Put $d=6-\epsilon$. The one-loop two-point bubble has symmetry factor $1/2$ and, after a Feynman parameter and momentum shift, its pole is proportional to
$$
-\frac{2}{(4\pi)^3\epsilon}\int_0^1dx\,[m^2+x(1-x)p_E^2]
=-\frac{2}{(4\pi)^3\epsilon}\left(m^2+\frac{p_E^2}{6}\right).
$$
Cancelling the pole in the <minimal subtraction scheme> gives, with $\epsilon=6-d$,
$$
\boxed{\delta_Z=-\frac{\lambda^2}{6(4\pi)^3\epsilon},\qquad
\delta_{m^2}=-\frac{\lambda^2m^2}{(4\pi)^3\epsilon}}.
$$
If one defines dimensional regularization by $d=6-2\epsilon$, these same poles are written with $2\epsilon$ in place of $\epsilon$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 304 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-304/3/b/solution