=
1
rσixi
r1
2
(σjσnxjpn+ (σjσn+ 2iǫnjkσk)xnpj) =1
rσixi
r(
1
2
(σjσnxjpn+σjσnxnpj) +iǫnjkσkxnpj)=1
rσixi
r(
1
2
(σjσnxjpn+σnσjxjpn) +iσkǫnjkxnpj) =1
rσixi
r(
1
2
(σjσn+σnσj)xjpn+iσkLk)=1
rσixi
r(
1
2
2 δjnxjpn+iσkLk) =1
rσixi
r(xjpj+iσkLk)~σ·~p =1
r~σ·~x
r(
−i ̄hr∂
∂r+i~σ·L~)
~σ·~p=1
r~σ·~x
r(
−i ̄hr∂
∂r
+i~σ·~L)
Note that the operators~σr·~xandi~σ·L~act only on the angular momentum parts of the state. There
are no radial derivatives so they commute with−i ̄hr∂r∂. Lets pick a shorthand notation for the
angular momentum eigenstates we must use. These have quantum numbersj,mj, andℓ.ψAwill
haveℓ=ℓAandψB must have the other possible value ofℓwhich we labelℓB. Following the
notation of Sakurai, we will call the state|jmjℓA〉 ≡ Y
mj
jℓA=αYℓA,mj−^12 χ++βYℓA,mj+^12 χ−. (Note
that our previous functions made use ofm=mℓparticularly in the calculation ofαandβ.)
c1
r~σ·~x
r(
−i ̄hr∂
∂r+i~σ·L~)(
ψB
ψA)
=
(
E−V(r)−mc^20
0 E−V(r) +mc^2)(
ψA
ψB)
c
1
r~σ·~x
r(
−i ̄hr∂
∂r+i~σ·~L)(
if(r)YjℓmBj
g(r)YjℓmAj)
=
(
E−V(r)−mc^20
0 E−V(r) +mc^2)(
g(r)YjℓmAj
if(r)YjℓmBj)
Theeffect of the two operators related to angular momentumcan be deduced. First,~σ·L~
is related toK. For positiveκ,ψAhasℓ=j+^12. For negativeκ,ψAhasℓ=j−^12. For either,ψB
has the opposite relation forℓ, indicating why the full spinor is not an eigenstate ofL^2.
K =(
~σ·~L+ ̄h 0
0 −~σ·L~+− ̄h)
Kψ = −κ ̄hψ=(
~σ·~L+ ̄h 0
0 −~σ·~L− ̄h)(
ψA
ψB)
=∓
(
j+1
2
)
̄h(
ψA
ψB)
(~σ·L~+ ̄h)ψA = −κ ̄hψA
~σ·~LψA = (−κ−1) ̄hψA
(−~σ·~L− ̄h)ψB = −κ ̄hψB
~σ·~LψB = (κ−1) ̄hψBSecond, ~σr·~xis a pseudoscalar operator. It therefore changes parity and theparity of the state is
given by (−1)ℓ; so it must changeℓ.
~σ·~x
rYjℓmAj=CYjℓmBjThe square of the operator
(~σ·~x
r) 2
is one, as is clear from the derivation above, so we know the effect
of this operator up to a phase factor.
~σ·~x
rYjℓmAj=eiδYjℓmBj