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I am trying to form intuition for the following `well-known' facts about spectrum of unbounded operators (Schrodinger/wave etc.) $L$ on $\mathbb{R}^n$.

Let $\lambda\in\mathbb{R}$ satisfy $Lf=\lambda f$, for some function $f$.

Roughly speaking: we call "eigenvalues" the things that give us $f$ in some $L_2$ space. If the eigenvalue equation is satisfied but the $f$ is not in $L_2$ (roughly, it doesn't decay at infinity), then we don't call $\lambda$ an eigenvalue, but consider it part of essential (or continuous ?) spectrum. Example of latter is the usual Laplacian.

Is there a resource where I can understand how generic is this "equivalence":

Existence of "bound state" <-> Existence of eigenvalue

I was motivated to ask this after reading the following : https://en.m.wikipedia.org/wiki/Bound_state_in_the_continuum

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    $\begingroup$ Not exactly sure what you are after, but you could simply look at a multiplication operator on $L^2$, $f\mapsto gf$ for some function $g$. By trying different $g$'s you will see by yourself the difference between an eigenvalue and a mere point in the spectrum. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Jul 12, 2021 at 16:32

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My intuition for the bound state/discrete spectrum relationship is that the discrete spectrum $\lambda_1,\lambda_2,\ldots$ of a Hermitian operator allows the construction of a set of eigenfunctions $f_1,f_2,\ldots$ which is orthonormal, $$\int_{\mathbb{R}}\bar{f}_n(x)f_m(x)dx=\delta_{nm},$$ and hence square integrable --- which is what we mean by a "bound state".

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  • $\begingroup$ In other words, the equivalence the OP asked about is extremely generic.... because it is by definition. $\endgroup$
    – Willie Wong
    Commented Jul 13, 2021 at 3:53
  • $\begingroup$ @williewong : does this also rule out bound states for values of $\lambda$ in the continous spectrum? $\endgroup$
    – mathamphetamine
    Commented Jul 13, 2021 at 4:52
  • $\begingroup$ See this: en.m.wikipedia.org/wiki/Bound_state_in_the_continuum $\endgroup$
    – mathamphetamine
    Commented Jul 13, 2021 at 4:57
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    $\begingroup$ there is no need to assume that $\lambda_n$ is distinct from a $\lambda$ in the continuous spectrum -- states which are localized in space at an energy that falls in the continuum band do exist. $\endgroup$
    – Carlo Beenakker
    Commented Jul 13, 2021 at 5:58

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