\rangle. \label{osc eq}\end{aligned}$$ In order to find the quantum states for Eq. (\[osc eq\]), we employ the quantum invariant method, which satisfies the Liouville-von Neumann equation $$\begin{aligned}
i \frac{\partial \hat{I} (t)}{\partial t} + [\hat{I}(t), \hat{H} (t) ] = 0. \label{ln eq}\end{aligned}$$ Then, the exact solution to Eq. (\[osc eq\]) is given by an eigenstate of the quantum invariant up to a time-dependent phase factor [@lewis-riesenfeld69] $$\begin{aligned}
\vert \Psi (t) \rangle = \sum_{\lambda} C_{\lambda} e^{- i \int^t dt' \langle \lambda, t' \vert \hat{H} (t')
- i \partial/ \partial t' \vert \lambda, t' \rangle} \vert \lambda, t \rangle, \label{lr sol}\end{aligned}$$ where $C_{\lambda}$ is a constant, and $\lambda$ is a constant eigenvalue corresponding to the eigenvalue problem of the invariant operator $$\begin{aligned}
\hat{I} (t)  \vert \lambda, t \rangle = \lambda  \vert \lambda, t \rangle.\end{aligned}$$ Considering the symmetric form of the Hamiltonian (\[q-ham\]) from two fields $\phi$ and $\phi^*$ and their conjugate momenta, we may search for the quantum invariants in the extended phase space (\[phase\]) of the form $$\begin{aligned}
\hat{I} (t) = (U(t), V(t)) \frac{\hat{Z}+ \hat{Z}^*}{\sqrt{2}}, \label{q-inv}\end{aligned}$$ where $U (t)$ and $V (t)$ are matrix-valued functions, carrying indices of $\alpha, \beta$, and are determined by Eq. (\[ln eq\]).

Invariant Operators in Time-Dependent Electric 