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Consider the following linear (Hamiltonian) operator on functions on the real line $\mathbb{R}$ $$H\psi(x)=-\frac{d^2}{dx^2}\psi(x)+V(x)\psi(x).$$

Assume that $V$ is a smooth function and $V(x)\to +\infty$ as $|x|\to +\infty$.

Is it true that the spectrum of $H$ is discrete? In my case $V$ is a polynomial of fourth degree. If the answer is negative in general, is it positive in my case?

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    $\begingroup$ Sure it is (the resolvent of your operator is compact). Also, you can argue along the lines of that questions mathoverflow.net/questions/278568/… $\endgroup$
    – Sascha
    Commented May 29, 2018 at 10:46
  • $\begingroup$ @Sascha: Thanks. It seems that an answer there contains the answer to my question too. $\endgroup$
    – asv
    Commented May 29, 2018 at 10:51
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    $\begingroup$ Possible duplicate of Harmonic oscillator discrete spectrum $\endgroup$
    – Christian Remling
    Commented May 29, 2018 at 21:07

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