Estimate the smallest eigenvalue of a Schrodinger operator

Question

There are several results on the estimate of the number of negative eigenvalues of a Schrodinger operator, see a recent paper of Grigor'yan-Nadirashvili-Sire and references therein. I wonder how to estimate the smallest eigenvalue $\lambda_1$ of a Schrodinger operator $-\Delta+h$ on a closed Riemannian manifold $(M^n,g)$? I am interested in the case that $h$ has some positivity, for example $h>0$ somewhere on $M$ or $\int_M hdv_g>0$, in particular under what condition $\lambda_1\geq0$. Thank you very much.

Answer 1

This question is actually quite subtle (at least I think so). The lowest eigenvalue of the Schrodinger operator $-\Delta_g+h$ on a closed Riemannian manifold, where the potential $h$ is as you described, can actually be positive, negative or zero, and the sign depends upon the "size" of the potential.

I'll say more about this in a moment, but first I'd like to remark that In all other cases of potential functions, i.e. when the potential has non-positive average or when it has positive average and is non-negative, the sign of the lowest eigenvalue of the associated Schrodinger operator is easy to determine. Also, while it is only in the case where the potential has negative average, and hence the sign of the lowest eigenvalue is easy to determine, that the result of Grigor'yan-Nadirashvili-Sire is applicable (note, there is a sign change between conventions), their work bounding the number of negative eigenvalues from below is highly non-trivial.

In a recent work joint with M. Dabkowski arxiv:1508.02755, we study the lowest eigenvalue question in the situation you bring up. More specifically, we show the following. On a closed Riemannian manifold $(M,g)$, let $h$ be a potential function that changes sign on the manifold and satisfies $\int_M hdV_g>0$ (the conditions you described). Then, consider the $1$-parameter family of Schrodinger operators \begin{align} -\Delta_g+t\cdot h, \end{align} where $0<t<\infty$. Note that, since $t>0$, the scaled potential $t\cdot h$ satisfies the conditions of the original potential $h$. This scaling of the potential can likewise be understood as the $1$-parameter family of Schrodinger operators obtained by fixing the potential $h$ and scaling the manifold since the Laplace-Beltrami operator of a scaled metric $tg$, for some $t>0$, is just $\Delta_{tg}=\frac{1}{t}\Delta_g$, and the corresponding Schrodinger operator $-\frac{1}{t}\Delta_g+h$.

For the $1$-parameter family of Schrodinger operators $-\Delta_g+t\cdot h$, it turns out that there exists a unique $t=t^*>0$ for which the lowest eigenvalue is zero, and that the interval $(0,\infty)$ decomposes as \begin{align} (0,\infty)=I^+\sqcup\{t^*\}\sqcup I^-, \end{align} where $I^+=(0,t^*)$ and $I^-=(t^*,\infty)$, and the lowest eigenvalue of $-\Delta_g+t\cdot h$ is \begin{align} &\bullet\text{ positive for $t\in I^+$}\\ &\bullet\text{ zero for $t=t^*$}\\ &\bullet\text{ negative for $t\in I^-$.} \end{align}

Furthermore, the lowest eigenvalue of $-\Delta_g+t\cdot h$ will be strictly positive for all $t$ where \begin{align} 0<t\leq\frac{\int_M hdV_g}{P||h||_{\infty}\big(\int_M hdV_g+4Vol(M)||h||_{\infty}\big)}, \end{align} where $P$ is the Poincare constant of the manifold. In other words, what we have here is a lower bound on the size of $I^+$.