= Solution
Distinguish in-operators and out-operators before identifying the symbols in the final <vacuum> condition. The full charged-field Fourier component before the pulse contains
$$
\frac1{\sqrt{2\omega}}
\left(a_{\mathbf k}^{\rm in}e^{-i\omega t}
+b_{-\mathbf k}^{{\rm in}\dagger}e^{i\omega t}\right).
$$
The conjugate incoming mode has the conjugate matched solution. Collecting the coefficients of $e^{-i\Omega t}$ and $e^{i\Omega t}$ after the pulse gives the real Bogoliubov map
$$
a_{\mathbf k}^{\rm out}=\alpha_{\mathbf k}a_{\mathbf k}^{\rm in}
+\beta_{\mathbf k}b_{-\mathbf k}^{{\rm in}\dagger},\qquad
b_{-\mathbf k}^{{\rm out}\dagger}
=\beta_{\mathbf k}a_{\mathbf k}^{\rm in}
+\alpha_{\mathbf k}b_{-\mathbf k}^{{\rm in}\dagger}.
$$
Since $\alpha^2-\beta^2=1$, the inverse map is
$$
a_{\mathbf k}^{\rm in}
=\alpha_{\mathbf k}a_{\mathbf k}^{\rm out}
-\beta_{\mathbf k}b_{-\mathbf k}^{{\rm out}\dagger}.
$$
The <Inverse Bogoliubov vacuum condition> therefore yields
$$
\boxed{u_{\mathbf k}=\alpha_{\mathbf k}
=\frac{\Omega+\omega}{2\sqrt{\Omega\omega}},
\qquad
v_{\mathbf k}=-\beta_{\mathbf k}
=\frac{\omega-\Omega}{2\sqrt{\Omega\omega}}.}
$$
Here the $a,b$ inside $(u_{\mathbf k}a_{\mathbf k}+v_{\mathbf k}b_{-\mathbf k}^\dagger)$ mean out-operators. Using the forward-map coefficient $\beta$ with the same plus sign would be inconsistent with this convention. Mode rephasings can change a displayed coefficient sign, but the map and squeezed-state exponent must change together.
The complete <in-vacuum> also obeys the <antiparticle> annihilation condition. In out-operators this is $(u_{\mathbf k}b_{-\mathbf k}^{\rm out}+v_{\mathbf k}a_{\mathbf k}^{{\rm out}\dagger})|0_{\rm in}\rangle=0$. The two conditions jointly select the <vacuum> sector used below; the <particle> condition alone would not exclude arbitrary incoming <antiparticle> excitations.
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