in Sec. \[numerical\], we go back to Eqs.  and . These equations define an instantaneous vacuum state ${\vert{0_t}\rangle}$ such that ${\hat{b}}_{\bm p}(t){\vert{0_t}\rangle}=0$ and ${\hat{d}}_{\bm p}(t){\vert{0_t}\rangle}=0$. It is different than the in-vacuum state, as $${\hat{b}}_{\bm p}(t){\vert{0_{-\infty}}\rangle}=\xi_{\bm p}(t){\hat{d}}_{-{\bm p}}^\dagger{\vert{0_{-\infty}}\rangle}, \quad
{\hat{d}}_{\bm p}(t){\vert{0_{-\infty}}\rangle}=\xi_{-{\bm p}}(t){\hat{b}}_{-{\bm p}}^\dagger{\vert{0_{-\infty}}\rangle}.
\label{extra}$$ In addition, the charge field operator can be derived from , $${\hat{Q}}=e\int\frac{{\mathrm{d}}^3{\bm p}}{(2\pi)^3}\Bigl({\hat{b}}_{\bm p}^\dagger(t){\hat{b}}_{\bm p}(t)-{\hat{d}}_{-{\bm p}}^\dagger(t){\hat{d}}_{-{\bm p}}(t)\Bigr),
\label{charge-op}$$ where the normal ordering of the creation and annihilation operators has been introduced. As it follows from , the mean value of ${\hat{Q}}$ in the in-vacuum state is zero. One can also show that the charge field operator is conserved during the time-evolution. Building upon the definition of , we can interpret $$\begin{aligned}
Q^{(0)}({\bm p},t)&=e{\langle{0_{-\infty}}\vert}{\hat{b}}_{\bm p}^\dagger(t){\hat{b}}_{\bm p}(t){\vert{0_{-\infty}}\rangle}\nonumber\\
&=e{\langle{0_{-\infty}}\vert}{\hat{d}}_{-{\bm p}}^\dagger(t){\hat{d}}_{-{\bm p}}(t){\vert{0_{-\infty}}\rangle}\nonumber\\
&=e|\xi_{\bm p}(t)|^2=e|c_{\bm p}^{(2)}(t)|^2\label{bf12}\end{aligned}$$ as the charge distribution of created bosons with momentum ${\bm p}$ and antibosons with momentum $-{\bm p}$ from the initial vacuum state by the pulsed electric field. Here, we have used Eq. . While this accounts for quasiparticles, the charge distribution of a real boson pair is obtained from  by taking the limit $t\rightarrow +\infty$. Moreover, when considered as a function of ${\bm p}$, it will be related to the momentum distribution of created bosons. We 