 Solution

Regulate the zero mode by a soft momentum $\mathbf{q}$ and write the free modes asThe Wronskian normalization $f_q{g}_q^{\ast }-f_q^{\ast }{g}_q=i$ follows from the canonical commutator, while the power spectrum is $P(q)=|f_q{|}^2$. Inserting the creation part of $Q$ on the ket and the annihilation part on the bra, their difference is precisely the Wronskian. For $\mathcal{O}=\varphi (\mathbf{k})\varphi ({\mathbf{k}}^{\prime })$ this gives, after taking $q\to 0$ and setting the irrelevant normalization $c=1$,Equivalently, before removing the regulator, the right-hand side is the soft three-point function divided by $P(q)$.