0
$\begingroup$

Is every Quantum Field theory, including CFTs, an Effective Field Theory?

Can every interacting CFT become the infra-red fixed point of some high energy QFT?

$\endgroup$
6
  • 4
    $\begingroup$ Do you want your UV theory to be complete? then the answer is no, there are many QFTs that do not admit any UV completions at all (e.g., if they have a suitable higher-form symmetry, cf. arxiv.org/abs/2108.00732). Do you allow for non-complete UV theories? then the answer is trivially yes: just add some irrelevant interaction. $\endgroup$ Commented Mar 26, 2023 at 21:57
  • $\begingroup$ physics.stackexchange.com/a/467869/133418 $\endgroup$
    – Avantgarde
    Commented Mar 27, 2023 at 11:33
  • $\begingroup$ Only two types of QFTs should ever be used in physics. Those which are UV complete and those which are not but still make sense as an EFT for some as yet undiscovered UV completion. Examples of the former type (like QCD) could be called "not EFTs" but this is artificial. QCD makes sense mathematically at high energies but in real life there will still be a scale of new physics. $\endgroup$ Commented Mar 27, 2023 at 12:58
  • $\begingroup$ The answer to the second question is yes. Take the CFT you're interested in and deform by an irrelevant operator. This produces a QFT whose IR fixed point is that CFT again. $\endgroup$ Commented Mar 27, 2023 at 13:00
  • $\begingroup$ @ConnorBehan Massless QCD with proper ratio of colors to flavors, has a UV fixed point. But what if I get interested in massive QCD where the beta function changes with scale as new flavors are included. Then massive QCD is just an EFT where it can have a UV comoletion. No? In this sense isn't quark masses a sign that QCD should be included in a higher symmetry group necessarily? $\endgroup$ Commented Mar 27, 2023 at 13:20

0

Your Answer

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