For small $\boldsymbol\pi$, expand $(1-\boldsymbol\pi^2)^{-1}=1+O(\boldsymbol\pi^2)$ and retain the quartic interaction $(\boldsymbol\pi\mathbin\cdot\partial_i\boldsymbol\pi)^2/(2g_0)$. Split $\boldsymbol\pi=\boldsymbol\pi^-+\boldsymbol\pi^+$ and contract the fast fields in the shell $\Lambda/\zeta<|q|<\Lambda$. To first order in
$$
I_d=\int_{\Lambda/\zeta}^{\Lambda}\frac{d^dq}{(2\pi)^d}\frac1{q^2},
$$
the contraction adds $I_d$ to the inverse coefficient of the slow-field kinetic term. The prescribed <wave-function renormalization>
$$
\boldsymbol\pi'=\frac{\boldsymbol\pi^-}{A},
\qquad
A=1-\frac12(N-1)g_0I_d,
$$
then multiplies that coefficient by $A^2$. Hence
$$
\left(\frac1{g_0}+I_d\right)A^2
=\frac1{g_0}-(N-2)I_d+O(g_0I_d^2).
$$
Finally the coordinate rescaling contributes $\zeta^{d-2}$, and thus
$$
\boxed{\frac1{g(\zeta)}=\zeta^{d-2}\left[\frac1{g_0}+(2-N)I_d\right]}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 303 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-303/3/c/solution