Since $dt=a\,d\tau$ and $\dot\phi=a^{-1}\phi'$, the interaction is
$$
S_{\rm int}=\frac\lambda{3!}\int d\tau\,d^3x\,a^2\phi'^2\phi,
\qquad
H_{\rm int}=-\frac\lambda{3!}\int d^3x\,a^2\phi'^2\phi
$$
to first order in $\lambda$. The tree-level <in-in formalism> gives
$$
\langle\phi_{\mathbf k_1}\phi_{\mathbf k_2}\phi_{\mathbf k_3}\rangle
=2\operatorname{Im}\int_{-\infty(1-i\epsilon)}^0d\tau\,
\langle0|\phi_{\mathbf k_1}(0)\phi_{\mathbf k_2}(0)
\phi_{\mathbf k_3}(0)H_{\rm int}(\tau)|0\rangle.
$$
There are two <Wick contractions> for each choice of the undifferentiated field at the vertex. With $K=k_1+k_2+k_3$, the stated <Bunch-Davies vacuum> mode obeys
$$
f_k(0)=\frac{H}{\sqrt{2k^3}},
\qquad
f_k^{*\prime}(\tau)=\frac{Hk^2\tau}{\sqrt{2k^3}}e^{ik\tau}.
$$
The two powers of $\tau$ from the differentiated modes cancel $a^2=1/(H^2\tau^2)$, and the remaining integral is
$$
\int_{-\infty(1-i\epsilon)}^0
(1-ik_i\tau)e^{iK\tau}\,d\tau
=-i\frac{K+k_i}{K^2}.
$$
Removing the momentum-conserving delta function, the <bispectrum from a time-derivative cubic scalar interaction> is therefore
$$
\boxed{
B(k_1,k_2,k_3)=
\frac{\lambda H^4}{12k_1^3k_2^3k_3^3K^2}
\left[
k_2^2k_3^2(K+k_1)+k_3^2k_1^2(K+k_2)
+k_1^2k_2^2(K+k_3)
\right]}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 312 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-312/1/b/solution