130_notes.dvi

The components of orbital angular momentum do not commute withH.

[H,Lz] =icγ 4 [γjpj,xpy−ypx] = ̄hcγ 4 (γ 1 py−γ 2 px)

The components of spin also do not commute withH.

Σz =

[γ 1 ,γ 2 ]

2 i

=

γ 1 γ 2

i

[H,Sz] = [H,

̄h

2

Σz] =c

̄h

2

[γ 4 γjpj,γ 1 γ 2 ] =c

̄h

2

pj[γ 4 γjγ 1 γ 2 −γ 1 γ 2 γ 4 γj]

= c

̄h

2

pj[γ 4 γjγ 1 γ 2 −γ 4 γ 1 γ 2 γj] =c

̄h

2

pjγ 4 [γjγ 1 γ 2 −γ 1 γ 2 γj] = ̄hcγ 4 [γ 2 px−γ 1 py]

However, thehelicity, or spin along the direction of motion does commute.

[H,S~·~p] = [H,S~]·~p= ̄hcγ 4 ~p×~γ·~p= 0

From the above commutators [H,Lz] and [H,Sz], thecomponents of total angular momentum
do commutewithH.

[H,Jz] = [H,Lz] + [H,Sz] = ̄hcγ 4 (γ 1 py−γ 2 px) + ̄hcγ 4 [γ 2 px−γ 1 py] = 0

The Dirac equation naturallyconserves total angular momentumbut not conserve the orbital
or spin parts of it.

We will need another conserved quantity for the solution to the Hydrogen atom; something akin to
the±inj=ℓ±^12 we used in the NR solution. We can show that [H,K] = 0 for

K=γ 4 Σ~·J~−

̄h

2

γ 4.

It is related to the spin component along the total angular momentum direction. Lets compute the
commutator recalling thatHcommutes with the total angular momentum.

H = icγ 4 ~γ·~p+mc^2 γ 4

[H,K] = [H,γ 4 ](Σ~·J~−

̄h

2

) +γ 4 [H,~Σ]·J~

[H,γ 4 ] = ic[γ 4 ~γ·~p,γ 4 ] = 2ic~γ·~p

[H,Σz] = 2cγ 4 [γ 2 px−γ 1 py]

[

H,~Σ

]

= − 2 cγ 4 ~γ×~p

[H,K] = 2ic(~γ·~p)(Σ~·J~−

̄h

2

)− 2 c~γ×~p·J~

~p·~L = 0

J~ = L~+ ̄h

2

Σ~

[H,K] = 2c

(

i(~γ·~p)(Σ~·J~)−~γ×~p·J~−i

̄h

2

(~γ·~p)

)