# Lower estimate of the minimal eigenvalue of a Hamiltonian

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.

• 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. – Mateusz Kwaśnicki Jan 7 at 12:59
• 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. – Christian Remling Jan 7 at 20:26

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).