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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.

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  • $\begingroup$ It's clear that the integral cannot be equal to $1/2$ since even/odd eigenfunctions are also eigenfunctions of the half-line problem with Neumann/Dirichlet boundary conditions, so every even eigenfunction is orthogonal in $L^2(0,\infty)$ to every other even eigenfunction and the same for odd. Finally, an even and an odd functions are obviously not multiples of one another. $\endgroup$ Commented Dec 12, 2022 at 18:40
  • $\begingroup$ I don't see any obvious reason why (in general, I don't know about this operator) the integral could not be close to $1/2$, though. $\endgroup$ Commented Dec 12, 2022 at 18:40
  • $\begingroup$ To slightly rephrase my first comment (and your question), you are asking if eigenfunctions of two different half-line operators (Dirichlet, Neumann) could be almost collinear, and I don't see why not. $\endgroup$ Commented Dec 12, 2022 at 18:43
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    $\begingroup$ @ChristianRemling, thanks for your comment, in the case $a=\frac12$, the eigenfunctions can be explicitly expressed by translations of Airy function or their derivatives, see e.g.eudml.org/doc/210776 (Theorem 2.6), By a direct the compution, I got that $|\int_{0}^{\infty}\phi_n\phi_{n+1}dx|\rightarrow \frac{1}{\pi}$. $\endgroup$
    – Tomas
    Commented Dec 13, 2022 at 12:26
  • $\begingroup$ In the above computation, the Airy kernel is involved (which is a little bit unexpected), and I also use the fact that $\frac{\pi}{2}\lambda_{n}^{-1/2}\ge\lambda_{n+1}-\lambda_n\ge\frac{\pi}{2}\lambda_{n+1}^{-1/2}$, the factor $\frac{1}{\pi}$ comes from this. $\endgroup$
    – Tomas
    Commented Dec 13, 2022 at 12:39

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