Is the energy always discrete? In the von Neumann axioms for quantum mechanics, the first postulate states that a quantum state is a vector in a separable Hilbert space. It means it is assumed the Hilbert space has a basis with at most infinite countable elements (cardinality). In other words, it states that the energy of an arbitrary system is discrete. Is it always true? If not, can you make an specific example in nature that its eigen-energies are uncountable?
 A: One counterexample is so simple as to be trivial:
The free particle has a completely continuous energy spectrum, since the Hamiltonian $H = \frac{p^2}{2m}$ has $[0,\infty)$ as its spectrum (this follows directly from $p$ having completely continuous spectrum $(-\infty,\infty)$. The reason this does not violate the spectral theorem/the countability of the basis of the Hilbert space is that this $H$ is an unbounded operator, and the "eigenstates" $\lvert p \rangle$ (which are $\psi_p(x) = \mathrm{e}^{\mathrm{i}px}$ in the position wavefunction representation) are not inside the Hilbert space (just note that $\psi_p(x)$ is not square-integrable on $\mathbb{R}$ to see that it is not in the canonical space of position wavefunctions $L^2(\mathbb{R},\mathrm{d}x)$). Only wavepackets, i.e. square-integrable superpositions of the plane waves $\psi_p(x)$, lie inside the Hilbert space of states.
In fact, the "eigenstates" belonging to continuous eigenvalues are never "normalizable", cf. this phys.SE question, and thus never are vectors in the actual Hilbert space, but only in the larger space of the so-called rigged Hilbert space, cf. this phys.SE question
Another counterexample would be the hydrogen atom, where the energies above a certain threshhold are continuous, and are again essentially free states.
