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An attractive, one-dimensional square well has depth $V_0$ as shown above. Which of the following best shows a possible wave function for a bound state?

  1. 9277_29_abcde

Quantum Mechanics}Bound State

Tunneling should show exponential decay for a finite-potential well, and thus choice (E) is eliminated. Choice (C) is eliminated because the wave function is not continuous. One eliminates choice (D) because the bound-state wave functions of a finite well isn't linear. The wave function for a bound state should look similar to that of an infinite potential well, except because of tunneling, the well appears larger---thus the energy levels should be lower and the wave functions should look more spread out. Choice (B) shows a more-spread-out version of a wave function from the infinite potential well.

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2015-10-22 13:22:39
\"An attractive, one dimensional square well\" I find it attractive too.NEC
2013-11-28 15:39:39
Is eliminating option A based on the fact that a bound state should go towards zero outside of the well correct?
2015-10-22 13:30:14
Read first sentence from the top.
2011-03-24 01:32:09
in xx2 , the wave decays as E<0 and in x1
2011-03-29 15:05:17
so sorry, I don't know why it is typed incomplete,I 'm not familiar with the format of Latex
2009-10-05 13:42:14
"Tunneling should show exponential decay for a finite-potential well, and thus choice (E) is eliminated"rnrnIs function E not demonstrating exponential decay for x < x1, x > x2? It looks to me like -e^{-Ax} \sin Bx. There must be a better explanation for eliminating E.
2009-10-05 14:32:34
The behavior in (E) isn't really exponential decay. It's oscillating, and although the oscillations seem to decay, that's not the behavior exhibited in tunneling.

If you're looking for more detail about the physics behind that, the gist of it is that in the region outside the well, the eigenstates have to satisfy \psi^{\prime\prime}(x) = [-2m(E - V)/\hbar^2]\psi(x) where E < V. That has solutions of the form Ae^{bx} where b is strictly real.
2007-09-22 16:38:58
Choice D is eliminated more rigorously by invoking the property that the derivative of a wavefunction need be continuous (except at infinite potentials). This wavefunction is not even differentiable at x_1 and x_2.
2015-10-22 13:18:57
2006-03-30 15:29:23
Choice (A) is easy to rule out because the wave function should tend to sero for x-> +/-infinity.
2009-03-13 19:23:52
you are right.
2010-04-03 16:28:46
I think that (A) is correct answer too. What differnce between (A) and (B). I think - no difference.
2010-04-08 13:40:31
Plantis, choice (A) does not go to zero at the "tails" (for \ x < \ x_{1} and \ x > \ x_{2})
2010-08-29 15:55:08
2010-11-12 18:39:13
On the other hand, wavefunctions are often drawn on pictures as if the "0" were the corresponding energy level. Given that, I think the question was poorly written.
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