Floquet quasienergy spectrum, continuous or discrete?

In the stationary Schrödinger equation, we can have a continuous or a discrete spectrum. How about Floquet quasienergies?

You can have both. In one sense it is trivial to show this, since any constant hamiltonian is also periodic, but presumably you want some more physical examples, so here's two.

• For a continuous spectrum, start with a nonrelativistic free charged particle and add an oscillating uniform electric field,so the hamiltonian is $$\hat H(t)=\frac12\hat p^2+\hat x\,E_0\cos(\omega t).$$ The cleanest solutions are Volkov states $|\Psi_p(t)⟩$, which are plane waves with canonical momentum $p$ but a kinematic momentum $p+A(t)=p+\tfrac{E_0}{\omega}\sin(\omega t)$ which follows the vector potential of the field, i.e. $$⟨x|\Psi_p(t)⟩=\frac{1}{\sqrt{2\pi}}e^{-\frac i2\int^t(p+A(\tau))^2\mathrm d\tau}e^{i(p+A(t))x}.$$ (Modulo constants and signs, which you should check yourself.) The Volkov states are Floquet states, with quasienergy $$\varepsilon_p=\frac{p^2}{2}+U_p=\frac{p^2}{2}+\frac{E_0^2}{4\omega^2},$$ where $U_p$ is the ponderomotive potential of the field, i.e. the mean energy of the oscillatory motion. They're also a complete set, with $\int |\Psi_p(t)⟩⟨\Psi_p(t)|\mathrm dp=1$ and $⟨\Psi_p(t)|\Psi_{p'}(t)⟩=\delta(p-p')$, which is nice, but it also means that they're not the only possible Floquet basis as any linear combination of $|\Psi_p(t)⟩$ and $|\Psi_{\pm\sqrt{p^2+2n\omega}}(t)⟩$, $n\in\mathbb Z$, is also a Floquet state. So the Floquet manifold is either one big continuum, or multiple overlapping continua, which are equivalent given the usual Floquet-ladder degeneracy.

• For a discrete spectrum, simply take any finite-dimensional initial Hilbert space $\mathcal{H}$ and add any periodic hamiltonian $H(t)=H(t+T)$. Then the quasienergies $\varepsilon$ (or rather, the exponentiated form $e^{i\varepsilon T}$) are the eigenvalues of the one-period propagator $U(t_0+T,t_0)$ for any starting time $t_0$, where the propagator obeys $i\partial_t U(t,t')=H(t)U(t,t')$ and $U(t,t')=1$. Since $U(t_0+T,t_0)$ is an operator on the finite-dimensional $\mathcal H$, it can only have a discrete set of eigenvalues.

  I can hear you grumble and say that that's cheating, and that one should take a "natural" discrete-spectrum problem and show that its Floquet quasienergies are still discrete. For some examples of this nature, see e.g. Commun. Math. Phys. 177 no. 2, 327 (1996).