Intuiting qualitatively the shapes of the eigenfunctions of a finite well-like potential, using the infinite well eigenfunctions as an inspiration

Question

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

Now consider the following potential:

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:

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?

Answer 1

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.