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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-301/3/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 301 / 3 / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

Varying the displayed gauge-fixed Lagrangian density gives the kinetic operator

D^{μ ρ} = η^{μ ρ} □ - (1 - \frac{1}{α}) \partial^{μ} \partial^{ρ} .

Its quantum field theory propagator should satisfy

D_{x}^{μ ρ} Δ_{ρ ν} (x - y) = i δ_{ν}^{μ} δ^{(4)} (x - y),

and inversion into transverse and longitudinal projectors gives the numerator

Π_{μν}^{correct} = - η_{μν} + (1 - α) \frac{k _{μ} k _{ν}}{k ^{2}} .

The paper instead prints

- η_{μν} + (α - 1) k_{μ} k_{ν} / k^{2}

. Except at

α = 1

, that is not the inverse of the displayed Lagrangian's kinetic operator. Taken literally, for

α \neq = 2

it is the Green function of

[η^{μ ρ} □ + \frac{α - 1}{2 - α} \partial^{μ} \partial^{ρ}] Δ_{ρ ν} = i δ_{ν}^{μ} δ^{(4)},

and at

α = 2

its longitudinal part is noninvertible. Thus the longitudinal sign in the printed propagator is a typographical error; the two forms coincide in Feynman gauge.

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