Particle in a box with absolutely continuous spectrum Let's consider a particle on a real line in a potential $V(x)$ which disappears at infinity. The Hamiltonian is:
$$
  H: W^{2,2}(\mathbb{R}) \subset L^2(\mathbb{R}) \to L^2(\mathbb{R}) \\
  \big( H \, \psi \big)(x) = -\psi''(x) + V(x) \, \psi(x)
$$
(We set $\frac{\hbar^2}{2m} = 1$.) Such Hamiltonians typically have a non-empty absolutely continuous spectrum – therefore there are energies for which there is no corresponding eigenstate. Since physicists (except mathematical physicists) often prefer to have eigenvalues and eigenvectors, they frequently use a finite box approximation (eg. this paper): instead of $L^2(\mathbb{R})$ they work on $L^2([-R, R])$ for an arbitrarily large $R$, with Dirichlet boundary conditions. Then, the corresponding Hamiltonian is:
$$
  H: \big\{ \; \psi \in W^{2,2}([-R, R]) \; \big| \; \psi(-R) = \psi(R) = 0 \; \big\} \to L^2([-R, R]) \\
  \big( H \, \psi \big)(x) = -\psi''(x) + V(x) \, \psi(x)
$$
The spectrum of this Hamiltonian is typically discrete, because the kinetic energy of a particle in a box is discrete. As pointed out in comments, instead of Dirichlet boundary conditions one can use periodic boundary conditions $\psi(-R) = \psi(R), \; \psi'(-R) = \psi'(R)$ to get a better approximation.
I want to know whether there are any “physically plausible” Hamiltonians for a particle in a box with either Dirichlet or periodic boundary conditions which have a non-empty continuous spectrum. By plausible I mean bounded from below, containing the kinetic term $P^2$ (or $(P-A)^2$ for a magnetic interaction) and a potential $V(x)$, which can be reasonably ill-behaved (eg. it may contain a point-interaction $\delta_{x_0}$).
 A: The answer is negative for smooth bounded below potentials: there is no continuous part of the spectrum.
That is because, as is well known, the heat semigroup  $e^{-tH}$, $t\geq 0$, for strictly positive $t$, is made of compact, Hilbert-Schmidt, trace class operators, where $H$ is the unique selfadjoint extension of your Hamiltonian (the operator  you consider  is essentially selfadjoint with standard boundary conditions or one may always consider the Friedrichs selfadjoint extension).
This result is more generally valid on compact manifold for operators Laplace-Beltrami + potential (see e.g. my
old paper and the references therein).
As is well known, compact operators have a spectrum like this.

*

*There is a  point spectrum with eigenspaces with finite dimension and at most $0$ as accumulation point of the whole spectrum.


*The unique point of the continuous spectrum is at most $0$. Actually, it is possible to prove that, in the considered case the sequence of eigenvectors is in fact accumulated by 0. There is a general estimate by Weyl.


*In the present case, no residual spectrum exists because the operator $e^{-tH}$ is selfadjoint.
From elementary functional calculus, it turns therefore out that the spectrum of $H$ is a pure point spectrum, made of isolated points tending to infinity, with eigenspaces with finite dimension.
I do not know what happens when using singular potentials as deltas, but I do not expect that the result changes substantially.
