2
$\begingroup$

Consider, for example, the third excited state of an infinite square well:

enter image description here

Now consider the following potential:

enter image description here

If we wanted to sketch the rough shape of the third excited eigenfunction of this potential, we shouldn't have to go through the full set of calculations, but we should be able to deduce what it will look like by considering how the potential differs from the infinite square well. How can we intuitively figure this out? To define the question further and resolve any ambiguity, I've included my attempt at this below.

My reasoning

This is my attempt at sketching the approximate shape of the third excited state:

enter image description here

Between B and C the particle would have less potential energy, and therefore more kinetic energy. Thus, there is a greater chance of the particle being found between A and B, so the central peak shifts to the left of the central dividing line, to the A-B section. The area under the A-B part of the function must be greater than the area under the B-C part for the same reason.

Since the potential is deeper in the B-C region, the exponential decay after the C boundary must be faster than that before the A boundary, because the taller C boundary is closer to an infinite well boundary than the A boundary.

Am I on the right lines?

Improve this question
$\endgroup$
4
  • $\begingroup$ Can you label the 4 energy levels on $V(x)$? It will make it easier to discuss. $\endgroup$
    – JEB
    Commented Jul 14, 2020 at 13:59
  • $\begingroup$ You mean E1 and E2? I'm not sure where I would place them. $\endgroup$
    – Pancake_Senpai
    Commented Jul 14, 2020 at 14:17
  • 1
    $\begingroup$ You may have a look at this paper from where problems as these probably originated- aapt.scitation.org/doi/abs/10.1119/1.1986336 The authors of this classic paper say that there is partial controversy on the solution of this problem, "even among professionals" which I think, is still true as of now. $\endgroup$
    – Manas Dogra
    Commented Jul 15, 2020 at 6:23
  • $\begingroup$ Thank you! That paper was really helpful. So in effect my answer is mostly correct, except that the central peak should be in the region B-C, because loosely speaking there is less momentum in region A-B and thus in that region the wavelength should be longer. $\endgroup$
    – Pancake_Senpai
    Commented Jul 15, 2020 at 9:56

1 Answer 1

Reset to default
0
$\begingroup$

So had you labeled the energy levels for $V(x)$, say $V_i$ for $i=(1,2,3,4)$ for the 4 levels from right to left, then you can actually talk about them clearly.

If you start with a pure finite square well:

$$ V_1=V_4 $$ $$ V_2=V_3$$

you know the solutions.

Likewise, setting $V_2=V_1$, you know have a different square well, for which you know the solutions.

If you adiabatically shift between the two, that is:

$$ V_2 = V_3 $$

and slowly make the transition:

$$ V_2 \rightarrow V_1 $$

it should be clear that the probability in the $V_2$ sector needs to slowly shrink, from "equal to $V_3$" to being an exponential decay. It should not be getting larger in regions of low available kinetic energy.

From there, raising so that $V_4 > V_1$ means a fast exponential decay, and maybe a smaller $\psi(x)$ at the right boundary versus the left boundary.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.