 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-304/3/e/solution

μ^{ϵ}

N_{0} = 2 i (1 - x) \neq p + 4 m, N_{1} = i (1 - x) \neq p + m .

Using

Γ (ϵ /2) = 2/ ϵ - γ + O (ϵ)

gives

Σ (\neq p) = - \frac{e ^{2}}{16 π ^{2}} \int_{0}^{1} d x {\frac{2 N _{0}}{ϵ} + N_{0} [lo g \frac{4 π μ ^{2}}{Δ} - γ] - 2 N_{1}} + O (ϵ) .

2/ ϵ - γ + lo g 4 π

leaves

Σ_{\overline{MS}} (\neq p) = - \frac{e ^{2}}{16 π ^{2}} \int_{0}^{1} d x [N_{0} lo g \frac{μ ^{2}}{Δ} - 2 N_{1}] .

On the tree-level mass shell,

p^{2} = - m^{2}

and

i \neq p = - m

, so

Δ = x^{2} m^{2}

δ m_{\overline{MS}} = - \frac{3 e ^{2} m}{16 π ^{2}} (\frac{2}{ϵ} - γ + lo g 4 π) .

The pole condition

m_{phys} = m - Σ_{\overline{MS}} ∣_{i \neq p = - m} + O (e^{4})

therefore gives

m_{phys} = m (μ) [1 + \frac{e ^{2}}{8 π ^{2}} \int_{0}^{1} d x ((1 + x) lo g \frac{μ ^{2}}{x ^{2} m ^{2}} - x)] + O (e^{4}) .

Evaluating the elementary parameter integral yields the equivalent expression

m_{phys} = m (μ) [1 + \frac{e ^{2}}{4 π ^{2}} (1 + \frac{3}{4} lo g \frac{μ ^{2}}{m ^{2}})] + O (e^{4}) .

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