Semiboundedness of Schrödinger operators Is the Schrödinger operator $$H~=~-\Delta+V$$
bounded from below? For example, I would like to analyze the case where $V\in L^{2}_\mathrm{loc}(\mathbb{R}^{n})$ is a locally square integrable function, but what about other situations, are there general results?
 A: The operator $H$ is bounded from below if all of its inner products $\langle\psi,H\psi\rangle$ are bounded from below. Thus, we can easily compute
$$\langle\psi,H\psi\rangle=\int_{\mathbb{R}^n}\mathrm{d}\textbf{x}\,\psi^*(\textbf{x})\left(-\Delta+V(\textbf{x})\right)\psi(\textbf{x}).$$
Integrating this expression by parts gives
$$\langle\psi,H\psi\rangle=\int_{\mathbb{R}^n}\mathrm{d}\textbf{x}\,\left(\left|\boldsymbol{\nabla}\psi\right|^2+\psi^*(\textbf{x})V(\textbf{x})\psi(\textbf{x})\right)=\langle\boldsymbol{\nabla}\psi,\boldsymbol{\nabla}\psi\rangle+\langle\psi,V\psi\rangle\geq\langle\psi,V\psi\rangle.$$
Thus, so long as $V$ is bounded from below, then $H$ is bounded from below. This, of course, is not a two way implication. There are cases where $V$ is unbounded from below but $H$ is still bounded from below.
I hope this helps!
A: A locally square integrable potential function $V\in {\cal L}^{2}_\mathrm{loc}(\mathbb{R}^{n})$ does not guarantee that $H$ is bounded from below. Consider e.g. a linear potential $$V(\vec{r}) ~=~ -\vec{E}\cdot \vec{r}, \qquad \vec{E}~\neq~\vec{0}.$$ 
