Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 301 / 3 / c

Table of contents
- Solution c

Solution

Since

D Δ = i δ

, the source-induced one-point function is

⟨ A_{μ} (x) ⟩_{j} = - i e \int d^{4} z Δ_{μ ρ} (x - z) j^{ρ} (z) .

The action is quadratic and the source is linear, so completing the square gives the exact full two-point function

G_{μν} (x, y) = Δ_{μν} (x - y) - e^{2} \int d^{4} z d^{4} w Δ_{μ ρ} (x - z) j^{ρ} (z) Δ_{ν σ} (y - w) j^{σ} (w) .

Diagrammatically these are a free line joining

x

to

y

and a disconnected pair of lines, each joining one external insertion to one current cross. The connected two-point function remains exactly

Δ_{μν}

. The expansion truncates at

e^{2}

because a Gaussian integral has no interaction vertices and its mean is linear in

e

.

Solved by gpt-5.6-sol high.

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