κ^2 ̄h^2 −̄h^2
4
= j(j+ 1) ̄h^2κ^2 = j^2 +j+1
4
κ = ±(j+1
2
)
Kψ = ±(j+1
2
)ψThe eigenvalues ofKare
κ=±(
j+1
2
)
̄h.We may explicitly write out the eigenvalue equation forKforκ=±
(
j+^12)
̄h.Kψ=−κ ̄hψ=(
~σ·L~+ ̄h 0
0 −~σ·~L− ̄h)(
ψA
ψB)
=∓
(
j+1
2
)
̄h(
ψA
ψB)
The difference betweenJ^2 andL^2 is related to~σ·L~.
L^2 =J^2 − ̄h~σ·L~−3
4
̄hWe may solve for the effect of~σ·~Lon the spinorψ, then, solve for the effect ofL^2. Note that since
ψAandψBare eigenstates ofJ^2 and~σ·L~, they are eigenstates ofL^2 but have different eigenvalues.
(
~σ·~L+ ̄h 0
0 −~σ·~L− ̄h
)(
ψA
ψB)
= ̄h(
(±(j+^12 )ψA
(±(j+^12 )ψB)
(
~σ·~L 0
0 −~σ·~L)(
ψA
ψB)
= ̄h(
(±(j+^12 ∓1)ψA
(±(j+^12 ±1)ψB)
~σ·~L(
ψA
ψB)
= ̄h(
(±(j+^12 ∓1)ψA
(∓(j+^12 ±1)ψB)
L^2
(
ψA
ψB)
= j(j+ 1) ̄h^2(
ψA
ψB)
− ̄h^2(
(±(j+^12 ∓1)ψA
(∓(j+^12 ±1)ψB)
−
3
4
̄h^2(
ψA
ψB)
= ̄h^2((
j(j+ 1)−3
4
)(
ψA
ψB)
−
(
(±(j+^12 ∓1)ψA
(∓(j+^12 ±1)ψB))
= ̄h^2(
(j(j+ 1)−^34 ∓(j+^12 ∓1))ψA
(j(j+ 1)−^34 ±(j+^12 ±1))ψB)
= ̄h^2(
(j^2 +j∓j∓^12 + 1−^34 )ψA
(j^2 +j±j±^12 + 1−^34 )ψB)
= ̄h^2(
(j^2 +j∓j∓^12 +^14 )ψA
(j^2 +j±j±^12 +^14 )ψB