# Can energy levels rise faster than $n^2$?

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?

• Are you restricting attention to 1D potentials? Jun 23 at 12:45
• Not necessarily. Jun 23 at 13:07
• @MariusLadegårdMeyer, I think OP means the principal quantum number "n" Jun 23 at 13:31
• possibly interesting: en.wikipedia.org/wiki/Weyl_law Jun 24 at 14:15

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 $$\sqrt{n}$$.

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.

• The WKB result from @Qmechanic would seem to indicate that there might be an impediment to realizing this Hamiltonian in the form $P^2 + V$. Jun 24 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 $V$. Jun 24 at 22:09