Particle number from Bogoliubov coefficients

Ifthen the expected out-particle number in mode $i$ in the in-vacuum is

If $b_s={\sum }_j(A_{sj}{a}_j+B_{sj}{a}_j^{†})$ and the state is the in-vacuum, put $n_s={\sum }_j|B_{sj}{|}^2$ and $c_s={\sum }_j{A}_{sj}{B}_{sj}$. The canonical commutation relations give $N_s^2=b_s^{†2}{b}_s^2+N_s$. Acting with $b_s^2$ on the vacuum produces $c_s|0⟩$ and a two-particle vector whose squared norm is $2n_s^2$, because its two possible vacuum contractions have equal contributions. This proves the displayed second moment and $\mathrm{Var}(N_s)=n_s(n_s+1)+|c_s{|}^2$. A thermal-mode formula requires $c_s=0$; a general squeezed vacuum state need not satisfy that restriction.