I may be wrong, but it seems that only logarithmic divergences need to be retained when using the Callan-Symanzik equation, finding running couplings, etc. Why is this the case? Is there some simple intuitive understanding for why the logarithmic divergences are most important for these applications?

# [Physics] Why are only logarithmic divergence relevant for the Callan-Symanzik equation? Intuitive understanding

quantum-field-theoryregularizationrenormalization

#### Related Solutions

I recently stumbled upon a good comment about this in Jared Kaplan's AdS/CFT notes

Any quantum field theory which has hope of having an UV-completion can be viewed as as effective theory at point in the RG flow from an UV complete theory.

Field theories at the UV fixed point are conformal.

Hence all 'well-defined' field theories are either CFTs or points in the RG flow from one (UV) CFT to another CFT.

So in a sense, in Kaplan's words:

studying the space of CFTs basically amounts to studying the space of all well-defined QFTs

So that's one way of seeing the place of CFTs in general QFTs.

1) Is this (the second type of counterterms) the only way that power divergences will be cancelled?

Yes, you need to add counter terms that look like the divergences you find. So in this case, you are exactly right, you need to add a counter term corresponding to $\phi^2 \partial(\phi)^2$.

2) If yes, are we always able to set counterterms with respect to the symmetries of the system in this way, so that power divergences can always be cancelled at all scales (then we do not have to worry about power divergences at all!)?

I'm not sure what symmetry you are worried about here, but often the power law divergences will break the symmetries you started with and will require you to add counterterms that break the symmetry you started with. In fact, this is precisely why people tend to focus on the logs, or use regularization methods like dim reg, where the power laws don't appear and you don't have this annoying problem.

The general principle is that you will need counterterms that violate the symmetry you started with, if your regularization method itself breaks the symmetry. An example is QED. Regulating with a hard cutoff violates gauge invariance, so the power law divergences also violate gauge invariance. (Just compute the one loop correction to the two point function for the photon, gauge invariance tells you the propagator should be proportional to $\eta_{\mu\nu}-p_\mu p_\nu$, but the power law divergences will give different relative contributions to those two terms).

The best solution is to use a regularization scheme that respects all the symmetries. Dim reg is usually a good choice. If you can't find such a regularization scheme, that may be a sign that the symmetry is anamolous and can't be maintained at the quantum level (eg--massless fermions have a chiral symmetry that is anamolous when you couple them to gauge fields).

If you insist on using a different regularization scheme, you can still get the right answer, but it will take a little more work. You will need to add counterterms that violate the symmetry. However the real question is that when you compute something physical (such as the S matrix), after you renormalize, will the physical quantities respect the symmetry? You will end up finding, if you do things correctly, that the failure of the counterterms to respect the symmetry will exactly cancel the failure of the divergences to respect the symmetry, and the final answer will be symmetric. I would consult Weinberg if you want to get a precise prescription on how to proceed.

The lesson is that to renormalize, you simply add the counterterms you need to cancel the divergences you find, without thinking about what they mean. Later on, you may need to do some re-interpreting to figure out exactly what the physics is.

Also, it is true that this interaction is renormalizable, so you will end up generating an infinite number of counterterms. However, depending on what you are trying to do, that's not necessarily as bad as it sounds.

## Best Answer

Of course, you can't just neglect power divergent terms. However, we have a regulator (dimensional regularization, DR) that automatically eliminates power divergent terms. To the extent that all consistent regularization schemes are equivalent, we would expect that nothing new can be learned by including power divergent terms. There are, however, cases where DR, while not wrong, obscures the physics or leads to apparently poorly converging expansions. See http://arxiv.org/abs/nucl-th/9802075 for an example in which the authors modify DR to retain power divergences (``PDS''), and solve the associated RG equations.