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Consider a linear operator $H\colon L^2(\mathbb{R}^3)\to L^2(\mathbb{R}^3)$ given by $$H(\psi)(x):=-\Delta\psi(x)+V(x)\cdot \psi(x),$$ where $V\colon \mathbb{R}^3\to \mathbb{R}$ is a continuous (or smooth) non-negative function. If necessary, one may assume $\lim_{|x|\to \infty}V(x)=+\infty$.

Are there known estimates from below of the minimal eigenvalue of $H$ in terms of $V$?

(Of course, the trivial estimate is by 0.)

I am not an expert and may not aware even of the most standard results.

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  • $\begingroup$ Such an estimate is equivalent to Poincaré's inequality for $-\Delta + V(x)$. I believe quite a lot is known about this problem, but unfortunately I do not know the references. I would start by looking into Robert Seiringer's notes. $\endgroup$ – Mateusz Kwaśnicki Jan 7 at 12:59
  • $\begingroup$ In this generality, of course one can't say more than the trivial statement you pointed out ($V$ could be zero on a large set). I suspect there has to be a lot of work on more specific situations. $\endgroup$ – Christian Remling Jan 7 at 20:26
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There is a generalization of the variational method already shown in Linus Pauling's "Introduction to Quantum Mechanics" that isn't mentioned much anymore (possibly because of limited practical use):

Defining both $E = \langle H\rangle $ and $D = \langle H^2 \rangle $ in any trial state, there exists an eigenvalue $W$ of $H$ satisfying either $E+\sqrt{D-E^2 } \geq W \geq E$ or $E \geq W \geq E-\sqrt{D-E^2 } $. If $W$ is the minimal eigenvalue of $H$, the second alternative applies, giving a bound on $W$ from both sides. If there is enough symmetry, it's often not so hard to construct a trial state close enough to the ground state such that $W$ is, in fact, the minimal eigenvalue (e.g., for spherically symmetric problems, look for s-waves and no nodes in the radial wave function).

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