Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 304 / 1 / b

Solution

For

n = 2

, the coupling is a mass squared,

m^{2} = g

. After integration by parts, the quadratic action has kernel

K = - (□ + g - i 0)

. Completing the square in the Gaussian integral and choosing the source-independent normalization so that

Z_{2} [0] = 1

gives

Z_{2} [J] = exp [- \frac{i}{2} \int d^{d} x d^{d} y J (x) K^{- 1} (x - y) J (y)] .

Equivalently, in momentum space,

Z_{2} [J] = exp [- \frac{i}{2} \int \frac{d ^{d} p}{( 2 π ) ^{d}} \frac{J ( - p ) J ( p )}{p ^{2} - g + i 0}] .

The

i 0

prescription selects the Feynman propagator.

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