\delta_{m, n-2}
- \sqrt{(n+1) (n+2)} \delta_{m, n+2} \Bigr]. \label{ll tr}\end{aligned}$$ Note that the coupling matrix is antisymmetric, $\Omega^T = - \Omega$.

Thus, omitting the longitudinal momentum for simplicity, we have the expansions $$\begin{aligned}
\phi_{k_z} ({\bf x}_{\perp}, t) &=& \sum_{n} \phi_{n} (t) \Phi_n ({\bf x}_{\perp}, t), \nonumber\\
\dot{\phi}_{k_z} ({\bf x}_{\perp}, t) &=& \sum_{n} \dot{\phi}_{n} (t) \Phi_n ({\bf x}_{\perp}, t) + \sum_{n l} \phi_{n} (t) \Omega_{nl} (t) \Phi_l ({\bf x}_{\perp}, t), \label{lan exp}\end{aligned}$$ where $ \Phi_n ({\bf x}_{\perp}, t) = \langle {\bf x}_{\perp} \vert n, t \rangle$ is the wave function for the $n$-th Landau level. Finally, we obtain the Hamiltonian of the form $$\begin{aligned}
H(t) = \int \frac{dk_{z}}{2\pi} \Bigl[ \sum_{n} \bigl( \pi_{n}^* \pi_{n} + \omega_n^2 \phi_n^* \phi_n \bigr) + \sum_{m n} \bigl( \pi_{m}^* \Omega_{mn}  \phi_{n}^* + \pi_{m} \Omega_{mn} \phi_{n}  \bigr) \Bigr], \label{B-ham}\end{aligned}$$ where the canonical momenta are $$\begin{aligned}
\pi_n &=& \dot{\phi}_n^* + \sum_{m} \phi_{m}^* \Omega_{mn} = \dot{\phi}_n^* - \sum_{m}  \Omega_{nm} \phi_{m}^*, \nonumber\\
\pi_n^* &=& \dot{\phi}_n + \sum_{m} \phi_{m} \Omega_{mn} = \dot{\phi}_n - \sum_{m} \Omega_{nm} \phi_{m},\end{aligned}$$ and the time-dependent Landau energies are $$\begin{aligned}
\omega_n^2 (t) = |qB(t)| (2n+1) + m^2 + k_z^2. \label{lan en}\end{aligned}$$ Note that other choice of the basis, for instance, the initial Landau states at $t = - \infty$ 