 Solution

ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2019/1/a/iv/solution

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

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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