GR 8677927796770177 | # Login | Register

GR9277 #29
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\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.

Alternate Solutions
 There are no Alternate Solutions for this problem. Be the first to post one!
Comments
RusFortunat
2015-10-22 13:22:39
\"An attractive, one dimensional square well\" I find it attractive too.
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?
 RusFortunat2015-10-22 13:30:14 Read first sentence from the top.
maryami
2011-03-24 01:32:09
in xx2 , the wave decays as E<0 and in x1
 maryami2011-03-29 15:05:17 so sorry, I don't know why it is typed incomplete,I 'm not familiar with the format of Latex
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.
 kroner2009-10-05 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.
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 $x_2$.
 RusFortunat2015-10-22 13:18:57 yep
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.
 zaijings2009-03-13 19:23:52 you are right.
 Plantis2010-04-03 16:28:46 I think that (A) is correct answer too. What differnce between (A) and (B). I think - no difference.
 raevyn2010-04-08 13:40:31 Plantis, choice (A) does not go to zero at the "tails" (for $\ x$ < $\ x_{1}$ and $\ x$ > $\ x_{2}$)
 shak2010-08-29 15:55:08 right
 Crand0r2010-11-12 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.

Post A Comment!
 Username: Password:
Click here to register.
This comment is best classified as a: (mouseover)
Mouseover the respective type above for an explanation of each type.

## Bare Basic LaTeX Rosetta Stone

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}\$ $\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...