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 Incoming links: Ghost-loop pole of the gluon self-energy

Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2019 / 1 / a / ii / Solution 2026-10-25

Route a directed ghost loop so that its internal four-momenta are

l

and

l + k

. At the external gluon with indices

(μ, a)

, let the incoming ghost carry colour

c

and the outgoing ghost carry colour

d

; the other Feynman vertex has the reverse order. The stated ghost-gluon Feynman rules supply

g_{D}^{2} f^{a c d} f^{b d c} (l + k)^{μ} l^{ν}

. Their colour factor is

f^{a c d} f^{b d c} = - C_{A} δ^{ab} .

The two ghost propagators supply

i^{2} = - 1

, while the closed ghost loop supplies another

- 1

. Thus, with the insertion convention stated above,

i Π_{gh}^{μν, ab} = - g_{D}^{2} C_{A} δ^{ab} I^{μν}, I^{μν} = \int \frac{d ^{D} l}{( 2 π ) ^{D}} \frac{( l + k ) ^{μ} l ^{ν}}{( l ^{2} + i 0 ) (( l + k ) ^{2} + i 0 )} .

This establishes every prefactor before integration. To isolate its ultraviolet pole, take generic off shell

k

, initially

k^{2} < 0

, so the massless integral has no overlapping infrared divergence. Introduce a Feynman parameter, shift

r = l + x k

, and set

Δ = - x (1 - x) k^{2} > 0

. Odd powers of

r

vanish, giving

I^{μν} = \int_{0}^{1} d x [\frac{g ^{μν}}{D} \int \frac{d ^{D} r}{( 2 π ) ^{D}} \frac{r ^{2}}{( r ^{2} - Δ + i 0 ) ^{2}} - x (1 - x) k^{μ} k^{ν} \int \frac{d ^{D} r}{( 2 π ) ^{D}} \frac{1}{( r ^{2} - Δ + i 0 ) ^{2}}] .

The needed dimensional regularization integrals follow from the supplied gamma function formula. In

D = 4 - ϵ

,

J_{1} (Δ) J_{2} (Δ) \int \frac{d ^{D} r}{( 2 π ) ^{D}} \frac{r ^{2}}{( r ^{2} - Δ + i 0 ) ^{2}} = \int \frac{d ^{D} r}{( 2 π ) ^{D}} \frac{1}{r ^{2} - Δ + i 0} = \frac{i Δ}{16 π ^{2}} \frac{2}{ϵ} + O (1), = \int \frac{d ^{D} r}{( 2 π ) ^{D}} \frac{1}{( r ^{2} - Δ + i 0 ) ^{2}} = \frac{i}{16 π ^{2}} \frac{2}{ϵ} + O (1), = J_{1} + Δ J_{2} = \frac{i}{16 π ^{2}} \frac{4Δ}{ϵ} + O (1) .

Here

Γ (ϵ /2) = 2/ ϵ + O (1)

, rather than

1/ ϵ

. Using

1/ D = 1/4 + O (ϵ)

and

\int_{0}^{1} x (1 - x) d x = 1/6

, obtain

I^{μν}_{UV} = - \frac{i}{96 π ^{2} ϵ} (k^{2} g^{μν} + 2 k^{μ} k^{ν}) .

Therefore the ghost-loop pole of the gluon self-energy is

Π_{gh}^{μν, ab}_{UV} = \frac{g ^{2} C _{A} δ ^{ab}}{96 π ^{2} ϵ} (k^{2} g^{μν} + 2 k^{μ} k^{ν}) .

The factor

μ^{ϵ}

in

g_{D}^{2}

affects only finite terms at this order. The isolated ghost loop is not transverse: contracting this result gives

g^{2} C_{A} δ^{ab} k^{2} k^{ν} / (32 π^{2} ϵ)

. Transversality is a condition on the complete gluon self-energy, including the other loop graphs.

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Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2019 / 1 / a / iv / Solution 2026-10-25

The Slavnov-Taylor identity implies transversality of the full perturbative gluon self-energy:

k_{μ} (Π_{3 g}^{μν, ab} + Π_{4 g}^{μν, ab} + Π_{gh}^{μν, ab} + Π_{q}^{μν, ab}) = 0.

The quark loop is separately transverse. This follows without evaluating its Feynman integral: contract its vector Feynman vertex with

k_{μ}

and use

\neq k = (\neq l + \neq k - m) - (\neq l - m)

between its two Dirac propagators. One obtains a difference of two loop integrals that cancel after a shift of

l

in dimensional regularization. The colour factor is proportional to

tr (T^{a} T^{b})

and does not obstruct this argument.

The four-gluon tadpole diagram vanishes after removal of

m_{g}

. Thus the nontransverse parts of the two-three-gluon-vertex loop and the ghost loop must cancel. Contracting the preceding ghost-loop pole of the gluon self-energy gives

k_{μ} Π_{3 g}^{μν, ab}_{UV} = - k_{μ} Π_{gh}^{μν, ab}_{UV} = - \frac{g ^{2} C _{A} δ ^{ab}}{32 π ^{2} ϵ} k^{2} k^{ν} .

C_{A} = 3

. No evaluation of the remaining three-gluon Feynman diagram is required. Its uncontracted transverse part is not fixed by this identity.

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