130_notes.dvi

Then we compute the energy shift in first order perturbation theory for s states.

∆E=

〈

e

mec

S~·B~

〉

=−

Ze^2 gN

2 meMNc^2

(

S~·~I

〈

∇^2

1

r

〉

−

〈

SiIj

∂

∂xi

∂

∂xj

1

r

〉)

The second term can be simplified because of the spherical symmetry of s states. (Basically the
derivative with respect to x is odd in x so when the integral is done, only the terms wherei=jare
nonzero). ∫

d^3 r|φn 00 (~r)|^2

∂

∂xi

∂

∂xj

1

r

=

δij

3

∫

d^3 r|φn 00 (~r)|^2 ∇^2

1

r

So we have

∆E=−

2

3

Ze^2 gN

2 meMNc^2

S~·I~

〈

∇^2

1

r

〉

.

Now working out the∇^2 term in spherical coordinates,
(
∂^2
∂r^2

+

2

r

∂

∂r

)

1

r

=

2

r^3

+

2

r

(

− 1

r^2

)

= 0

we find that it is zero everywhere but we must be careful atr= 0.

To find the effect atr= 0 we will integrate.

∫ε

r=0

∇~^21

r

d^3 r=

∫ε

r=0

~∇·(∇~^1

r

)d^3 r=

∫

(∇~

1

r

)·dS~ =

∫

∂

∂r

1

r

dS

=

∫ε

r=0

− 1

r^2

dS= (4πε^2 )(

− 1

ε^2

) =− 4 π

So the integral is nonzero for any region including the origin, which implies
(
∇^2

1

r

)

=− 4 πδ^3 (~r).

We can now evaluate the expectation value.

∆E=−

2

3

Ze^2 gN

2 meMNc^2

S~·~I(− 4 π|φn 00 (0)|^2 )

4 π|φn 00 (0)|^2 =|Rn 0 (0)|^2 =

4

n^3

(

Zαmec

̄h

) 3

∆E=

2

3

Ze^2 gN

2 meMNc^2

S~·I~^4

n^3

(

Zαmec

̄h

) 3

Simply writing thee^2 in terms ofαand regrouping, we get

∆E=

4

3

(Zα)^4

(

me

MN

)

(mec^2 )gN

1

n^3

S~·~I

̄h^2

.