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classes:2009:fall:phys4101.001:lec_notes_1125 [2009/11/29 12:26] – x500_razi0001classes:2009:fall:phys4101.001:lec_notes_1125 [2009/11/29 22:20] (current) – yk
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 ===Part 2: How to add angular momenta=== ===Part 2: How to add angular momenta===
 +The basic idea of the following section is to define <math>J^2</math> and then show how to find the eigenvalues of <math>J^2</math> for the basic spin-up and spin-down eigenfunctions. In the lecture, and in what follows, we just found the eigenvalues for the combined state <math>\uparrow\,_1\uparrow\,_2</math>. At the end, Yuichi just wrote out a matrix that included the other eigenvalues for the other combined spin states. They can be found in a similar manner to what is shown below. I hope my writeup makes some sense.
 +
 First of all, we know that  First of all, we know that 
 <math>\left|S_1\right|^2=\frac{1}{2}(\frac{1}{2}+1)\hbar^2 = \frac{3}{4}\hbar^2</math> and by the same method <math>\left|S_1\right|^2=\frac{1}{2}(\frac{1}{2}+1)\hbar^2 = \frac{3}{4}\hbar^2</math> and by the same method
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 For each of the two Hilbert spaces (one for S1 and one for S2) we have a pair of spin-up and spin-downs represented as follows: For each of the two Hilbert spaces (one for S1 and one for S2) we have a pair of spin-up and spin-downs represented as follows:
 <math>\chi_1^+ = \begin{pmatrix} 1\\ 0\end{pmatrix}_1 \:\:</math> <math>\chi_1^+ = \begin{pmatrix} 1\\ 0\end{pmatrix}_1 \:\:</math>
-<math>\chi_1^- = \begin{pmatrix} 0\\ 1\end{pmatrix}_1</math>+<math>\chi_1^- = \begin{pmatrix} 0\\ 1\end{pmatrix}_1</math> \\
 <math>\chi_2^+ = \begin{pmatrix} 1\\ 0\end{pmatrix}_2 \:\:</math> <math>\chi_2^+ = \begin{pmatrix} 1\\ 0\end{pmatrix}_2 \:\:</math>
-<math>\chi_2^- = \begin{pmatrix} 0\\ 1\end{pmatrix}_2</math>+<math>\chi_2^- = \begin{pmatrix} 0\\ 1\end{pmatrix}_2</math> \\ 
 +Remember these can also be represented with up and down arrows, which will be done below.\\ 
 + 
 +We have to be careful with these since each pair is associated with a different Hilbert space. If we want to be able to represent combined states of S1 and S2, one way to do it is to use a column 4-vector and define it like this...\\ 
 +<math>\uparrow\uparrow = \begin{pmatrix} 1\\ 0 \\0\\0\end{pmatrix} \:\: 
 +\uparrow\downarrow = \begin{pmatrix} 0\\ 1 \\0\\0\end{pmatrix} \:\: 
 +\downarrow\uparrow = \begin{pmatrix} 0\\ 0 \\1\\0\end{pmatrix} \:\: 
 +\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> when they are both in the spin up state....\\ 
 +<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.) 
 + 
 +Now, how do we find this sum? \\ 
 +Well, we already know that \\ 
 +<math>\vec{S_1}^2 \uparrow\;_1\uparrow\:_2 = (\frac{3}{4}\hbar^2\uparrow\;_1)\uparrow\;_2</math>  and, similarly,\\ 
 +<math>\vec{S_2}^2 \uparrow\;_1\uparrow\:_2 = (\frac{3}{4}\hbar^2\uparrow\;_2)\uparrow\;_1</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>.\\ 
 + 
 +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> 
  
-Me or the other guy will finish the notes later; happy Thanksgiving!+\\ 
 +\\ 
 +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_1125.1259519213.txt.gz · Last modified: 2009/11/29 12:26 by x500_razi0001

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