1_\xi \rangle.\end{aligned}$$

A crucial difference between the single photon state and the vacuum state is that the later state factorizes $\vert 0 \rangle =\vert 0_{t]} \rangle
\otimes \vert 0_{(t} \rangle $ with respect to the temporal factorization $\mathfrak{F}=\mathfrak{F}_{t]} \otimes \mathfrak{F}_{(t}$ of the Fock space, with $\vert 0_{t]} \rangle \in \mathfrak{F}_{t]} $ and $\vert
0_{(t} \rangle \in \mathfrak{F}_{(t}$, while the former does not. Rather, we have $$\vert 1_\xi \rangle = B^\ast(\xi) \vert 0 \rangle = \vert {1_\xi}_{t]} \rangle
\otimes \vert 0_{(t} \rangle + \vert 0_{t]} \rangle \otimes \vert
{1_\xi}_{(t} \rangle ,  \label{eq:factor-additive-1}$$ where $$\vert {1_\xi}_{t]} \rangle = B^{-\ast}_t(\xi) \vert 0_{t]} \rangle, \ \text{and} \ \vert {1_\xi}_{(t} \rangle = B^{+\ast}_t(\xi) \vert 0_{(t} \rangle ,$$ and $$B^-_t(\xi) = B(\xi \chi_{[0,t]}), \ \ B^+_t(\xi) = B(\xi \chi_{(t,\infty]}),
\ \ B(\xi) = B^-_t(\xi) + B^+_t(\xi) .  
\label{eq:B-xi-decomp-1}$$ Here, $\chi_{[0,t]}$ is the indicator function for the time interval $[0,t]$. Note that while $\vert 1_\xi \rangle$ has unit norm, we have $$\parallel \vert  {1_\xi}_{t]} \rangle \parallel^2 = \int_0^t \vert \xi(s) \vert^2
ds, \ \text{and} \ \parallel \vert {1_\xi}_{(t} \rangle \parallel^2 =
\int_t^\infty \vert \xi(s) \vert^2 ds .$$

A consequence of the additive decomposition (\[eq:factor-additive-1\]) and the definitions (\[eq:B-xi-decomp-1\]) is the following. Let $K(t)$ be a bounded operator 