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GR9677 #52
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Alternate Solutions |
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Comments |
WoolfianOperator
2009-11-05 19:11:27 | i think im losing it. im trying to find the expectation value of  .  . Using =-i  . applying the operator to  one gets a cos( ) term. But then one has an integral from 0 to 2pi of sin()*cos), which is zero. I understand the other method, but why is this not appear to work?
WoolfianOperator
2009-11-06 15:31:48 |
ok. so the problem wants to know the possible outcomes. What i calculated above is the expectation value, which is the average of the possible outcomes. In order to find the possible outcomes we are really going to find the eigenvalues of the operator. One can do this using Yosun's method which involves Eulers formula. I guess my point is that one should remember that the expectation value and eigenvalues are not the same thing.
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|  | wittensdog
2009-11-03 19:53:17 | The more I think about this problem, the more I'm convinced that Void is getting at what ETS is looking for. For spherical harmonics, all of the phi dependence comes in the form of Fourier modes exp(i * m * phi). So if you can break a state down into its Fourier decomposition in phi, you can immediately read off what the values of m are. This is pretty easy to do with sin ( phi ), since you can just use Euler's formula. It would also work very easily for something like sin ( 2 * phi ), in which case you would have m = +/- 2. | | sullx
2009-11-03 18:44:16 | Why is it that the given rigid rotator state is not even included in the above solution(s)?
One could just apply the Lz operator to the given function and equate it with the function multiplied by its eigenvalue m*h-bar.
This results in m depending on sin(PHI) which is a superposition of TWO complex functions (see the last line of void's explination below). | | tin2019
2007-10-18 07:50:16 | Well I solved it this way: Apply the z-component operator twice to the given function  . Now this is obviously not an eigenfunction of the z component, but a superposition of two such functions. Nonetheless it is an eigenfunction of  operator which yields that  and nothing else. Square root will give correct answers (although I think that strictly speaking this is not entirely justified). | | yosun
2005-11-10 11:40:22 | void: surround your LaTeX commands with  ... not \[slash-brackets]\... i only programmed the parser to recognize $ signs---as the LaTeX Rosetta Stone below explains. (your message has been manually re-parsed. but, use $ signs in the future!)
| | Void
2005-11-10 06:41:12 |
Let's say you forget the first few spherical harmonics. (Happens to the best of us.) Hopefully you could remember that:
1) The eigenstates of the z-component of angular momentum are given by
2) You'd remember the form of the angular momentum operator in the z-direction as well:
3) Combining the two, you see that the part of the spherical harmonics that contains m (besides the normalization) is proportional to  . I think this is pretty easy and important to memorize.
The trick here is to realize in the given wave function, the information about m is contained in the  term. Ignore the others. Then remember back from complex analysis that
 Thus, the wave function is in a superposition of states m = 1 and m = -1. The answer is C. | | Void
2005-11-10 06:19:29 | | |
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