Let $O=\varphi(\mathbf k_1)\varphi(\mathbf k_2)$. Since $Q_S=-2\omega_{ij}\Pi_{ij}(\mathbf0)$,
$$
i\langle[Q_S,O]\rangle
=4\omega_{ij}\operatorname{Im}\langle\Pi_{ij}(\mathbf0)O\rangle.
$$
For a soft mode $\gamma^s=f_q a_s+f_q^*a_s^\dagger$ and $\Pi^s=g_q a_s+g_q^*a_s^\dagger$, the vacuum obeys
$$
\langle0|\Pi^s(\mathbf q)
=\frac{g_q}{f_q}\langle0|\gamma^s(\mathbf q).
$$
The mode-function <Wronskian normalization>
$$
f_qg_q^*-f_q^*g_q=i
$$
implies
$$
\operatorname{Im}\frac{g_q}{f_q}
=-\frac1{2|f_q|^2}=-\frac1{2P_\gamma(q)},
$$
where $P_\gamma(q)$ is the <graviton power spectrum> per polarization. The equal-time bispectrum is real in this parity-even configuration, and therefore
$$
\boxed{
i\langle[Q_S,O]\rangle'
=-2\omega_{ij}\lim_{\mathbf q\to0}
\frac{\langle\gamma_{ij}(\mathbf q)
\varphi(\mathbf k_1)\varphi(\mathbf k_2)\rangle'}{P_\gamma(q)}}.
$$
This turns the charge insertion into the soft-graviton insertion used in the <Soft graviton theorem>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 312 / 2 / iii / Solution

{title2=$P_\gamma(k)$}

The graviton power spectrum is the two-point spectrum of a tensor metric perturbation. A convention may quote it per polarization or summed over both polarizations, so that choice must accompany its normalization.


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 Incoming links: Graviton power spectrum