How limiting a potential affects the energy eigenvalues?

Question

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} $

Or:

$ 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.

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.