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

Table of contents
- Solution d

Solution

The operator

\nabla^{2 m} ϕ^{2 n}

has engineering dimension

2 m + 2 n \frac{d - 2}{2} = 2 m + n (d - 2) .

The action integral is dimensionless, so

[α] = d - 2 m - n (d - 2) = 2 n - 2 m - (n - 1) d .

Thus

α

is a relevant coupling, marginal coupling, or irrelevant coupling according as

d < \frac{2 ( n - m )}{n - 1}, d = \frac{2 ( n - m )}{n - 1}, d > \frac{2 ( n - m )}{n - 1}

respectively.

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