Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 304 / 2 / a

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- Solution a

Solution

Completing the square in the Gaussian functional integral and choosing

N^{- 1} = \int D ϕ exp (i \int d^{4} x L_{b})

normalizes

Z_{0} [0] = 1

. The result is

Z_{0} [J] = exp [- \frac{1}{2} \int d^{4} x, d^{4} y J (x) D_{F} (x - y) J (y)],

where the Feynman propagator in the convention of the question is

D_{F} (x - y) = \int \frac{d ^{4} p}{( 2 π ) ^{4}} \frac{i , e ^{- i p \cdot (x - y)}}{p ^{2} - M ^{2} + i ϵ} .

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