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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 $P(q)$.

The positive-frequency solution for $u_k=a{f}_k$ is proportional to $e^{-ick\tau }$. Since $a=-1/(H\tau )$, its scalar-field mode function has the stated form $f_k=\mathcal{N}\tau e^{-ick\tau }$. The canonical momentum in conformal time is $\pi =a^2{\phi }^{\prime }$, so the canonical commutation relation requires the Wronskian normalizationSubstitution gives $2ick{a}^2|\mathcal{N}{|}^2{\tau }^2=i$. Hence, up to an irrelevant constant phase,The choice $e^{-ick\tau }$ is the Bunch-Davies vacuum condition at early conformal time.

Let $O=\phi ({\mathbf{k}}_1)\phi ({\mathbf{k}}_2)$. Since $Q_S=-2{\omega }_{ij}{\Pi }_{ij}(0)$,For a soft mode ${\gamma }^s=f_q{a}_s+f_q^{\ast }{a}_s^{†}$ and ${\Pi }^s=g_q{a}_s+g_q^{\ast }{a}_s^{†}$, the vacuum obeysThe mode-function Wronskian normalizationimplieswhere $P_{\gamma }(q)$ is the graviton power spectrum per polarization. The equal-time bispectrum is real in this parity-even configuration, and thereforeThis turns the charge insertion into the soft-graviton insertion used in the Soft graviton theorem.