Mon., Nov. 29:
#  Flux quantization in superconductors: In a superconductor the Meissner effect forces
the magnetic field to vanish except within a small distance, the penetration depth, of the
surface. This also implies that the electric current vanishes. The current is expressible in
terms of the superconducting order parameter field Psi = rho exp(i theta). Roughly speaking,
there is a Bose-Einstein condensate of Cooper pairs, and rho^2 measures the density of
Cooper pairs, and psi is proportional to the pair wavefunction. The current is thus 2e j
where j is the probability current density,
j = Re Psi* v Psi = (hbar/m) rho^2 [grad theta - (2e/hbar c)A].
(The coefficient of the A-term is 2e rather than e since the Cooper pairs have charge 2e, not e.)
Then j = 0 implies A = (hbar c/2e) grad theta, so the magnetic flux around any loop is
Phi = \oint A.dl = (hbar c/2e) 2pi n = n Phi_0/2, (we used the fact that the continuity of exp(i theta)
implies that theta can only change around the loop by an integer multiple of 2pi.)
# In a superconducting ring geometry, if we take the loop around the ring we learn that the
flux through the ring is quantized. As a ring is cooled, eddy currents will be set up in the
surface of the superconductor to enforce this quantization.
# In the bulk of a superconductor, there may be vortex configurations in the core of which
the superconducting order parameter vanishes and the magnetic field need not vanish.
Taking the loop to encircle such a vortex we learn that the flux is quantized.
# Anyons: A brief remark about one way to conceptualize anyons. Think of the 2d quasiparticles
as flux lines in the 3rd dimension, carrying a pseudo-magnetic flux Phi, with vanishing
pseudo-magnetic field outside. Then each anyon has a pseudo- vector potential circulating around it
with \oint A.dl = Phi. Now imagine physically interchanging the position of two anyons, and then
doing it again. In the process, one anyon goes 360 degrees around the other. The wavefunction
of the anyon thus accumulates a phase factor exp(i \oint A.dl) = exp(i Phi). This means (I suppose)
that the interchange, which is the square root of the double interchange, entails a phase factor
exp(i Phi/2). The pseudo-flux Phi can be anything, hence the quasiparticles are "anyons".
I have not read it, but there is a Scientific American article that might be good: Anyons,
F. Wilczek (Princeton, Inst. Advanced Study). 1991. Sci.Am.264:24-31,1991 (No.5) (Issue no.5).
And a technical book by the same author:  Wilczek, Frank, Fractional statistics and anyon
superconductivity (World Scientific, 1990).
# Relativistic effects:
[Nice treatments of this topic are in Schwabl and in Cohen-Tannoudji et.al., vol 2.]
The non-rel. H-atom involves the dimensionful quantities (hbar, m, e), where m is the electron
mass. It also involves the proton mass, but that only makes for a small correction since m_e/m_p <<1.
The energies are E_n = - (me^4/2hbar^2) (1/n^2), and the Bohr radius is a_B = hbar/me^2.
When we bring in the speed of light c we get a dimensionless constant, the fine structure constant:
 
alpha = e^2/hbar c = 1/137.036...
To begin with let's write the ground state energy,  Bohr radius, and typical electron velocity,
of the non-relativistic H-atom in relativistic terms:
 
E_1 =  (1/2) alpha^2 mc^2,           a_B = alpha^-1 (hbar/mc),           v/c = alpha
Note that dimensional analysis gives this, since we put in whatever power of alpha is needed
to cancel the c-dependence since the non-relativistic H-atom doesn't involve c.
The length (hbar/mc) is the Compton wavelength of the electron. The typical speed of the electron in the ground state of hydrogen is given by KE = |E_1|, i.e. v/c = alpha. This is darn fast, but not very relativistic. There will be small relativistic corrections.
Note that for a nucleus of charge Z|e|, alpha in the above formulae is replaced by (Z alpha).
# The form of the lowest order corrections comes from the Dirac equation. Here we just quote the
result, and motivate the form of these terms.
 
Name term in Hamiltonian  origin
relativistic  kinetic energy H1 = - p^4/8m^3c^2 KE + mc^2 = 
sqrt[p^2c^2 +m^2c^4]
spin-orbit coupling H2 = (e^2/2m^2c^2)(1/r^3) S.L Interaction of spin magnetic moment with induced magnetic field 
B = - (v/c) x E
Darwin term H3 = 
(e^2 pi/2)(hbar/mc)^2 delta^3(r)		 Zitterbewegung: electron feels potential averaged over a volume of size hbar/mc
The ratios of H1, H2, and H3 to the ground state energy, or, equivalently,
H1/(p^2/2m), H2/(e^2/a_B), and H3/(e^2/a_B),
are all of order (alpha)^2 ~ 5 10^-5.
In showing this we set S=L=hbar, r=a_B, and interpret the Dirac delta
function in H3 in terms of its expectation value, |psi(0)|^2 =~ (a_B)^-3.

Wed. Dec. 1:
# Explained origin of the spin-orbit term, and hand-waved about the Darwin term.
On the spin-orbit term, we go to the rest frame of the electron, and compute the magnetic dipole
energy in the indeuced magnetic field. This energy not equal to that in the rest frame of the atom,
but the corrections are of higher order in v/c, and we are already looking at a mall correction.
Also, the rest frame of the electron is not inertial, which explains why we end up missing a factor of 1/2.
In section 11.8 of Jackson's Electrodynamics book, this is explained classically as a consequence of
Thomas precession, which is an extra rotation of the spin due to the rotational acceleration of the
electron's rest frame.
# It is possible to solve the Coulomb problem exactly for the Dirac equation, but at this stage we
treat the relativisitc effects using perturbation theory. This is simpler, it reveals the basic physics,
it shows what happens even for many-electron atoms for which the potential is not Coulomb but
rather the screened nuclear potential, and it gives us practice with perturbation theory.
# For a hamiltonian H = H_0 + H_1, the first order shift of the energy of a state |n> is given by
the expectation value of H_1 in the state |n>. This can be seen as follows.
Let H(v) = H_0 + v H_1, and H(v)|n,v> = E_n(v)|n,v>. Using the identity
dE/dv = <n,v| dH/dv |n,v> = <n,v| H_1 |n,v>
in a Taylor expansion around v=0 we find
E(1)-E(0) = <n,v| H_1 |n,v> + O(v^2).
# There is an important caveat about this: it is only true
provided that the eigenstate does not change discontinuously when the perturbation is turned on,
i.e. provided |n,v> is differentiable with respect to v at v=0.
This condition is automatically satisfied if |n> is non-degenerate.
It is also true even if |n> is degenerate,
provided that H_1 is diagonal in the degenerate subspace,
i.e., <n'|H|n>=0 for any  states |n'> with which |n> is degenerate.
# A simple example illustrates the need for this caveat:
H_0 = E_0 I, and H_1 = v sigma_x.
The exact eigenvalues are E_0 +/- v, corresponding to eigenvectors (1,1) and (1,-1).
If we happened to start with the unperturbed eigenvectors (1,0) and (0,1), they would jump
discontinuously when v is turned on. The first order shift in this case is evidently NOT given by the
expectation value of H_1 in the unperturbed eigenstates, since that is ZERO!

 Fri. Dec. 3:
# The relativistic corrections give rise to the fine structure of the H atom spectrum.
In particular, part of the degeneracy is lifted.
# History: The first observation of the H fine structure
of the H_alpha line corresponding to n=3 -> n=2 transitions (according to my sources) was
by Michelson and Morley, using an interferometer. It is difficult (but I think possible)
to observe spectroscopically using a grating. In 1916 Sommerfeld, using the old quantum theory
in which the phase integrals for both angular and radial motion were quantized in
units of Planck's constant,  accounted for this fine structure splitting!
(For an account see H.E. White, Introduction to Atomic Spectra (McGraw-Hill, New York, 1934).)
In this theory, the ratio of the major and minor axes of the orbital ellipse must be a ratio
of integers. This allows a discrete set of orbits for each energy, with the energy being
proportional to the square of  the sum of the two quantum numbers, quite like the
exact nonrelativistic quantum result. For the second level, there were two orbits, one circular
and one elliptical. The relativistic corrections for these orbits are different, hence the level
splits into two. Sommerfeld worked out the exact relativistic result, and it looks quite
similar to the exact result from the Dirac equation! The splitting comes out to agree with
observation at lowest order, although this is partly accidental, since both the Darwin
term and the spin-orbit interaction are not included.
# We now compute the first order relativistic corrections to the H atom energy levels.
H1 and H3 are spin-independent scalar operators, hence diagonal in the basis {|nlmm_s>}
of degenerate states for fixed principal quantum number n. Thus we can just take their
expectation value.
# A trick: p^4 = (p^2)^2 =   [2m (H0 + e^2/r)]^2, so H1=(-1/2mc^2)[H0 +e^2/r]^2.
Hence <H1> = (-1/2mc^2)<(E0+e^2/r)^2>, where E0 is the unperturbed energy.
So we need <1/r> and <1/r^2>. Tricks for computing these in H-atom eigenstates are
given in various books.
# H2 is a scalar operator (1/r^3) times a number times  2 L.S = J^2 - L^2 - S^2.
So we need <1/r^3>, and we must diagonalize the spin-orbit coupling in the degenerate
subspaces. L.S commutes with J, L^2, and S^2,  where J=L+S is the total angular momentum.
(S^2  is uninteresting since all states of the one electron atom have the same value for S^2.)
The eigenvalues of 2 L.S are [j(j+1) - l(l+1) - s(s+1)] hbar^2, where j = l +/- 1/2 is the
total angular momentum quantum number (see below for details).
# Putting it all together, the result for the first order energy shift is

Delta E^(1)_n,j = (Ry/n^2)(alpha^2/n^2)[3/4 - n/(j+1/2)].

The exact result for the energy levels from the Dirac equation is

E_n,j  = mc^2 {1 + alpha^2 [n - (j+1/2) + Sqrt[(j+1/2)^2 - alpha^2]]^-2}^-1/2.

Comments:
1. The first order shift is proportional to alpha^2.
2. Both the shift and the exact eigenvalues depend on n and j, but not on l separately.
Thus the degeneracy of the Coulomb spectrum is only partly lifted. For a given n, the
possible j's range from 1/2 to n-1/2, and they all have different energy, however each j
can arise from two l's, namely j+1/2 and j-1/2.
3. The states are labeled as L_J. Thus the ground state is 1s_1/2, and the first excited
state is 2s_1/2 or 2p_1/2.
4. These are split from the 2p_3/2 state by alpha^2 Ry/16 = 10,950 MHz = 4.5 x 10^-5 eV.
5. In a many electron alkalai atom, such as sodium, the ground state has the configuration
(1s)^2 (2s)^2 (2p)^6 (3s). Because of the different screening of the nuclear charge seen
by the 3p and 3d states, these are not degenerate with 3s or with each other. But
we still have a fune structure splitting of the degenerate 3p states, for example, into
3p_1/2 and 3p_3/2 states.

# Switching  from the basis of eiegenstates of L^2, L_z, S_z to that of L^2, J^2, J_z:
|l j m_j>  =   Sum_{m_l+m_s=m_j}   C^{j,m_j}_{m_l,m_s} |l m_l m_s>. The coefficients C are
Clebsch-Gordon coefficients.
Let's work them out explicitly for l=1.
1) Divide up the |m_l>|m_s> states into the possible eigenvalues of J_z, m_l+m_s.
2) The top value is l+s= 1 + 1/2 = 3/2, and there is a unique such state. This must be the top state of
a spin 3/2 rep. space. Work down by applying J- till you get to m_j = -3/2. This fills out
a spin 3/2 irrep.
3) What's left? The top m_j now remaining is the other linear combination of
m_j = 1/2 states, so it is the top state of a spin 1/2 irrep. Since it is an eigenvector of J^2 with
different eigenvalue, it must be orthogonal to the spin 3/2 state.  We thus take the state that is
orthogonal to the one used in the spin 3/2 rep. Then lowering this we get the other state in the
spin 1/2 rep.
4) In carrying out this procedure we use J+/- |j,m> = Sqrt[j(j+1) - m(m +/- 1)] |j,m p/m 1>.
For j=1 the Sqrt is always Sqrt[2]. For j=1/2 the Sqrt is always 1.
5) The count: 1 x 1/2 = 3/2 + 1/2. In dimensions: 3 x 2 = 4 + 2:
 

J_z L_z,  S_z eigenstates j = 3/2 j = 1/2
3/2  |1> |1/2> |1> |1/2> ---
1/2  |0> |1/2>  ,   |1> |-1/2> Sqrt[2/3] |0> |1/2>  + Sqrt[1/3] |1> |-1/2> Sqrt[1/3] |0> |1/2>  -  Sqrt[2/3] |1> |-1/2>
-1/2  |0> |-1/2>  ,   |-1> |-1/2> Sqrt[2/3] |0> |-1/2>  + Sqrt[1/3] |-1> |1/2>  - Sqrt[1/3] |0> |-1/2>  +  Sqrt[2/3] |-1> |1/2> 
-3/2  |-1> |-1/2> |-1> |-1/2> ---