= Solution
Suppress the mode label temporarily and let $a,b$ be out-operators. With $q=v/u$, the commuting-mode identity
$$
a\,e^{-qa^\dagger b^\dagger}
=e^{-qa^\dagger b^\dagger}(a-qb^\dagger)
$$
shows that $(a+qb^\dagger)e^{-qa^\dagger b^\dagger}|0_{\rm out}\rangle=0$. The analogous relation for $b$ gives the <antiparticle> condition. Thus in a finite regulated collection of modes,
$$
\boxed{|0_{\rm in}\rangle
=\mathcal N\prod_{\mathbf k}
\exp\!\left[-\frac{v_{\mathbf k}}{u_{\mathbf k}}
a_{\mathbf k}^{{\rm out}\dagger}
b_{-\mathbf k}^{{\rm out}\dagger}\right]
|0_{\rm out}\rangle.}
$$
For one particle-antiparticle mode pair, $(a^\dagger b^\dagger)^n|0,0\rangle=n!|n,n\rangle$. The exponential's factorial therefore cancels:
$$
|0_{\rm in}\rangle_{\mathbf k}
=\sqrt{1-|q_{\mathbf k}|^2}
\sum_{n=0}^{\infty}(-q_{\mathbf k})^n|n,n\rangle_{\rm out}.
$$
The resulting <particle-antiparticle two-mode squeezed vacuum> has the <geometric pair-number distribution>
$$
\boxed{P_{\mathbf k}(n)
=(1-r_{\mathbf k})r_{\mathbf k}^{\,n},
\quad n=0,1,\ldots,\qquad
r_{\mathbf k}=\left|\frac{v_{\mathbf k}}{u_{\mathbf k}}\right|^2
=\left(\frac{\omega-\Omega}{\omega+\Omega}\right)^2.}
$$
Here one pair means one <particle> in canonical mode $\mathbf k$ and one <antiparticle> in mode $-\mathbf k$. They are distinct species, and the state has equal <particle> and <antiparticle> occupation. In particular,
$$
\langle n_{\mathbf k}\rangle
=\frac{r_{\mathbf k}}{1-r_{\mathbf k}}=|v_{\mathbf k}|^2,\qquad
\operatorname{var}n_{\mathbf k}
=|v_{\mathbf k}|^2(1+|v_{\mathbf k}|^2).
$$
For a finite set, $|\mathcal N|=\prod_{\mathbf k}\sqrt{1-r_{\mathbf k}}=\prod_{\mathbf k}|u_{\mathbf k}|^{-1}$. An overall <phase> is arbitrary. With no pulse, $\Omega=\omega$, so $v=0$ and only the zero-pair state remains.
\Image[/past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2016/paper-1-pair-creation.png]
{title=Original sudden-pulse calculation: normalized frequency-matching coefficients and the resulting geometric pair-number distributions}
The <ultraviolet divergence of an instantaneous vector-potential quench> qualifies this formal product. For an exactly instantaneous vector-potential step in three spatial dimensions, at large $k$ the mixing behaves as $v_{\mathbf k}\sim e\mathbf A_0\cdot\widehat{\mathbf k}/(2\hbar k)$. Thus $\int\mathrm d^3k\,|v_{\mathbf k}|^2$ diverges in the ultraviolet. In a finite quantization box, a finite momentum cutoff makes the product a normalized Fock-space vector; a sufficiently smooth finite-duration pulse removes this particular instantaneous-quench pathology. The per-mode distribution remains the result above, while the unregulated infinite product should not be asserted to have a finite nonzero global normalization.
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