The in-vacuum satisfies $a_j^{\rm in}|0_{\rm in}\rangle=0$. For the out-mode number operator $N_i^{\rm out}=a_i^{{\rm out}\dagger}a_i^{\rm out}$, substitute the operator transformation and use the <canonical commutation relations>. Only the $a_j^{\rm in}a_k^{{\rm in}\dagger}$ contraction survives, giving the <particle number from Bogoliubov coefficients>
$$
\boxed{
\langle0_{\rm in}|N_i^{\rm out}|0_{\rm in}\rangle
=\sum_j|\beta_{ij}|^2}.
$$
For a continuous mode label, the sum becomes the corresponding integral with Dirac-delta normalization.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 311 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-311/4/b/solution