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

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