Is the eigenvalue decomposition of the Sturm-Liouville operator $$ Lf(x)=-f''(x)+h\sin(x)f'(x),\quad h>0, $$ with Neumann boundary conditions $f'(-\pi)=f'(\pi)=0$ on the Hilbert space $L^2([-\pi,\pi],\mu_h)$ known? Here $d\mu_h(x)=Z^{-1}e^{h\cos x}dx$ and $Z>0$ is chosen such that $\mu_h$ is a probability measure. I suspect it involves Bessel functions.

N.B. The Sturm-Liouville problem is unitarily equivalent to the Schrödinger operator $$ Hf(x)=-f''(x)+\frac{h}{4}(h\sin^2(x)-2\cos(x))f(x) $$ on $L^2([-\pi,\pi])$.

Edit: By the hint of Sascha, we transform $H$ into a Schrödinger operator with Whittaker-Hill potential. Set $y=x/2$, then $$ Hf(y)=-\frac14 f''(y)+\left(\frac{h^2}{8}-\frac{h^2}{8}\cos(4y)-\frac{h}{2}\cos(2y)\right)f(y). $$ Hence, $4H-h^2/2$ is a Schrödinger operator with Whittaker-Hill potential with parameters $\alpha=h/2$, $s=1$ (in the convention of the paper by Hemery and Veselov).

Is anything known about the spectral gap of this operator (as noted by Hemery and Veselov, the ground state eigenvalue is (not very surprisingly) $-h^2/2$)?

  • $\begingroup$ The operator $L$ is not self-adjoint, right? Could you thus specify what exactly you mean by "eigenvalue decomposition"? $\endgroup$ Feb 27, 2019 at 19:07
  • $\begingroup$ My bad. I missed some crucial information. I edited the question accordingly. $\endgroup$
    – julian
    Feb 27, 2019 at 19:16
  • $\begingroup$ I'd be surprised if you could find the eigenvalues explicitly here, but I might be wrong of course. $\endgroup$ Feb 27, 2019 at 19:42
  • $\begingroup$ So what do precise information on the spectral gap do you need? $\endgroup$
    – Sascha
    Feb 28, 2019 at 14:51
  • $\begingroup$ I need the second lowest eigenvalue. $\endgroup$
    – julian
    Feb 28, 2019 at 14:52

1 Answer 1


I think this is one of the quasi exactly solvable potentials for the Schrödinger equation, see this paper.

To say a bit more: For particular choices of $h$ certain eigenfunctions are explicit but not all of them.

  • $\begingroup$ Thank you for this hint. It works out in fact (see the edited question). $\endgroup$
    – julian
    Feb 28, 2019 at 14:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.