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--- classes:2009:fall:phys4101.001:lec_notes_1125 [2009/11/29 11:30] – x500_razi0001
+++ classes:2009:fall:phys4101.001:lec_notes_1125 [2009/11/29 22:20] (current) – yk
@@ Line 90: / Line 90: @@
 <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 J to play with <math>\vec{L} \cdot \vec{S}</math> in a simple useful way that will come up in Chapter 6.
+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.
 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.
-Me or the other guy will finish the notes later; happy Thanksgiving!
+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 \\
+\vec{S_1}\cdot\vec{S_2}\downarrow\uparrow = -\frac{1}{2}\hbar^2\downarrow\uparrow + \hbar^2\downarrow\uparrow \\
+\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]]**\\

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