130_notes.dvi

(Frankie) #1
= −

̄h^2
2
〈n|e^2 δ^3 (~x)|n〉

= −

e^2 ̄h^2
2

|ψn(0)|^2

Only thes states will have a non-vanishing probability to be at the originwith|ψn 00 (0)|^2 =
1
πn^3 a^30 anda^0 =


h ̄
αmc. Therefore, only the s states will shift in energy appreciably. The shift will be.

∆En(obs) = −

2 α ̄h
3 πm^2 c^2

log

(

|ω ̄nj|
ωcut−off

)

e^2 ̄h
2

1

πn^3

(αmc
̄h

) 3

=

α^4 e^2 mc
3 π^2 ̄hn^3
log

(

ωcut−off
| ̄ωnj|

)

=

4 α^5 mc^2
3 πn^3

log

(

ωcut−off
|ω ̄nj|

)

∆E 2 (obss ) =

α^5 mc^2
6 π

log

(

mc^2
8. 9 α^2 mc^2

)

ν =

∆E

(obs)
2 s
2 π ̄h

=

α^5 mc^2 c
12 π^2 ̄hc

log

(

1

8. 9 α^2

)

= 1. 041 GHz

Thisagrees far too wellwith the measurement, considering the approximations made and the
dependence on the cut-off. There is, however, justification in therelativistic calculation. Typically,
the full calculation was made by using this non-relativistic approach up to some energy of the
order ofαmc^2 , and using the relativistic calculation above that. Therelativistic free electron
self-energy correction diverges only logarithmically anda very high cutoff can be used
without a problem. The mass of the electron is renormalized as above. TheLamb shift does not
depend on the cutoffand hence it is well calculated. We only need the non-relativistic part of
the calculation up to photon energies for which the E1 approximationis OK. Therelativistic part
of the calculation down toωminyields.


∆En=

4 α^5
3 πn^3

(

log

(

mc^2
2 ̄hωmin

)

+

11

24


1

5

)

mc^2

Thenon-relativistic calculationgave.


∆En=
4 α^5
3 πn^3

log

(

ωmin
|ω ̄nj|

)

mc^2

So thesum of the two gives.


∆En(obs)=

4 α^5
3 πn^3

(

log

(

mc^2
2 ̄hω ̄nj

)

+

11

24


1

5

)

mc^2

Thedependence onωmincancels. In this calculation, themc^2 in the log is the outcome of the
relativistic calculation, not the cutoff. The electric dipole approximation is even pretty good since
we did not need to go up to large photon energies non-relativistically and no E1 approximation is
needed for the relativistic part. That’s how we (and Bethe) got about the right answer.


TheLamb shift splits the 2 S 12 and 2 P 12 states which are otherwise degenerate. Its origin
ispurely from field theory. The experimentalmeasurement of the Lamb shift stimulated

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