quantum mechanics - A series of bound states covering an interval

Generally, the bound states (normalizable eigenvectors) of a Hamiltonian have discrete eigenvalues.

Is it possible for the eigenvalues to cover an interval? Say, $(a,b)$?

That is, for each $E \in (a,b)$, there is a corresponding bound state?

Answer

A self-adjoint operator $T$ on $L^2(\mathbb{R})$ has in its spectrum three different kinds of subspectra: A discrete point spectrum, a continuous spectrum, and a singular spectrum. The latter is physically discarded.

The point spectrum consists of the eigenvalues of $T$, that is, the spectral values for which true eigenvectors in $L^2(\mathbb{R})$, and hence normalizable eigenstates, exist.

The continuous spectrum may intersect with the point spectrum, but except for these discrete intersections, the continuous spectral values do not have eigenvectors in $L^2(\mathbb{R})$, but only in a larger rigged Hilbert space, which are consequently non-normalizable (for example because the Hermitian product of $L^2(\mathbb{R})$ is not a proper inner product for them).

Thus, continuous bound states do not exist in the usual quantum mechanical setting.

A way to see that the true eigenvalues cannot form a continuum is to notice that they would have to be uncountably many, but the separable Hilbert spaces of quantum mechanics have only countably many basis vectors, and hence there are only countably many possible independent eigenvectors.