Suppose that a RG (renormalization group) fixed point of some RG trajectory (or flow) is a CFT. Then do theories in this trajectory have to be CFTs as well?


1 Answer 1


No. The points along the RG trajectory represent the theory’s behavior at some scale. The most baby example of this is a massive free field $\mathcal{L} = (\partial \phi)^2 + m^2 \phi^2$. This can be viewed as a perturbation of the Gaussian fixed point by the relevant operator $\phi^2$. In the deep UV the mass is negligible and the theory approaches massless free fields. In the deep IR the mass is effectively infinite and we get the trivial conformal theory, with no low-lying excitations. The spectrum is gapped above the vacuum. For generic points along the flow, the theory is of course non-conformal.

Families of CFTs correspond to manifolds of fixed points in the coupling constant space.

  • $\begingroup$ I'm curious, what does the phrase "manifolds of fixed points in the coupling constant space mean exactly"? which coupling constant space would you be talking about? The space of ALL possible Lagrangians minimally coupled to a relevant operator as $\lambda\int dx \mathcal{O}(x)$?? $\endgroup$ Jun 7, 2019 at 0:33
  • $\begingroup$ The coupling constant space is just the space of all Lagrangians, the space that the RG flow takes place in. Coordinates are the coefficients of the operators in the Lagrangian. $\endgroup$
    – user145190
    Jun 8, 2019 at 14:44
  • $\begingroup$ Could you please explain why the theory is conformal in the IR? $\endgroup$ Jun 9, 2019 at 16:26
  • 1
    $\begingroup$ The answer is kind of trivial—in the deep IR, the mass can be taken as infinite (this has the same effect as scaling up all length scales in the problem) and all the correlation functions go to zero. The only finite-energy state left is the Fock vacuum with nothing in it. So the only thing left is the vacuum, which furnished the trivial representation of the conformal algebra. $\endgroup$
    – user145190
    Jun 11, 2019 at 14:46

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