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This question is related to these others: mostly this one, but also this one and this one.

Do we care about CFTs in particle physics? Let me explain. Suppose we don’t know anything about string theory, holography, etc. And suppose we don’t know about the Wick rotation, so that we are unable to relate QFTs and statistical field theories. In this hypothetical world the only application of QFT is the Standard Model and its different extensions or possible minor changes. Would we be interested in CFTs? Why?

I'm asking this because I want to motivate the study of CFTs from the point of view of standard QFTs. The best argument I've found is that all "reasonable" QFTs are actually points in the RG flow from a fixed point at high energies $\text{CFT}_\text{UV}$ (UV completion) and a fixed point at low energies $\text{CFT}_\text{IR}$ (see for instance these lectures by Simmons-Duffin). Is the Standard Model an example of this? Looking at the running couplings of the SM, my answer would be no. Look for instance at this article by Rychkov. There he says there are basically three types of IR phases:

  • A CFT.

  • A theory with a mass gap. The example is QCD, because the beta function is negative and the coupling blows at low energies.

  • A theory with massless particles. The example is QED, because it is free in the IR, and at energies $E \ll m_e$ you're only left with photons. Someone said in the comments that this is a CFT (an empty CFT, with just the vacuum state for electrons) but Rychkov doesn't consider it as such.

So at least QCD is a counterexample of the above notion: it has no CFT$_\text{IR}$. On the other hand, for the UV part QED gives a counterexample: its coupling constant grows towards the UV, so it has no $\text{CFT}_\text{UV}$. Is this reasoning correct? If it is, then why are the only real-world QFTs (QED and QCD) a counterexample of such an extended notion (that all QFTs are points in the RG flow between two fixed points)? The "good" (conformal) behaviour of QCD at high energies (and arguably QED at low energies) seems accidental, not fundamental. So why do we care about CFTs in particle physics?

MBolin
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  • "...why are the only QFTs that work (experimentally, if you wish) a counterexample of such an extended notion?" That question can be turned around: Why would we expect those QFTs that are most directly relevant to the real world to have UV completions that are also QFTs (CFTs), if we don't expect the most fundamental theory of the real world to be a QFT? – Chiral Anomaly Aug 15 '19 at 02:00
  • I'm not sure if I understand your point. Are you saying that my argument is right? – MBolin Aug 15 '19 at 09:12
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    Any QFT flows to a CFT in the IR. QCD is not a counterexample, but in that case the IR theory is trivial: it is an empty theory (or a free theory of Goldstone bosons if you forget about the quark masses). – M.Jo Aug 15 '19 at 13:05
  • @M.Jo, What does this empty CFT consist of? A vacuum state and nothing else? – octonion Aug 15 '19 at 13:40
  • @octonion just the vacuum state and nothing else, precisely. Of course it's a bit of a special CFT, but it's still a CFT by all means. – M.Jo Aug 15 '19 at 13:53
  • @M.Jo Then why Rychkov says it is not a CFT? Instead he says it is "a theory with a mass gap". – MBolin Aug 15 '19 at 15:10
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    @MBolin The "theory with a mass gap" is the QFT. But in the deep IR, below the mass gap, there is only an empty theory (which happens to be a CFT): if you don't have enough energy to create at least one particle, the only state that exists is the vacuum. I think all of what I'm saying is pretty standard, but often people use the word "CFT" to talk about theories that are neither free nor empty. This might be why Rychkov makes a distinction between a flow to a non-trivial CFT and a flow to a theory with a mass gap ($\equiv$ an empty CFT). – M.Jo Aug 15 '19 at 15:28
  • I don't know, maybe you are right, and it's just that it is trivial. But he says exactly: "One can ask which IR phases are possible. A priori, there are three possibilities: a theory with a mass gap, a theory with massless particles in the IR and a Scale Invariant theory with a continuous spectrum. It is the last class that we will call CFT and will mostly study in these lectures. [...] Non-abelian Yang-Mills theory in 4 dimensions belongs to the first type". Moreover, if it is what you say, is there any example of a theory which is not a CFT in the IR? – MBolin Aug 15 '19 at 15:47
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    @MBolin My first comment only meant that I don't know of any compelling argument that every perturbatively-defined QFT should have a UV completion which is also a QFT. I don't know if the Standard Model has such a completion or not. I don't even know of any non-perturbative construction of the Standard Model on a lattice. – Chiral Anomaly Aug 16 '19 at 00:05
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    Some points: 1) QED is simply not UV-complete, at least I don't know any UV-completions. Same for Standard Model, but in principle you can run its coupling constants to extremely high energies (even up to Planck scale maybe? -- don't remember) before encountering any problems of this sort. 2) Even if you don't know about Wick-rotation, Lorentzian CFTs still describe long-distance physics of quantum critical points. ... – Peter Kravchuk Aug 24 '19 at 02:26
  • Even if a theory is not CFT, it might be close to being a CFT, and then you can use conformal perturbation theory to describe it. E.g., before Higgs discovery there were (maybe still are?) things like walking technicolor.
    • – Peter Kravchuk Aug 24 '19 at 02:26
  • Also, I would argue that your question is extremely artificial. No, Standard Model is not conformal, it does not posses a non-trivial IR fixed point (it is IR-free), and we do not think it has a CFT UV fixed point. End of story. But this is not why we care about CFTs. For example, you discard connection to statistical phase transitions, but QCD itself is likely to exhibit a second-order phase transition associated to chiral symmetry breaking. – Peter Kravchuk Aug 24 '19 at 02:35
  • I am not saying the SM has any problem because it's not UV complete or because it's not conformal at long distances. All I'm saying is that I've often read that a "reasonable" QFT must be a point in the flow between two CFTs (see the lectures by Simmons-Duffin I just linked), and I don't see that in the SM. 2) If quantum critical points are an application of CFTs in particle physics then I am interested in that.
    • – MBolin Aug 24 '19 at 10:43
  • SM is not conformal, end of story? OK, but then what is the meaning of "We can think of a UV-complete QFT as an RG flow between CFTs"? (from https://arxiv.org/abs/1602.07982 ) – MBolin Aug 24 '19 at 10:46
  • (I don't get notified of your responses unless you @ me.) 1) You are asking "Is the Standard Model an example of this?" and I am answering: no, it is not. As a whole at least -- QCD by itself is an example of this. Simmons-Duffin is taking only about UV-complete QFTs, obviously. You can generalize a bit and allow non-QFT UV-completions (lattice, for example, or whatever UV-completes SM), but then the UV fixed point may not be a CFT or even exist. – Peter Kravchuk Aug 27 '19 at 05:49
  • Experimentally established theoretical particle physics = SM. SM is not a conformal theory. It is not UV complete, its IR fixed point is not particularly exciting. You can make use of CFTs in particle physics, but that seems to not fit in your question. I gave an example: there likely is a chiral critical point in QCD phase diagram that is described by some CFT in Euclidean signature. Is this particle physics? But it is Wick-rotated, so you don't want it?
    • – Peter Kravchuk Aug 27 '19 at 05:57