When Schrodinger operator has discrete spectrum?

Question

On $L^{2}(\mathbb{R})$ we have a linear operator $S=-\frac{d^{2}}{dx^{2}}+u(x)$. As I understand for some choices of potential $u$ (like harmonic oscillator $u(x)=\frac{\omega x^2}{2}$) Schrodinger operator will have only a countable set of $\lambda_{n} \in \mathbb{R}, f_{n} \in L^{2}(\mathbb{R}), (f_{n} \neq 0)$ such that $S f_{n}=\lambda_{n} f_{n} \\$.
My question is for what other choices (if any) of a linear subspace $V$ (possibly without a non-trivial norm) of the space of set-theoretical functions $Map(\mathbb{R},\mathbb{R})$ Schrodinger operator will have only countably many real eigenvalues (say $u$ has finitely many poles on $\mathbb{R}$ and outside poles is infinitely differentiable. Probably a satisfactory answer can be achieved with weaker regularity assumptions on the potential)?

Answer 1

This is a nice result that holds in $\mathbb{R}^n$ for $n\ge 1$ and strikes a nice balance between generality and applicability. The potential $V$ is real, and $H=-\Delta + V$. The requirement on $V$ is that there is finite area in the given region (or finite volume in $\mathbb{R}^n$ for $n > 1$.)