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 Incoming links: Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2025 / 2 / a

Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2025 / 2 / c / Solution 2026-10-25

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

. It is a one-particle-irreducible Feynman diagram with no closed fermion loop sign. In Feynman gauge,

- 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},

or

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 ]} .

For a direct ultraviolet divergence extraction, use a Feynman parameter

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} .

As a check on the same fermion self-energy calculation,

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

This is the ultraviolet pole of QED fermion self-energy. It is safe to extract it at generic external momentum; an on shell or massless limit must distinguish any infrared divergence from this ultraviolet divergence.

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Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2025 / 2 / d / Solution 2026-10-25

In the right panel of part a,

p

enters the fermion line,

p^{'} = p + q

leaves it, and a virtual photon of momentum

k

bridges the external photon vertex. Write

Γ^{μ} = γ^{μ} + Λ^{μ}

and

F_{1} = 1 + F_{1}^{(1)} + \dots

. The one-loop amplitude is

- i e Λ^{μ} = \int_{k} (- i e γ^{α}) \frac{i ( \neq p ^{'} - \neq k + m )}{( p ^{'} - k ) ^{2} - m ^{2} + i 0} (- i e γ^{μ}) \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}, Λ^{μ} = - i e^{2} \int_{k} \frac{γ ^{α} ( \neq p ^{'} - \neq k + m ) γ ^{μ} ( \neq p - \neq k + m ) γ _{α}}{[ k ^{2} + i 0 ] [( p ^{'} - k ) ^{2} - m ^{2} + i 0 ] [( p - k ) ^{2} - m ^{2} + i 0 ]} .

This Feynman integral has a logarithmic superficial degree of divergence. Its ultraviolet pole is independent of the external momenta and

m

, because every such numerator insertion or denominator correction lowers the large-

k

degree. Preserve an auxiliary positive mass

M

while extracting the ultraviolet pole; setting everything to zero would create a scaleless integral and conceal the cancellation of ultraviolet and infrared divergences.

The leading numerator is

γ^{α} \neq k γ^{μ} \neq k γ_{α}

. Symmetric integration replaces

k_{ρ} k_{σ}

by

g_{ρ σ} k^{2} / d

. Two contractions of gamma matrices then give

γ^{α} \neq k γ^{μ} \neq k γ_{α} ⟼ \frac{( d - 2 ) ^{2}}{d} k^{2} γ^{μ}, tr [γ^{α} \neq k γ^{μ} \neq k γ_{α} γ_{μ}] = 4 (d - 2)^{2} k^{2} .

For the auxiliary-mass extraction of ultraviolet poles,

J_{M} = \int_{k} \frac{k ^{2}}{( k ^{2} - M ^{2} + i 0 ) ^{3}} = \int_{k} [\frac{1}{( k ^{2} - M ^{2} + i 0 ) ^{2}} + \frac{M ^{2}}{( k ^{2} - M ^{2} + i 0 ) ^{3}}], J_{M}_{UV} = \frac{i}{16 π ^{2} ϵ} .

The second term is ultraviolet convergent. Thus

\frac{1}{4 d} tr (Λ^{μ} γ_{μ})_{UV} = - i e^{2} \frac{( d - 2 ) ^{2}}{d} J_{M}_{UV}, F_{1}^{(1)} (q^{2})_{UV} = \frac{e ^{2}}{16 π ^{2} ϵ} .

The ultraviolet pole of the QED vertex is therefore

Λ^{μ} ∣_{UV} = e^{2} γ^{μ} / (16 π^{2} ϵ)

. The Pauli structure has zero trace against

γ_{μ}

and no ultraviolet pole at this order.

Form-factor qualification. The two electron electromagnetic vertex form factors in the source describe matrix elements between external on shell Dirac spinors. A general off shell vertex has more structures, so the unprojected trace is not a projector for the complete finite on-shell

F_{1}

. It does correctly isolate the divergent local

γ^{μ}

coefficient, which is all that is required here.

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