Consider the Schrödinger operator $H=-\Delta+|x|^a$ on $\mathbb{R}$, where $a>0$. Since the potential is growing at $\infty$, we have compact resolvent thus the eigenvalues are discrete and tend to $\infty$. We denote the eigenvalues by $\{\lambda_n\}_{n=1}^{\infty}$. Let $\{\phi_n\}$ be the correspoding $L^2$-normalized eigenfunctions. It's trivial that eigenfunctions corresponding to different eigenvalues are orthogonal, i.e., $$ \int_{R}\phi_n\phi_m=0,\quad\text{whenever $m\ne n$.} $$ But if we only consider a subregion of $\mathbb{R}$, for instance, let $$ I_{mn}(\Omega):=\int_{\Omega}\phi_n\phi_m,\quad\text{where $\Omega\subset \mathbb{R}$,} $$
then this quantity may not vanish. Actually, I'm asking the opposite. How large can $I(\Omega)$ be? To make it more precise, let $\Omega=(0, \infty)$ be the half line. Note that all eigenfunctions are either odd or even, thus one has $\int_0^{\infty}\phi_n^2\,dx=\frac12$ for all $n$. Thus by Cauchy inequalty one has the upper bound \begin{align}\label{eq} \left\lvert\int_0^{\infty}\phi_n\phi_m\right\rvert\leq \frac12. \end{align} Is this upper bound the sup of $I_{mn}((0, \infty))$? or can we find a sequence of eigenfunctions such that $$I_{mn}((0, \infty))\rightarrow \frac12?$$ Moreover, is there some physical explanation to this? I think it's related to something which is opposite to orthogonality. Note that the equality holds in Cauchy inequality iff $\phi_n$ and $\phi_m$ are linearly dependent, which seems "impossible" at first sight. I'm trying to prove this when $a=\frac12$, since the spectrum of the unharmonic oscillator grows like $(\frac{3\pi}{4}n)^{\frac{2}{3}}$, in this case, the spectral gap tends to zero when $n\rightarrow \infty$, then intuitively, the eigenfunctions of two consecutive eigenvalues may be almost linearly dependent in this special subregion.
