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| classes:2009:fall:phys4101.001:lec_notes_1125 [2009/11/29 11:17] – x500_razi0001 | classes:2009:fall:phys4101.001:lec_notes_1125 [2009/11/29 22:20] (current) – yk |
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| Notice that we are going to ignore the <math>(\frac{p^2}{2m} + V)</math> term of the hamiltonian since that part isn't changing from the magnetic field; we're not interested in that part of it right now. | Notice that we are going to ignore the <math>(\frac{p^2}{2m} + V)</math> term of the hamiltonian since that part isn't changing from the magnetic field; we're not interested in that part of it right now. |
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| Anyway, if the magnetic field we apply is | Anyway, if the magnetic field we apply is in the <math>\hat{z}</math> direction, then we have |
| | <math>\vec{B}=B_z</math> and <math>\hat{H}=-\alpha B_z S_z</math> |
| | In that case, which is associated with the Zeeman Effect, <math>S_z</math> is an eigenfunction of <math>\hat{H}</math> |
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| Me or the other guy will finish the notes later; happy Thanksgiving! | If the magnetic field is not on, then the <math>\vec{L} \cdot \vec{S}</math> becomes important, and |
| | <math>[\vec{L} \cdot \vec{S},S_z] \neq 0</math> |
| | |
| | So, we want to find eigenfunctions of <math>\vec{L} \cdot \vec{S}</math>, and in some situations, we may also want to find <math>\vec{S_1} \cdot \vec{S_2}</math> eigenfunctions. |
| | |
| | Well, if we let <math>\vec{J} = \vec{L}+\vec{S}</math>, we'll have the following... |
| | <math>[J^2,\vec{L} \cdot \vec{S}] = 0</math> and |
| | <math>[J_z,\vec{L} \cdot \vec{S}] = 0</math>, |
| | and then we can use <math>\vec{J}</math> to play with <math>\vec{L} \cdot \vec{S}</math> in a simple useful way that will come up in Chapter 6. |
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| | In short, let's just trust that this addition of angular momentum technique we develop in this chapter will be very useful later, and move on to the technique itself. |
| | |
| | ===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. |
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| | 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_2\right|^2 = \frac{3}{4}\hbar^2</math>. |
| | |
| | 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} 0\\ 1\end{pmatrix}_1</math> \\ |
| | <math>\chi_2^+ = \begin{pmatrix} 1\\ 0\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> |
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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]]**\\ |