Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 303 / 2

A momentum-shell renormalization group transformation consists of three steps. First split $\varphi ={\varphi }_{-}+{\varphi }_{+}$ into slow modes with $|k|<\Lambda /\zeta$ and fast modes with $\Lambda /\zeta <|k|<\Lambda$, and integrate out ${\varphi }_{+}$ to obtain a Wilsonian effective action for ${\varphi }_{-}$. Next rescale momenta by $k^{\prime }=\zeta k$, equivalently coordinates by $x^{\prime }=x/\zeta$, to restore the cutoff to $\Lambda$. Finally rescale the field to restore the chosen normalization of the gradient term. Repeating these operations produces a renormalization-group flow of every permitted mass and coupling.

The free energy is dimensionless, while $d^{d}x$ has engineering dimension $-d$ and each derivative has dimension one. Requiringto be dimensionless gives $2+2[\varphi ]-d=0$. Thus the engineering dimension of the field is

A scalar field of scaling dimension ${\Delta }_{\varphi }$ has a scale-invariant two-point correlation function proportional to $|x{|}^{-2{\Delta }_{\varphi }}$. Comparison with the stated form givesThe difference $\eta /2$ from the engineering value is the field's anomalous dimension, generated by fluctuations and interactions at a non-Gaussian renormalization-group fixed point.

Ignoring interactions, every field has dimension $(d-2)/2$, and every Laplacian contributes two derivatives. The operator ${\varphi }^n({\nabla }^2\varphi {)}^m$ therefore has dimensionBecause the integrated interaction $\int d^{d}xg\mathcal{O}$ is dimensionless, the coupling has scaling dimensionIt is a relevant coupling when ${\Delta }_g>0$, a marginal coupling when ${\Delta }_g=0$, and an irrelevant coupling when ${\Delta }_g<0$.

Expand the quintic interaction after the slow-fast split. Its term with three slow fields and two fast fields isThe first term of the cumulant expansion therefore containsSince the coincident fast-mode propagator isthe lowest-order correction isIts Feynman diagram is one five-valent vertex with three external slow-field legs and the remaining two legs contracted into a tadpole diagram.

One cubic vertex cannot leave four external legs. Two cubic vertices can be joined by one contracted fast-field line, leaving four slow-field legs, so the first correction to ${\lambda }_0$ involving ${\alpha }_0$ isIt arises from the second term of the cumulant expansion.

A correction with three external legs can be made from one cubic and one quartic vertex by contracting four of their seven fields into internal lines. Consequently the first correction to ${\alpha }_0$ involving ${\lambda }_0$ isA purely quartic interaction cannot generate an odd interaction because its ${\mathbb{Z}}_2$ symmetry forbids odd powers of $\varphi$.