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For a 1D particle in a box, energy levels are exactly proportional to $n^2$.

For the harmonic oscillator, $E_n\sim n$. And for a particle in an $|x|^\alpha$ potential, the energies are $\sim n^\beta$ with $\beta <2$ from the WKB approximation.

I also vaguely recall that the eigenvalues for a regular Sturm-Liouville problem are $\sim n^2$.

All of this suggests that the eigenvalues cannot rise faster than $n^2$. Is my intuition correct here?

Brendan Darrer
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FusRoDah
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2 Answers2

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  1. In this answer we generalize OP's 1D semiclassical WKB estimate for a power law potential $\Phi(x) \propto |x|^{\alpha}$ to an arbitrary potential $\Phi(x)$. If $\ell(V)$ denotes the accessible length function at potential energy-level $V$, then the number of bound states $N(E)$ below energy-level $E$ is $$ N(E) ~\approx ~\frac{\sqrt{2m}}{h} \int_{V_0}^E \frac{\ell(V)~dV}{\sqrt{E-V}},\tag{2} $$ cf. my Phys.SE answer here. Since $V\mapsto \ell(V)$ is a monotonically (weakly) increasing non-negative function, we can pick a potential energy-level $V_1>V_0$ such that $$ N(E) ~\gtrsim~\frac{\sqrt{2m}}{h} \int_{V_1}^E \frac{\ell(V_1)~dV}{\sqrt{E-V}} ~=~\frac{2\sqrt{2m}}{h} \ell(V_1)\sqrt{E-V_1},$$ i.e. the number of bound states $N(E)$ grows at least as fast as $\sqrt{|E|}$.

    Or equivalently: $E_n$ grows no faster than $n^2$.

  2. On the other hand, from regular Sturm-Liouville theory, it is known that the resolvent is a compact operator, which implies that $$\sum_{n\in\mathbb{N}}|E_n|^{-2}~<~\infty$$ is convergent, and which shows that $E_n$ must grow faster than $\sqrt{n}$.

Qmechanic
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It sort of depends on what restrictions you impose on $H$.

If we let $|n\rangle$ be the eigenstates of the harmonic oscillator, then you can surely define $$ H:=\sum_{n\ge0}E_n|n\rangle\langle n| $$ where $E_n$ is any function you want, say $E_n=n^3$. This Hamiltonian has eigenvalues $E_n$ and eigenfunctions $\psi_n(x)\sim H_n(x)e^{-x^2}$, which are perfectly well-defined wave-functions.

You may even be able to write the above as $P^2+V(x)$ for some nasty $V(x)$, although I haven't tried.

So unless you specify which precise type of Hamiltonian you want, the answer is that you can have any spectrum you want.

AccidentalFourierTransform
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  • The WKB result from @Qmechanic would seem to indicate that there might be an impediment to realizing this Hamiltonian in the form $P^2 + V$. – Michael Seifert Jun 24 '22 at 16:58
  • @MichaelSeifert QM's answer applies to sufficiently nice $V$. Although I don't have any particularly convincing argument, I suspect that it is still possible with a sufficiently nasty $V$. – AccidentalFourierTransform Jun 24 '22 at 22:09