So I am looking at this paper (Instability and control of a periodically-driven Bose-Einstein condensate) and I am interested in solving equation 8, which reads $$i\frac{d}{dt}\begin{bmatrix} u(t)\\v(t) \end{bmatrix}=L(q,t)\begin{bmatrix} u(t)\\v(t) \end{bmatrix}$$ It is then stated

To find the corresponding Floquet states, we numerically evolve Eq. 8 over one period of driving, using the 2x2 identity matrix as the initial state. The result of this procedure is the single-period propagator U. The eigenstates of U are then the excitation Floquet states, while its eigenvalues are related to the excitation quasienergies via $\lambda_i=\exp[−\epsilon_iTi]$.

I am confused as to how to do this. I have tried just solving for u(t) and v(t) through coupled differential equations, but from there I am not sure if I just say that I can solve the earlier equation

we now introduce a perturbation $\alpha_n(t) > =\alpha_n^{(0)}(t) (1+u(t)\exp[iqn]+v^*(t)\exp[−iqn])$

For $\alpha(t)$ and say that $\alpha_n(t)=\alpha_n(0)\exp[-i\lambda_n(t)T]$. Since when I numerically solve the matrix equation, i get a number (for n=0) for $\alpha(T)$ that I can just invert and solve for.

I appreciate any advice on how to perform the time evolution and find the U matrix!


The single-period propagator $U$ is defined as follows: $$ \begin{pmatrix} u(T) \\ v(T) \end{pmatrix} = U \begin{pmatrix} u(0) \\ v(0) \end{pmatrix} , $$ the eigenvalues of $U$ are related to the Floquet quasienergies as stated in the paper. To determine the matrix $U$, take the differential equation $$ \mathrm i \frac{\mathrm d}{\mathrm dt} \begin{pmatrix} u_1(t) & u_2(t) \\ v_1(t) & v_2(t) \end{pmatrix} = \mathcal L(q,t) \begin{pmatrix} u_1(t) & u_2(t) \\ v_1(t) & v_2(t) \end{pmatrix} . $$ Solve it numerically (for fixed $q$ and other parameters) with the initial conditions $u_1(0) = v_2(0) = 1$, $u_2(0) = v_1(0) = 0$.* Then, $$ U = \begin{pmatrix} u_1(T) & u_2(T) \\ v_1(T) & v_2(T) \end{pmatrix} .$$

* Which amounts to solving equation (8) once for the initial condition $(1,0)^T$ and once for $(0,1)^T$.


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