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Quantum Mechanics}Symmetry

One recalls the simple harmonic oscillator wave functions to be symmetric about the vertical-axis (even) for the 0th energy level, symmetric about the origin (odd) for the first energy level, and so on.

If there is a wall in the middle of the well, then all the 0th energy level wave function would disappear, as would all even wave functions.

Recall the formula for SHO E=\hbar \omega \left(n+\frac{1}{2}\right). The first few odd states (the ones that remain) are E_1 = 3/2 \hbar \omega, E_3=7/2 \hbar \omega, etc. This is choice (D).

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Comments
mike
2009-11-06 17:00:34
Look at the 1992 test, problem 89...

I didn't know the ETS gave the same problems over
NEC
anmuhich
2009-04-02 12:16:25
You can also immediately eliminate all but C and D because the energy can't be zero.
segfault
2009-08-19 20:32:07
Also (C) is just the energy levels of an unperturbed harmonic oscillator, so it has to be D.
cilginfizikci
2010-03-09 02:36:30
why the energy can not be "0" ??? dont tell me that this is harmonic oscillator... coz u already change the system... ????????????
NEC

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