n_{\alpha}, t \rangle = \frac{(\hat{A}^{\dagger}_{\alpha} (t))^{n_{\alpha}} }{\sqrt{n_{\alpha}!}} \vert 0, t \rangle
= \sum_{k_{\alpha}=0}^{n_{\alpha}}  \sqrt{\frac{n_{\alpha}!}{2^{\alpha} (n_{\alpha} - k_{\alpha})! k_{\alpha}!}}
\vert n_{\alpha} - k_{\alpha}, \bar{k}_{\alpha}, t \rangle, \label{n-p-a}\end{aligned}$$ in which the statistical weight comes from the number of ways for arranging indistinguishable $n_{\alpha}$ particle-antiparticles. Any general particle-antiparticle state is a linear combination of (\[n-p-a\]).

It is one of the merits of the invariant operators that the exact quantum state can also be given by the eigenstates of another invariant $\hat{N}_{\alpha} (t) = \hat{A}_{\alpha}^{\dagger} (t) \hat{A}_{\alpha} (t)$ [@kim-page01]. Hence, the total number of initial particle-antiparticle with the quantum $\alpha$ contained in the vacuum state at $t$ is the same as that of particle-antiparticle at $t$ contained in the initial vacuum state at $t_0$: $$\begin{aligned}
{\cal N}_{\alpha} (t) = \langle 0, t \vert \hat{A}_{\alpha}^{\dagger} (t_0) \hat{A}_{\alpha} (t_0) \vert 0, t \rangle = \langle 0, t_0 \vert \hat{A}_{\alpha}^{\dagger} (t) \hat{A}_{\alpha} (t) \vert 0, t_0 \rangle =  |\dot{\varphi}_{\alpha} (t)|^2 |\varphi_{\alpha} (t_0)|^2 + |\varphi_{\alpha} (t)|^2|\dot{\varphi}_{\alpha} (t_0)|^2.\end{aligned}$$ Here, $\hat{A}_{\alpha} (t_0)$ and $\hat{A}_{\alpha}^{\dagger} (t_0)$ are the operators (\[pa op\]) and (\[an op\]) evaluated at $t_0$. Note that $\hat{A}_{\alpha}^{\dagger} (t) \hat{A}_{\alpha} (t)$, $(\hat{A}_{\alpha}^{\dagger} (t))^2$ and $(\hat{A}_{\alpha} (t))^2$ form an S(1,1) algebra, which leads to the quantum master equation [@kim-schubert11], and 