Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 306 / 3 / b / Solution

After integration by parts, the quadratic operator in the action is

K = - π^{- 1} \partial \overset{ˉ}{\partial}

. Its Green function

G (z, w) = ⟨ X (z, \overset{z}{ˉ}) X (w, \overset{w}{ˉ})⟩

therefore satisfies

- \frac{1}{π} \partial \overset{ˉ}{\partial} G (z, w) = δ^{(2)} (z - w) .

Using the given distributional identity gives

⟨ X (z, \overset{z}{ˉ}) X (w, \overset{w}{ˉ})⟩ = - \frac{1}{2} lo g ∣ z - w ∣^{2} + C,

where the additive constant depends on the infrared convention and cancels from neutral correlators.

Solved by gpt-5.6-sol high.

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