 Solution

Besides the Gaussian fixed point, solve the nonzero-coupling fixed-point equations. The quartic equation gives ${\lambda }_{\ast }=2ϵ(1+{\rho }_{\ast }{)}^2/(3K_D)$. Substitute into the mass equation to get $2{\rho }_{\ast }+ϵ(1+{\rho }_{\ast })/3=0$, hence ${\rho }_{\ast }=-ϵ/(6+ϵ)$ within the displayed one-loop truncation. Since $K_4=2{\pi }^2/(2\pi {)}^4=1/(8{\pi }^2)$,These are the Wilson-Fisher fixed point coordinates in this cutoff scheme. The higher terms obtained by solving the truncated equations exactly should not be regarded as a full higher-order epsilon calculation.

The attraction claim in the PDF needs a thermal qualification. The stability matrix of a renormalization-group fixed point at this solution has entriesThe lower-left entry is $O({ϵ}^2)$, so its eigenvalues areIncreasing $\ell$ means approaching the infrared. The quartic scaling direction is attractive, but a perturbation in the thermal scaling direction grows as $e^{y_t\ell }$. Therefore the Wilson-Fisher fixed point is an infrared attractor on its tuned critical surface, not an attractive fixed point of the entire two-dimensional coupling plane. In the plane it is a saddle; an arbitrary small thermal perturbation is a concrete counterexample to unconditional attraction. This is the thermal relevant direction at the Wilson-Fisher fixed point.