In the space of field theories, Conformal field theories are fixed points in the RG flow. However, a lot of literature on CFT usually talks about a QFT being the RG flow between two CFTs: one UV and the other IR. However, I do not understand what UV and IR CFTs are. Suppose there are only 2 fixed points, with characteristic lengths $R_c$ of $0$ and $\infty$, the trajectory from $R_c = \infty$ to $R_c = 0$ is a trajectory through QFTs. We have the following relationship on the trajectory,

$$ Z^{-1/2}(L)\langle \phi(x_1/L)...\phi(x_n/L) \rangle_{A'} = \langle\phi(x_1)...\phi(x_n) \rangle_A$$

where, $RG_L[A] = A'$. But I don't understand if the fixed point at $L = \infty$ would be considered UV or the other way around and why.

Clearly, there are QFTs that do not lie on this trajectory or any trajectory between 2 fixed points. In this case why do people associate CFTs with the UV and IR limits of QFTs?

  • $\begingroup$ Can you explain what the symbol $RG_L[\bullet]$ is supposed to mean? $\endgroup$ Jun 29, 2022 at 18:22
  • $\begingroup$ By $RG_L[]$, I meant following the RG flow with the scale factor going from 1 to L. $\endgroup$
    – Chandrahas
    Jun 30, 2022 at 9:46

1 Answer 1

  • The theory with characteristic length $R_\text{c}=0$ is the UV CFT, since it is probing things at the smallest possible length scale, i.e. at the largest possible energy. In other words, its characteristic energy scale is $\Lambda_{\text{UV}}=\infty$. Similarly the $R_\text{c}=\infty$ is the IR CFT, with $\Lambda_{\text{IR}}=0$.

  • The non-trivial statement here is the existence of a UV CFT. If there exists such a theory, there exists an IR CFT too. It might be trivial, but it's there. Namely, just turn on a relevant deformation in your UV CFT and start flowing down. In the deep IR there are the following possible scenarios:

    1. The theory is gapless. There are excitations which you can reach without spending energy. It is described either by a free or by an interacting CFT, depending on the details.
    2. The theory is trivially gapped. There is a unique ground state, and then an energy gap. If you go below that energy gap, the theory is empty. It is the trivial theory containing only the operator $\mathbf{1}$ corresponding to the unique vacuum state.
    3. The theory is topological. It is a theory of multiple degenerate vacua. Its spectrum is a set of topological operators corresponding to the various vacua.

    Hence, starting from a UV CFT, you will eventually reach an IR CFT, passing through QFTs in the middle. This is the way to understand QFTs as intermediate points between CFTs. However, if you just start with a QFT, at some energy scale $0<\Lambda<\infty$, you can only flow down, to its IR CFT. There is no unambiguous way to run up the RG flow, making sure that you remain with a local theory (and hence a QFT). Turning on an irrelevant deformation, usually results in non-local theories (cf. $\text{T}\overline{\text{T}}$-deformed QFTs, and related stories).

  • $\begingroup$ I don't know if I am missing something but the RG flow always decreases the characteristic length right? So if we flow from UV to IR, shouldn't UV have $R_c = \infty$ and IR have $R_c = 0$? $\endgroup$
    – Chandrahas
    Jun 30, 2022 at 10:51
  • $\begingroup$ No it doesn't. Why would it? $\endgroup$ Jun 30, 2022 at 15:47
  • $\begingroup$ I might be misremembering but I think we derived an expression for the characteristic length when flowing along an RG flow in the our course and we obtained $R_c(l) = e^{-l}R_c$ for $l = log L$ and $R_c(l)$ is the characteristic length after flowing down the flow from 0 to $l$. $\endgroup$
    – Chandrahas
    Jul 1, 2022 at 10:08
  • $\begingroup$ Well, this equation is clearly wrong on dimensional analysis grounds, so I can't trust it as is. Furthermore if $R_c(0)$ is 0 or $\infty$, then $e^{-l}R_c(0)$ is also 0 or $\infty$, so this doesn't help much. $\endgroup$ Jul 1, 2022 at 11:48
  • 1
    $\begingroup$ Still if $R_c(1)$ is 0 or $\infty$ $R_c(l)$ given by this equation remains 0 or $\infty$, do you agree? And well, are there any notes or further context for these equations? Can you reproduce their derivation? Or are we relying on possibly misremembered equations with undefined symbols? $\endgroup$ Jul 1, 2022 at 12:20

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