Let $-\frac{\mathrm{d}^2}{\mathrm{d} x^2} + V(x)$ be an one-dimensional Schrödinger operator for a given potential $V(x)$. Is it possible to know if such operator has a discrete spectrum (or not) without solving the eigenvalue problem? Thanks in advance.
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4$\begingroup$ you could start here: ma.utexas.edu/mp_arc/c/08/08-191.pdf $\endgroup$– Carlo BeenakkerCommented Oct 13, 2017 at 10:24
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$\begingroup$ A sufficient condition is for the Schrodinger operator to be compactly resolved. $\endgroup$– NealCommented Oct 13, 2017 at 11:29
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$\begingroup$ The grandmaster on this topic is probably Barry Simon. Rather than me passing on references it would probably be easier for you to look him up yourself, e.g., on mathscinet. $\endgroup$– saoneCommented Oct 13, 2017 at 13:41
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3$\begingroup$ There is a characterization of when the spectrum is purely discrete (quick version: exactly when $V\to\infty$ in a suitable sense). See my answer here: mathoverflow.net/questions/278568/… $\endgroup$– Christian RemlingCommented Oct 13, 2017 at 16:09
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1$\begingroup$ On the other hand, if you are asking if it's possible to detect if the operator has some discrete spectrum, then there's a lot on that too but no very general complete answer (there can't be one because existence of eigenvalues may depend on small details). $\endgroup$– Christian RemlingCommented Oct 13, 2017 at 16:12
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