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| classes:2009:fall:phys4101.001:lec_notes_1202 [2009/12/10 21:11] – x500_vinc0053 | classes:2009:fall:phys4101.001:lec_notes_1202 [2009/12/10 21:45] (current) – x500_vinc0053 |
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| Well, this problem involves looking at finding a matrix corresponding to the operator <math>\vec{S}\cdot\vec{L}</math>. | 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> | First off, we know that <math>\vec{S}\cdot\vec{L} = {S_x}{L_x}+{S_y}{L_y}+{S_z}{L_z}</math> |
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| | <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>. |
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| To work with this more easily, let's recall that (as on p.174), we have \\ | 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>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>L_x =\frac{1}{2}(L_+ + L_-) \qquad\qquad L_y = \frac{1}{2i}(L_+ - L_-) </math> |
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| | <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> |
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| 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). \\ | 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). \\ |
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| The rules for how this works for any values of <math>\vec{L_1}</math> and <math>\vec{L_2}</math> are that | 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. | j can range from (L1+L2) to (L1-L2) in integer steps: |
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| | <math>j = (l_1 + l_2), (l_1 + l_2 - 1), ... ,(l_1-l_2)</math> |
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| 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). | 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). |