GR9677 #98



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mpdude8 20120420 19:35:55  Here's some "it's problem 98, it can't be that easy" logic, even if you don't see the solution right off the bat.
As long as you know the energy eigenvalues for the S.H.O. you know A, B, and E are impossible.
C would be the energy eigenvalues for the straight up S.H.O. without the wall. Let's go out on a limb and say that the wall messes up the energy states. Pretty unlikely that a giant wall in your potential wouldn't affect the energy in any way. So, I went with D.   mike 20091106 17:00:34  Look at the 1992 test, problem 89...
I didn't know the ETS gave the same problems over   anmuhich 20090402 12:16:25  You can also immediately eliminate all but C and D because the energy can't be zero.
segfault 20090819 20:32:07 
Also (C) is just the energy levels of an unperturbed harmonic oscillator, so it has to be D.

cilginfizikci 20100309 02:36:30 
why the energy can not be "0" ??? dont tell me that this is harmonic oscillator... coz u already change the system... ????????????

carle257 20100405 19:09:54 
If the energy were zero, then this would imply that the position is zero because the harmonic oscillator potential goes as , and the kinetic term would also have to be zero to make the total energy zero () But because of the uncertainty principle, we cannot simultaneously define the momentum and position to be zero.
Put another way, if the potential and thus position is zero, you would have a huge uncertainty in the momentum and thus the kinetic term would be nonzero.

 

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