$$\begin{aligned}
\langle 0_{\alpha}, t \vert \hat{a}^{\dagger}_{\alpha} (t)  \hat{a}_{\alpha} (t) \vert 0_{\alpha}, t \rangle = \langle 0_{\alpha}, t \vert \hat{b}^{\dagger}_{\alpha} (t)  \hat{b}_{\alpha} (t) \vert 0_{\alpha}, t \rangle = 0. \label{zer pa}\end{aligned}$$ The zero particle and antiparticle content (\[zer pa\]) excludes another null state (\[zer st\]) with one particle or antiparticle $$\begin{aligned}
\vert 1_{\alpha}, t \rangle_{\rm spurious}
= \sqrt{\frac{1}{2}} \Bigl(\vert 1_{\alpha}, \bar{0}_{\alpha}, t \rangle - \vert 0_{\alpha}, \bar{1}_{\alpha}, t \rangle\Bigr).\end{aligned}$$ Indeed, $\vert 0_{\alpha}, t \rangle := \vert 0_{\alpha}, \bar{0}_{\alpha},t \rangle$ is the unique state satisfying Eqs. (\[zer st\]) and (\[zer pa\]) and constructs the time-dependent vacuum state for the field $$\begin{aligned}
\vert 0, t \rangle = \prod_{\alpha} \vert 0_{\alpha}, t \rangle. \label{vac st}\end{aligned}$$ The first excited state is obtained by acting $\hat{A}^{\dagger}_{\alpha} (t)$ on the vacuum $$\begin{aligned}
\vert 1_{\alpha}, t \rangle := \hat{A}^{\dagger}_{\alpha} (t) \vert 0, t \rangle
= \sqrt{\frac{1}{2}} \Bigl(\vert 1_{\alpha}, \bar{0}_{\alpha}, t \rangle + \vert 0_{\alpha}, \bar{1}_{\alpha}, t \rangle\Bigr). \label{1-p-a}\end{aligned}$$ The one particle-antiparticle state (\[1-p-a\]) is the superposition of a particle with ${\bf k}$ and an antiparticle with $- {\bf k}$ from Eq. (\[fourier\]) but is not $\vert 1_{\alpha}, \bar{1}_{\alpha}, t \rangle$, as expected naively. Similarly, acting $n$-times $\hat{A}^{\dagger}_{\alpha} (t)$ excites $n$ particle-antiparticle state for the quantum $\alpha$ $$\begin{aligned}
\vert 