# Floquet quasienergy spectrum, continuous or discrete?

I haven't got a feeling about Floquet quasienergy, although it is talked by many people these days.

Floquet theorem:

Consider a Hamiltonian which is time periodic $$H(t)=H(t+\tau)$$. The Floquet theorem says that there exists a basis of solutions to the Schrödinger equation of the form $$\psi(r,t)=e^{-i\varepsilon t}u(r,t)\ ,$$ where $$u(r,t)$$ is a function periodic in time.

We can rewrite the Schrödinger equation as

$$\mathscr{H}u(r,t)=\left[H(t)-\mathrm{i}\hbar\frac{\partial}{\partial t}\right]u(r,t)=\varepsilon u(r,t)\ ,$$

where the Floquet hamiltonian $$\mathscr{H}$$ can be thought as a Hermitian operator in the Hilbert space $$\mathcal{R}\otimes\mathcal{T}$$, where $$\mathcal{R}=L_2(\mathbb R^3)$$ is the Hilbert space of square-integrable functions of $$\vec r$$, and $$\mathcal{T}$$ is a Hilbert space with all square integrable periodic functions with periodicity $$\tau$$. Then the above equation can be thought of as an analogue to the stationary Schrödinger equation, with the real eigenvalue $$\varepsilon$$ defined as Floquet quasienergy.

My question is, since for the stationary Schrödinger equation we can have both continuous or discrete spectra, how about the Floquet quasienergy?

Another thing is, is this a measurable quantity? If it is, in what sense it is measurable? (I mean, in the stationary case, the eigenenergy difference is a gauge invariant quantity, what about quasienergy?)

• Evolution equations with time-dependent generators are difficult to treat in a rigorous way. One standard source is the book by Pazy. A reference that seems more tailored on your question is this book. – yuggib Oct 5 '14 at 14:06

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).

• Thanks for your edit and instructive answer. Do you have an idea how the floquet quasi-energy can be measured? That is my second part of the question, since E and E+nhw are the same states, it is a little bit confussing that when you measure the same state, you got two energies as in the experiment linked in the other answer. Can the floquet quasi-energies represent the same state be distinguished within the floquet formalism? I think no. Do you agree? – an offer can't refuse Jan 5 '16 at 9:13
• I'm not completely sure, but the answer is probably "it depends what you mean". In a purely monochromatic case, maybe, but real experiments take finite time so some of the Floquet energy surfaces can couple (example). These are subtle questions and they're not easy to answer, and they're often covered in ambiguities: does the molecule go through a light-induced conical intersection, or does it simply absorb a photon? If you ask a more precise question I may be able to help. – Emilio Pisanty Jan 5 '16 at 18:30
• (If you do, it would be good if you can ask it separately and in a way that Xcheckr's answer can be migrated there. You're asking two completely different questions in the current version.) – Emilio Pisanty Jan 5 '16 at 18:32

You can think of a Floquet energy in a similar way to a Bloch state. In the latter case, because space is periodic, the momentum states are repeated at every reciprocal lattice vector, $\textbf{G}$. For a Floquet state, because time is periodic, energy states are repeated every $n\hbar \omega$ where $n$ is an integer and $\hbar \omega$ depends on the time, $\tau$ (where $\tau$ in the experiment is the time between laser pulses).

Here is an image from the attached paper in case you cannot view it, but I highly recommend reading the below paper if you are interested in Floquet states. You can see (barely) in the image below that the Dirac Cone (which was chosen to be the system studied here for no particular reason), is repeated at several values of $\hbar \omega$ above and below the "actual" Dirac Cone at $n=0$. You can see the $n=1$, $n=2$, and $n=-1$ states pretty clearly. See the paper here:

https://www.sciencemag.org/content/342/6157/453?related-urls=yes&legid=sci;342/6157/453

• I happened to see this paper before. So your answer seem to say quasi-energy is a gauge invariant in the same sense as Bloch energy. Is that what you mean? Also, what about my first question? – an offer can't refuse Oct 6 '14 at 5:28
• @luming Yes, the quasi-energy must be gauge-invariant or it wouldn't be measurable! Also, perhaps I don't understand in what sense you mean discrete vs. continuous? It seems to me that the quasi-energies are not very different than Bloch states with different bands, except separated by constant energy steps. – Xcheckr Oct 6 '14 at 13:45