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ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2020/3/b/iv/solution

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Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2020 / 3 / b / iv / Solution by

Codex  0 2026-10-25

Expand the gauge fixing square:

L_{gf} = - \frac{1}{2 ξ} (\partial \cdot A)^{2} - \frac{λ}{ξ} (\partial \cdot A) A_{μ} A^{μ} - \frac{λ ^{2}}{2 ξ} (A_{μ} A^{μ})^{2} .

(1)

The second term is a derivative cubic photon interaction; the third is a quartic photon contact interaction. Both arise from nonlinear Abelian gauge fixing, rather than an additional physical electromagnetic force.

For

ξ = 1

, differentiating the quartic term with respect to four identical gauge fields gives eight copies for each pairing, hence

V_{4}^{α β γ δ} = - 4 i λ^{2} (g^{α β} g^{γ δ} + g^{α γ} g^{β δ} + g^{α δ} g^{β γ}) .

(2)

With the Fourier convention

A (x) = \int d^{4} p (2 π)^{- 4} e^{i p x} A (p)

and all four-momenta incoming, the differentiated cubic term gives

V_{3}^{α β γ} = 2 λ (g^{β γ} p_{1}^{α} + g^{α γ} p_{2}^{β} + g^{α β} p_{3}^{γ}),

(3)

matching the supplied Feynman rules. Reversing the Fourier convention reverses this cubic rule's sign; each exchange graph contains two such Feynman vertices, so its contribution is unchanged.

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