Hamiltonian (\[diag2\]) with the aid of the quantum invariant method and then improve the unperturbed states by the Hamiltonian (\[off2\]). The unperturbed states consist of the time-dependent vacuum state $$\begin{aligned}
\hat{a}_{\alpha} (t) \vert 0_{\alpha}, t \rangle_{(0)} = \hat{b}_{\alpha} (t) \vert 0_{\alpha}, t \rangle_{(0)} =0, \label{lan qst}\end{aligned}$$ and the excited states $$\begin{aligned}
\vert n_{\alpha}, t \rangle_{(0)} = \frac{(\hat{\cal A}_{\alpha} (t))^{n_{\alpha}}}{\sqrt{n_{\alpha}!}} \vert 0_{\alpha}, t \rangle_{(0)}.\end{aligned}$$ For each $\alpha$ the time-dependent vacuum state (\[lan qst\]) leads to the Landau state and thereby the time-dependent vacuum state for the field $$\begin{aligned}
\vert 0, t \rangle_{(0)} = \prod_{\alpha} \vert 0_{\alpha}, t \rangle_{(0)}. \label{B-vac}\end{aligned}$$ Furthermore, the general excited states are [@kim-khanna00] $$\begin{aligned}
\vert {\cal N}, t \rangle_{(0)} = \prod_{{\cal N}} \frac{\hat{\cal A}^{\dagger {\cal N}} (t)}{\sqrt{ \{ {\cal N} \}!}} \vert 0, t \rangle_{(0)}\end{aligned}$$ where $\{ {\cal N} \} = (n_0, \cdots, n_{\alpha}, \cdots)$, $\hat{\cal A}^{\dagger {\cal N}} = \hat{\cal A}_{0}^{\dagger n_0} \cdots \hat{\cal A}_{\alpha}^{\dagger n_{\alpha}} \cdots$, and $ \{{\cal N} \}! = \prod_{\alpha} n_{\alpha}!$.

The quantum state of the field in a constant magnetic field $B_0$, though looks trivial because of $\Omega_{\alpha \beta} = 0$, requires a careful understanding in contrast to that of a charged particle in the non-relativistic theory. Each Landau level has the 