# See Landau & Lifshitz,  "The Quasi-classical Case", for the best treatment of WKB.

 hbar^2 k^2(x)/2m + V(x) = E

#  WKB tunneling formula   P =~ exp(-2 \int |k| dx), where the integral goes between
the classical turning points.

# WKB quantization condition: counting nodes of the wf gives number n of bound state:

\int k dx =~ n pi,   integral between classical turning points, or with p = hbar k,

\int p dx =~ n h/2,  or, integrating over a full classical cycle

\oint p dx =~ n h.

This selects certain values of E, E_n.

# More precise treatment takes into account the breakdown of WKB at the turning points.
The result of this depends on the nature of these points, e.g. smooth potential, infinite
square barrier, infinite well... The most elegant treatment of the matching is
to analytically continue in x and go around the turning point inthe complex x=plane
so WKB remains good everywhere! The brute force approach is to solve the eqn
in the neighborhood of the turning point, which is the Airy eqn since the potential
is locally approximately linear. For the quantization cndn, this results in the modification

\oint p dx = (n + gamma) h,   where gamma depends on the boundary conditions,
and is 0 for infinite sq. well, 1/2 for smooth potentials.
 
 
# Number of bound states with energy less than E is (\oint p(x,E) dx)/h = area of classical
orbit in phase space in units of h, i.e. number of "cells" in phase space.

# For systems with N degrees of freedom, number of states =~ volume in phase space
in units of h^N.

Addition of angular momenta:

# Rotations in space are implemented on QM systems by unitary transformations
U(R)=exp(-i theta.J/hbar), where J^i are the hermitian generators of rotation.

# J^i are also the angular momentum operators, and are conserved
if the Hamiltonian is invariant under rotations.

# Rotation group structure implies [J^i,J^j] = ihbar epsilon^ijk J^k.

# representations: can simultaneously diagonalize J_z and J^2, since [J_z,J^2]=0.
We analyzed this last semester.
Call the eigenstates |jm>,
where J_z |jm> = m |jm>,  J^2|jm> = j(j+1) |jm>,  with hbar=1 from now on.
The possible values of j are
0, 1/2, 1, 3/2, 2, ... and the possible values of m, for a given j, are j, j-1, j-2, ..., -j.
The representation with a given j is called the "spin-j" representation,
and it is 2j+1 dimensional.
These representations are irreducible, in the sense that there is no subspace
that is  invariant (mapped into itself) under all rotations.
We can see this from the fact that
J_+ |jm> = Sqrt[j(j+1) - m(m+1)]  |j,m+1> and  J_- |jm> = Sqrt[j(j+1) - m(m-1)]  |j,m-1>,
where J_+ = J_x + i J_y,   J_- = J_x - i J_y,
from which it is clear that by acting with rotations we move through all the states.

# spin-0 : singlet, trivial, scalar
   spin-1/2: spinor
   spin-1: vector
   spin-3/2: e.g., gravitino of supergravity
   spin-2: e.g. graviton

#example: 3d vectors V^i  form the spin-1 rep. The tensor product of two of these
is the rank two tensors like V^i W^j, or more generally, T^ij.  These are not irreducible.
Rather the antisymmetric part is by itself irreducible, and three dimensional,
hence another spin-1 rep. The symmetric part is reducible into the part proportional to
the Kronecker delta (trace) and the rest (symmetric trace-free part).
The trace part is the j=0 rep, the symm tracefree part is j=2
(since then 2j+1=5=number of independent components of a symmetric tracefree tensor).

# example: 1/2 x 1/2 = 1 + 0, example: 1 x 1 = 2 + 1 + 0 (this is equivalent to the example above).

# Note three different examples of spin-1 rep:
vector, antisymmetric tensor, |2p, m=-1,0,1> states of H-atom.
I.e., the rep is the abstract structure. Many things can realize it.

# general scheme:   j1xj2 spanned by basis  {|j1m1>|j2m2>}. Decomposes into irreducibles.
Find by starting with top J_z state and working down with lowering operator J_-.
When fill out a rep, go back and start with the next highest top J_z state, which is the other
linear combination of the two second to two top J_z states.
This results in
j1 x j2 = (j_1+j_2) + (j_1+j_2 - 1) + ... + |j_1 -j_2|.

# The largest spin rep, j1+j2, starts with top state equal to the  product of the two top states |j1j1>|j2j2>.
To see that the smallest spin rep is  |j1-j2|, suppose first that j1>=j2.
The argument I gave in class, cleaned up a bit here,  was that the largest degeneracy
that occurs for fixed total m is   2j2+1, so there must be 2j2+1 different irreps in the decomposition.
Working our way down from the j1+j2 rep the last one must therefore be the j1-j2 rep.
A (sort of) different argument goes as follows.  Each state |j1m1> must occur in every
rep, since acting with J_+ and J_- will eventually introduce it. In particular, |j1j1> must occur.
The smallest total m the state |j1j1>|j2m2> can have is if m2 is as small as possible, m2=-j2.
In this case, the total m is j1-j2, hence the smallest rep we have is spin- (j1-j2).
If j2>j1 then reverse the roles, and the smallest rep is spin-(j2-j1).
In general, we have that the smallest is spin-|j1-j2|. You can check that the total dimension
(2j1+1)(2j2+1) is equal to the sum over integer steps from j = |j1-j2| to j1+j2 of (2j+1).

# |jm> = |m1m2><m1m2|jm> ,  sum on m1,m2 with m=m1+m2.
Similarly, |m1m2>= |jm><jm|m1m2>, where the sum is over j with m=m1+m2 fixed.
The expansion coefficients are the Clebsch-Gordan coefficients. The construction above
shows that they can always be taken to be real, so <m1m2|jm>=<jm|m1m2>*=<jm|m1m2>.
There is still an overall sign ambiguity of the CG coeffs, that is typically fixed by requiring that
the coefficient of |m1=j1>|m2=j-j1> in the the expansion of the top state |jj> of the spin-j rep. is positive,
i.e. <j1,j-j1|jj> is positive. (There is a typo in Baym in the fourth line after (15-40), where it reads
m1=j instead of m1=j1.)

# Baym works out the case of   j x 1/2. There is a typo in eqn (15-44), which should have m2 = -/+1/2.)

# The CG coeffs can be computed by:
-brute force
-Mathematica: ClebschGordan[{j1,m1},{j2,m2},{j,m}]   (Note:  I mis-spelled it "Gordon" in class.)
-tables
-recursion relations
-a projection operator method
-amazingly enough, a CLOSED FORM formula has been found by Wigner for all the CG coeffs,
which was given in a more symmetrical form by Racah. See (106.14) of Landau & Lifshitz.
It is so complicated as to be unusable.