Past exam of the quantum field theory course of the University of Oxford / 2018 / 3 / a / Solution

A phi-fourth theory interaction has mass dimension $4-D$, so its bare coefficient must contain a renormalization scale factor ${\mu }^{4-D}=({\mu }^2{)}^{2-D/2}$ when expressed through a dimensionless renormalized coupling. The tree-level four-leg interaction vertex is linear in that coupling. At the next order, two quartic vertices form the three one-loop bubble diagram graphs shown in c, with external pairings $12|34$, $13|24$ and $14|23$.

For each pairing there are two identical internal lines and Feynman-diagram symmetry factor 2. The loop integral has the large-momentum behaviourIt is logarithmically divergent in four dimensions. Dimensional regularization represents that local divergence by a simple pole proportional to $1/(D-4)$. Its residue is independent of external momenta, so a local quartic counterterm cancels it. The three graphs sum to one constant coefficient; this argument establishes its structure without requiring its value.

The relation consequently has the formThe sign of $C$ incorporates the chosen pole denominator and vertex convention. In particular, a negative $C$ here corresponds to a positive coefficient of $1/(4-D)$ in the bare coupling. Finite pieces depend on the renormalization scheme; a pure-pole relation is natural in minimal subtraction scheme. One-loop two-point tadpole diagrams contribute mass renormalization, but not the momentum-dependent field renormalization that could change this coupling relation at order $g^2$. The four-point object relevant here is the amputated one-particle-irreducible vertex; external-leg self-energy insertions are dealt with separately.