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Let us act intentionally stupid and assume we do not know that we can solve for the spectrum of the harmonic oscillator

$$-\frac{d^2}{dx^2}+x^2$$

explicitly.

Is there an abstract argument why the spectrum of this operator is discrete and tends to infinity?

It almost looks like a result that would follow from the compactness of some operator involved, yet we are here on an unbounded domain $\mathbb{R}.$

So does it follow from spectral theory only that the spectrum is discrete?

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    $\begingroup$ Indeed, it follows from compactness of the resolvent of the Schrödinger operator. This is proved, if I remember correctly, in Kato's "Perturbation Theory for Linear Operators". $\endgroup$
    – José Figueroa-O'Farrill
    Commented Aug 12, 2017 at 9:16
  • $\begingroup$ Being stupid, what's the way you know in short ? And $T$ commutes with the Fourier transform so I would look at Hermite functions and things like $\exp(-x^2/2-2xt-t^2)$ @JoséFigueroa-O'Farrill $\endgroup$
    – reuns
    Commented Aug 12, 2017 at 11:21
  • $\begingroup$ is the algebraic argument using ladder operators to your liking? (see, for example, section 3.1.1 of these lecture notes ) $\endgroup$
    – Carlo Beenakker
    Commented Aug 12, 2017 at 12:23
  • $\begingroup$ I think semiclassical approach (also known as VKB approximation) can explain why the spectrum is discrete and tends to infinity. $\endgroup$
    – Nemo
    Commented Aug 12, 2017 at 13:02
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    $\begingroup$ The appropriate notion from semiclassics is called the Bohr-Sommerfeld rule. $\endgroup$
    – Igor Khavkine
    Commented Aug 12, 2017 at 14:34

2 Answers 2

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As already pointed out in the comments, $V\to\infty$ does imply discrete spectrum in general. In fact, this becomes an equivalence if a somewhat more general version of this condition is used:

Theorem: Suppose that $V(x)$, $x\ge 0$, is bounded below (and locally integrable, as usual). Then the spectrum of $-d^2/dx^2+V(x)$ on $L^2(0,\infty)$ (with arbitrary boundary condition at $x=0$) is purely discrete if and only if $\lim_{x\to\infty} \int_x^{x+d} V(t)\, dt = \infty$ for all $d>0$.

This is usually called the Molchanov criterion, you can try a search for this if you want more information.

I've formulated this for half line problems, but by the decomposition method $\sigma_{ess}(H)=\sigma_{ess}(H_-)\cup\sigma_{ess}(H_+)$ you also get a corresponding result about whole line problems.

Addendum: A few words on the proof perhaps, especially since "compact resolvent" has been mentioned a few times already as a possible method. This is an indeed an obvious approach and there's nothing wrong with it, but my personal preference here in one dimension would be to use oscillation theory. More precisely, $\sigma_{ess}=\emptyset$ is equivalent to the claim that any non-trivial solution $y$ of $-y''+Vy=Ey$ has only finitely many zeros for any $E\in\mathbb R$ (the number of zeros counts the number of eigenvalues below $E$, give or take one).

Now it's fairly straightforward to show, using Prufer variables, that this is indeed equivalent to $\lim\int_x^{x+d}V\, dt=\infty$.

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The physical idea is that $-\Delta+q$ for a sufficiently-growing "confining potential" $q$ on $\mathbb R^n$ should have compact resolvent, which then proves discreteness of its spectrum. In many explicit situations, one can prove this directly by imitating proofs of Rellich's compactness lemma(s).

Not an abstract version, but maybe non-trivial, is illustrated for automorphic forms in http://www.math.umn.edu/~garrett/m/v/simplest_afc_schrodinger.pdf

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