Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 312 / 4 / a / i / Solution

Use the comoving curvature perturbation mode normalization. Take $H,ϵ,c_s$ to be effectively constant at leading slow-roll approximation order, with $H,ϵ,c_s>0$, and use the Bunch-Davies vacuum selected by the early-time $i\epsilon$ contour. The regulator $\epsilon$ is distinct from the Hubble slow-roll parameter $ϵ$.

Let $C=ϵ+1-c_s^2$ and use the Fourier transform convention $\zeta (\mathbf{x})=\int d^{3}k\zeta (\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{x}}/(2\pi {)}^3$. Conversion to conformal time givesThe last sign follows from substituting $a=-1/(H\tau )$, which is positive for $\tau <0$; dropping the explicit minus sign in that substitution would reverse the final primordial bispectrum.

At the order needed by the in-in formalism, the unequal-time Wick contraction isIt uses the Gaussian quantum vacuum two-point function with the displayed operator ordering; an arbitrary classical unequal-time covariance cannot replace it. For the connected three-point function, each external field contracts with one of the three fields at the vertex. There are $3!=6$ such Wick contractions, all equal. The spatial integral conserves total momentum and leaves one factor $(2\pi {)}^3{\delta }^{(3)}({\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3)$. The other nine pairings are disconnected tadpole terms supported on a zero external momentum; they are excluded from the connected primordial bispectrum, equivalently by subtracting the one-point function or normal ordering the vertex.

The free Gaussian odd correlator is zero. Applying the stated first-order in-in formalism therefore givesBy definition, the full connected correlator is $(2\pi {)}^3{\delta }^{(3)}({\sum }_r{\mathbf{k}}_r)B$. The assumptions are a Gaussian Bunch-Davies vacuum, the given leading cubic vertex, tree order, the late-time limit $\tau \to 0^{-}$, effectively constant background parameters, and connected nonzero external modes. No symmetry factor $1/3!$ is inserted into the given Hamiltonian.