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classes:2009:fall:phys4101.001:lec_notes_1125 [2009/11/29 20:18] – x500_razi0001classes:2009:fall:phys4101.001:lec_notes_1125 [2009/11/29 22:20] (current) – yk
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 \downarrow\downarrow = \begin{pmatrix} 0\\ 0 \\0\\1\end{pmatrix}</math> \downarrow\downarrow = \begin{pmatrix} 0\\ 0 \\0\\1\end{pmatrix}</math>
  
-Now, let's try adding <math>\vec{S_1}</math> and <math>\vec{S_2}</math>....\\+Now, let's try adding <math>\vec{S_1}</math> and <math>\vec{S_2}</math> when they are both in the spin up state....\\
 <math>\vec{J}=\vec{S_1} + \vec{S_2}</math> so \\ <math>\vec{J}=\vec{S_1} + \vec{S_2}</math> so \\
 <math>J^2\uparrow\uparrow=(\vec{S_1} + \vec{S_2})^2\uparrow\uparrow = (S_1^2+S_2^2+2S_1 \cdot S_2)\uparrow\uparrow</math> (This last bit works because S1 and S2 commute.) <math>J^2\uparrow\uparrow=(\vec{S_1} + \vec{S_2})^2\uparrow\uparrow = (S_1^2+S_2^2+2S_1 \cdot S_2)\uparrow\uparrow</math> (This last bit works because S1 and S2 commute.)
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 but what about <math>2\vec{S_1}\cdot\vec{S_2}\uparrow\,_1\uparrow\,_2</math> ?? \\ but what about <math>2\vec{S_1}\cdot\vec{S_2}\uparrow\,_1\uparrow\,_2</math> ?? \\
  
 +Well,\\
 +<math>2\vec{S_1}\cdot\vec{S_2}\uparrow\,_1\uparrow\,_2 = 2(\vec{S_{1x}}\,\vec{S_{2x}}+\vec{S_{1y}}\,\vec{S_{2y}}+\vec{S_{1z}}\,\vec{S_{2z}})\uparrow\,_1\uparrow\,_2</math> \\
  
 +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>.\\
 +//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>.\\
  
-Me or the other guy will finish the notes later; happy Thanksgiving!+Using a similar technique, we can also find (with corrections)... 
 + 
 +<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 \\ 
 +<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>  
 + 
 +\\ 
 +\\ 
 +Happy Thanksgiving!
 ------------------------------------------ ------------------------------------------
 **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_1125.1259547486.txt.gz · Last modified: 2009/11/29 20:18 by x500_razi0001

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