# A strange quantum potential: $V(x) = \frac{x^2}{5}+\mu \left(\left\lfloor x+\frac{1}{2}\right\rfloor \right).$

So I have a strange quantum potential I have been playing with:

$$V(x) = \frac{x^2}{5}+\mu \left(\left\lfloor x+\frac{1}{2}\right\rfloor \right).$$

where $\mu$ is the Möbius function. This is what it looks like.

I wanted to see what the bound states looked like, but $V$ is discontinuous, so I changed it into the following:

$$\frac{x^2}{5}+\sum _{n=-M}^M\mu(n)\left[\tanh\left(b\left(x-\frac{1}{2}-n+1)\right)\right)-\tanh\left(b\left(x+\frac{1}{2}-n-1\right)\right)\right]$$

This $\tanh$ method is a common way to approximate step functions in numerical quantum mechanics. The limit of the above expression as $M$ and $b$ go to infinity is $V$.

I picked a reasonably high value for $b$ and took $M$ out to $10$. The bound states ended up looking like this: (LHS is $\psi_n$, RHS is $|\psi_n|$ overlayed on $V$)

and so on. I don't have anything much smarter to say about these than that they are pretty weird looking.

Can anything rigorous be said about the eigenfunctions of this potential?

By this, I mean the potential in the limit as $M,b\to\infty$.

Mainly I am wondering whether including higher $M$ would have given me more bound states between the above energies. The energies seem to be at very strange intervals, like there could perhaps should have been more between them. In particular, if I were to take $M$ out to infinity, would I get a dense spectrum of energies?

-
Where does that come from? Just curious... –  vonbrand Feb 17 '13 at 23:44

Now as for taking $M\to\infty$. You haven't yet reached any high enough states so that the wavefunction would be localized far from center to hit the hills/wells far from center. You'll only get visible influence of M only on highly excited states (with threshold energy depending on M).
Also, you'll never get a continous spectrum of energies because your basic part of potential is a harmonic oscillator $\frac{x^2}{5}$, which has purely discrete spectrum (because it has finite width at arbitrary height, so the particle doesn't go to infinity) and addition $\mu$ is bounded.