Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 304 / 2

The first three terms are the renormalized kinetic, mass, and quartic interaction terms. The remaining three are counterterms: $\delta Z$ gives wave-function renormalization, $\delta m^2$ renormalizes the mass, and $\delta \lambda$ renormalizes the four-point coupling. With all momenta entering a vertex, the momentum-space Feynman rules aretogether with momentum conservation at every vertex and an integral $\int d^{d}p/(2\pi {)}^d$ for each independent loop momentum.

The amputated connected four-point function through one loop contains the tree quartic vertex, the quartic-counterterm vertex, and three bubble diagrams in the $s$, $t$, and $u$ channels. The unamputated connected function also has one-loop self-energy and two-point-counterterm insertions on each external propagator, which do not determine $\delta g$.

For channel momentum $P$, the bubble has symmetry factor $1/2$ and the Feynman parameter identity gives an integral proportional toIn dimensional regularization, $\Gamma (ϵ/2)=2/ϵ+O(1)$, so the pole from one channel is $g^2/(16{\pi }^2ϵ)$. Summing the three crossing channels gives $3g^2/(16{\pi }^2ϵ)$. The minimal subtraction scheme cancels only this pole, hence

The bare coupling does not depend on the renormalization scale. Expressing it through the renormalized field and coupling gives, to the stated order,At one loop $\delta Z=0$. Put $a=3/(16{\pi }^2)$ and $\delta g=a{g}^2/ϵ$. If ${\beta }_g=\mu dg/d\mu$, differentiation and expansion through $g^2$ yieldSince $\lambda ={\mu }^{ϵ}g$,and in four dimensions