Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 48 / 3 / i / Solution

Let $p$ be the total external Euclidean momentum crossing one four-point channel. The two internal scalar propagators produce the scalar bubble integralUsing the Feynman parameter identity $1/(ab)={\int }_0^{1}dx[xa+(1-x)b{]}^{-2}$ and shifting the loop momentum yieldsIn phi-fourth theory the bubble diagram has two quartic vertices and a symmetry factor $1/2$ from exchanging its two identical internal lines. There are three partitions of four labelled external legs into two pairs, the $s,t,u$ channels. The connected amputated loop diagram has positive coefficient ${\lambda }^{2}B/2$, whereas the Euclidean one-particle-irreducible vertex in the effective action has the opposite sign. ThusOne can check this sign without external-line conventions by expanding the one-loop quantum effective action $\frac{1}{2}\mathrm{Tr}\log (K+{\mu }^{ϵ}\lambda {\varphi }^2/2)$. Its second logarithmic term is $-{\mu }^{2ϵ}{\lambda }^2\mathrm{Tr}(K^{-1}{\varphi }^2{K}^{-1}{\varphi }^2)/16$; four field derivatives give precisely the three negative bubble terms above.

The four-dimensional pole of $I_d$ is independent of its mass argument, so each channel contributes $2/[(4\pi {)}^2ϵ]$. The divergence of the proper four-point vertex is $-3{\lambda }^2/[(4\pi {)}^2ϵ]$. The minimal subtraction scheme therefore requiresFor a Minkowski interaction $-{\mu }^{ϵ}\lambda {\varphi }^4/4!$, the corresponding local interaction counterterm is $-{\mu }^{ϵ}\delta \lambda {\varphi }^4/4!$. No momentum-dependent four-point counterterm is needed for these logarithmic poles. This calculation concerns the requested four-point bubble contributions; a massive two-point tadpole and vacuum diagrams require their own mass and vacuum-energy subtractions if those functions are also renormalized.