GR 8677927796770177 | # Login | Register

GR9677 #98
Problem
 GREPhysics.NET Official Solution Alternate Solutions
This problem is still being typed.
Quantum Mechanics$\Rightarrow$}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).

Alternate Solutions
 There are no Alternate Solutions for this problem. Be the first to post one!
mpdude8
2012-04-20 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
2009-11-06 17:00:34
Look at the 1992 test, problem 89...

I didn't know the ETS gave the same problems over
anmuhich
2009-04-02 12:16:25
You can also immediately eliminate all but C and D because the energy can't be zero.
 segfault2009-08-19 20:32:07 Also (C) is just the energy levels of an unperturbed harmonic oscillator, so it has to be D.
 cilginfizikci2010-03-09 02:36:30 why the energy can not be "0" ??? dont tell me that this is harmonic oscillator... coz u already change the system... ????????????
 carle2572010-04-05 19:09:54 If the energy were zero, then this would imply that the position is zero because the harmonic oscillator potential goes as $\frac{1}{2}kx^2$, and the kinetic term would also have to be zero to make the total energy zero ($\Rightarrow v=0$) 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 non-zero.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$