In this paper "Photonic Floquet Topological Insulators" the authors calculate the bandstructure of a time-periodic Hamiltonian. They create a time-dependent tight-binding Hamiltonian via the Peierl's substitution.

Now my first question would be about equation (3) of this paper. Since it is in lattice site basis, would i need to FT in order to get to transverse crystal momentum basis so i can create a bandstructure? And what would happen to the phase factor aquired through the Peierls substitution?

Secondly, how would I calculate the Floquet quasi-energies from there? From what I have read so far they are usually calculated by calculating the Eigenvalues of the stroboscopic time-evolution operator $U(T)=\mathrm{exp}\left[\int_{0}^{T}{H(t)dt}\right]$. This has not worked for me so far. Is this even the right approach?

I hope this question is not too "do-my-homework-for-me"-y. Its a completely new topic for me and I have been trying to solve this for what feels like and eternity. Any help would be greatly appreciated.


In that paper, the authors are looking at a honeycomb lattice, which is well-known as the lattice of graphene. This is not a simple Bravais lattice, but has a two atom basis which gives rise to the two bands. There is a lot of literature treating how to obtain the bandstructure for this sort of lattice, which as you say, corresponds to making a Fourier transform to to momentum space, so I would like to concentrate on the second part of your question, how to obtain the quasienergy spectrum.

You are correct to say that the quasienergies can be obtained from the time-evolution operator for a single period, $U(T)$. As $U$ is unitary, its eigenvalues are pure phases, which are related to the quasienergies $\epsilon_j$ via $\lambda_j = \exp[-i T \epsilon_j ]$. So a typical procedure is to prepare $U(T)$ by time-evolving the identity matrix over a single period of driving, finding its eigenvalues, and then taking their logarithm (and dividing by a factor of $T$).

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  • $\begingroup$ First of all, thank you so much. So I would now proceed by transforming equation (3) in transverse k-space and use it to evolve the identity matrix over one period, then take the logarithm and divide by T. I will try this first thing tomorrow and report back if it worked. $\endgroup$ – JanJasper May 4 at 17:12
  • $\begingroup$ M = [0,exp(-1i*((kx+R*sin(omega*t))*a*sin(120/360*2*pi)-(ky-R*cos(omega*t))*a*cos(120/360*2*pi)))+exp(-1i*(-(kx+R*sin(omega*t))*a*sin(120/360*2*pi)-(ky-R*cos(omega*t))*a*cos(120/360*2*pi)))+exp(+1i*(ky-R*cos(omega*t))*a);exp(1i*((kx+R*sin(omega*t))*a*sin(120/360*2*pi)-(ky-R*cos(omega*t))*a*cos(120/360*2*pi)))+exp(1i*(-(kx+R*sin(omega*t))*a*sin(120/360*2*pi)-(ky-R*cos(omega*t))*a*cos(120/360*2*pi)))+exp(-1i*(ky-R*cos(omega*t))*a),0]; So I've tried using this Hamiltonian, but it seems it's still not quite right. Now I'm not sure if the Hamiltonian is incorrect or my time-evolution is faulty $\endgroup$ – JanJasper May 5 at 13:23
  • $\begingroup$ It's a bit hard to read, but is $M$ the Hamiltonian matrix? Are you trying to get the topological edge states, as shown in Fig. 2 of the paper, or do you want the bandstructure (like Fig. 1)? If you are trying to get the full bandstructure, I would expect the Hamiltonian to be a 2x2 matrix, which is a function of $k_x$ and $k_y$. But if you want the edge states, then you should work in real-space using a tight-binding Hamiltonian, with periodic boundary conditions in$x$ and closed boundaries in $y$. And how are you doing the time-evolution? $\endgroup$ – Clara Diaz Sanchez May 5 at 17:17
  • $\begingroup$ Yes, it is supposed to be a 2x2-Matrix. I tried doing the evolution both by calculating the TE-Operator as I defined it above, except I approximated the integral by dividing T into steps (this in matlab). Also I used mathematica's NDSolve to solve for the solutions of eq. (3), then calculated the eigenvalues of the resulting monodromy matrix. Currently I am trying a different approach, by Fourier decomposing the Hamiltonian into its Floquet eigenmodes (as seen in this paper journals.aps.org/prb/pdf/10.1103/PhysRevB.97.035422 , eq. (18)). The last one actually showed results yesterday. $\endgroup$ – JanJasper May 6 at 12:22

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