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

The in-vacuum satisfies $a_j^{\mathrm{in}}|0_{\mathrm{in}}⟩=0$. For the out-mode number operator $N_i^{\mathrm{out}}=a_i^{\mathrm{out}†}{a}_i^{\mathrm{out}}$, substitute the operator transformation and use the canonical commutation relations. Only the $a_j^{\mathrm{in}}{a}_k^{\mathrm{in}†}$ contraction survives, giving the particle number from Bogoliubov coefficientsFor a continuous mode label, the sum becomes the corresponding integral with Dirac-delta normalization.