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Consider a Schrodinger operator $H=H_0+V$, where $H_0$ is a $L_2$ realization of the negative Laplace operator $-\Delta$ with homogeneous Neumann boundary condition on a bounded, smooth domain $\Omega\subset\mathbb{R}^n$ and $V$ is a smooth real-valued function on $\overline{\Omega}$. I am interested in the following

Problem

Find nontrivial necessary and/or sufficient conditions for $V$ under which $H$ is nonnegative i.e.

\begin{align} \int_{\Omega}|\nabla u|^2+Vu^2\geq0 \end{align} for every smooth function $u$ satisfying the Neumann boundary condition.

It is known that the operator $H$ is self-adjoint and has a compact resolvent, hence by the appropriate version of the spectral theorem the spectrum of $H$ is real and discrete and the above inequality is equivalent to the claim that the smallest eigenvalue of $H$ is nonnegative.

A trivial sufficient condition is that $V$ is nonnegative. A trivial necessary condition is that $\int_{\Omega}V\geq 0$ or more generally that \begin{align*} \int_{\Omega}(V+\lambda)\phi^2\geq0 \end{align*} for any $\lambda$ - eigenvalue of $H_0$ and $\phi$ - corresponding eigenfunction.

What happens when there exists $x_0\in\Omega$ such that $V(x_0)<0$? Can $H$ be still nonnegative? I will appreciate any comments/references even for the one-dimensional case $\Omega=(0,1)$.

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    $\begingroup$ I don't think this can have simple neat final answers. As for the dependence on the dimension, maybe my answer here is of interest to you: mathoverflow.net/questions/180846/… $\endgroup$ – Christian Remling Feb 27 '18 at 17:34
  • $\begingroup$ Thank You for the comment. Does this result hold also for a bounded domain? What happens when $V$ changes its sign? $\endgroup$ – Marcin Malogrosz Mar 1 '18 at 21:37
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    $\begingroup$ I think the term $Vu$ in your definition of non-negativity of $H$ should actually be $Vu^2$, shouldn't it? $\endgroup$ – Jochen Glueck Mar 2 '18 at 16:03
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    $\begingroup$ About your last question: there always exist potentials which are negative somewhere, but the associated Schrödinger operator is non-negative. Indeed, start from a function $V$, which is non-negative, vanishes somewhere, but is not identically equal to $0$. Then the smallest eigenvalue is strictly positive, because the operator has no kernel, so we can subtract this eigenvalue to get a silly example. $\endgroup$ – Mateusz Wasilewski Mar 2 '18 at 20:27

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