GREPhysics.NET: GR9277 #29

GR9277 #29

Problem

GREPhysics.NET Official Solution

\prob{29}

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?

Quantum Mechanics $\Rightarrow$ }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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Comments

RusFortunat
2015-10-22 13:22:39

\"An attractive, one dimensional square well\" I find it attractive too.

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psychonautQQ
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?

RusFortunat
2015-10-22 13:30:14

Read first sentence from the top.

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maryami
2011-03-24 01:32:09

in xx2 , the wave decays as E<0 and in x1

maryami
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

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e2ka
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.

kroner
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.

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eshaghoulian
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 $x_2$ .

RusFortunat
2015-10-22 13:18:57

yep

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Ingrid
2006-03-30 15:29:23

Choice (A) is easy to rule out because the wave function should tend to sero for x-> +/-infinity.

zaijings
2009-03-13 19:23:52

you are right.

Plantis
2010-04-03 16:28:46

I think that (A) is correct answer too. What differnce between (A) and (B). I think - no difference.

raevyn
2010-04-08 13:40:31

Plantis, choice (A) does not go to zero at the "tails" (for $\ x$ < $\ x_{1}$ and $\ x$ > $\ x_{2}$ )

shak
2010-08-29 15:55:08

right

Crand0r
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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