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

Solution

Substitution of

δ N = Cφ

and conversion to conformal time turn the interaction into

S_{int} = - \int d η d^{3} x a^{2} Cφ (φ^{'})^{2} .

Treat

C

and

H

as constant at leading slow-roll order and use the massless de Sitter mode

f_{k} (η) = \frac{H}{2 k ^{3}} (1 + ik η) e^{- ik η} .

For

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

i

undifferentiated is proportional to

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

Summing the three choices of undifferentiated leg gives the primordial bispectrum

⟨ φ_{k_{1}} φ_{k_{2}} φ_{k_{3}} ⟩ = (2 π)^{3} δ^{(3)} (k_{1} + k_{2} + k_{3}) B_{φ},

B_{φ} = - \frac{C H ^{4}}{2 ( k _{1} k _{2} k _{3} ) ^{3}} cyc \sum k_{j}^{2} k_{l}^{2} (\frac{1}{K} + \frac{k _{i}}{K ^{2}}) .

The sign follows from the interaction sign displayed in the question and

H_{I} = - L_{int}

at this order.

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