# Is every classical field theory with dimensionless couplings conformally invariant?

I'm trying to learn conformal field theory and getting rather frustrated, because I can't find any source that gives decent examples or straightforward logic.

In most sources I have found, conformal invariance for a classical theory is established by showing that the theory is scale invariant, then performing some handwaving. However, this reasoning can't be remotely correct, as every theory with only dimensionless couplings is scale invariant, as long as we give each field a scaling dimension equal to its usual mass dimension.

For example, under this reasoning, the Standard Model without a Higgs mass term is a conformal field theory at the classical level, but it can't possibly be, or else I would have heard about that already. Similarly, massless $$\phi^4$$ theory in $$d = 4$$ would be conformally invariant under this logic, but it isn't, as its stress-energy tensor isn't traceless. Yet another example is a free scalar field in a curved spacetime background, which requires a non-minimal coupling term $$- R \phi^2 / 12$$ to achieve conformal invariance.

What's going on? Are all CFT sources just being incredibly sloppy, and if so, why are they being this sloppy? Is there some reason that this fuzzy logic actually works in most examples they consider, e.g. is the situation different in $$d = 2$$, perhaps? And what's the real way to show a theory is classically conformally invariant?

• Maybe useful: arxiv.org/abs/1101.5385 – Avantgarde Mar 12 '19 at 0:03
• @Avantgarde From that paper: "First, it provides a clearcut counterexample to the often assumed conjecture that any unitary and scale invariant theory is conformal. It is pretty amazing that this counterexample escaped the attention of the recent literature on the subject." Phew, I guess that answers my question. Except for why everybody is so sloppy... – knzhou Mar 12 '19 at 0:15

The question whether a classical theory is conformal or not can be answered by looking at the trace of the energy-momentum tensor $$T^{\mu\nu}$$. If $$T^\mu_\mu = \partial_\mu K^\mu$$ for some operator $$K^\mu$$, then the theory is invariant under scale transformations. If moreover $$T^\mu_\mu = \partial_\mu \partial_\nu L^{\mu\nu}$$ for some $$L^{\mu\nu}$$, then the theory is also invariant under special conformal transformations. This is nicely explained in a paper by Polchinski.

Not every classical field theory with dimensionless coupling is conformal. A counterexample is Maxwell theory in dimension $$d \neq 4$$ (as mentioned in the comments). Another would be the theory of elasticity in 2 dimensions. But the Standard Model without a Higgs mass term is definitely a conformal field theory at the classical level. So is the massless $$\phi^4$$ theory (see below for a detailed explanation).

The probable reason why you have never heard that the Standard Model without a Higgs mass term is classically conformal is that nobody cares. After all it is a quantum theory, and conformal invariance is broken at the quantum level in that theory. Also it is not immediately obvious what you gain by knowing that a classical theory is conformal. (On the other hand if a theory is conformal at the quantum level, then you gain a lot because you can use tools like the state/operator correspondence to construct the Hilbert space of the theory explicitly, but this goes beyond the scope of the question.)

## Conformal invariance of the classical massless $$\phi^4$$ theory

Consider a theory in $$d$$ spacetime dimensions defined by the Lagrangian $$\mathscr{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{\lambda}{4!} \phi^4$$ The equation of motion is $$\square \phi + \frac{\lambda}{3!} \phi^3 = 0$$ and the canonical energy-momentum tensor $$T_c^{\mu\nu} = \frac{\partial \mathscr{L}}{\partial(\partial_\mu \phi)} \partial^\nu \phi - \eta^{\mu\nu} \mathscr{L} = \partial^\mu \phi \partial^\nu \phi - \frac{1}{2} \eta^{\mu\nu} \partial_\rho \phi \partial^\rho \phi + \frac{\lambda}{4!} \eta^{\mu\nu} \phi^4$$ The trace of this tensor is non-zero (even when $$\lambda = 0$$ in $$d \neq 2$$), but this does not mean that the theory is not invariant under conformal transformations. A theory is conformal if its energy-momentum tensor can be "improved" with a term of the form $$\partial^\mu \partial^\nu L - \eta^{\mu\nu} \square L$$ into a traceless tensor. In our case we can take $$T^{\mu\nu} = T_c^{\mu\nu} - \frac{d-2}{4(d-1)} (\partial^\mu \partial^\nu - \eta^{\mu\nu} \square) \phi^2$$ which gives $$T^{\mu\nu} = \frac{1}{2(d-1)} \left[ d \partial^\mu \phi \partial^\nu \phi - \eta^{\mu\nu} \partial_\rho \phi \partial^\rho \phi - (d-2) \phi \partial^\mu \partial^\nu \phi + (d-2) \eta^{\mu\nu} \phi \square \phi \right] + \frac{\lambda}{4!} \eta^{\mu\nu} \phi^4$$ The trace of this tensor is $$T^\mu_\mu = \frac{d-2}{2} \phi \square \phi + \frac{d \lambda}{4!} \phi^4$$ and you see that it vanishes by the equation of motion exactly when $$d = 4$$ (that is when $$\lambda$$ is dimensionless).

• This is very helpful, thanks! So to explain the residual sloppiness: people don't include $-R \phi^2/12$ because they're usually concerned with CFTs on flat spacetime where this term vanishes, and they give fuzzy logic in the classical case because they just want to dispose of it quickly to get to the quantum. – knzhou Mar 12 '19 at 13:49
• Nice answer. Note that in the classical context, conformal transformations aren't entirely useless — they're essential for defining things like null infinity in curved-spacetime theories. But GR isn't conformally invariant, and it's also unclear to me whether knowing that a classical model is conformally invariant gains you anything other than a warm fuzzy feeling. – Michael Seifert Mar 12 '19 at 13:50
• And the reason that I never see people actually show conformal invariance (in $d > 2$) by explicitly showing the action doesn't change under a conformal transformation is because it's complicated, and showing the tracelessness of $T^{\mu\nu}$ is a much easier route. Is that all right? – knzhou Mar 12 '19 at 13:51
• Finally, classically, the "conformal/scaling dimension" is always equal to the ordinary dimension, right? Sorry for all the questions, I am just completely unable to find any sources where any of this is stated explicitly. – knzhou Mar 12 '19 at 13:52
• @knzhou Computing $T^{\mu\nu}$ or the variation of the action are really the same thing at the end of the day... – M.Jo Mar 12 '19 at 13:56