 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-312/1/d/solution

H_{I} (η) = - \frac{λ}{4 !} \int d^{3} x a^{4} (η) φ^{4} (η, x) .

⟨ φ_{k_{1}} φ_{k_{2}} φ_{k_{3}} φ_{k_{4}} ⟩_{c} = 2 λ (2 π)^{3} δ^{(3)} (a \sum k_{a}) \times Im [a = 1 \prod 4 f_{k_{a}}^{*} (τ) \int_{- \infty (1 - i ϵ)}^{τ} d η a^{4} (η) a = 1 \prod 4 f_{k_{a}} (η)] .

The factor

4!

from the Wick contractions cancels the vertex factor. Writing

K = k_{1} + k_{2} + k_{3} + k_{4}

, the powers of

η

cancel because

a^{4} \prod_{a} f_{k_{a}}

is constant apart from its phase, and the i-epsilon prescription gives

\int_{- \infty (1 - i ϵ)}^{τ} e^{- i cKη} d η = \frac{i}{cK} e^{- i cK τ} .

Consequently

⟨ φ_{k_{1}} φ_{k_{2}} φ_{k_{3}} φ_{k_{4}} ⟩_{c} = (2 π)^{3} δ^{(3)} (a \sum k_{a}) \frac{λ H ^{4} τ ^{4}}{8 c ^{5} K k _{1} k _{2} k _{3} k _{4}} .

The full four-point function also contains the three disconnected products of the free two-point function found in part c.

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