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Please try to include the following

• main points understood, and expand them - what is your understanding of what the points were.
  • expand these points by including many of the details the class discussed.
• main points which are not clear. - describe what you have understood and what the remain questions surrounding the point(s).
  • Other classmates can step in and clarify the points, and expand them.
• How the main points fit with the big picture of QM. Or what is not clear about how today's points fit in in a big picture.
• wonderful tricks which were used in the lecture.
  

The original plan for today's lecture was to discuss the following things:
1) Clebsch-Gordon coefficients
2) Taking another look at the spherical harmonics (aka <math>Y_\ell^m</math>)
3) 2-particle system

However, we didn't get to any of these items. Instead the lecture covered two main areas:
First, we went over some details about Discussion Problem #13,
Second, we talked about addition of angular momenta (aka <math>\vec{J}=\vec{L_1}+\vec{L_2}</math>)

Yuichi asked for student input on the discussion problem, since the TA's said a lot of students found it challenging. Class comments ranged from 'it was pretty straightforward' to 'I have no idea how to build the H matrix'.

Well, this problem involves looking at finding a matrix corresponding to the operator <math>\vec{S}\cdot\vec{L}</math>. First off, we know that <math>\vec{S}\cdot\vec{L} = {S_x}{L_x}+{S_y}{L_y}+{S_z}{L_z}</math>

<math>L^2\begin{pmatrix} *
*
*\end{pmatrix} = l(l+1)\hbar^2\begin{pmatrix} *
*
*\end{pmatrix} = 2\hbar^2, s^2\begin{pmatrix} *
*
*\end{pmatrix} = \frac{3}{4}\hbar^2\begin{pmatrix} *
*
*\end{pmatrix}</math>.

To work with this more easily, let's recall that (as on p.174), we have
<math>S_x =\frac{1}{2}(S_+ + S_-) \qquad\qquad S_y = \frac{1}{2i}(S_+ - S_-)</math> , and similarly,
<math>L_x =\frac{1}{2}(L_+ + L_-) \qquad\qquad L_y = \frac{1}{2i}(L_+ - L_-) </math>

<math>S_+ \downarrow or S_- \downarrow from \sqrt{(l \mp m)(l\pm m+1)}\hbar</math> <math>l = \frac{1}{2}, m = \pm \frac{1}{2}</math>

Using these expressions, we can rewrite <math>\vec{S}\cdot\vec{L}</math> in terms of just the + and - operators and the z direction operators. This is useful because we know exactly how these operators affect the vector they act on. They pull out certain constants, and the + and - operators also change the vector. (The constants for the + and - operators are given by eq 4.136 on p.172).

Now, you can use this information to find the Hamiltonian by operating on each of our 6 orthogonal basis vectors. Each one we operate on will give us one row of the Hamiltonian (and one column too because the hermitian H matrix is symmetric about its diagonal).

Consider any two arbitrary angular momenta <math>\vec{L_1}</math> and <math>\vec{L_2}</math>.

Let <math>\vec{J}=\vec{L_1}+\vec{L_2}</math>.

When <math>L_1=L_2=\frac{1}{2}</math>, the two vectors can either be parallel, giving you j=1, or they can be antiparallel, which gives you j=0.

By the same reasoning, when <math>L_1=1</math> and <math>L_2=\frac{1}{2}</math>, then <math>j = \frac{3}{2}</math> or <math>j=\frac{1}{2}</math>

Notice that in this case, when <math>j = \frac{3}{2}</math>, then <math>j_z</math> can range from <math>\frac{3}{2}</math> to <math>-\frac{3}{2}</math> in integral steps, (4 possible values)

and when <math>j = \frac{1}{2}</math>, then <math>j_z</math> can range from <math>\frac{1}{2}</math> to <math>-\frac{1}{2}</math> in integral steps (i.e. jz can be + or - 1/2) (2 possible values)

The rules for how this works for any values of <math>\vec{L_1}</math> and <math>\vec{L_2}</math> are that j can range from (L1+L2) to (L1-L2) in integer steps:

<math>j = (l_1 + l_2), (l_1 + l_2 - 1), … ,(l_1-l_2)</math>

Here's another example. Let L1=L2=1. Then the total spin is J=2 or J=1 or J=0. For J=2, jz={2,1,0,-1,-2} (5 possibilities). For J=1, jz={1,0,-1} (3 possibilities), and for J=0, jz=0 (1 possibility).

Notice that there are 5 + 3 + 1 = 9 distinct possible states. This makes sense, since there are 3 possible states of L1, and 3 possible states of L2, so there are 3 * 3 = 9 possible states total.

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