Campuses:
Return to Q&A main page: Q_A
Q&A for the previous lecture: Q_A_1106
Q&A for the next lecture: Q_A_1111
If you want to see lecture notes, click lec_notes
Main class wiki page: home
Does anyone have any idea why E14 is a “interesting case” in part c of problem 4.2? I am guessing cause it has a high degeneracy, but I think I am missing something…
It's hard to give a direct answer without giving away the homework, so I'll do my best to get the point across.
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.
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).
Can anyone tell me how to derive px, py, and pz in spherical coordinates?
Are you asking how to derive Lx, Ly, lz in spherical coordinates ? I didn't find the expression for px,py,pz in spherical coordinates system in the book.
No, I know how what Lx, Ly, and Lz are in spherical coordinates, I was just curious what px, py, and pz was. I saw this question asking for the px, py and pz in spherical coordinates, and couldn't figure it out. I found that if you just the expressions found for Lx, Ly, and Lz, you can just use L=r x p=rpsin(theta), and solve for p, which I think should work out.
This was tricky for me, too, just interpreting the problem. The algebra/calc isn't so bad, but it's extensive. Without giving away the entire problem, I'll help you with the major concept: The key is to use the product rule and substitute all of your (many) defined values into the definition for Lz. From the product rule we know that
<math>\frac{\partial}{\partial x}= \frac{\partial \rho}{\partial x} \frac{\partial}{\partial \rho} + \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta} + \frac{\partial \phi}{\partial x} \frac{\partial}{\partial \phi} </math>
Just find each of the derivatives above from the Cartesian to spherical conversion equations and work through the algebra. If your answer is very simple you've done it right. (I guess it's actually in the book)
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?
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>.
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?
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
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?
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.
This might be a bit late to ask for homework help, but I'm having some trouble with 4.6. First I assumed that l=l' (assuming that this yields the non-zero solution from the Kronecker delta),so I have <math>\int_{-1}^{1}P_{l}^{2}(x)\; dx </math> but then what do we do with the <math>\frac{d}{dx}</math> in the integral? Any hint would be appreciated!
This was a tricky problem… What I did was began by assuming that l and l' weren't necessarily equal. If you set up the integral you can perform Integration By Parts, giving you the derivatives to the power of l+1 and l' (the polynomials in the integral aren't affected by IBP, just the derivative operators) and a boundary term. If you perform IBP “l” times you end up with the integral of the polynomial term times the derivative to the (l'+l) power and “l” number of boundary terms. Each bdry term must be zero (which you'll have to show) and you can solve the remaining integral described above. You'll see (subtly) that l = l' for the final integral. It's odd looking, having a derivative operator in the middle of two polynomials each to a variable power, and tricky to solve. It took me a long time to figure it out so I'll give you a hint to use a substitution for <math> x = \cos(\theta) </math> when you get to the final integral. There may be an easier way, but at least it works. Keep track of the constants and just remember when it gets tedious (toward the end) that you know what result you're trying to get. I guess the main key to this problem is recognizing the patterns to set up a formula for performing these operations a variable amount of times.
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.
Can a multidimensional hermitian operator be expressed as a non-square matrix? Such as a 3×27 matrix?
er.. chicken…. I believe that in fact the operator has to be square – the rules of matrix multiplication say that a 1×4 * 4×4 give 1×4. We need to preserve the space coordinates – if it were a 4×3… we'd get a 1×3 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 1×4, 4×4 etc. or… maybe there's a space creating operator… 4×27 … STring theory'd be happy.
I'm also pretty sure it needs to be square in order to take it's determinant.