Let us consider the Schrödinger operator $$ H_hf(x)=-\frac{d^2}{dx^2}f(x)+h(h\sin^2(x)-\cos(x))f(x) $$ on $L^2[-\pi,\pi]$ with Neumann boundary conditions $f^\prime(\pm\pi)=0$. Here, $h\geq 0$ is a parameter.

It is easy to see that (up to normalization) $$ \psi_0^h(x)=e^{h\cos(x)} $$ is an eigenfunction for the eigenvalue $0$.

Running some numerics suggests that the mapping $h\mapsto\lambda_n(H_h)$ where $\lambda_n$ denotes the $n^\text{th}$ eigenvalue of $H_h$, $n=0,1,2,\dots$, is monotone non-decreasing. The question is how to prove this.

My approach was the following: By a famous result, $$ \frac{d}{dh}\lambda_n(H_h)=\langle \psi_n^h,H_h^\prime\psi_n^h\rangle $$ where $\psi_n^h$ is the $n^\text{th}$ eigenfuntion of $H_h$ and $H_h^\prime f(x)=(2h\sin^2(x)-\cos(x))f(x)$. However, I am struggling to prove that $\langle \psi_n^h,H_h^\prime\psi_n^h\rangle\geq 0$.


Maybe we should first focus on the second eigenvalue $\lambda_1(H_h)$. By general theory, $\psi_1^h$ is a differentiable odd function. Furthermore, $\psi_1^h$ has only one zero (at $x=0$). To obtain $\frac{d}{dh}\lambda_n(H_h)\geq 0$, it would thus be sufficient to prove $$ \max_{x\in[0,x_0]}|\psi_1^h(x)|\leq\min_{x\in[x_0,\pi]}|\psi_1^h(x)|, $$ where $x_0=2\arctan\left(\sqrt{\sqrt{16h^2 + 1} - 4 h}\right)$ is the zero of $2h\sin^2(x)-\cos(x)$ in $[0,\pi]$, since $$ \int_{0}^\pi 2h\sin^2(x)-\cos(x)\,dx=h\pi\geq 0. $$

Update 2

This is based on the comment by Carlo Beenakker. We transform the initial Neumann problem into a Dirichlet problem. A computation gives that the non-zero spectrum of $H_h$ coincides with the spectrum of \begin{align} K_hf(x)&=-f''(x)+h(h\sin^2(x)+\cos(x))f(x)\\&=-f''(x)+\left(h\cos(x)-\frac{h^2}{2}\cos(2x)+\frac{h^2}{2}\right)f(x) \end{align} acting on $L^2[-\pi,\pi]$ with Dirichlet boundary conditions $f(\pm\pi)=0$.


1 Answer 1


As requested in the comment, let me explain what I could extract from the literature on this problem. (This is not the solution asked for in the OP, but what I have would be too long for a comment.)

The potential in the OP is known as the "Razavy potential" or "double cosine potential". A recent study is Exact Solutions of the Razavy Cosine Type Potential (2018). The Schrödinger equation is $$-\psi''(x)+V(x)\psi(x)=E\psi(x),\;\;V(x)=\tfrac{1}{4}\xi^2\sin^2 x-(a+1)\xi\cos x,$$ on $-\pi\leq x\leq\pi$ with $\psi(\pm\pi)=0$. The differential equation in the OP is for $\xi=2h$ and $a=-3/2$. (I note that the cited paper assumes $\xi,a>0$, but that does not seem to be an essential condition on the solution.)

The substitution $\psi(x)=\exp(\tfrac{1}{2}\xi\cos x)\phi(x)$ and the change of variables $z=\cos^2(x/2)$ produces a confluent Heun differential equation with solution given by the Heun function: $$\psi(x)=\exp(\tfrac{1}{2}\xi\cos x)H\bigl(2\xi,-1/2,-1/2,-(2a+1)\xi,2(a+1)\xi+3/8-E;\cos^2(x/2)\bigr).$$ The energy $E$ should then be obtained from the boundary condition $x=\pm\pi$, so at the origin for the Heun function, but the cited paper does not succeed in obtaining a closed-form solution.

  • 1
    $\begingroup$ I contend myself with the fact that there is no explicit spectral decomposition. Thank you for your effort. $\endgroup$
    – julian
    Mar 28, 2019 at 14:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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