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| classes:2009:fall:phys4101.001:lec_notes_1125 [2009/11/29 20:09] – x500_razi0001 | classes: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. Follow along, and hopefully it will make sense. | 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. |
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| First of all, we know that | First of all, we know that |
| \downarrow\downarrow = \begin{pmatrix} 0\\ 0 \\0\\1\end{pmatrix}</math> | \downarrow\downarrow = \begin{pmatrix} 0\\ 0 \\0\\1\end{pmatrix}</math> |
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| 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.) |
| 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> ?? \\ |
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| | 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> \\ |
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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 \\ |
| | <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>.\\ |
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| Me or the other guy will finish the notes later; happy Thanksgiving! | Using a similar technique, we can also find (with corrections)... |
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| | <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> |
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| | 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> |
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| | //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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| | 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]]**\\ |