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?
 A: *

*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 $n^2$.


*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 $\sqrt{n}$.
A: 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.
