Defining a CFT using beta-functions

Won't it be correct to define a CFT as a QFT such that the beta-function of all the couplings vanish?

But couldn't it be possible that the beta-function of a dimensionful coupling vanishes but it does so at a non-zero value of it - then the scale invariance is not generated though the renormalization flow is stopped? Is this possible?

(..it is obviously true that a theory with no intrinsic scale or dimensionfull parameter can still not be a CFT - like a marginal deformation of a CFT may not keep it a CFT and then this deformation parameter has to flow to a fixed point for a new CFT to be produced at that fixed point value of the marginal coupling..)

-
Consider a 4d theory with a Yukawa coupling and a $\phi^4$ interaction for the scalar. This has IR fixed points for a particular value of $\frac{\lambda}{g^2}$, and the two couplings can be nonzero. –  Siva Apr 2 '13 at 1:15
@Siva What is your definition of $\lambda$ and $g$? And how does your example fit the type that I am looking for? I would think in your suggested theory all interactions are still marginal. (see here - physics.stackexchange.com/questions/62070/… ) –  user6818 Apr 24 '13 at 0:17
The couplings are defined by the following terms in the lagrangian: $\lambda \phi^4$ and $g \phi \psi \psi$. You asked if there can be a theory with beta functions that are zero (i.e. couplings are fixed and don't renormalize) but the couplings are not zero. I gave an example of that. That example essentially happens to have N=1 SUSY (hidden in the background) relating the scalar and the fermion (and you needn't even be aware of that) which gives you such a fixed point. And yes, in my example, all interactions are marginal. –  Siva Apr 24 '13 at 4:32
@Siva Precisely my point - that it is NOT surprising if the RG flow of a marginal coupling has a non-zero fixed point. The issue is very interesting (and non-intuititve!) if it happens for non-marginal couplings - thats what I asked in my question. –  user6818 Apr 28 '13 at 0:17
Oh, I'm very sorry. I misunderstood that part. –  Siva Apr 28 '13 at 4:30

Your definition is quite good and works almost always. I'm quite sure it is rigorously true in 2D. You'll actually find it in some lecture notes. Remember that a theory is conformal if the trace of the stress tensor vanishes: $T \equiv T_\mu^{\mu} = 0.$ Indeed there is a folk theorem that states that

$T = \sum \beta_I \mathcal{O}^I$

where the sum runs over those operators $O^I$ in the theory with their beta functions $\beta_I$ (up to terms generating the conformal anomaly in curved space).

However, this is not completely true, and there are important classes of counterexamples where additional terms appear. Recently, these examples have led to some confusion in the literature (in the search for scale but not conformally invariant theories). All of this is well understood now and a good starting point for your studies would be 1204.5221 [hep-th].

Edit: don't forget that operator dimensions aren't protected and change under the RG flow.

-
When you say "T=0" what are you thinking about the usual conformal anomaly that goes proportional to the scalar curvature? What exactly do you mean by "counterexamples with additional terms"? They are counterexamples to what? And can you give a reference to this claim "T=\sum \beta_I O_I" ? Can one prove that a theory when sitting at a zero of all its beta-functions necessarily has conformal invariance? –  user6818 Feb 13 '13 at 3:15
As to the first remark: in even dimensions there will be terms proportional to those scalar densities, but since we're probably thinking about QFT in flat space let's forget about them. The literature about them is abundant. Those counterexamples have extra terms: $T = \ldots + \text{tensor with gauge indices},$ since they occur in gauge theories. So to have $T = 0$, $\beta_I = 0$ is not the right criterion in those cases. As to your last remark: my post meant to say that often the answer is yes, but in some cases there is some additional complexity (that is fully understood now). –  Vibert Feb 13 '13 at 8:47
@Vibert So you are saying that in general it is known that for CFTs one will always have $T = \sum \beta_I O_I + (\text{tensor with gauge indices})$? Does your reference prove this fact? Also can you comment on the other aspect as to if for a dimensionful coupling is it enough for the beta-function to go to $0$ or do you also want the zero to be at the the value of the coupling being 0? –  user6818 Feb 13 '13 at 19:46
That is true for general QFT's. In CFT, $T = 0$. The reference explains the fact, the proof is given in an old paper by Jack and Osborn, I think the original resource is 'Analogs of the c-theorem...' NPB 343. For dimensionful couplings it should suffice that $\beta(g) = 0$, you don't need the coupling itself to vanish. –  Vibert Feb 13 '13 at 21:06
@Vibert So when you say that $T=0$ do you mean $\langle T \rangle =0$ ? Because I would think that in general even if classically for something $T=0$, it would become anomalous by quantum corrections..right? So where is a reference which will prove this thing that $<T^\mu_\mu = \beta_I O_I \rangle = 0$ implies $\beta_I = 0$ ? Or is this still an open question? –  user6818 Feb 14 '13 at 1:33