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Can You Take the Cutoff To Infinity At A Conformal Fixed Point?

+ 5 like - 0 dislike
147 views

A conformal fixed point is defined by

$$\beta(g)=0$$

We hence know that couplings, masses and dimensions of operators do not flow in the effective Lagrangian when we change the renormalization scale $\mu$.

My question then is, can we take $\mu$ to infinity and retain finite results for all separated correlation functions at this point? Intuitively that seems sensible to me, but perhaps there is a technical obstruction.

If no, could you provide me with an example where separated correlation functions diverge in a CFT?

This question is related, but doesn't seem to answer my specific question.

First Edit

The answer is definitely no. Here is an example of a UV divergent correlation function in planar $\mathcal{N}=4$ SYM, a theory which is conformal at all values of the coupling!

But I still don't understand where this divergence comes from! Surely if $\beta=0$ then nothing flows in the effective Lagrangian, so there can't be a divergence? Why is this simplistic viewpoint erroneous?

Second Edit

Actually I've changed my mind. My example above had a divergence before renormalization. But after renormalization it's fine! How come you need to renormalize, even though the coupling constants and anomalous dimensions are fixed? Well, you still have the wavefunction renormalization $Z$ which is unaffected by $\beta(g)=0$. So a CFT can have UV divergences which must be cancelled in $Z$ so that the correlation functions are finite.

To quote from that paper

The UV divergences, on the other hand, require renormalisation. In $\mathcal{N} = 4$ SYM theory, the appropriate combinations of the self-energies of the elementary fields and the one-particle-irreducible (1PI) corrections to the elementary vertices are UV finite, ensuring the vanishing of the $\beta$-function. The only sources for UV divergences are the insertions of composite operators as external states, which hence need to be renormalised.

If anyone disagrees with my reasoning let me know, otherwise I'll write this up into a Q and A!

This post imported from StackExchange Physics at 2015-05-24 09:00 (UTC), posted by SE-user Edward Hughes
asked May 12, 2015 in Theoretical Physics by Edward Hughes (130 points) [ no revision ]
retagged May 24, 2015
I think you are correct: from the Euler characteristic of the graph, you can derive that only wave function renormalizations may be required in supersymmetric models. I also agree that all infinite contributions to the correlator can be interpreted as contact terms, since integrated correlation functions can have UV divergences.

This post imported from StackExchange Physics at 2015-06-01 09:10 (UTC), posted by SE-user giulio_hep
@Edwardhughes Maybe its is time to write that answer, I don't want the bounty going to waste. Very good question nevertheless.

This post imported from StackExchange Physics at 2015-06-01 09:10 (UTC), posted by SE-user Prathyush
Yes - I shall do over the weekend. Thanks for the bounty btw!

This post imported from StackExchange Physics at 2015-06-01 09:10 (UTC), posted by SE-user Edward Hughes
@Prathyush - find my answer below. Hopefully it's a lucid collection of thoughts! Many thanks once again for bringing more community attention to this via your bounty. It's nice to know there are people checking my reasoning!

This post imported from StackExchange Physics at 2015-06-01 09:10 (UTC), posted by SE-user Edward Hughes
I did not use this site for a while, I found this questions after some searching. I do like your answer, as far as I can see it seems accurate. But I am only learning the subject, so it will take some time to fully understand it. It only take 7 hours for questions to disappear on the main page now, Perhaps that is a reason for a lack of attention. Do check out physicsoverflow, maybe you will find a larger communitybase, interested in your questions.

This post imported from StackExchange Physics at 2015-06-01 09:10 (UTC), posted by SE-user Prathyush

1 Answer

+ 3 like - 0 dislike

For clarity, I shall always assume that the theory we are considering is renormalizable, conformal at the quantum level, and well-defined at least at some finite energy scale.

The answer is yes, I believe. Here's my reasoning.

  1. Correlation functions cannot have direct dependence on the renormalization scale $\mu$, by definition.

  2. Therefore they can only depend on $\mu$ through physical parameters like the coupling, masses and dimensions.

  3. In a conformal theory $\beta(g)=0$ implies that all physical parameters are fixed independent of $\mu$.

  4. Therefore we may extend our well-defined theory at finite $\mu$ to $\infty$ by assuming a trivial group flow in the opposite direction.

So why was I confused? Well, people still talk about UV divergences in conformal field theories. But these are the divergences in the unrenormalized theory. Obviously many of these must still cancel out to ensure $\beta =0$. But not all of them!

In particular, the wavefunction renormalization $Z$ can (and often does) absorb a UV divergent factor when correcting the bare fields to interacting ones. Intuitively this makes sense to me, at least. If you want an example, check out this paper in which a certain form factor require field strength renormalization in $\mathcal{N}=4$ super-Yang-Mills theory.

To quote from that paper

The UV divergences, on the other hand, require renormalisation. In $\mathcal{N} = 4$ SYM theory, the appropriate combinations of the self-energies of the elementary fields and the one-particle-irreducible (1PI) corrections to the elementary vertices are UV finite, ensuring the vanishing of the $\beta$-function. The only sources for UV divergences are the insertions of composite operators as external states, which hence need to be renormalised.

So to conclude, you can take $\mu$ to $\infty$ in conformal field theories, provided that you go about field strength renormalization correctly when doing it!

This post imported from StackExchange Physics at 2015-06-01 09:10 (UTC), posted by SE-user Edward Hughes
answered May 31, 2015 by Edward Hughes (130 points) [ no revision ]

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