130_notes.dvi

= −

2 α ̄h

3 πm^2 c^2

∑

j

ωcut∫−off

0

ωnj

ω−ωnj

|〈n|~p|j〉|^2 dω

= −

2 α ̄h

3 πm^2 c^2

∑

j

ωnj[log(ω−ωnj)]ω 0 cut−off|〈n|~p|j〉|^2

=

2 α ̄h

3 πm^2 c^2

∑

j

ωnj[log(|ωnj|)−log(ωcut−off−ωnj)]|〈n|~p|j〉|^2

≈

2 α ̄h

3 πm^2 c^2

∑

j

ωnj[log(|ωnj|)−log(ωcut−off)]|〈n|~p|j〉|^2

=

2 α ̄h

3 πm^2 c^2

∑

j

ωnjlog

(

|ωnj|

ωcut−off

)

|〈n|~p|j〉|^2

The log term varies more slowly than does the rest of the terms in thesum. We canapproximate
it by an average. Bethe used numerical calculations to determine that the effectiveaverage of
̄hωnjis 8. 9 α^2 mc^2. We will do the same and pull the log term out as a constant.

∆E(nobs)=

2 α ̄h

3 πm^2 c^2

log

(

|ω ̄nj|

ωcut−off

)∑

j

ωnj|〈n|~p|j〉|^2

This sum can now be reduced further to a simple expression proportional to the|ψn(0)|^2 using a
typical clever quantum mechanics calculation. The basic Hamiltonian for the Hydrogen atom

isH 0 =p

2

2 m+V(r).

[~p,H 0 ] = [~p,V] =

̄h

i

∇~V

〈j|[~p,H 0 ]|n〉 =

̄h

i

〈j|∇~V|n〉

∑

j

〈n|~p|j〉〈j|[~p,H 0 ]|n〉 =

̄h

i

∑

j

〈n|~p|j〉·〈j|∇~V|n〉

∑

j

(Ei−En)〈n|~p|j〉〈j|~p|n〉 =

̄h

i

∑

j

〈n|~p|j〉〈j|∇~V|n〉

Thismust be a real number so we may use its complex conjugate.



∑

j

(Ei−En)〈n|~p|j〉〈j|~p|n〉





∗

=

∑

j

(Ei−En)〈n|~p|j〉〈j|~p|n〉

= −

̄h

i

∑

j

〈n|∇~V|j〉〈j|~p|n〉

∑

j

(Ei−En)〈n|~p|j〉〈j|~p|n〉 =

̄h

2 i





∑

j

〈n|~p|j〉〈j|∇~V|n〉−〈n|∇~V|j〉〈j|~p|n〉





=

̄h

2 i

〈n|[~p,∇~V]|n〉

= −

̄h^2

2

〈n|∇^2 V|n〉