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--- classes:2009:fall:phys4101.001:lec_notes_1125 [2009/11/29 20:38] – x500_razi0001
+++ classes:2009:fall:phys4101.001:lec_notes_1125 [2009/11/29 22:20] (current) – yk
@@ Line 130: / Line 130: @@
 Now, we know all of the spin matrices in the above expression, so we can just plug them all in and solve, and we will get \\
 <math>J^2\uparrow\uparrow = \frac{1}{2}\hbar^2\uparrow\uparrow</math>.\\
-Using a similar technique, we can also find...
+//Yuichi// I mis-stated here.  It should have been:<math>2\vec{S_1}\cdot\vec{S_2}\uparrow\uparrow = \frac{1}{2}\hbar^2\uparrow\uparrow</math>.\\
-<math>J^2\uparrow\downarrow = -\frac{1}{2}\hbar^2\uparrow\downarrow + \hbar^2\uparrow\downarrow \\
-J^2\downarrow\uparrow = -\frac{1}{2}\hbar^2\downarrow\uparrow + \hbar^2\downarrow\uparrow \\
+Using a similar technique, we can also find (with corrections)...
-J^2 \downarrow\downarrow = \frac{1}{2}\hbar^2\downarrow\downarrow \\</math>
+<math>2\vec{S_1}\cdot\vec{S_2}\uparrow\downarrow = -\frac{1}{2}\hbar^2\uparrow\downarrow + \hbar^2\uparrow\downarrow \\
+\vec{S_1}\cdot\vec{S_2}\downarrow\uparrow = -\frac{1}{2}\hbar^2\downarrow\uparrow + \hbar^2\downarrow\uparrow \\
+\vec{S_1}\cdot\vec{S_2} \downarrow\downarrow = \frac{1}{2}\hbar^2\downarrow\downarrow \\</math>
 We can use our 4 element column vector notation described above to represent <math>J^2</math> using a single matrix, which is \\
-<math>J^2 = \frac{1}{2}\hbar^2 \begin{pmatrix} 1 \ 0 \ 0 \ 0 \\ 0 \ -1 \ 2 \ 0 \\ 0 \ 2 \ -1 \ 0 \\ 0 \ 0 \ 0 \ 1 \end{pmatrix} </math>
+<math>J^2 = \frac{1}{2}\hbar^2 \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} </math>
+//Yuichi// and this should have been <math>2\vec{S_1}\cdot\vec{S_2} = \frac{1}{2}\hbar^2 \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} </math>
 \\
 \\

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