Dimensional regularisation in $phi^4$ theory

Question

My question is in regards to the 1-loop corrections of phi 4 theory.

The question is in regards to these notes: http://www.damtp.cam.ac.uk/user/dbs26/AQFT/chap5.pdf

On page 15 of these notes (PDF page 15, actual page 96) I don't understand the argument with the integral measure as to why we only get

$$
g^2 mu^{epsilon}
$$
as the integral prefactor and not
$$
g^2 mu^{2epsilon}
$$
The argument pretty much says that is makes the whole integral dimensionless (or is it simply making that comment in passing im not sure), but then why not also do this on PDF page 13 where the integral is not dimensionless? Would this not all have a bearing in the beta function at the end?

Hans Moleman · Answer

This is just bookkeeping. The correction to $lambda$ is
$$delta lambda sim lambda^2 times text{loop integral} = g^2 (mu^{4-d})^2 times text{loop integral}$$
so
$$delta g sim g^2 mu^{4-d} times text{loop integral}.$$

Dimensional regularisation in $phi^4$ theory

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