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 $\langle \psi_n^h,H_h^\prime\psi_n^h\rangle\geq 0$.

**Update**

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 $f(\pm\pi)=0$.