opposite for energy eigenstates. We already know the relationℓ=j±^12 from NR QM. We simply
check that it is the same here.
ℓ(ℓ+ 1) = (j±1
2
)(j+ 1±1
2
) =j^2 +j±1
2
j±1
2
j±1
2
+
1
4
=j^2 +j±j±1
2
+
1
4
It is correct. SoψAandψBareeigenstates ofL^2 but with different eigenvalues.
L^2
(
ψA
ψB)
= ̄h^2(
ℓ∓(ℓ∓+ 1)ψA
ℓ±(ℓ±+ 1)ψB)
ℓ±=j±1
2
Now weapply the Dirac equationand try to use our operators to help solve the problem.
(
γμ
∂
∂xμ+γμie
̄hcAμ+mc
̄h)
ψ= 0
(
γi∂
∂xi+γ 4∂
∂x 4+γ 4ie
̄hciA 0 +mc
̄h)
ψ= 0
(
cγi∂
∂xi−iγ 4∂
∂t−γ 4e
̄he
r+
mc^2
̄h)
ψ= 0
(
̄hcγ 4 γi∂
∂xi−i ̄h∂
∂t+V(r) +mc^2 γ 4)
ψ= 0̄hcγ 4 γi∂
∂xiψ=(
i ̄h∂
∂t−V(r)−mc^2 γ 4)
ψ= 0̄hc(
1 0
0 − 1
)(
0 −iσi
iσi 0)
∂
∂xiψ=(
i ̄h∂
∂t−V(r)−mc^2 γ 4)
ψ= 0c(
0 −i ̄hσi
−i ̄hσi 0)
∂
∂xiψ=(
i ̄h∂
∂t−V(r)−mc^2 γ 4)
ψ= 0c(
0 σipi
σipi 0)(
ψA
ψB)
=
(
i ̄h∂t∂ −V(r)−mc^20
0 i ̄h∂t∂ −V(r) +mc^2)(
ψA
ψB)
c(
0 σipi
σipi 0)(
ψA
ψB)
=
(
E−V(r)−mc^20
0 E−V(r) +mc^2)(
ψA
ψB)
The Dirac Equation then is.
c~σ·~p(
ψB
ψA)
=
(
E−V(r)−mc^20
0 E−V(r) +mc^2)(
ψA
ψB)
We can use commutation and anticommutation relations towrite~σ·~pin terms of separate
angular and radial operators.
σixˆiσjxˆjσnpn = σiσjxˆixˆjσnpn=
1
2
(σiσj+σjσi)ˆxiˆxjσnpn=1
2
2 δijxˆixˆjσnpn=σnpnσnpn = σixˆiσjxˆjσnpn=1
rσixi
r(σjσnxjpn) =1
rσixi
r1
2
(σjσnxjpn+σnσjxnpj)