A Eigenvectors of a Floquet Hamiltonian

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1. Aug 24, 2017

DeathbyGreen

I'm trying to recreate some results from a paper:

https://arxiv.org/pdf/1406.1711.pdf

Basically they take the Hamiltonian of graphene near the Dirac point (upon irradiation by a time periodic external field) and use Floquet formalism to rewrite it in an extended Hilbert space incorporating time periodic functions; in doing this they effectively remove the time dependence of the periodically modulated Hamiltonian. Skipping a few steps, they rewrite this Floquet Hamiltonian in a subspace of two potentially degenerate eigenvector branches near the Dirac point (eigenvalues of this matrix are equation (15)). The matrix is (for only the m=0 and m=1 modes)

$H_f = \left[ \begin{array}{cc} \hbar\Omega -\hbar v_fk_p& \frac{v_fe}{2c}A_0e^{i\theta}\\ \frac{v_fe}{2c}A_0e^{-i\theta}& \hbar v_f k_p \end{array} \right]$

The eigenvectors of this matrix are quasienergy states $|\phi^\alpha\rangle$ from the equation $H_F |\phi^\alpha\rangle = \epsilon_\alpha|\phi^\alpha\rangle$. My problem is from going from equation (9)
$|\phi^\alpha\rangle = (|u^\alpha_1\rangle, |u^\alpha_0\rangle)^T$to (8) $|\phi^\alpha(t)\rangle = |u^\alpha_0\rangle + |u^\alpha_1\rangle e^{i\Omega t}$ in which the coefficients $|u^\alpha_m\rangle$ are 1x2 kets. Expanding $|\phi^\alpha\rangle$:

$|\phi^\alpha\rangle = \left[ \begin{array}{cc} u^\alpha_{1a}&u^\alpha_{1b}\\ u^\alpha_{0a}&u^\alpha_{0b} \end{array} \right]$

I need to apply this to the 2x2 Hamiltonian and solve for the $u^\alpha_m$coefficients. Then take those coefficients and put them into

$|\phi^\alpha(t)\rangle = \left[ \begin{array}{c} u^\alpha_{0a}\\ u^\alpha_{0b} \end{array} \right]+ \left[ \begin{array}{c} u^\alpha_{1a}\\ u^\alpha_{1b} \end{array} \right]e^{i\Omega t}$

to get the final answer they have in equation (17). In doing so I get a system of 2 equations and 4 unknowns. When I work this out my answer doesn't come out like theirs. Is this the right way to go about going from the form in the extended Hilbert space
$|\phi^\alpha\rangle = (|u^\alpha_1\rangle, |u^\alpha_0\rangle)^T$to the form in the standard Hilbert space (8) $|\phi^\alpha(t)\rangle = |u^\alpha_0\rangle + |u^\alpha_1\rangle e^{i\Omega t}$?

2. Aug 29, 2017

PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Aug 29, 2017

DeathbyGreen

For any people from the future who come across this and are curious about the answer, I figured it out. So after finding the eigenvectors of the 2x2 Hamiltonian

$H_f = \left[ \begin{array}{cc} \hbar\Omega-\hbar v_vk&\frac{A_0ev_f}{c2}e^{i\theta}\\ \frac{A_0ev_f}{c2}e^{-i\theta}&\hbar v_fk \end{array} \right]$

you get the solution

$|\phi^\alpha\rangle = \left[ \begin{array}{c} -\sin(\psi_k^\pm/2)e^{-i\theta}\\ \cos(\psi_k^\pm/2) \end{array} \right]$

where the psi angle is specified in the paper (a combination of variables). You then take the upper element of this matrix, corresponding to the u1 mode and multiply it by the corresponding eigenvector of the 4x4 matrix to convert it back to the original basis. Also, my initial statement of $|\phi_\alpha\rangle$ is incorrect and was due to some confusing notation used in the paper. It should be a 2x1 for the 2x2 and a 4x1 for the 4x4.