\Bigl(\frac{qB(t)}{2} \Bigr)^2 {\bf x}_{\perp}^2 - qB(t) L_z \label{tr-ham}\end{aligned}$$ is the Hamiltonian transverse to the magnetic field, and $L_z$ is the angular momentum from the orbital motion of the charge. Following Ref. [@kim14a], we diagonalize the Hamiltonian (\[tr-ham\]) as $$\begin{aligned}
\hat{H}_{\perp} (t) = q B(t) \bigl[2 \hat{c}_{-}^{\dagger} (t) \hat{c}_{-} (t) +1 \bigr], \label{lan ham}\end{aligned}$$ where the time-dependent annihilation and the creation operators $$\begin{aligned}
\hat{c}_{-} (t) = \frac{1}{\sqrt{2}} \bigl( \hat{a}_{x} (t) - i \hat{a}_{y} (t)\bigr), \quad \hat{c}_{-}^{\dagger} (t) = \frac{1}{\sqrt{2}} \bigl( \hat{a}^{\dagger}_{x} (t) + i \hat{a}^{\dagger}_{y} (t)\bigr)\end{aligned}$$ are constructed by the annihilation operators $\hat{a}_{x} (t)$ and $\hat{a}_{y} (t)$ for the $x$-component and the $y$-component of the oscillator $$\begin{aligned}
H_{\perp 0} (t) = {\bf p}_{\perp}^2 + \Bigl(\frac{qB(t)}{2} \Bigr)^2 {\bf x}_{\perp}^2.\end{aligned}$$ The number states $\vert n, t \rangle$ in the oscillator representation (\[lan ham\]), which are the instantaneous Landau states, constitute a basis. Thus, the rate of the change of each number state can be expressed by themselves, which has been first derived in Ref. [@kim92], and then Eqs. (12) and (13) of Ref. [@kim14a] and Eq. (13) of Ref. [@kim13], which reads as $$\begin{aligned}
\langle m, t \vert (\frac{\partial}{\partial t} \vert n, t \rangle) := \Omega_{mn} (t) = \frac{\dot{B}}{4 B} \Bigl[ \sqrt{n (n-1)} 