Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 2 / a / ii

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- Solution ii

Solution

Use dimensional regularization with

d = 4 - ϵ

and choose Feynman gauge. Combining the electron and photon denominators with a Feynman parameter, shifting the loop momentum, and discarding the odd term leaves the numerator

(2 - d) (1 - x) γ^{μ} p_{μ} + d m .

The pole of the rotationally symmetric integral consequently gives

Σ_{pole} (p) = \frac{e ^{2}}{8 π ^{2} ϵ} \int_{0}^{1} d x [- 2 (1 - x) γ^{μ} p_{μ} + 4 m] = \frac{e ^{2}}{8 π ^{2} ϵ} (- γ^{μ} p_{μ} + 4 m) .

With the inverse-propagator convention

S^{- 1} (p) = γ^{μ} p_{μ} - m + δ_{2} γ^{μ} p_{μ} - δ_{m} - Σ (p),

the minimal subtraction scheme chooses the counterterm pole to equal

Σ_{pole}

. Hence

δ_{2} = - \frac{e ^{2}}{8 π ^{2} ϵ}, δ_{m} = - \frac{m e ^{2}}{2 π ^{2} ϵ} .

Solved by gpt-5.6-sol high.

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