while $\hat{A}^{\dagger}_{\alpha} (t)$ creates the pair with the same quantum $\alpha$, and the field has the canonical representation $$\begin{aligned}
\hat{\phi} (t, {\bf x}) = \sum_{\alpha} \Bigl[ \varphi_{\alpha} (t) \hat{a}_{\alpha} (t) + \varphi_{\alpha}^{*} (t) \hat{b}_{\alpha}^{\dagger} (t) \Bigr].\end{aligned}$$

Using the commutation relations (\[com\]), we may construct the time-dependent particle-antiparticle states $$\begin{aligned}
\vert k_{\alpha}, \bar{l}_{\beta}, t \rangle = \frac{(\hat{a}_{\alpha}^{\dagger} (t))^{k_{\alpha}}}{\sqrt{k_{\alpha}!}}  \frac{(\hat{b}_{\beta}^{\dagger}(t))^{l_{\beta}}}{\sqrt{l_{\beta}!}} \vert 0_{\alpha}, \bar{0}_{\beta}, t \rangle,\end{aligned}$$ where bars denote antiparticles, and the vacuum state is given by $$\begin{aligned}
\hat{a}_{\alpha} (t) \vert 0_{\alpha}, \bar{0}_{\beta}, t \rangle = \hat{b}_{\beta} (t) \vert 0_{\alpha}, \bar{0}_{\beta}, t \rangle = 0.\end{aligned}$$ These states are orthonormal among themselves at the equal time $$\begin{aligned}
\langle m_{\alpha}, \bar{n}_{\beta}, t \vert k_{\gamma}, \bar{l}_{\delta}, t \rangle = \delta_{mk} \delta_{\alpha \gamma} \delta_{nl} \delta_{\beta \delta}.\end{aligned}$$ However, each multiple particle-antiparticle state is not a solution to Eq. (\[osc eq\]), because $\hat{a}_{\alpha} (t)$ and $\hat{b}_{\alpha} (t)$ cannot separately become quantum invariants for Eq. (\[ln eq\]). Instead, the particle-antiparticle pair constitutes the quantum invariants $\hat{A}_{\alpha} (t)$ and $\hat{A}^{\dagger}_{\alpha} (t)$. Hence, the vacuum state for a given $\alpha$ should be annihilated by the zero particle-antiparticle operator $$\begin{aligned}
\hat{A}_{\alpha} (t) \vert 0_{\alpha}, t \rangle = 0, \label{zer st}\end{aligned}$$ and at the same time have the zero particle and antiparticle number, respectively, 