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  What does it mean for a QFT to not be well-defined?

+ 8 like - 0 dislike

It is usually said that QED, for instance, is not a well-defined QFT. It has to be embedded or completed in order to make it consistent. Most of these arguments amount to using the renormalization group to extrapolate the coupling up to a huge scale, where it formally becomes infinite.

First of all, it seems bizarre to me to use the renormalization group to increase resolution (go to higher energies). The smearing in the RG procedure is irreversible, and the RG doesn't know about the non-universal parts of the theory which should be important at higher energies.

I have read in a note by Weinberg that Kallen was able to show that QED was sick using the spectral representation of the propagator. Essentially, he evaluated a small part of the spectral density and showed that it violated the inequalities for the field strength $Z$ that are imposed by unitarity + Lorentz invariance, but I do not have the reference.

So what I would like to know is :

What does it take to actually establish that a QFT is not well-defined?

In other words, if I write down some arbitrary Lagrangian, what test should I do to determine if the theory actually exists. Are asymptotically free theories the only well-defined ones?

My initial thought was: We would like to obtain QED (or some other ill-defined theory) using the RG on a deformation of some UV CFT. Perhaps there is no CFT with the correct operator content so that we obtain QED from the RG flow. However, I don't really know how to make that precise.

Any help is appreciated.

This post imported from StackExchange Physics at 2014-08-29 16:51 (UCT), posted by SE-user Dan
asked Sep 12, 2013 in Theoretical Physics by UnknownToSE (505 points) [ no revision ]
retagged Aug 29, 2014
More on rigor in QFT: physics.stackexchange.com/q/27569/2451 , physics.stackexchange.com/q/6530/2451 , physics.stackexchange.com/q/27665/2451 and links therein.

This post imported from StackExchange Physics at 2014-08-29 16:51 (UCT), posted by SE-user Qmechanic
Also, doesn't this just indicate that the free-particle description of QED breaks down at high energies, just like the quarks and gluons description of QCD breaks down at low energies?

This post imported from StackExchange Physics at 2014-08-29 16:51 (UCT), posted by SE-user Jerry Schirmer

1 Answer

+ 5 like - 0 dislike

Usually if someone says a particular QFT is not well-defined, they mean that effective QFTs written in terms of these physical variables don't have a continuum limit. This doesn't mean that these field variables aren't useful for doing computations, but rather that the computations done using those physical variables can only be entirely meaningful as an effective description/approximation to a different computation done with a different set of variables. (Think of doing an expansion in a basis and only keeping terms that you know make large contributions to the thing you're trying to compute.)

A funny and circular way of saying this is that a QFT is well-defined if it can be an effective description of itself at all length scales.

You're right that it's slightly perverse to talk about renormalization flow in terms of increasing resolution. But there's a(n inversely) related story in the terms of decreasing resolution. Suppose you have a set of fields and a Lagrangian (on a lattice for concreteness, but you can cook up similar descriptions for other regularizations). You can try to write down the correlation functions and study their long distance limits. Your short distance QFT is not well defined if it only flows out to a free theory.

When people talk about having coefficients blow up, they're imagining that they've chosen a renormalization trajectory which involves the same basic set of fields at all distance scales. Flowing down to a non-interacting theory means that -- along this trajectory -- the interactions have to get stronger at shorter distance scales. Testing this in practice can be difficult, especially if you insist on a high degree of rigor in your arguments.

Not all well-defined theories are asymptotically free. Conformal field theories are certainly well-defined, but they can be interacting and these ones are not asymptotically free.

This post imported from StackExchange Physics at 2014-08-29 16:51 (UCT), posted by SE-user user1504
answered Sep 12, 2013 by user1504 (1,110 points) [ no revision ]
I like this, but I’m still a little confused. In order to ever talk about RG, I have to make the explicit assumption that my theory is near a fixed point. If I do not assume approximate scale invariance resulting from a nearby critical point, I don’t know my operator dimensions and there is no way to organize the RG so that I know which terms to throw away as I smear (in other words I cannot rank operators). Saying “a QFT is well-defined if it can be an effective description of itself at all length scales” sounds to me like requiring that the QFT lives near a fixed point.

This post imported from StackExchange Physics at 2014-08-29 16:51 (UCT), posted by SE-user Dan
Approximate scale invariance says when I smear an operator, I get it back to first order, plus sub-leading corrections. This means that I can use the same degrees of freedom to describe the theory at all scales by integrating out and generating EFT’s. So it seems like any QFT which I am going to use RG on has already been assumed to satisfy your definition. Of course we can have QFT’s which flow down to another non-trivial fixed point, where the operator content is different and we cannot use the same degrees of freedom. But I’ve never heard that such theories are ill-defined.

This post imported from StackExchange Physics at 2014-08-29 16:51 (UCT), posted by SE-user Dan

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