I am asking myself, how constraining a potential changes the energy eigenvalues.

With the WKB-Approximation-Method one can derive that the dependence of the eigenenergies regarding a potential $V(x) \propto x^{\lambda}$ is:

$$ E_n \propto sgn(\lambda)\cdot n^{\frac{1}{1/\lambda + 1/2}}, $$

see e.g. Dependence of energy $E$ with principal quantum number $n$.

But that is for an infinite potential.

What would be the effect on the difference between the energy levels, if the potential vanishes? For example:

$ V_1(x) = \begin{cases} |x|-a \, , if \ x\le |a|\\ 0, \ otherwise \end{cases} $


$ V_2(x) = \begin{cases} x^2 -a^2 \ if \ x\le |a|\\ 0, \ otherwise \end{cases} $

For the harmonic oscillator, the difference between the energie-levels is constant. I suppose this wouldn't be the case anymore for the finite potential.

  • $\begingroup$ Did you mean $|x|<a$, and are you sure you want the potential to be 0 otherwise, and not some positive value like $a^2$. If not, the answer is really simple - there are no bound states and there is only a continuum (if you draw the potential this should be clear). I'd also guess you're more likely to want $|x|$ rather than $x$ in the first potential. I think you should draw your potentials and think more carefully about what question you want to ask. If you like all the changes I suggest I can answer the question. $\endgroup$
    – AXensen
    Commented Mar 25, 2023 at 16:17
  • $\begingroup$ I'm sorry for the bad formulation. I've fixed it. $\endgroup$
    – Kubrik
    Commented Mar 25, 2023 at 16:38

1 Answer 1


There are a few ways to approach this. You might also want to look at the "finite square well" quantum solution to get a better sense of the qualitative behavior of this system.

1-Both of these potentials are analytically solvable. $V_1$ is going to be really hard, and $V_2$ is going to be pretty hard. The solution to the Schrodinger equation in a linear potential is the Airy functions, and you probably know the solution in constant or quadratic potentials. You just need to find a bunch of coefficients to make this piecewise wavefunction continuous and differentiable. I don't really recommend it, it won't be fun. If only a small number of energy states fit in your well this is the only good way to find the energy levels.

2-Qualtitative statements about what will happen in different limits. First off there will always be finitely many bound states. You can use the "infinite harmonic oscillator/|x|" solution to approximately find how many bound states there are. So by solving $a^2=(n+1/2)\hbar\omega$ for $n$ you find roughly how many bound states there are (at least for the harmonic oscillator).

For low lying states you can use perturbation theory to find a good approximation of their new energy. So the nominal potential is just $V_0=x^2-a^2$, and the perturbing Hamiltonian is $$ V_1=\begin{cases} 0\text{ if }|x|<a\\ -x^2+a^2\text{ if }|x|>a \end{cases} $$ So to find the energy perturbation, you'd calculate $\int_x \psi_iV_1\psi_i^*$ ($\psi_i$ is the $i$th energy level wavefunction of the unperturbed harmonic oscillator). This will be a negative number since $V_1$ is a negative perturbation (the energy levels will be slightly shifted to be more negative). The approximation becomes poor at higher energy states, but at that point you can return to using the WKB method.

The perturbation will also be exponentially small in $-E$ (the energy E in your system stars at $-a^2$ and goes up to $0$), because the wavefunction starts exponentially decreasing at the classical turning points. So the wavefunction is exponentially suppressed where the perturbing potential is nonzero. So only states with energy closer to zero will be significantly affected, and the closer they are to zero the more they are affected, so you will get more energy levels closer together near zero. Of course beyond zero you will have a continuum of unbound states.

  • 1
    $\begingroup$ Nice answer! Thanks! Is it reasonable to say, that the approximation with the harmonic oscillator eigenenergies for the lower states is better than for the higher ones? $\endgroup$
    – Kubrik
    Commented Mar 26, 2023 at 12:44
  • $\begingroup$ Yes that's correct. $\endgroup$
    – AXensen
    Commented Mar 26, 2023 at 12:46

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