Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 303 / 2

A Wilsonian renormalization group step:

The resulting effective free energy defines the transformed couplings and their flow.

Solved by gpt-5.6-sol high.

At the Gaussian fixed point, set $a_d=12{\Omega }_{d-1}{\Lambda }^{d-2}/(2\pi {)}^d$. The linearized flow isThe mass direction, corresponding to ${\varphi }^2$, has coupling dimension $y_t=2$. For $d\ne 2$, the second eigendirection isand has $y_g=4-d$; it is the tadpole-subtracted mixture of ${\varphi }^4$ and ${\varphi }^2$. The corresponding operator dimensions are $d-2$ and $2d-4$.

Solved by gpt-5.6-sol high.

For $d<4$, $y_g=4-d>0$, so the quartic coupling is a relevant direction of a fixed point. Generic Ising-like systems therefore flow away from the Gaussian point toward the interacting Wilson-Fisher fixed point, which controls the Ising universality class below four dimensions.

Solved by gpt-5.6-sol high.

Put $r={\mu }^2/{\Lambda }^2$ and $\tilde{g}={\Lambda }^{d-4}g$. At $d=4-ϵ$, ${\Omega }_3/(2\pi {)}^4=1/(8{\pi }^2)$ givesThe nonzero fixed point iswhich is the stated Wilson-Fisher fixed point.

Solved by gpt-5.6-sol high.

Linearization at the fixed point givesThe relevant eigendirection is $(\delta r,\delta \tilde{g})=(1,O({ϵ}^2))$. An irrelevant eigendirection isThus the mass-like combination is relevant and the quartic-like combination is irrelevant; neither is marginal for $ϵ>0$.

Solved by gpt-5.6-sol high.

The coefficient $A$ comes from one sextic vertex with two external slow legs and two fast tadpole loops. The coefficient $B$ comes from one sextic vertex with four external slow legs and one fast tadpole. The coefficient $C$ comes from one quartic and one sextic vertex joined by two fast propagators, leaving six external slow legs.

Solved by gpt-5.6-sol high.

Choosing two slow legs in $h({\varphi }_{<}+{\varphi }_{>}{)}^6$ and contracting four fast legs gives $\left(\genfrac{}{}{0ex}{}{6}{2}\right)(3)h{I}_1^2{\varphi }_{<}^2$. Since the mass term is ${\mu }^2{\varphi }_{<}^2/2$,Choosing four slow legs and contracting the remaining pair gives $B=\left(\genfrac{}{}{0ex}{}{6}{4}\right)=15$. For the connected quartic-sextic cumulant, the factor is $\left(\genfrac{}{}{0ex}{}{4}{2}\right)\left(\genfrac{}{}{0ex}{}{6}{2}\right)2!=180$ and the cumulant has a minus sign, soAs a check, two quartic vertices give $-\frac{1}{2}{\left(\genfrac{}{}{0ex}{}{4}{2}\right)}^{2}2!=-36$, matching the supplied coefficient.

Solved by gpt-5.6-sol high.

The canonical eigenvalue of $h$ isThus ${\beta }_h=(-2+O(ϵ))h+O(g^3)+O(h^2)$. At a fixed point this forces $h_{\ast }=O(g_{\ast }^3)=O({ϵ}^3)$. Its feedback into the mass and quartic flows begins only at order ${ϵ}^3$, so the order-$ϵ$ fixed point from part (b)(iii) is unaffected.

Solved by gpt-5.6-sol high.