Put $r=\mu^2/\Lambda^2$ and $\widetilde g=\Lambda^{d-4}g$. At $d=4-\epsilon$, $\Omega_3/(2\pi)^4=1/(8\pi^2)$ gives
$$
\beta_r=2r+\frac{3}{2\pi^2}\frac{\widetilde g}{1+r},
\qquad
\beta_{\widetilde g}=\epsilon\widetilde g-\frac9{2\pi^2}
\frac{\widetilde g^2}{(1+r)^2}.
$$
The nonzero fixed point is
$$
\widetilde g_*=\frac{2\pi^2}{9}\epsilon+O(\epsilon^2),
\qquad
r_*=-\frac\epsilon6+O(\epsilon^2),
$$
which is the stated <Wilson-Fisher fixed point>.

Solved by gpt-5.6-sol high.

Solution

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

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.

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

For $d>4$, the quartic coupling has $y_g=4-d<0$, but setting it to zero removes the stabilizing term. It is a <dangerously irrelevant coupling>. At $t=0$, minimizing $g\phi^4-B\phi$ gives $\phi\sim(B/g)^{1/3}$ and $\delta=3$. This is consistent with Gaussian scaling because
$$
\frac{\Delta_B-y_g}{3}
=\frac{(d+2)/2-(4-d)}3
=\frac{d-2}{2}=\Delta_\phi.
$$
For $d<4$, $g$ flows to a nonzero <Wilson-Fisher fixed point>, so no dangerously irrelevant zero-coupling limit spoils the ordinary <scaling relation for critical exponents>[scaling relations].

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 303 / 3 / a / v / Solution

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{wiki=Wilson–Fisher_fixed_point}

Below four dimensions, the scalar quartic theory has an interacting Wilson-Fisher fixed point that controls the Ising universality class.


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 Incoming links: Wilson-Fisher fixed point