 Solution

ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2020/1/v/solution

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Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2020 / 1 / v / Solution by

Codex  0 2026-10-25

Keep

m > 0

and set

p = p^{'} = 0

, which is an off shell choice. The numerator is

N^{μ} = γ^{ν} (- \neq k + m) γ^{μ} (- \neq k + m) γ_{ν} .

Terms odd in

k

integrate to zero. Lorentz covariance and dimensional regularization give, inside this integral,

k_{ρ} k_{σ} ⟼ \frac{k ^{2}}{D} g_{ρ σ}, \neq k γ^{μ} \neq k ⟼ \frac{2 - D}{D} k^{2} γ^{μ} .

Applying the gamma matrices identity from part (ii) once more yields

N^{μ} ⟼ [\frac{( D - 2 ) ^{2}}{D} k^{2} + (2 - D) m^{2}] γ^{μ} .

Consequently

i V_{loop}^{μ} (0, 0) = - e_{D}^{3} γ^{μ} \int \frac{d ^{D} k}{( 2 π ) ^{D}} \frac{α k ^{2} + β m ^{2}}{( k ^{2} + i 0 ) ( k ^{2} - m ^{2} + i 0 ) ^{2}}, α = \frac{( 2 - ϵ ) ^{2}}{4 - ϵ}, β = ϵ - 2.

This massive zero-momentum QED vertex integral uses an integrated tensor identity for the replacement of

N^{μ}

, not an equality for each fixed

k

. The physical photon remains massless. The retained fermion mass supplies a scale and makes this zero-external-four-momentum calculation suitable for extracting the ultraviolet pole without a scaleless integral.

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