Eigenvectors of a Floquet Hamiltonian

In summary, the conversation discusses the process of recreating results from a paper that uses Floquet formalism to rewrite the Hamiltonian of graphene near the Dirac point in an extended Hilbert space. The conversation also mentions the steps needed to rewrite the Floquet Hamiltonian in a subspace of two potentially degenerate eigenvector branches and the difficulties encountered in the process. The conversation also includes the solution to the problem and clarifies some confusing notation used in the paper.
  • #1
DeathbyGreen
84
16
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)

[itex]
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]
[/itex]

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

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

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

[itex]
|\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}
[/itex]

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
[itex] |\phi^\alpha\rangle = (|u^\alpha_1\rangle, |u^\alpha_0\rangle)^T [/itex]to the form in the standard Hilbert space (8) [itex]|\phi^\alpha(t)\rangle = |u^\alpha_0\rangle + |u^\alpha_1\rangle e^{i\Omega t}[/itex]?
 
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  • #2
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

[itex]
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]
[/itex]

you get the solution

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

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 [itex]|\phi_\alpha\rangle[/itex] 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.
 

1. What are eigenvectors of a Floquet Hamiltonian?

Eigenvectors of a Floquet Hamiltonian are the special set of vectors that do not change direction when multiplied by the Floquet Hamiltonian. They represent the stationary states of the system and determine the evolution of the system over time.

2. How are eigenvectors of a Floquet Hamiltonian different from regular eigenvectors?

Eigenvectors of a Floquet Hamiltonian are different from regular eigenvectors in that they are time-dependent. This means that they can vary over time, unlike regular eigenvectors which are constant. This is because the Floquet Hamiltonian describes the dynamics of a system that is subject to periodic driving forces.

3. What is the significance of eigenvectors of a Floquet Hamiltonian in quantum mechanics?

Eigenvectors of a Floquet Hamiltonian are important in quantum mechanics as they provide a way to analyze and understand the behavior of a system that is subject to periodic driving forces. They allow us to determine the energy levels and dynamics of the system, and are crucial in predicting the behavior of quantum systems.

4. How are eigenvectors of a Floquet Hamiltonian related to the concept of Floquet states?

Eigenvectors of a Floquet Hamiltonian are closely related to Floquet states, which are solutions to the time-dependent Schrödinger equation. Floquet states are obtained by expanding the wavefunction in terms of the eigenvectors of the Floquet Hamiltonian. This allows for the time-dependence of the system to be taken into account, resulting in a more accurate description of the system's behavior.

5. Can the eigenvectors of a Floquet Hamiltonian be used to predict the behavior of a system over a long period of time?

Yes, the eigenvectors of a Floquet Hamiltonian can be used to predict the long-term behavior of a system subject to periodic driving forces. This is because they represent the stationary states of the system, and the time evolution of the system can be described as a linear combination of these states. However, the accuracy of these predictions may decrease if the system is subject to strong perturbations or if the driving forces are not periodic.

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