The early and late stationary regions define distinct positive-frequency mode bases and hence an in-vacuum and out-vacuum. Write their <Bogoliubov transformation> as
$$
u_i^{\rm out}=\sum_j
\left(\alpha_{ij}u_j^{\rm in}+\beta_{ij}u_j^{{\rm in}*}\right).
$$
The corresponding operators satisfy
$$
\widehat b_i^{\rm out}=\sum_j
\left(\alpha_{ij}^*\widehat a_j^{\rm in}
-\beta_{ij}^*\widehat a_j^{{\rm in}\dagger}\right).
$$
Using $\widehat a_j^{\rm in}|0_{\rm in}\rangle=0$ and the canonical commutator gives the <particle number from Bogoliubov coefficients>:
$$
\boxed{\langle0_{\rm in}|\widehat N_i^{\rm out}|0_{\rm in}\rangle
=\sum_j|\beta_{ij}|^2}.
$$
Time dependence in the sandwich region can make $\beta_{ij}\neq0$, so the in-vacuum contains out-particles.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 311 / 4 / b / i / Solution

 Solution

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