 Solution

A momentum-shell renormalization group step has three parts. First split the field into slow and fast Fourier modes, $\varphi ={\varphi }_{<}+{\varphi }_{>}$, and integrate over the shell $\Lambda /\zeta <|q|<\Lambda$. Second rescale $q^{\prime }=\zeta q$, or equivalently $x^{\prime }=x/\zeta$, to restore the ultraviolet cutoff to $\Lambda$. Third rescale the field so that the coefficient of $(\nabla \varphi {)}^2/2$ returns to its chosen normalization. The resulting free energy has the same operator expansion but new couplings; iteration traces a renormalization-group flow in coupling space.

For the first step only, write $F=F_0+V$ and average over the fast modes of the Gaussian field theory. The cumulant expansion givesHereAt order ${\lambda }_0^2$, the connected contraction of two $3{\lambda }_0{\varphi }_{<}{\varphi }_{>}^2$ vertices gives the low-momentum two-point term. Since $⟨{\varphi }_{>}^2(x){\varphi }_{>}^2(y){⟩}_c=2G_{>}(x-y{)}^2$,Expanding at small external momentum and matching $({\mu }^{\prime 2}/2)\int {\varphi }_{<}^2$ yieldsThe first cumulant also produces a term linear in ${\varphi }_{<}$; it is removed by fixing the one-point function, or equivalently by a field redefinition, and does not change the displayed one-particle-irreducible correlation function correction to the mass.

The leading vertex correction is order ${\lambda }_0^3$. Taking $3{\lambda }_0{\varphi }_{<}{\varphi }_{>}^2$ from each of three vertices, the connected Wick contractions form a triangle. There are eight contractions, so the third cumulant contributes $27\times 8/3!=36$ times the triangle integral. At zero external momentum,For nonzero external momenta the three propagators carry the corresponding shifted loop momenta, with every internal line restricted to the fast shell.