6.4. ENERGY LEVELS OF SUBATOMIC PARTICLES 363
where~σ= (σ^1 ,σ^2 ,σ^3 )is the Pauli matrix operator,~Dis as in (6.4.13).
We now derive spectral equations for massive bound states from (6.4.14). Let the solu-
tions of (6.4.14) be in the form
Ψk=e−i(λ+mkc(^2) )t/ ̄h
ψk.
Then equations (6.4.14) become
(λ−gA 0 )
(
ψk 1
ψk 2)
=−ic ̄h(~σ·~D)(
ψk 3
ψk 4)
(6.4.15) ,
(λ−gA 0 + 2 mkc^2 )(
ψ 3 k
ψ 4 k)
=−ich ̄(~σ·~D)(
ψ 1 k
ψ 2 k)
(6.4.16) ,
for 1≤k≤N. The equation (6.4.16) can be rewritten as
(6.4.17)
(
ψ 3 k
ψ 4 k)
=
−i ̄h
2 mkc(
1 +
λ−gA 0
2 mc^2)− 1
(~σ·~D)(
ψ 1 k
ψ 2 k)
.
In physics,λis the energy, andλ−gA 0 is the kinetic energy
λ−gA 0 =1
2
mkν^2.For massive particles,ν^2 /c^2 ≃0. Hence, (6.4.17) can be approximatively expressed as
(
ψ 3 k
ψ 4 k)
=
−i ̄h
2 mkc
(~σ·~D)(
ψ 1 k
ψ 2 k)
.
Inserting this equation into (6.4.15), we deduce that
(6.4.18) (λ−gA 0 )
(
ψ 1 k
ψ 2 k)
=−
̄h^2
2 mk(~σ·~D)^2(
ψ 1 k
ψ 2 k)
.
Now, we need to give the expression of(~σ·~D)^2. To this end, note that the Pauli matrices
satisfy
(σk)^2 = 1 , σkσj=−σjσk=iεljkσl.Hereεljkis the arrange symbal:
εijk=
1 as(jkl) the even arrange,
−1 as(jkl) the odd arrange,
0 otherwise.Hence we obtain
(6.4.19) (~σ·~D)^2 = (
3
∑
k= 1σkDk)^2 =D^2 +i~σ·(~D×~D).