= Solution
Applying the scalar transformation from part i to both fields gives
$$
\langle\Delta O\rangle
=\omega_{ij}\sum_{a=1}^2k_{a i}\frac{\partial}{\partial k_{a j}}
\left[(2\pi)^3\delta^{(3)}(\mathbf k_1+\mathbf k_2)P_\varphi(k_1)\right].
$$
The derivative of the <Dirac delta distribution> is proportional to $\delta_{ij}$ and vanishes after contraction with traceless $\omega_{ij}$. Removing that delta function leaves
$$
\langle\Delta O\rangle'
=\omega_{ij}k_i\frac{\partial}{\partial k_j}P_\varphi(k).
$$
Equating the two sides of the <Ward-Takahashi identity> and resolving the soft graviton into a polarization $s$ yields the <cosmological soft-graviton consistency relation>
$$
\boxed{
\lim_{q\to0}
\frac{\langle\gamma^s(\mathbf q)
\varphi(\mathbf k)\varphi(-\mathbf k-\mathbf q)\rangle'}{P_\gamma(q)}
=-\frac12\epsilon_{ij}^s(\mathbf q)
k_i\frac{\partial}{\partial k_j}P_\varphi(k)}.
$$
For an isotropic power spectrum this is equivalently
$$
-\epsilon_{ij}^s k_i k_j\frac{\partial P_\varphi}{\partial k^2}.
$$
The long-wavelength <adiabatic tensor mode> acts on the short two-point function as an anisotropic rescaling of its momentum.
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