= Solution
In conformal time the interaction Hamiltonian is
$$
H_I(\eta)=-\frac{\lambda}{4!}\int d^3x\,a^4(\eta)\varphi^4(\eta,\mathbf x).
$$
The first-order <in-in formalism> formula and <Wick theorem> give the connected <primordial trispectrum>
$$
\begin{aligned}
\langle\varphi_{\mathbf k_1}\varphi_{\mathbf k_2}
\varphi_{\mathbf k_3}\varphi_{\mathbf k_4}\rangle_c
={}&2\lambda(2\pi)^3\delta^{(3)}\!\left(\sum_a\mathbf k_a\right)\\
&\times\operatorname{Im}\left[
\prod_{a=1}^4f_{k_a}^*(\tau)
\int_{-\infty(1-i\epsilon)}^\tau d\eta\,
a^4(\eta)\prod_{a=1}^4f_{k_a}(\eta)
\right].
\end{aligned}
$$
The factor $4!$ from the <Wick contractions> cancels the vertex factor. Writing $K=k_1+k_2+k_3+k_4$, the powers of $\eta$ cancel because $a^4\prod_af_{k_a}$ is constant apart from its phase, and the <i-epsilon prescription> gives
$$
\int_{-\infty(1-i\epsilon)}^\tau e^{-icK\eta}\,d\eta
=\frac{i}{cK}e^{-icK\tau}.
$$
Consequently
$$
\boxed{
\langle\varphi_{\mathbf k_1}\varphi_{\mathbf k_2}
\varphi_{\mathbf k_3}\varphi_{\mathbf k_4}\rangle_c
=(2\pi)^3\delta^{(3)}\!\left(\sum_a\mathbf k_a\right)
\frac{\lambda H^4\tau^4}{8c^5K\,k_1k_2k_3k_4}}.
$$
The full four-point function also contains the three disconnected products of the free two-point function found in part c.
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