Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 304 / 3 / i / Solution

For a connected Feynman diagram with cubic vertices with four external legs and one loop, $3V=2I+4$ and $1=I-V+1$ give $V=I=4$. In a one-particle-irreducible Feynman diagram, every internal edge lies on the single cycle. Each of the four cubic vertices therefore has two internal and one external leg, giving a box Feynman diagram. With labelled external momenta there are three inequivalent cyclic orderings, modulo rotation and reversal, represented by $1234$, $1243$ and $1324$:

Take all momenta incoming with ${\sum }_i{p}_i=0$. The Feynman rules for the Euclidean action give a propagator $(q^2+m^2{)}^{-1}$ and a cubic insertion $-g$. For order $1234$, the amputated connected insertion iswhere $\chi (q)=1_{\{\Lambda <|q|<{\Lambda }_0\}}$ is the high-mode projector for a strict Wilsonian effective action. Each labelled box has Feynman-diagram symmetry factor one. A common fixed-loop-cutoff convention instead restricts only the chosen $q$ integration to the shell; both prescriptions give the same zero-external-momentum value. The effective-action vertex has the opposite sign to this connected insertion. Thus $\boxed{\text{three labelled boxes, with one cyclic topology}.}$