# What are renormalons from a physics point of view?

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This is again a question in the context of this paper about the Exact Renormalization Group. On p 23 and the following few pages, it is explained that for a $\lambda \phi^4$ bare action at the bare scale $\Lambda_0$, after integrating out degrees of freedom and assuming a small coupling $\lambda$, the effective action at the larger scale $\Lambda$ can be written as the sum of a perturbative series plus nonperturbative power corrections (Eq. 2.6)

$$S_{\Lambda,\Lambda_0}[\phi] = \sum\limits_{i=0}^{\infty}\lambda^{i-1}S_i[\phi] +O(\Lambda/\Lambda_0)$$

Naively taking $\Lambda_0 \rightarrow \infty$ makes the second part disappear and leaves the first part which is self-similar which means that the theory is renormalizable. However, as stated in the paper, UV renormalons can make the perturbation series ambiguous, such that the power corrections and therefore the bare scale can not be removed and the theory would then not be renormalizable.

The issue is explained by the mathematical argument that in order to obtain a unique finite value for the perturbative series, one makes use of the so-called Borel transformation to define a function which has the same power series expansion defined as an integral in the complex Borel plane. This integral exists only of there are no poles on the real axis, otherwise the contour of integration around these poles can be deformed which makes the integral and the function defined by it ambiguous. If this is the case, the power correction terms have to be kept to restore the uniqueness of the integral which means that the bare scale can not be removed.

In the paper it is said, that the poles on the real axis are for example due to UV renormalons, that arise from large loop momenta in certain Feynman diagrams.

My question now is:

What are these renormalons from a physics point of view? How do they enter the Lagrangian of the theory? Are they some kind of unphysical auxilliary fields that appear in the mentioned Feynman diagrams? And what do the Feynman diagrams that contain them look like?

retagged Mar 9, 2014
Have you done a Google search first? Within a minute, I found a Wiki and a paper with section 3 on renormalons from Feynman diagrams.

This post imported from StackExchange Physics at 2014-03-09 16:16 (UCT), posted by SE-user Larry Harson
@LarryHarson thanks for the paper, why dont you post an answer and explain it a little bit?

This post imported from StackExchange Physics at 2014-03-09 16:16 (UCT), posted by SE-user Dilaton
I wish I could, but I'm not competent enough to explain any part of it.

This post imported from StackExchange Physics at 2014-03-09 16:16 (UCT), posted by SE-user Larry Harson

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Larry Harson is right; you should read Beneke's Physics Report. But I think I can make your reading easier by clearing up a misconception:

The name 'renormalon' is a bit misleading. Renormalons (like instantons) aren't real physical things. They don't appear in the Lagrangian, and they don't correspond to any physical state. They are not auxiliary fields. What they are is divergences that show up when you use a particular approximation scheme. You could say, I suppose that the renormalons are the field configurations which lead to the divergences, but that's not quite the right spirit. If you do pure non-perturbative calcuations, you never see renormalons. They're an artefact of the perturbative approach.

This post imported from StackExchange Physics at 2014-03-09 16:16 (UCT), posted by SE-user user1504
answered Jul 11, 2013 by (1,110 points)
Thanks for these clarifications, they are very helpful since the term renormalon mislead me indeed ...

This post imported from StackExchange Physics at 2014-03-09 16:16 (UCT), posted by SE-user Dilaton

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