Is it possible to write the time evolution of a generic quantum state in terms of a floquet basis? It seems that the answer is yes, although this seems wrong.

In general, time evolution in a system with a time dependent hamiltonian is complex, and the propagator involves a time ordered exponential. Of course, if the Hamiltonian is periodic, then the hamiltonian will commute with itself after multiples of the period. Using group theory, one can see that if we symmetrize with respect to the group of finite time translations (i.e. take advantage of the discrete time translational symmetry) then there is a quasi-energy, which lives on a circle, and describes how the phase evolves under a 'stroboscopic' time evolution of multiples of the driving period.

The answer to my original question is apparently yes, as this is essentially the floquet theorem that can be widely found from a google search. However, I struggle to understand how to make the final leap to the actual floquet theorem. That is, symmetrizing with respect to the finite time translations doesn't appear to suggest that the Hamiltonian commutes with itself in any way, just that if we only consider the state in a stroboscopic manner then we can know it, based on the original state.

To summarize, it is not immediately apparent to me why a generic time evolution can be written, when it is only generally understood to have a 'stroboscopic' evolution that is well-defined. Would anyone be able to offer an explanation as to why infinitesimal-times wave function evolution should also be understood in a floquet picture?

  • $\begingroup$ I don't understand your question. I could show you how to derive the Floquet theorem, but it seems you already know the derivation. Why do you think that the time evolution is well-defined only for multiples of the period length? The time evolution is well-defined for arbitrary times (just difficult to calculate due to the time ordering, note that there is usually also no practical way to actually calculate the Floquet energies and basis exactly). $\endgroup$
    – Noiralef
    Commented Aug 31, 2018 at 13:57
  • $\begingroup$ Could you point me in the right direction? I found the representations of the discrete time translation group, but this doesn't really cover general times, just the stroboscopic evolution. $\endgroup$
    – speeze
    Commented Sep 1, 2018 at 15:38
  • $\begingroup$ "Right direction" towards what? The proof of the Floquet theorem? $\endgroup$
    – Noiralef
    Commented Sep 1, 2018 at 21:31
  • $\begingroup$ Yep! I appreciate that there is no way to exactly diagonalize the floquet Hamiltonian, but I am just concerned with general time evolution. Does the floquet 'picture' help at all with the issue of the time-ordered exponential propogator business? I suppose it doesn't, but the way floquet's theorem is presented (e.g. page 33 iht.univ.kiev.ua/nanoqt2016/materials/presentations/… ) make it seem as if this is no longer a concern, which seems wrong to me. $\endgroup$
    – speeze
    Commented Sep 2, 2018 at 22:37

1 Answer 1


The Floquet theorem only tells you one thing: that the propagator $U(t) = \mathcal{T} e^{-i \int_0^t H(\tau)\, d\tau}$ ($\hbar=1$) can be written in the factorized form $$ U(t) = Q(t)\, e^{-i G t} $$ where $Q(t)$ is periodic and $G$ a Hermitian operator (assuming that the Hamiltonian $H(t)$ is periodic itself, $H(t+T) = H(t)$).

The proof of this is surprisingly simple: We define $G = \frac{i}{T} \log U(T)$ (so that $U(T) = e^{-i G T}$, this gives a Hermitian operator due to the spectral theorem) and define $Q(t) = U(t)\, e^{i G t}$. It is only left to show that $Q(t)$ is periodic, that can be seen from the fact that $U(t+T) U(T)^{-1} = U(t)$.

Let now $|n\rangle$ be an eigenbasis of $G$, $G | n \rangle = E_n | n \rangle$, and define $|n(t) \rangle = Q(t) |n \rangle$. Then $$ U(t)\, | n(0) \rangle = e^{-i E_n t}\, | n(t) \rangle , $$ this is the periodic "Floquet basis".

Some notes:

  • Yes, we can write an arbitrary state in terms of this basis. Let $|\psi(0)\rangle = \sum_n \psi_n |n\rangle$, then $\psi(t) = \sum_n \psi_n e^{-i E_n t}\, |n(t)\rangle$.
  • This does not "help with the time-ordered exponential business" because normally it is just as hard to calculate $G$ as it is to calculate $U(t)$. It does help in some simple cases, and with perturbation theory.
  • The energies $E_n$ are only defined up to multiples of $2\pi / T$ because the logarithm is only defined up to multiples of $2\pi i$. This is analogous to Bloch waves, where the wave vector is only defined up to reciprocal lattice vectors.
  • From the definitions it is easy to see that $$ [ H(t) - i \partial_t ]\, |n(t) \rangle = E_n |n(t)\rangle . $$ In this sense, we have reduced the problem to finding the eigenvectors of $\tilde H = H(t) - i \partial_t$ on the extended Hilbert space $\mathcal H \otimes \mathcal P$ (where $\mathcal H$ is the original Hilbert space and $\mathcal P$ the space of $T$-periodic functions). In this formulation, the "eigenvector" $|n(t)\rangle$ already contains the full time-evolution over one period of the basis state, the eigenenergies that appear are all of $E_n \pm 2\pi m / T$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.