Substitution of $\delta N=C\varphi$ and conversion to <conformal time> turn the interaction into
$$
S_{\rm int}=-\int d\eta\,d^3x\,a^2C\varphi(\varphi')^2.
$$
Treat $C$ and $H$ as constant at leading slow-roll order and use the massless de Sitter mode
$$
f_k(\eta)=\frac{H}{\sqrt{2k^3}}(1+ik\eta)e^{-ik\eta}.
$$
For $K=k_1+k_2+k_3$, the <in-in formalism> time integral at a vertex with leg $i$ undifferentiated is proportional to
$$
\int_{-\infty(1-i\epsilon)}^0(1-ik_i\eta)e^{iK\eta}\,d\eta
=-i\left(\frac1K+\frac{k_i}{K^2}\right).
$$
Summing the three choices of undifferentiated leg gives the <primordial bispectrum>
$$
\langle\varphi_{\mathbf k_1}\varphi_{\mathbf k_2}\varphi_{\mathbf k_3}\rangle
=(2\pi)^3\delta^{(3)}(\mathbf k_1+\mathbf k_2+\mathbf k_3)B_\varphi,
$$
$$
\boxed{B_\varphi
=-\frac{CH^4}{2(k_1k_2k_3)^3}
\sum_{\rm cyc}k_j^2k_l^2
\left(\frac1K+\frac{k_i}{K^2}\right)}.
$$
The sign follows from the interaction sign displayed in the question and $H_I=-L_{\rm int}$ at this order.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 312 / 2 / v / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-312/2/v/solution