# Are fixed points of RG evolution really scale-invariant?

It is often stated that points in the space of quantum field theories for which all parameters are invariant under renormalisation – that is to say, fixed points of the RG evolution – are scale-invariant field theories. Certainly this should be a necessary condition, since the theory must look the same at all scales. However, I have doubts whether this is sufficient.

In the classical theory, there is no notion of renormalisation, but nevertheless we can talk about scale-invariant theories; these theories possess a global dilatation symmetry. Theories with inherent mass scales, such as $\phi^4$ theory with a non-zero quadratic term, are not classically conformal because they lack such a symmetry. It seems to me that the vanishing (or perhaps blowing up?) of such dimensionful couplings must also be a condition to impose on a quantum field theory, if we wish for it to be scale-invariant.

My question is then: are the dimensionful couplings of all fixed points in RG flow necessarily trivial? Is it impossible for an RG fixed point to have some mass scale $M \neq 0$ in its Lagrangian?

As I see it, either the answer is yes, in which case being at a fixed point is enough to guarantee scale invariance, or the answer is no, in which case we also need to make an additional requirement of our theories if we wish for them to be conformal, beyond the vanishing of all beta functions

• 1. Concerning the last sentence (v1): If you are really only asking about scale invariance (as opposed to conformal invariance), consider to edit the post accordingly. 2. The issue of scale vs. conformal symmetry is covered in e.g. this and this Phys.SE posts. – Qmechanic Oct 4 '17 at 15:32

No, dimensionful couplings do not have to be all set to zero at an RG fixed point. An RG fixed point is one where all of the beta functions vanish, and beta functions generally have the form $$\beta(g_i) = (d_i - d) g_i + \hbar A_{ij} g_j + \ldots$$ where $d_i$ is the dimension of the corresponding operator. If one truncates the series at $O(\hbar^0)$ then the only possible solution is to have $g_i = 0$ if $d_i \neq 0$, so in classical field theory the only fixed points are the massless free theory and massless $\phi^4$ theory.

In a quantum field theory we must account for loop diagrams, which give terms that are higher order in $\hbar$. Then the zeroes of the beta functions are completely different; the massless free theory remains a fixed point, called the Gaussian fixed point, but massless $\phi^4$ theory acquires a mass scale by dimensional transmutation. But this process can also work in reverse. In this case there's a new fixed point, the Wilson-Fisher fixed point, where the classical mass term is nonzero. This is dimensional transmutation running in reverse; the mass renormalizes to exactly zero.

• Your last sentence is particularly interesting. Are you saying that just as theories with no scales in the Lagrangian can nevertheless develop a scale in the quantum theory (as in QCD), so theories with scales in the Lagrangian can lose that scale in the quantum theory? – gj255 Mar 1 '18 at 19:38
• @gj255 Yes, and this is exactly what happens at, e.g. the Wilson-Fisher fixed point. There is a mass term in the Lagrangian, but after renormalization the mass of the quantum particles is exactly zero! – knzhou Mar 1 '18 at 19:41

If all the couplings are zero then you are sitting on the trivial Gaussian fixed point. Being a fixed point is characterized by the vanishing of the beta functions (some derivative of the couplings as functions of scale), not that of the couplings themselves.

Also, in general, scale invariance does not imply conformal invariance. You need something extra like a traceless energy-momentum tensor. The study of this issue is an active area of research, see the review "Scale invariance vs conformal invariance" by Nakayama.

• My question is whether all the dimensionful couplings are zero. If a RG fixed point has some non-zero dimensionful coupling, then I claim it is not conformal. – gj255 Oct 4 '17 at 15:32
• Talking about dimensionful couplings is a bit meaningless here because you have to take a singular limit and really the only possible values for that limit are $0$ and $\infty$. Take Ising/$\phi^4$ in 3d then the dimensionful $\phi^4$ coupling is zero at trivial Gaussian fixed point but $\infty$ at the nontrivial Wilson-Fisher fixed point. – Abdelmalek Abdesselam Oct 4 '17 at 15:37
• Right, so the dimensionful coupling is always trivial in some sense. Is this true of all theories at fixed points? – gj255 Oct 4 '17 at 15:41
• I would say yes but again we are talking about a meaningless quantity. Siri has a nice answer to "what is zero divided by zero?" maybe Siri also has one for "what is the dimensionful coupling of an RG fixed point?" – Abdelmalek Abdesselam Oct 4 '17 at 15:46