Are large logarithms tied to dimensional regularization?

Question

When introducing the renormalization group many textbooks start by stating that scattering amplitudes often contain logarithms of the form $\log\frac{\mu^2}{p^2}$, where $p^2$ is the characteristic energy scale of the process and $\mu^2$ is a parameter related to renormalization. $\mu$ either comes from dimensional regularization where it is introduced to keep the right dimensions or it could be a momentum cut-off (then usually called $\Lambda$). It appears to me (for example from reading Schwartz QFT chapter 23) that these large logarithms always arise, independently of the regularization scheme used. I can see how they would enter in dimensional regularization, since loop integrals will always be proportional to $\mu^{4-\epsilon}$ in $4-\epsilon$ dimensions which gives a term containing $\log \mu$ if expanded in $\epsilon$. However, if I used a momentum cutoff $\Lambda$ on a quadratically divergent integral the result would of course be proportional to $\Lambda^2$ and no $\log \Lambda$ enters. So my question is: Is the "problem of large logarithms" tied to dimensional regularization or am I missing something?

Answer 1

Large logarithms actually occur in different ways depending on the regularization scheme. With a hard cutoff they do appear as $\log \Lambda$, but actually $\log \mu$ is not a divergent logarithm in dimreg. That is because, when working with dimreg, $\mu$ is not a cutoff, but rather could be thought of as an auxiliary scale. The divergence in dimensional regularization is actually the term $\frac{1}{\epsilon}$. However, dimensional regularization will not be able to express power divergences that would occur on a hard cutoff regularization, which are often not required for practical calculations anyway.

As you pointed out, though, not every divergence is logarithmic. You can also have power divergences which, as I mentioned, are actually not captured by dimensional regularization. Dimreg can only pick up the large logarithms, but for many applications that is enough. I expect that the integral you computed will be convergent in dimreg as you take the limit $\epsilon \to 0$. $\mu$ is just an auxiliary scale, and hence $\log\mu$ is not really a large logarithm.

In short, large logarithms are not an artifact of dimensional regularization.