Spectrum of Laplace operator with potential acting on $L^2(\mathbb R)$ is discrete

Asked May 09 '16 at 18:14

Active May 09 '16 at 18:19

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Consider an operator $H=-\Delta +U(x)$ on $L^2(\mathbb R)$ for a function $U(x): \mathbb R \to \mathbb R$ that tends to $+\infty$ as $|x|$ grows. These kinds of operators appear all over non-relativistic quantum mechanics as Hamiltonians.

A statement that I have read is that such an operator has a discrete spectrum, and it was presented as being a standard result.

How can this be proven?

(I'm sorry if this is a lazy question, it seems to be common knowledge so one would expect proofs to abound like sand at the beach, but I couldn't find them on this website.)

edited May 09 '16 at 18:19

asked May 09 '16 at 18:14

s.harp

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This paper of Barry Simon is a good one : https://www.ma.utexas.edu/mp_arc/c/08/08-191.pdf . – Disintegrating By Parts May 09 '16 at 19:29
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For a more classical source, you could consider Titchmarsh, E. C. (1962), Eigenfunction expansions associated with second order differential equations, Vol. I, second edition, Oxford University Press, ISBN 0-608-08254-6 . You might also be helped by the information in (https://en.wikipedia.org/wiki/Spectral_theory_of_ordinary_differential_equations) – Frits Veerman May 10 '16 at 09:11

Spectrum of Laplace operator with potential acting on $L^2(\mathbb R)$ is discrete

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