TITLE: Lower estimate of the minimal eigenvalue of a Hamiltonian
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: 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).