Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2019 / 1 / a / iii

Table of contents
- Solution iii

Solution

There is one four-gluon tadpole diagram. Temporarily replace its internal gluon propagator by

- i g_{ρ σ} δ^{c d} / (l^{2} - m_{g}^{2} + i 0)

. Contract the supplied four-gluon Feynman vertex with

g_{ρ σ} δ^{c d}

. Its first colour factor vanishes because

f^{cce} = 0

. Each of the other two terms contributes

C_{A} δ^{ab} (D - 1) g^{μν}

, so

δ^{c d} g_{ρ σ} V_{4}^{μν ρ σ, ab c d} = - 2 i g_{D}^{2} C_{A} δ^{ab} (D - 1) g^{μν} .

Include the symmetry factor

1/2

and the factor

- i

from the internal gluon propagator:

i Π_{4 g}^{μν, ab} = - g_{D}^{2} C_{A} δ^{ab} (D - 1) g^{μν} J_{1} (m_{g}^{2}) .

The exact regulated tadpole integral is

J_{1} (m_{g}^{2}) = - \frac{i}{( 4 π ) ^{D /2}} Γ (1 - D /2) (m_{g}^{2})^{D /2 - 1} .

In particular,

i Π_{4 g}^{μν, ab}_{UV} = - \frac{i g ^{2} C _{A} δ ^{ab}}{8 π ^{2} ϵ} (D - 1) m_{g}^{2} g^{μν} .

The entire expression is proportional to

(m_{g}^{2})^{1 - ϵ /2}

and vanishes as the auxiliary mass tends to zero near four dimensions. Equivalently, its expanded terms are proportional to

m_{g}^{2} / ϵ

,

m_{g}^{2} lo g m_{g}^{2}

and

m_{g}^{2}

, all of which disappear when the regulator is removed. Hence

Π_{4 g}^{μν, ab} = 0 (m_{g} \to 0) .

This is the massless scaleless integral result obtained directly in dimensional regularization. The artificial mass is only an intermediate device for this graph; retaining it in the full theory would modify the gauge symmetry and cannot be used when applying the massless transversality identity in part (iv).

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