130_notes.dvi

Since this does not depend on eithermℓorj, totaljstates and the product states give the same
answer. We will choose to use the totaljstates,ψnjmjℓs, so that we can combine this correction
with the spin-orbit correction.

23.4.4 Perturbation Calculation for H2 Energy Shift

We now calculate the expectation value ofH 2. We will immediately use the fact thatj=ℓ±^12.

〈

ψnjmjℓs|H 2 |ψnjmjℓs

〉

=

ge^2 ̄h^2

4 m^2 c^2

1

2

[j(j+ 1)−ℓ(ℓ+ 1)−s(s+ 1)]

〈

1

r^3

〉

nl

=

ge^2 ̄h^2

8 m^2 c^2

[

(ℓ±

1

2

)(ℓ+ 1±

1

2

)−ℓ(ℓ+ 1)−

3

4

]

1

a^30

(

1

n^3 ℓ(ℓ+^12 )(ℓ+ 1)

)

= −En

g ̄h^2

4 m^2 c^2 a^20

[

ℓ^2 +ℓ±ℓ±

1

2

+

1

4

−ℓ^2 −ℓ−

3

4

](

1

nℓ(ℓ+^12 )(ℓ+ 1)

)

=

(g

2

)(−En

2 mc^2

)

̄h^2

ma^20 n^2

[

ℓ

−(ℓ+ 1)

](+)

(−)

n

ℓ(ℓ+^12 )(ℓ+ 1)

=

(g

2

)(−E

n

2 mc^2

)

̄h^2 α^2 m^2 c^2

m ̄h^2 n^2

[

ℓ

−(ℓ+ 1)

](+)

(−)

n

ℓ(ℓ+^12 )(ℓ+ 1)

=

(g

2

)E(0)

n

2

2 mc^2

2

[ n

(ℓ+^12 )(ℓ+1)

−n

ℓ(ℓ+^12 )

]

j=ℓ+^12

j=ℓ−^12

Note that in the above equation, we have canceled a termℓℓwhich is not defined forℓ= 0. We will
return to this later.

23.4.5 The Darwin Term

We get a correction at the origin from theDirac equation.

HD=

πe^2 ̄h^2

2 m^2 ec^2

δ^3 (~r)

When we take the expectation value of this, we get the probability for the electron and proton to
be at the same point.

〈ψ|HD|ψ〉=

πe^2 ̄h^2

2 m^2 ec^2

|ψ(0)|^2

Now,ψ(0) = 0 forℓ >0 andψ(0) =√^14 π 2

(

z

na 0

) 3 / 2

forℓ= 0, so

〈HD〉n 00 =

4 e^2 ̄h^2

8 n^3 a^30 m^2 c^2

=

e^2 ̄h^2 α^2 m^2 c^2

2 n^3 a 0 m^2 c^2 ̄h^2

=

2 nEn^2

mc^2

This is the same asℓ= 0 term that we got for the spin orbit correction. This actually replaces the
ℓ= 0 term in the spin-orbit correction (which should be zero) making the formula correct!