# Clebsch table
# Spectroscopic notation
# L.S = (J^2 -L^2 - S^2)/2
# Example of Beryllium ion 9Be+. Nuclear spin I=3/2, so S_1/2 ground state
has two hyperfine levels F = 2,1 coming from 1/2 x 3/2 = 2 + 1.
Same goes for the P_1/2 excited state, while the P_3/2 state has
F=3,2,1,0 hyperfine levels.
# filled shells: There is only one totally antisymmetric tensor of rank n
in a n n-dimensional space. If the same linear transformation is applied
to each of it's indices, the result is still totally antisymmetric, hence
proportionl to the original tensor. The proportionality factor is the
determinant of the linear transformation. Apply this to a state of
identical fermions. There is a unique totally antisymmetric (2j+1)-particle
state in a spin-j representation, called a filled shell.  Under a rotation,
each of the vectors is transformed by a unitary transformation with unit determinant,
hence the state is invariant. Thus,  a filled shell has zero angular momentum.
A simple example of this is the spin singlet state of two spin-1/2 particles.
(An alternate method of proof is to just write out the Slater determinant
and show that it is annihilated by J_z and J_+\-. With J_z, the eigenvalues
of the individual J_z's add up to zero. With J_+\-, every terms involves
two particles in the same state, which is antisymmetrized, and hence vanishes.)
# A partly filled shell with a number of holes acts for many purposes like the same
number of particles in an otherwise empty shell.
See Supplement on tensor operators for a more precise statement of this
equivalence of holes and particles.

Fri., Feb. 25

# Baym typos: eqns 17-86,87,88 are all missing minus signs, I believe,
according to the usual Clebsh convention.
# isospin: read it in Baym, which was written before QCD. Now we understand
isospin symmetry as arising from the approximate equivalence of up and down
quarks as far as nuclear forces go. The proton is uud and the neutron is udd,
so they differ by u vs. d for one quark. The quarks differ in their electric
charge (+2/3 vs. -1/3) and their mass. But the electromagnetic interaction
is small compared to the nuclear one inside a nucleus, so the charge doesn't
make much difference, and both masses are very small compared to nuclear
energies.
# Vector operators: e.g. r^i, L^i, p^i.    ``Spherical components" of a vector:
V_0 = V_z,
V_+\-1 = (-\+ V_x  - iV_y)/Sqrt[2].
In terms of these the commutation relations [J^a,V^b] = i epsilon^abc V^c become
[J_z,V_q] = q V_q
[J_+\-, V_q] = Sqrt[2] V_q+/-1,
where it is understood that V_q=0 if |q|>1.
These are just like the l=1 spherical harmonics
Y_1m = Sqrt[4Pi/3](1/r) ( z   [if  m =0], -/+ x + iy [if m=+/-1]),
# The Y_lm also satisfy the above commutation relations when Y_lm is treated
as an operator that acts by multiplication, with Sqrt[2] replaced by
Sqrt[l(l+1)-m(m+\-1)]. These are irreducible tensor operators.
Generally, such an op will be denoted T_kq, where k=0,1/2,1,3/2,...
and q = k, k-1, ..., -k.
Example: r^i r^j is a tensor op. but not irreducible. It's irreducible parts
are  the tracefree part (r^i r^j - 1/3 r^2 d^ij) and the trace 1/3 r^2 d^ij.
The tracefree part has 5 independent components and hence is  Y_2m times
a function of r, while the trace is a scalar operator.
# Note ``vector operator" is always defined relative to a particular set of angular
momentum generators. Hence, for example, S^i is not a vector operator for L^i,
but it is for S^i or J^i=L^i+S^i.
# Wigner-Eckart theorem: all the matrix elements of a tensor operator
< a' j' m' | T_kq | a j m> = C(a',a,j',j) <j'm'|kj, qm>
taken between two irreducible subspaces are proportional to the CG coeffs,
hence they are determined by one number, the reduced matrix element
<a'j'|| T_k ||aj> := Sqrt[2j'+1] C(a',a,k).
See Supplement on tensor operators for more on this.