When you solve Schrödinger equation for a free particle with no boundary conditions your eigen states are indexed by quantum number $k \in \mathbb R $ and $\mathbb R$ isn't countable but if you add a harmonic oscillator potential, your eigen states are indexed by harmonic number $n \in \mathbb N$ which is countable.

More interestingly for the hydrogen atom, your potential is finite as it goes to $\infty$. So for $E>0$ you have an uncountable number of states but for $E<0$ there are only a countable number of eigenstates/values.

I can't think of any counter example but I have no idea how to go about proving the number of eigenstates for a general bound system is countable.

I find this interesting because if you ignore your ability to measure position and momentum it would appear that you can change the dimensionality of your system by changing your potential.

  • $\begingroup$ Keep in mind that the Schrodinger equation for the harmonic oscillator has an uncountable infinity of solutions but just countably infinite normalizable solutions. (Come to think of it, the free particle solutions aren't normalizable either). $\endgroup$ – Alfred Centauri Aug 29 '15 at 0:40
  • $\begingroup$ @AlfredCentauri There are uncountable many normalizable superpositions between just the two lowest energy states. I agree that there is a countable maximal orthonormal set within some Hilbert Space but it's not clear to me what you are trying to say. $\endgroup$ – Timaeus Aug 29 '15 at 2:28
  • $\begingroup$ @Timaeus, the point of your comment is unclear to me. $\endgroup$ – Alfred Centauri Aug 29 '15 at 2:31
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    $\begingroup$ Sturm-Liouville theory $\endgroup$ – DanielSank Aug 29 '15 at 14:14
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    $\begingroup$ I think the correct way to phrase the question would be something like, let $H = -(\hbar^2/2m) \nabla^2 + V(x)$, acting in a suitable rigged Hilbert space. When are there at most countably many negative eigenvalues of $H$? $\endgroup$ – Robin Ekman Aug 29 '15 at 14:52

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