Continue the incoming positive-frequency mode through $\eta=0$ as
$$
u_-(\eta)=\alpha u_+(\eta)+\beta u_+^*(\eta)
\qquad(\eta>0).
$$
Continuity of the mode and its first derivative gives the <sudden frequency quench> coefficients
$$
\alpha=\frac{\omega_++\omega_-}{2\sqrt{\omega_+\omega_-}},
\qquad
\beta=\frac{\omega_+-\omega_-}{2\sqrt{\omega_+\omega_-}}.
$$
The associated <Bogoliubov transformation> says that the incoming vacuum has expected outgoing occupation number
$$
\boxed{\langle N_{k_z\mathbf k}^{(+)}\rangle
=|\beta|^2
=\frac{(\omega_+-\omega_-)^2}{4\omega_+\omega_-}}.
$$
It vanishes when $a_+=a_-$, as required when there is no change of geometry.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 311 / 4 / b / iii / Solution

For an oscillator whose positive frequency jumps from $\omega_-$ to $\omega_+$, continuity of the normalized mode and its first derivative gives the Bogoliubov coefficients
$$
\alpha=\frac{\omega_++\omega_-}{2\sqrt{\omega_+\omega_-}},
\qquad
\beta=\frac{\omega_+-\omega_-}{2\sqrt{\omega_+\omega_-}}.
$$
The initial vacuum therefore contains $|\beta|^2$ final particles in that mode.


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 Incoming links: Sudden frequency quench