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| classes:2009:fall:phys4101.001:q_a_1116 [2009/11/16 18:09] – kuehler | classes:2009:fall:phys4101.001:q_a_1116 [2009/11/20 14:08] (current) – myers |
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| ===== Nov 16 (Mon) ===== | ===== Nov 16 (Mon) 4.3 Angular momentum ===== |
| **Return to Q&A main page: [[Q_A]]**\\ | **Return to Q&A main page: [[Q_A]]**\\ |
| **Q&A for the previous lecture: [[Q_A_1113]]**\\ | **Q&A for the previous lecture: [[Q_A_1113]]**\\ |
| **Main class wiki page: ** [[home]] | **Main class wiki page: ** [[home]] |
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| ====Anaximenes - 19:25 - 11/13/09==== | ====Anaximenes - 19:25 - 11/13/09==== |
| On the last page, Esquire asked whether a non-square matrix could represent a hermitian operator. A hermitian matrix must be square and, more-over, must be equal to its "transpose conjugate" (which is the same as a hermitian conjugate). The transpose conjugate is attained by transposing the matrix (flipping it along the diagonal) and taking the complex conjugate of each element. See pages 443-444. To get some insight into why this is so, try finding <math><\alpha |\hat{Q} \beta></math> and <math><\hat{Q} \alpha | \beta ></math> where <math>\displaystyle \hat{Q}= \left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 1 \end{array} \right)</math>, <math>\displaystyle \vec{\alpha} = \left(\begin{array}{c} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{array} \right)</math>, and <math>\displaystyle \vec{\beta} = \left(\begin{array}{c} \beta_1 \\ \beta_2 \\ \beta_3 \end{array} \right)</math>. | On the last page, Esquire asked whether a non-square matrix could represent a hermitian operator. A hermitian matrix must be square and, more-over, must be equal to its "transpose conjugate" (which is the same as a hermitian conjugate). The transpose conjugate is attained by transposing the matrix (flipping it along the diagonal) and taking the complex conjugate of each element. See pages 443-444. To get some insight into why this is so, try finding <math><\alpha |\hat{Q} \beta></math> and <math><\hat{Q} \alpha | \beta ></math> where <math>\displaystyle \hat{Q}= \left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 1 \end{array} \right)</math>, <math>\displaystyle \vec{\alpha} = \left(\begin{array}{c} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{array} \right)</math>, and <math>\displaystyle \vec{\beta} = \left(\begin{array}{c} \beta_1 \\ \beta_2 \\ \beta_3 \end{array} \right)</math>. |
| ====Anaximenes - 19:10 - 11/14/09==== | ====Anaximenes - 19:10 - 11/14/09==== |
| I guess might as well ask what people thought of the test. I didn't know what to expect going in seeing as how the material we covered was mostly just the solution to the hydrogen atom, which doesn't lend itself well to a test format. The first problem seemed a little out of place to me; the practice test is the only place I recall seeing anything like it so far in the course. (I knew the units of <math>\eps_0</math>, so I didn't have any problems, but I'm curious if that was the kind of thing the proctors would answer; did anyone ask during the test?) All of the problems were straight-forward, though, and I felt like I had enough time. I don't know what to think of problem 3; using the spherical gradient made the problem trivial, so I tried to use the other method as well and made some errors. As a result, though, I now have a somewhat better understanding of partial vs. regular derivatives. Does anyone else have any comments? Agreement/disagreement? Something completely different? | I guess might as well ask what people thought of the test. I didn't know what to expect going in seeing as how the material we covered was mostly just the solution to the hydrogen atom, which doesn't lend itself well to a test format. The first problem seemed a little out of place to me; the practice test is the only place I recall seeing anything like it so far in the course. (I knew the units of <math>\eps_0</math>, so I didn't have any problems, but I'm curious if that was the kind of thing the proctors would answer; did anyone ask during the test?) All of the problems were straight-forward, though, and I felt like I had enough time. I don't know what to think of problem 3; using the spherical gradient made the problem trivial, so I tried to use the other method as well and made some errors. As a result, though, I now have a somewhat better understanding of partial vs. regular derivatives. Does anyone else have any comments? Agreement/disagreement? Something completely different? |
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| | ===Captain America 11-18 10:21=== |
| | I'll answer this a few days late, maybe someone will read it. I thought the test was fairly straight-forward and followed the practice exam well. I actually really liked question 1 because it connected quantum physics with more real-world type physics. It's easy to forget what all of these things (specifically h) mean in a physical sense and then just pretend like this is a math course, but it is good to remember that these things can (more or less) be related to physics we already know. I don't know if you saw, but in the last 5-10 minutes of class he put the units of <math>\eps_0</math> on the board, so I assume the proctors would have answered that question. Problem 3 was a bit annoying, if you did it the quick and easy way I fear you won't get full credit because I'm pretty sure they wanted us to show our work plugging in spherical stuff, but if you did it the second way then I fear most people (myself for sure) won't get full credit because it either took too long or some small mistake slipped into the solution. |
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| | But overall I'd say I enjoyed this test much more than the second, in terms of what I was expecting. |
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| ====Jake22 - 15:32 - 11/15/09==== | ====Jake22 - 15:32 - 11/15/09==== |
| If we look at the probability density of the spherical harmonic <math>\|Y({\theta},{\phi})\| ^2</math>, we can see that it is always expressed only in terms of a polynomial of degree 2l in <math>cos\theta</math>. How can or has this relationship been exploited? | If we look at the probability density of the spherical harmonic <math>\|Y({\theta},{\phi})\| ^2</math>, we can see that it is always expressed only in terms of a polynomial of degree 2l in <math>cos\theta</math>. How can or has this relationship been exploited? |
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| | ===Esquire (age of concurrence)=== |
| | I was also curious toward this issue. |
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| ====Blackbox 10:42 - 11/16/09==== | ====Blackbox 10:42 - 11/16/09==== |
| Could you explain how to derive from 4.97 to 4.98? | Could you explain how to derive from 4.97 to 4.98? |
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| | ===prest121 8pm 11/16/2009=== |
| | In 4.97 we have <math>[L_x,L_y] = [yp_z,zp_x] - [yp_z,xp_z] - [zp_y,zp_x] + [zp_y,xp_z]</math>. |
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| | The second term <math>[yp_z,xp_z] = y\frac{\partial}{\partial z}(x\frac{\partial f}{\partial z}) - x\frac{\partial}{\partial z}(y\frac{\partial f}{\partial z})</math> for a test function f. We apply the chain rule here and since <math>\frac{\partial x}{\partial z} = \frac{\partial y}{\partial z} = 0</math>, we get that this term is equal to: |
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| | <math>yx\frac{\partial^2 f}{\partial z^2} - xy\frac{\partial^2 f}{\partial z^2}</math>. This is equal to zero since x and y commute (<math>xy = yx</math>). The same argument applies for the third term in equation 4.97. To get the rest of equation 4.98, you just expand the remaining terms (1st and 4th terms). |
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| ====Can 11/16 11am==== | ====Can 11/16 11am==== |
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| ===Pluto 4ever 11/16 6:02pm=== | ===Pluto 4ever 11/16 6:02pm=== |
| From what I can tell <math>L_z ~{\hbar}m</math> as shown in equation [4.118]. Here m is a quantum value that goes from -l, -l+1,...,l-1, l. In equation [4.115] he set m = -l, since l_bar = -1, to find the angular momentum for the bottom rung. In equation [4.111] Griffiths does this again, except this time it is for the top rung (m = l) of the angular momentum states. | From what I can tell <math>L_z</math> is proportional to <math>{\hbar}m</math> as shown in equation [4.118]. Here m is a quantum value that goes from -l, -l+1,...,l-1, l. In equation [4.115] he set m = -l, since l_bar = -l, to find the angular momentum for the bottom rung. In equation [4.111] Griffiths does this again, except this time it is for the top rung (m = l) of the angular momentum states. |
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