12
$\begingroup$

This is claimed by Jared Kaplan in his Lectures on AdS/CFT from the Bottom Up.

He writes:

It seems that all QFTs can be viewed as points along an Renormalization Flow (or RG flow, this is the name we give to the zooming process) from a ‘UV’ CFT to another ‘IR’ CFT. Renormalization flows occur when we deform the UV CFT, breaking its conformal symmetry. [...] Well-defined QFTs can be viewed as either CFTs or as RG flows between CFTs. We can remove the UV cutoff from a QFT (send it to infinite energy or zero length) if it can be interpreted as an RG flow from the vicinity of a CFT fixed point. So studying the space of CFTs basically amounts to studying the space of all well-defined QFTs.

Why is this true?

Especially, how can we see that we can only remove the cutoff (i.e. renormalize) if the QFT "can be interpreted as an RG flow from the vicinity of a CFT fixed point"?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
11
$\begingroup$

Although I disagree with the definition of well-behaved QFT (why are these people always insisting on taking the continuum limit ?), the reason is the following.

If one wants to take the continuum limit (that is, take the limit of an infinite cut-off non-perturbatively), while having to specify only a finite number of coupling constant at a given (finite) scale, then one needs to have a UV fixed point of the RG flow. Furthermore, to control the flow in the IR, one also needs a IR fixed point.

However, there are not that many theories that have this property. One famous example is the trajectory that links the gaussian fixed point to the Wilson-Fisher (WF) fixed point in scalar field theories for dimension less than four. Note however that it is a very special theory that does not describe any real system, even though the WF fixed point does describe second order phase transitions of a lot of systems. (That's why I don't understand why some people insists on having a "well-defined" QFT... "Ill-defined" QFTs are also useful (and I would even dare to say, more useful), since they allow to describe real systems, and compute real quantities (such as critical temperatures).)

Another remark is that the standard $\phi^4$ theory in 4D is not well-defined if interacting, and the only such theory that is a CFT is the trivial one (no interactions), which is quite boring. However, this of course does not mean that $\phi^4$ in 4D is useless (and that's why people spent decades studying it), only that insisting on a continuum limit is meaningless.

To me, the must read reference on that is arXiv:0702365, section 2.6. See also Why do we expect our theories to be independent of cutoffs?

Improve this answer
$\endgroup$
2
  • $\begingroup$ So, if you want a theory that can be described at low scales with a finite number of parameters and is valid up to arbitrarily high energies (continuum limit), you need a theory at a critical point (a CFT)?! I.e. a UV complete theory that is described by a finite number of parameters at low scales is necessarily a CFT? Can we describe a theory that isn't at a critical point, but on a lattice, equally be described by only a finite number of parameters at low energies? $\endgroup$
    – jak
    Commented Jun 14, 2017 at 14:08
  • $\begingroup$ For the continuum limit, you need a UV and a IR attractive fixed point, which might be a CFT or not (note that scale invarience does not in principle imply conformal invarience). This is the idea behind the asymptotic safety scenario of quantum gravity (i.e. find a (non-perturbative) UV fixed point controlled by a few relevant directions). I would say yes to the last question, but it is too vague for a definite answer... $\endgroup$
    – Adam
    Commented Jun 14, 2017 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.