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Depending on the dimension and the symmetry and form of the potential, the energy eigenvalues of a quantum mechanical system have different functional forms. Eg. The particle in the 1D-box gives rise to $E_n \propto n^2$, the hydrogen atom $E_n \propto -\frac{1}{n^2}$ the harmonic oscillator $E_n \propto n + \frac{1}{2}$, and so on. My question now is, could there possibly exist a system with energy eigenvalues $E_{n,m}\propto f(n m)$, for example like $\propto -\frac{1}{nm}$ with $n,m\in \mathbb{N}$?

I suspect the way the Hamilton function is translated to QM operators might prohibit such a product formation, since a product of say momenta ($\vec{p}$) is translated into a successive action of momentum operators rather than the product of the action of two operators. Which in turn will result in a summation of terms depending on different quantum numbers.

The question arises from thinking about how prime numbers could occur in physics. Since say, there would be a system with $E_{n,m} = \hbar(n+1)(m+1)$ with $n,m\in\mathbb{N}^+$ it would mean exactly non-prime number multiples of $\hbar$ would be allowed energy levels. And the prime numbers would appear as gaps in its eigenvalue spectrum.

Raphael J.F. Berger
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Let $H_1$ be the Hamiltonian of a Harmonic oscillator, and let $m=\hbar=\omega=1$, that is, $$ H_1=\frac{1}{2}P_1^2+\frac{1}{2}X_1^2-\frac{1}{2} $$

Let $|n_1\rangle$ be the eigenvectors of $H_1$, i.e., $$ H_1|n_1\rangle=n_1|n_1\rangle $$

If we define $H=H_1 H_2$ with$^1$ $[H_1,H_2]=0$ we get multiplicative eigenvalues: $$ H|n_1,n_2\rangle=n_1n_2|n_1,n_2\rangle $$

$^1$ this is equivalent to $[X_1,X_2]=[P_1,P_2]=0$ and $[X_i,P_j]=i\delta_{ij}$, that is, the usual commutation relations.

AccidentalFourierTransform
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  • What would be a physical representation (i.e. what would it mean)? – Raphael J.F. Berger Apr 10 '16 at 13:57
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    @Franky_GTH nothing at all. This Hamiltonian doesnt model any physical system. Its a mathematical example that does the job. AFAIK, all the physical systems with multiplicative eigenvalues correspond to discrete symmetries, not continuous ones (see here). – AccidentalFourierTransform Apr 10 '16 at 14:02
  • @AccidentialFourierTransform: Do we really have to be that pessimistic about the realization of discrete symmetries? Think about lattice positions, like "holes" only being able to localize on crystal lattice positions, or the like, maybe? – Raphael J.F. Berger Apr 10 '16 at 14:26
  • Your Hamitonian looks suspiciously similar to the Hamiltonian from here. What I mean is that in both appear products of momentum and space operators. Is there something hidden which corresponds to attempting to measure "noise" (bearing in mind the uncertainty principle)? – Raphael J.F. Berger Mar 21 '18 at 13:46