130_notes.dvi

(~γ·~p)(~Σ·J~) = γipiΣjJj=

−i

2

piJjγiγmγnǫmnj

γiγmγnǫmnj = 2(δijγ 5 γ 4 +ǫijkγk)

(~γ·~p)(~Σ·J~) = −ipiJj(δijγ 5 γ 4 +ǫijkγk) =−iγ 5 γ 4 ~p·J~−i~γ×~p·J~

[H,K] = 2c

(

γ 5 γ 4 ~p·J~+~γ×~p·J~−~γ×~p·J~−i

̄h

2

(~γ·~p)

)

[H,K] = 2c

(

γ 1 γ 2 γ 3 γ 4 γ 4 ~p·J~−i

̄h

2

(~γ·~p)

)

[H,K] = 2c

(

γ 1 γ 2 γ 3 ~p·J~−i

̄h

2

(~γ·~p)

)

[H,K] = 2c

(

γ 1 γ 2 γ 3 ~p·(L~+

̄h

2

Σ)~ −i ̄h

2

(~γ·~p)

)

[H,K] = 2c

̄h

2

(

γ 1 γ 2 γ 3 ~p·~Σ−i(~γ·~p)

)

[H,K] = 2c

̄h

2

(i~p·~γ−i(~γ·~p)) = 0

It is also useful to show that [K,J~] = 0 so that we have a mutually commuting set of operators to
define our eigenstates.

[K,J~] = [γ 4 Σ~·J~−

̄h

2

γ 4 ,J~] = [γ 4 ,J~]~Σ·J~+γ 4 [~Σ·J,~J~]−

̄h

2

[γ 4 ,J~]

This will be zero if [γ 4 ,J~] = 0 and [Σ~·J,~J~] = 0.

[

γ 4 ,J~

]

= [γ 4 ,~L+

̄h

2

~Σ] = ̄h

2

[γ 4 ,Σ] = 0~
[
~Σ·J,~J~

]

= [~Σ·L,~ J~] + [~Σ·~Σ,J~] = [~Σ·L,~ J~] + [3,J~] = [~Σ·~L,J~]

[

~Σ·L,~J~

]

= [Li+

̄h

2

Σi,ΣjLj] = [Li,ΣjLj] +

̄h

2

[Σi,ΣjLj] = Σj[Li,Lj] +

̄h

2

[Σi,Σj]Lj

= i ̄hǫijkΣjLk+ 2i

̄h

2

ǫijkΣkLj=i ̄h(ǫijkΣjLk+ǫijkΣkLj) =i ̄h(ǫijkΣjLk−ǫikjΣkLj) =i ̄h(ǫijkΣjLk−ǫijkΣjLk)

So for the Hydrogen atom,H,J^2 ,Jz, andKform a complete set of mutually commuting operators
for a system with four coordinatesx,y,zand electron spin.

36.13The Relativistic Interaction Hamiltonian

Theinteraction Hamiltonianfor the Dirac equation can be deduced in several ways. The simplest
for now is to just use the same interaction term that we had for electromagnetism

Hint=−

1

c

jμAμ

and identify the probability current multiplied by the charge (-e) as the current that couples to the
EM field.
jμ(EM)=−eicψγ ̄μψ