The operator $\nabla^{2m}\phi^{2n}$ has <engineering dimension>
$$
2m+2n\frac{d-2}{2}=2m+n(d-2).
$$
The action integral is dimensionless, so
$$
\boxed{[\alpha]=d-2m-n(d-2)=2n-2m-(n-1)d}.
$$
Thus $\alpha$ is a <relevant coupling>, <marginal coupling>, or <irrelevant coupling> according as
$$
\boxed{
d<\frac{2(n-m)}{n-1},\qquad
d=\frac{2(n-m)}{n-1},\qquad
d>\frac{2(n-m)}{n-1}}
$$
respectively.

Solution

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

 Solution

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