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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)?

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    in general, in an unbounded potential (from above) $u(\pm\infty)=+\infty$ there exist only discrete boundstates. As soon as this condition is violated there exist also so called scattering states which have a contious spectrum. – 2017-01-11
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    @tired, thanks for the criterion. I just wonder if bound in "boundstates" can be meaningfully replaced by another condition. – 2017-01-11
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    Boundstates are usually exponentially localized wihich is in sharp constrast to scattering states which are (in principle) infinitly extended – 2017-01-11
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    But you see scattering states usually come in continuum (say for free particle there is one for every frequency) – 2017-01-11

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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 $n > 1$.) enter image description here

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    do you have a reference for that? – 2017-01-12
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    @AknazarKazhymurat The reference is [this paper by Barry Simon](https://www.ma.utexas.edu/mp_arc/c/08/08-191.pdf). – 2017-01-12
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    @s.harp : That's the paper. Thanks for posting a link. I couldn't find one. – 2017-01-12
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    @TrialAndError it was you that showed me that paper in [this question](http://math.stackexchange.com/questions/1778441/spectrum-of-laplace-operator-with-potential-acting-on-l2-mathbb-r-is-discre) :) – 2017-01-12
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    @s.harp : And so I did. Thanks for the laugh. :) – 2017-01-12