 Solution

ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2025/2/c/solution

The middle panel in part a has an internal photon of momentum

k

joining two points on a fermion line; the internal fermion has momentum

p - k

- i Σ (p) = \int_{k} (- i e γ^{α}) \frac{i ( \neq p - \neq k + m )}{( p - k ) ^{2} - m ^{2} + i 0} (- i e γ^{β}) \frac{- i g _{α β}}{k ^{2} + i 0},

Σ (p) = - i e^{2} \int_{k} \frac{( 2 - d ) ( \neq p - \neq k ) + d m}{[ k ^{2} + i 0 ] [( p - k ) ^{2} - m ^{2} + i 0 ]} .

The decomposition in the source gives

tr (\neq p Σ) = 4 p^{2} S (p^{2})

. In particular,

S (p^{2}) = - \frac{i e ^{2} ( d - 2 )}{p ^{2}} \int_{k} \frac{p \cdot k - p ^{2}}{[ k ^{2} + i 0 ] [( p - k ) ^{2} - m ^{2} + i 0 ]} .

x

multiplying the massive denominator, shift

r = k - x p

Δ = x m^{2} - x (1 - x) p^{2}

. Then

\neq p - \neq k = (1 - x) \neq p - \neq r

and

\int_{0}^{1} d x \int_{r} \frac{1}{( r ^{2} - Δ + i 0 ) ^{2}}_{UV} = \frac{i}{16 π ^{2} ϵ} \int_{0}^{1} d x .

Odd

r

terms integrate to zero. Taking the trace explicitly supplies the weight

(2 - d) (1 - x)

, whose integral tends to

- 1

. Hence

tr (\neq p Σ)_{UV} = - \frac{4 e ^{2} p ^{2}}{16 π ^{2} ϵ}, S (p^{2})_{UV} = - \frac{e ^{2}}{16 π ^{2} ϵ} .

Σ (p)_{UV} = \frac{e ^{2}}{16 π ^{2} ϵ} (- \neq p + 4 m) .

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