130_notes.dvi

QM. We have already seen that (even with no applied fields), while thetotal angular momentum
operator commutes with the Dirac Hamiltonian, neither the orbital angular momentum
operator nor the spin operators do commute withH. The addition of a spherically symmetric
potential does not change these facts.

We have shown in the section on conserved quantities that the operator

K=γ 4 ~Σ·J~−γ 4

̄h

2

also commutes with the Hamiltonian and withJ~.Kis a measure of the component of spin
along the total angular momentum direction. We will useKto help solve problems with spherical
symmetry and ultimately the problem of hydrogen. We therefore havefour mutually commuting
operatorsthe eigenvalues of which can completely label the eigenstates:

H, J^2 , Jz, K→nr, j, mj, κ.

TheoperatorKmay be written in several ways.

K = γ 4 ~Σ·J~−γ 4

̄h

2

=γ 4 Σ~·~L+

̄h

2

γ 4 Σ~·~Σ−γ 4

̄h

2

=γ 4 ~Σ·L~+γ 4

3 ̄h

2

−γ 4

̄h

2

= γ 4 ~Σ·~L+ ̄hγ 4 =

(

~σ·~L+ ̄h 0

0 −~σ·~L− ̄h

)

Assume that theeigenvalues ofKare given by

Kψ=−κ ̄hψ.

We now compare theK^2 andJ^2 operators.

K^2 = γ 4 (Σ~·~L+ ̄h)γ 4 (~Σ·~L+ ̄hγ 4 ) = (Σ~·~L+ ̄hγ 4 )^2

= ~Σ·L~~Σ·L~+ 2 ̄h~Σ·L~+ ̄h^2 = ΣiLiΣjLj+ 2 ̄h~Σ·L~+ ̄h^2

Σ 1 Σ 1 = −γ 2 γ 3 γ 2 γ 3 =γ 2 γ 2 γ 3 γ 3 = 1

Σ 1 Σ 2 = −γ 2 γ 3 γ 3 γ 1 =−γ 2 γ 1 =γ 1 γ 2 =iΣ 3

ΣiΣj = δij+iǫijkΣk

K^2 = LiLj(δij+iǫijkΣk) + 2 ̄h~Σ·L~+ ̄h^2

= L^2 +i~Σ·(~L×~L) + 2 ̄hΣ~·~L+ ̄h^2

(~L×~L)o=xipjxmpnǫijkǫmnlǫklo = 0 +

̄h

i

xiδjmpnǫijkǫmnlǫklo=

̄h

i

xipnǫijkǫjnlǫklo

= −

̄h

i

xipnǫjikǫjnlǫklo=−

̄h

i

xipn(δinδkl−δilδkn)ǫklo

= −

̄h

i

xipiǫllo+

̄h

i

xipkǫkio=−

̄h

i

xipkǫiko=i ̄hLo

K^2 = L^2 − ̄h~Σ·L~+ 2 ̄h~Σ·~L+ ̄h^2 =L^2 + ̄h~Σ·~L+ ̄h^2

J^2 = L^2 + ̄h~Σ·L~+

3

4

̄h^2 =K^2 −

̄h^2

4