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classes:2009:fall:phys4101.001:lec_notes_1125 [2009/11/29 20:37] x500_razi0001classes:2009:fall:phys4101.001:lec_notes_1125 [2009/11/29 22:20] (current) yk
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 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 \\ 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\uparrow\uparrow</math>.\\ +<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 \\ 
 +2\vec{S_1}\cdot\vec{S_2}\downarrow\uparrow = -\frac{1}{2}\hbar^2\downarrow\uparrow + \hbar^2\downarrow\uparrow \\ 
 +2\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 \\ 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 -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 -1 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} </math>  
-Me or the other guy will finish the notes later; happy Thanksgiving!+ 
 +//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 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} </math>  
 + 
 +\\ 
 +\\ 
 +Happy Thanksgiving!
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 **To go back to the lecture note list, click [[lec_notes]]**\\ **To go back to the lecture note list, click [[lec_notes]]**\\
classes/2009/fall/phys4101.001/lec_notes_1125.1259548625.txt.gz · Last modified: 2009/11/29 20:37 by x500_razi0001