Ph235-08
F. Merritt1
27-Oct-2008
Lecture 13 (October 27, 2008):
(version 1.0; 24-Oct-2008::00:00)
Fine Structure of Hydrogen (summary)
The last lecture was mainly spent on deriving the fine-structure splitting of
hydrogen. We found that the total FS shift in the energy levels is the sum of two separate
pieces, one due to the relativistic correction to the electrons energy, and one due to the
coupling between the electron magnetic moment and the B-field that the electron sees in
its own rest frame. The sum of these is
E
1
FS
E E
1
rel
1
LS
E 

0 2
n
mc 2
 1 
 j ( j  1)  l (l  1)  34  
4n 

n
 3 



1
1
 l (l  1)(l  2 )
 
 2  l  2 
En0  

4n 

3

2 
2mc 
j  12 
2
(13.1)
This is a rather striking result. Due to the partial cancellation between these two terms,
the final result is rather simple; in fact it is the same as the relativistic correction alone,
but with j replacing l.
It is worth noting that we found earlier that we could use either of two bases for
hydrogen, the nl; m, ms basis or the nl; j, jz . In the first, j is not a good quantum
number, and the second m and ms are not good quantum numbers. With the spin-orbit
effect included, the energy eigenstates are proportional only to basis vectors in the second
of these. Although l is still a good quantum number, m and ms are not (because both L
and S precess around J ).
The Simple Harmonic Oscillator and other examples
(Shankar section 17.2)
Shankar uses the 1D simple harmonic oscillator as the first example in discussing
perturbation theory. I skipped this in lecture, but it is a very good and instructive
example, since it can be solved exactly as well as by using perturbation theory. Also, it
gives a good review of the SHO raising and lowering operators. I urge you to work
through this in detail, and make sure you understand everything. (And it might make a
really good problem on the midterm or the final).
Deriving the Thomas-Reiche-Kuhn sum rule (Shankar Exercise 17.2.3) is also a
good exercise, and I have put the solution to this under “Examples” for week 5.
The next exercise in Shankar (17.2.5) asks you to find the first-order correction to
the helium ground state, treating the ee interaction term as a perturbation. Shankar says
this is “hard”. In fact, we have already calculated all the integrals in Lecture 9 as an
example of the “Variational Method” (and it was hard).. We assumed a wavefunction of
the form  (r )  A exp(Zr / a0 ) , and derived the expectation value of the total
Ph235-08
F. Merritt2
27-Oct-2008
Hamiltonian with the interaction term included. (You should convince yourself that this
is equivalent to calculating the first-order perturbative correction). The result was
1
EHe
 [4Z 2  4Z ( Z  2)  5Z / 4]EH0
(13.2)
In fact, we can use this as two examples of perturbation theory. If we assume Z=1
(hydrogen nucleus) and sum both the coulomb field of the second proton and the eeinteractions as H  , then we obtain the formula (13.2) for Z=1. If we use Z=2 and treat
only the ee interaction as H  , we get (13.2) for Z=2. Finally, if we minimize (13.2) with
respect to Z, then we get Z=1.69. This is the same as using this value of Z for the nuclear
charge, and treating the remaining charge of 0.31 e plus the ee interaction as a
perturbation. The results of each of these three perturbative calculations are:
Z=1
 EHe  194 EH0  64.6 eV
Z=2
 EHe 
22
4
EH0  74.8 eV
(13.3)
Z  1.69  EHe  77.46 eV
Each of these can be regarded as a first-order perturbative calculation; the variational
method tells us that the 3rd is the best one, since it gives the lowest limit on the ground
state.
The Zeeman Effect (A) [Strong-field case]
Back in Ph234 last spring we derived the Zeeman Effect by considering a spin ½
particle in a magnetic field. Let’s revisit that, now in terms of a perturbative correction to
the hydrogen energy levels for a hydrogen atom in a strong magnetic field. We take the
Coulomb Hamiltonian to be H 0 , and treat the interaction with the B-field as a
perturbation. Since each n-shell has an n2-fold degeneracy, we want to choose a basis in
which H  is diagonal. Since the electron has its own intrinsic magnetic moment, and the
orbit has an orientation energy of   B (see Shankar, page 287-289),
H     L B  e B
(13.4)
where
e
ge
 e 
L
and
e  g 
 B
S  
2mc
4mc
 2mc 
(Recall that g=2 for the electron.) Then
H   L B  e B  B  ml  2ms  BZ .ext
L 
(13.5)
(13.6)
where  L  e / 2mc , e  ge / 2mc  e / mc , and  B  e / 2mc . Then H  is diagonal
in the nl; ml , ms eigenbasis, and
(13.7)
EZ1  B  ml  2ms  BZ .ext
The n=1 subshell (which has only s-waves) is split into two components, one for
each of the 2 values of ms . One of these will increase with increasing B, and the other
will decrease.
But let us focus on the n=2 shell, which contains 8 states (2 s-wave and 6 pwave). For B large, these will be split into 5 components, 3 of which are doubly
degenerate. If we define N  ml  2ms in equation (13.7), these 5 levels correspond to
Ph235-08
F. Merritt3
27-Oct-2008
N  2,1, 0, 1, 2 . The doubly degenerate levels are those with N  1, 0, 1 . If we
include the fin-structure splitting, then each of these 3 levels will be split into 2
components corresponding to the 2 possible values of j  12 , 23 .
The Zeeman Effect (B) [Weak-field case]
But we have already seen, in deriving the fine structure of hydrogen, that in the
absence of external fields the relativistic and spin-orbit terms split the n=2 subshell into 2
levels, corresponding to j  32 and j  12 . Moreover, these are not basis vectors in the
nl; ml , ms basis. Instead, they are basis vectors of the nl; j, jz basis.
When we add a small magnetic field, then we would expect j to remain a good
quantum number, with the vector J precessing about B . More formally, we can use the
nl; j, jz basis as the eigenbasis of
and treat
H 0  H 0  H FS
(13.8)
H   HZ
(13.9)
as a perturbation.
(to be continued)