# What happens with the energies of the boundstates in an infinite square well if we put a small potential step in the middel?

I am wondering how (qualitatively) the energies of the boundstates in an infinite square well with a small potential step in the middle changes if we change that potential step. The problem is actually pretty similar to this post, but I especially would like to know how the gaps between the energies changes, if we change the height or the width of the potential step.

I tried to find a formula for the energies by solving the time independent problem, but couldn't find a clear solution (it looked more like a transcendantal equation but I might have made some mistakes).

• NB: This is a classic exercise in perturbation theory. Commented Dec 17, 2020 at 16:01
• Can you suggest me where I can find this particular exercice or something similar? Commented Dec 21, 2020 at 11:03

Solving for the energies of this system analytically does involve solving a trancendental equation numerically, if memory serves. There's nothing wrong with that, but it can be a bit difficult to clearly see the influences of the various parameters on the result.

A different approach is to treat this problem with perturbation theory. Since you're assuming that the step height is small$$^\dagger$$, a good start would be to calculate the first order corrections to the energy eigenvalues.

Explicitly, let your Hamiltonian be $$\hat H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+ \lambda V(x), \qquad V(x)=\cases{1 & x\in \left[\frac{L}{2}-\frac{a}{2},\frac{L}{2}+\frac{a}{2}\right]\\0 & else}$$

This is the Hamiltonian for an infinite potential well with a potential step of width $$a$$ and height $$\lambda$$ in the center. To first order in $$\lambda$$, the corrected energies are simply $$E_n \simeq E_n^{(0)}+ \lambda \left<\psi_n^{(0)}|\hat V |\psi_n^{(0)}\right> = E_n^{(0)} + \lambda \int_{L/2-a/2}^{L/2+a/2}\psi_n^{(0)*}\psi_n^{(0)} dx$$ where $$E_n^{(0)}$$ and $$\psi_n^{(0)}$$ are the uncorrected energies and (normalized) eigenvectors, respectively. We already know what those are from the elementary solution of the infinite potential well, so by evaluating that integral you can see how those energies will change when you introduce the step - at least as long as the step height is small.

$$^\dagger$$ What it means for an operator to be small can be a subtle issue. In this case, we'd want that $$\lambda$$ be much smaller than the expected value of the unperturbed Hamiltonian in any state of interest. In this case, that would be accomplished if

$$\lambda \ll \frac{\pi^2\hbar^2}{2mL^2}$$

If $$\lambda$$ exceeds this limit, then the first order correction will no longer be a good approximation of how the energy will have changed.

• Thank you! I calculated the first order energy correction with the advice you gave me and found that the energy is higher for a higher potential step (there is a factor height^2 in the first energy correction). For the influence of the length of the potential step it is a bit less clear because the energy correction is proportional to a/L - (2(-1)^n * sin(npia/L))/(n*pi). I plotted this function and it looks like sin(a) + a * const so it normally gets bigger when a is bigger, but sometimes it gets a bit smaller. Commented Dec 21, 2020 at 13:59
• @NicolasSchmid Hi Nicolas. Remember that the way that I formulated this problem, the step height is contained within $\lambda$, not $V$. As a result, your correction to the energy should be linearly proportional to the step height. I haven't done the integral myself, but one way to check it is to simply let $a=L$; in that case, you've simply shifted the "floor" of the potential well up by $\lambda$, so all of the energy levels should be shifted up by the same amount. It looks like that's what happens with your result, which is good. Commented Dec 21, 2020 at 16:42