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--- classes:2009:fall:phys4101.001:lec_notes_1202 [2009/12/06 15:53] – x500_razi0001
+++ classes:2009:fall:phys4101.001:lec_notes_1202 [2009/12/10 21:45] (current) – x500_vinc0053
@@ Line 21: / Line 21: @@
 ) Clebsch-Gordon coefficients \\
 ) Taking another look at the spherical harmonics (aka <math>Y_\ell^m</math>)\\
-) z-particle systems (I don't quite know what that means,sorry)\\
+) 2-particle system\\
 However, we didn't get to //any// of these items. Instead the lecture covered two main areas: \\
@@ Line 28: / Line 28: @@
 ====Part 1: Discussion Problem #13====
+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). \\
 ====Part 2: Adding Angular Momenta of two particles to get J====
+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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