wave functions , $$\begin{aligned}
\langle\Phi_{\bm p}^{(\beta)}|\Phi_{{\bm p}'}^{(\beta')}\rangle&=(2\pi)^3\delta^{(3)}({\bm p}-{\bm p}')\Bigl(\frac{{\mathrm{i}}}{c}[\Phi_{\bm p}^{(\beta)}(t)]^*
\dot{\Phi}_{\bm p}^{(\beta')}(t)-\frac{{\mathrm{i}}}{c}[\dot{\Phi}_{\bm p}^{(\beta)}(t)]^*\Phi_{\bm p}^{(\beta')}(t)\Bigr).\label{kg15}\end{aligned}$$ Using Eqs.  and , one can show that $$\frac{{\mathrm{d}}}{{\mathrm{d}}t}\Bigl(\frac{{\mathrm{i}}}{c}[\Phi_{\bm p}^{(\beta)}(t)]^*
\dot{\Phi}_{\bm p}^{(\beta')}(t)-\frac{{\mathrm{i}}}{c}[\dot{\Phi}_{\bm p}^{(\beta)}(t)]^*\Phi_{\bm p}^{(\beta')}(t)\Bigr)=0,
\label{kg16}$$ where the dot stands for time derivative. It follows from this equation that the quantity in the brackets is constant in time. Setting its value at $t\rightarrow-\infty$ gives $$\begin{aligned}
\underset{t\rightarrow-\infty}{\lim}\Bigl(\frac{{\mathrm{i}}}{c}[\Phi_{\bm p}^{(\beta)}(t)]^*
\dot{\Phi}_{\bm p}^{(\beta')}(t)-\frac{{\mathrm{i}}}{c}[\dot{\Phi}_{\bm p}^{(\beta)}(t)]^*\Phi_{\bm p}^{(\beta')}(t)\Bigr)
=\frac{2\omega_{\bm p}}{c}\beta\delta_{\beta\beta'}.
\label{kg17}\end{aligned}$$ Hence, the one-particle solutions of the Klein-Gordon equation $\Phi_{\bm p}^{(\beta)}(x)$ can be normalized such that $${\mathrm{i}}\int{\mathrm{d}}^3{\bm x}\,[\Phi_{\bm p}^{(\beta)}(x)]^*\overleftrightarrow{\partial_0}\Phi_{{\bm p}'}^{(\beta')}(x)=2mc\beta(2\pi)^3\delta^{(3)}({\bm p}-{\bm p}')\delta_{\beta\beta'}.
\label{kg18}$$ Finally, we write down the completeness relation for these wave functions, $$\frac{1}{2mc}\sum_{\beta}\int\frac{{\mathrm{d}}^3{\bm p}}{(2\pi)^3}\,{\mathrm{i}}\beta\Phi_{\bm p}^{(\beta)}(x)\overleftrightarrow{\partial_0}[\Phi_{{\bm p}'}^{(\beta')}(x)]^*
=\delta^{(3)}({\bm x}-{\bm x}').
\label{kg19}$$ Since the one-particle solutions of the Klein-Gordon equation  form a complete and orthonormal set of functions (see, also Refs. [@Grib1; @Grib2]), one can use them to construct the boson field operator .

### Bogolyubov transformation for the boson field {#bosons}

The instantaneous Hamiltonian of the bosonic field in the presence of an external time-dependent electric field is given by [@Greiner1] $$\begin{aligned}
{\hat{H}}(t)=\frac{1}{2m}\int{\mathrm{d}}^3{\bm x}\,\Bigl[\frac{1}{c^2}\dot{{\hat{\Phi}}}^\dagger(x)\dot{{\hat{\Phi}}}(x)
+{\hat{\Phi}}^\dagger(x)\Bigl(({\hat{\bm p}}-e{\bm A}(t))^2+(mc)^2\Bigr){\hat{\Phi}}(x)\Bigr].
\label{bf3}\end{aligned}$$ Substituting here , we arrive at $$\begin{aligned}
{\hat{H}}(t)=\int\frac{{\mathrm{d}}^3{\bm p}}{(2\pi)^3}\Bigl[\gamma^{(++)}_{\bm p}(t){\hat{b}}_{\bm p}^\dagger{\hat{b}}_{\bm p}+
\gamma^{(+-)}_{\bm p}(t){\hat{b}}_{\bm p}^\dagger{\hat{d}}_{-{\bm p}}^\dagger
+\gamma^{(-+)}_{\bm p}(t){\hat{d}}_{-{\bm p}}{\hat{b}}_{\bm p}+\gamma^{(--)}_{\bm p}(t){\hat{d}}_{-{\bm p}}{\hat{d}}_{-{\bm p}}^\dagger\Bigr],
\label{bf4}\end{aligned}$$ where 