The free energy is dimensionless in units with $k_BT=1$, so the integrand has momentum dimension $d$. Since a <derivative> has dimension one, the kinetic term gives
$$
2+2[\phi]_{\rm eng}=d,
\qquad
\boxed{[\phi]_{\rm eng}=\frac{d-2}{2}}.
$$
The mass term then gives
$$
[\mu_0^2]+2[\phi]_{\rm eng}=d,
\qquad
\boxed{[\mu_0^2]=2}.
$$
These are <engineering dimensions>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 303 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-303/2/b/solution