The bare coupling does not depend on the <renormalization scale>. Expressing it through the renormalized field and coupling gives, to the stated order,
$$
0=\mu\frac d{d\mu}
\left[\mu^\epsilon(g+\delta g)-2\mu^\epsilon g\delta Z\right].
$$
At one loop $\delta Z=0$. Put $a=3/(16\pi^2)$ and $\delta g=ag^2/\epsilon$. If $\beta_g=\mu\,dg/d\mu$, differentiation and expansion through $g^2$ yield
$$
\boxed{\beta_g=-\epsilon g+\frac{3g^2}{16\pi^2}+O(g^3)}.
$$
Since $\lambda=\mu^\epsilon g$,
$$
\beta_\lambda=\mu\frac{d\lambda}{d\mu}
=\epsilon\lambda+\mu^\epsilon\beta_g
=\frac{3\mu^{-\epsilon}\lambda^2}{16\pi^2}+O(\lambda^3),
$$
and in four dimensions
$$
\boxed{\beta_\lambda=\frac{3\lambda^2}{16\pi^2}+O(\lambda^3)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 304 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-304/2/c/solution