 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-46/4/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

(+ - - -)

and the Fourier transform convention

e^{- i p x}

. Introduce an even source

J

for the gauge field and odd sources

η, \overset{η}{ˉ}

Z [J, \overset{η}{ˉ}, η] = \frac{1}{Z [ 0 ]} \int D A D \overset{c}{ˉ} D c exp [i S + i \int d^{4} x (J^{a μ} A_{μ}^{a} + \overset{η}{ˉ}^{a} c^{a} + \overset{c}{ˉ}^{a} η^{a})] .

At tree level the two-point functions come from the quadratic action. Integrating the gauge kinetic term by parts gives its momentum-space kernel

K_{μν}^{ab} (p) = δ^{ab} [- p^{2} g_{μν} + (1 - ζ^{- 1}) p_{μ} p_{ν}] .

Its inverse follows from transverse and longitudinal projectors:

K = - p^{2} (P_{T} + ζ^{- 1} P_{L})

, so

K^{- 1} = - p^{- 2} (P_{T} + ζ P_{L})

. Completing the bosonic Gaussian square gives

Z_{0} [J] = exp [- i J K_{F}^{- 1} J /2]

, with the Feynman i-epsilon prescription in the inverse. Two source derivatives and their insertion factors give the gauge-boson propagator

D_{F} = i K_{F}^{- 1}

. Hence

⟨ Ω∣ T A_{μ}^{a} (x) A_{ν}^{b} (y) ∣Ω ⟩ = δ^{ab} \int \frac{d ^{4} p}{( 2 π ) ^{4}} e^{- i p (x - y)} \frac{- i}{p ^{2} + i 0} [g_{μν} - (1 - ζ) \frac{p _{μ} p _{ν}}{p ^{2} + i 0}] .

For

ζ = 1

this reduces to the Feynman-gauge adjoint propagator. The longitudinal double pole is understood with the compatible Feynman boundary prescription.

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