In my blog post Why riemannium? , I introduced the following idea. The infinite potential well in quantum mechanics, the harmonic oscillator and the Kepler (hydrogen-like) problem have energy spectra, respectively, equal to

1) $$ E\sim n^2$$ 2) $$ E\sim n$$ 3) $$ E\sim \dfrac{1}{n^2}$$

Do you know quantum systems with general spectra/eigenvalues given by

$$ E(n;s)\sim n^{-s}$$

and energy splitting

$$ \Delta E(n,m;s)\sim \left( \dfrac{1}{n^s}-\dfrac{1}{m^s}\right)$$

for all $s\neq -2,-1,2$?


1 Answer 1


Here we will not address the full quantum mechanical problem, but only discuss the semi-classical limit $n \gg 1$, i.e. only the highly excited part of the energy spectrum far away from the ground state energy.

If we are in one dimension with a power law potential

$$\Phi(x)~\sim~|x|^{p}, \qquad p>-2, $$

for $|x|$ sufficiently large, then we can use the semi-classical method of this Phys.SE answer to estimate the classically accessible length as

$$ \ell(V)~\sim~V^{\frac{1}{p}}, $$

where $V$ is the available potential energy. The number of states $N(E)$ below energy-level $E$ then goes as

$$ N(E)~\sim~E^{\frac{1}{p}+\frac{1}{2}}, $$

and therefore the semi-classical discrete energies also obey a power law

$$ E_n ~\sim~n^{\frac{2p}{p+2}} \quad\text{for}\quad n ~\gg~ 1. $$

The values $p=-1$, $p=2$, and $p=\infty$ correspond to the (radial) hydrogen atom, the harmonic oscillator, and the infinite potential well, respectively.

  • 1
    $\begingroup$ For completeness: If the power $p<-2$, then there will only be a finite number of bound states, so our semi-classical analysis does not work in that case. $\endgroup$
    – Qmechanic
    Apr 13, 2013 at 18:15
  • $\begingroup$ That's cool! And I am wondering what kind of fully quantum (not semiclassical) "field/system/potential" could produce them too. Your answer is a great help Qmechanic. I am also very interested in physical and real/factual systems with that asymptotical spectrum... It is quite interesting for many reasons for my current "thoughts"... $\endgroup$
    – riemannium
    Apr 13, 2013 at 21:49
  • $\begingroup$ And one more question, what if your "p" is complex? Can the 1D Schrödinger equation be solved for that potential?$ (T+V)\Psi=E\Psi$? There T is the usual (free) kinetic energy operator in QM and $V\sim \vert \Phi\vert^p$ with $p\in \mathbb{C}$ $\endgroup$
    – riemannium
    Apr 13, 2013 at 21:54
  • 1
    $\begingroup$ In standard QM, the Hamiltonian $H$ should be a Hermitian operator, and hence the potential $\Phi$ (and therefore $p$) should be real. You might try to look into PT-symmetric QM. $\endgroup$
    – Qmechanic
    Apr 13, 2013 at 22:03
  • 1
    $\begingroup$ p=1 is used as a model of quarkonium. See arxiv.org/abs/hep-ph/0608103 $\endgroup$
    – user4552
    Apr 14, 2013 at 0:27

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