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| classes:2009:fall:phys4101.001:q_a_1109 [2009/11/09 20:42] – kuehler | classes:2009:fall:phys4101.001:q_a_1109 [2009/11/30 09:13] (current) – x500_bast0052 |
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| ===== Nov 09 (Mon) ===== | ===== Nov 09 (Mon) Legendre closure, Radial WF and spherical Bessel ===== |
| **Return to Q&A main page: [[Q_A]]**\\ | **Return to Q&A main page: [[Q_A]]**\\ |
| **Q&A for the previous lecture: [[Q_A_1106]]**\\ | **Q&A for the previous lecture: [[Q_A_1106]]**\\ |
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| Take for instance the energy level after the ones it asks us to solve for in the question. <math>E_7</math> has states of n = 3,2,2 respectively. This would give us 3 degenerate states, of energy <math>17 {\pi^2\hbar^2}/{2ma^2}</math>. For <math>E_8</math> we have an energy of <math>18 {\pi^2\hbar^2}/{2ma^2}</math>. The next energy level is <math>19 {\pi^2\hbar^2}/{2ma^2}</math>, then, however, the energy is <math>21 {\pi^2\hbar^2}/{2ma^2}</math>, as no combination of <math>n_x^2+n_y^2+n_z^2</math> can add to equal 20. As the energy level increases, there are some energy levels that cannot be reached, and some energy levels that may be able to be reached by more than one configuration of <math>n_x^2+n_y^2+n_z^2</math>s. Think about that for a while and it should come to you. | Take for instance the energy level after the ones it asks us to solve for in the question. <math>E_7</math> has states of n = 3,2,2 respectively. This would give us 3 degenerate states, of energy <math>17 {\pi^2\hbar^2}/{2ma^2}</math>. For <math>E_8</math> we have an energy of <math>18 {\pi^2\hbar^2}/{2ma^2}</math>. The next energy level is <math>19 {\pi^2\hbar^2}/{2ma^2}</math>, then, however, the energy is <math>21 {\pi^2\hbar^2}/{2ma^2}</math>, as no combination of <math>n_x^2+n_y^2+n_z^2</math> can add to equal 20. As the energy level increases, there are some energy levels that cannot be reached, and some energy levels that may be able to be reached by more than one configuration of <math>n_x^2+n_y^2+n_z^2</math>s. Think about that for a while and it should come to you. |
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| | === Mercury 11/10/2009 7:30 pm === |
| | E14 is interesting because it has a degeneracy of 4, whereas all the previous energies have degeneracies of 1, 3, or 6. The different values for n are (3,3,3), (5,1,1), (1,5,1), and (1,1,5). |
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| ==== Schrodinger's Dog 11/7/09 7:51am ==== | ==== Schrodinger's Dog 11/7/09 7:51am ==== |
| ==== Schrodinger's Dog 11/7/09 8:00am ==== | ==== Schrodinger's Dog 11/7/09 8:00am ==== |
| Can someone tell me how Griffths found an expression for theta/phi hat in Cartesian components on page 168, equations 4.125(6). Does anyone know of quick tricks in finding these transformation of unit vectors from one coordinate system to another? | Can someone tell me how Griffths found an expression for theta/phi hat in Cartesian components on page 168, equations 4.125(6). Does anyone know of quick tricks in finding these transformation of unit vectors from one coordinate system to another? |
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| | ===chavez 11/10/08 11:00am=== |
| | You need to use the relations <math>x=r cos\theta sin\phi</math>, <math>y=r sin\theta sin\phi</math>, and <math>z = r cos\phi</math>. Then you can use <math>\vec{r} = \<x,y,z\> = \<r cos\theta sin\phi,r sin\theta sin\phi,r cos\phi\></math> and the fact that the unit vectors are give by <math>\hat{r}=\frac{\frac{d\vec{r}}{dr}}{\|\frac{d\vec{r}}{dr}\|}</math>, <math>\hat{\theta}=\frac{\frac{d\vec{r}}{d\theta}}{\|\frac{d\vec{r}}{d\theta}\|}</math>, <math>\hat{\phi}=\frac{\frac{d\vec{r}}{d\phi}}{\|\frac{d\vec{r}}{d\phi}\|}</math>. |
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| ==== nikif002 11/9/09 12:17am ==== | ==== nikif002 11/9/09 12:17am ==== |
| I have a very fundamental question about the material we have been covering in this class. I would like to know more about the time-independent potential assumption that we have been making. Since so many important results follow directly from it, just how different is quantum dynamics from time-independent QM? And how useful is quantum dynamics in applications? Of course, in reality, all potentials are time dependent, but I am guessing you can make a quasi-static assumption for time-dependence that is slow enough in comparison to the wavefunction time-dependence. Are there any phenomena for which you absolutely need to solve the time-dependent Schrodinger equation (whether numerically or analytically) to get a reasonably accurate result? And if so, what are they? | I have a very fundamental question about the material we have been covering in this class. I would like to know more about the time-independent potential assumption that we have been making. Since so many important results follow directly from it, just how different is quantum dynamics from time-independent QM? And how useful is quantum dynamics in applications? Of course, in reality, all potentials are time dependent, but I am guessing you can make a quasi-static assumption for time-dependence that is slow enough in comparison to the wavefunction time-dependence. Are there any phenomena for which you absolutely need to solve the time-dependent Schrodinger equation (whether numerically or analytically) to get a reasonably accurate result? And if so, what are they? |
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| | ===chavez 11/10/09 7:30pm=== |
| | If you're talking about things like the hyrdogen atom in free space then time-independent assumption is acceptable because the standard coulomb potential has no time-dependence. Time-dependence becomes important when you are no longer talking about free space and there are other fields that can affect the particle. As far as applications for a time-dependent potential go, I think lasers/masers would probably be one of the more interesting ones. |
| | I found the following link to be a pretty good read: http://galileo.phys.virginia.edu/classes/752.mf1i.spring03/Time_Dep_PT.htm |
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| ==== Cuthulhu Food 11/9/09 9:30am ==== | ==== Cuthulhu Food 11/9/09 9:30am ==== |
| On page 147 of Griffiths, he uses the phrase "peel off the asymptotic behavior" when solving for the radial wave function for the hydrogen atom. Can anyone lend some insight as to what this phrase means? | On page 147 of Griffiths, he uses the phrase "peel off the asymptotic behavior" when solving for the radial wave function for the hydrogen atom. Can anyone lend some insight as to what this phrase means? |
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| | === David Hilbert's Hat 11/10 12:50pm === |
| | It seems like you get two forms that solve 4.56: one when <math> \rho </math> approaches infinity and one when it approaches zero, and this is found by looking at the 4.56 differential equation "asymptotically." And then in 4.60 Griffiths just multiplies these two functions, which should give you a solution that solves u(<math> \rho </math>) at each asymptote but may not be equal to it everywhere, like in the middle ground. So you want to find the term that dominates by taking away the asymptotic behavior: you want to find v(<math> \rho </math>). It's easiest like this: you know that the solution must have these two asymptotic parts, but they will dominate some other "hidden" function at the asymptotes and you can't find a general solution without the "hidden" function, so you build a general form of u(<math> \rho </math>) by taking any function v(<math> \rho </math>) and multiplying it by the asymptotes, then solve for a function v that should be more simple than u. When he says peel off the behavior he must mean to take what we already know about u as it approaches 0 and infinity, then find whatever the "hidden" function is, which should be easier than just finding an abstract u function. |
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| ====Hydra 11/9 6pm==== | ====Hydra 11/9 6pm==== |
| ===Pluto 4ever 11/9 8:33PM=== | ===Pluto 4ever 11/9 8:33PM=== |
| I basically did the same thing by using integration by parts l times that way you can get the <math>(\frac{d}{dx})^l^'(x^2-1)^l^'</math> to reduce to (2l')!. From there I expanded out <math>(x^2-1)^l</math> and noticed that when I took the integral of the first term from -1 to 1 I got <math>\frac{2}{2l+1}</math>. From there I used some reasoning to get a final answer. | I basically did the same thing by using integration by parts l times that way you can get the <math>(\frac{d}{dx})^l^'(x^2-1)^l^'</math> to reduce to (2l')!. From there I expanded out <math>(x^2-1)^l</math> and noticed that when I took the integral of the first term from -1 to 1 I got <math>\frac{2}{2l+1}</math>. From there I used some reasoning to get a final answer. |
| | ====Esquire 11/12/09 AD (Information Age) 8:15pm==== |
| | Can a multidimensional hermitian operator be expressed as a non-square matrix? Such as a 3X27 matrix? |
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| | ===Spherical Harmonic 11/12/09=== |
| | er.. chicken.... |
| | I believe that in fact the operator has to be square -- the rules of matrix multiplication say that a 1x4 * 4x4 give 1x4. We need to preserve the space coordinates -- if it were a 4x3... we'd get a 1x3 and completely wipe out a space coordinate.... so unless you have an operator that shrinks space down a dimension... I do believe we need square operators of 1x4, 4x4 etc. or... maybe there's a space creating operator... 4x27 ... STring theory'd be happy. |
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| | ===Devilin=== |
| | I'm also pretty sure it needs to be square in order to take it's determinant. |
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| | ===Esquire (Age of Aquarius) 11/12/09=== |
| | I believe that such operators need to be square so that eigenvalues and eigenvectors can be found. This is done via the determinent method, which only is doable with square matrices. |
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