 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-312/4/b/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

Write

A_{k} = H / 4 ϵ M_{Pl}^{2} k^{3}

, so

u_{k} (0) = A_{k}

. For the cyclic term with the first field undifferentiated, insertion into the in-in formalism gives

a^{2} \prod_{r} u_{k_{r}} (0) u_{k_{1}}^{*} u_{k_{2}}^{*'} u_{k_{3}}^{*'} = \frac{H ^{4}}{64 ϵ ^{3} M _{Pl}^{6} ( k _{1} k _{2} k _{3} ) ^{3}} k_{2}^{2} k_{3}^{2} (1 - i k_{1} τ) e^{i K τ},

where

K = k_{1} + k_{2} + k_{3} > 0

. The two differentiated modes contribute

τ^{2}

, cancelling the

τ^{- 2}

a^{2}

Along the vacuum-selected contour, the lower boundary vanishes. The elementary integrals are

\int_{- \infty (1 - i ε)}^{0} e^{i K τ} d τ = - \frac{i}{K}, \int_{- \infty (1 - i ε)}^{0} τ e^{i K τ} d τ = \frac{1}{K ^{2}},

and therefore

\int_{- \infty (1 - i ε)}^{0} (1 - i k_{1} τ) e^{i K τ} d τ = - i (\frac{1}{K} + \frac{k _{1}}{K ^{2}}) .

Multiplying by

4 i M_{Pl}^{2} ϵ^{2}

⟨ ζ_{1} ζ_{2} ζ_{3} ⟩_{c} B^{sf} = (2 π)^{3} δ^{(3)} (k_{1} + k_{2} + k_{3}) B^{sf}, = \frac{H ^{4}}{16 ϵ M _{Pl}^{4} ( k _{1} k _{2} k _{3} ) ^{3}} cyc \sum k_{2}^{2} k_{3}^{2} (\frac{1}{K} + \frac{k _{1}}{K ^{2}}) .

This matches the final normalization. The coefficient requires all six connected Wick contractions.

The calculation assumes the Gaussian Bunch-Davies vacuum, tree order in the specified cubic interaction, effectively constant

H

and

ϵ

during the integral, the approximate de Sitter scale factor, and observation after the modes freeze as

τ \to 0^{-}

. The contour tilt is kept until the early boundary has been eliminated; an undamped real-axis integral cannot simply discard that boundary. The reduced Planck mass convention is the one in the supplied De Sitter curvature mode function. Other cubic interactions and nonlinear field redefinitions are outside the specified model, so this is not by itself the complete bispectrum of a general single-field action.

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