# A question about defining a classical CFT

This is kind of related to this, Defining a CFT using beta-functions

So what would be the right definition of a CFT even classically?

• Is it true that classically one will call a theory scale invariant only if the action is invariant under scale transformations? (and not by the Lagrangian density)

For example under the scale transformations $x' = \lambda x$, in $3+1$ the scalar field goes as $\phi'(x') = \lambda \phi(x)$ and in $1+1$ it goes as $\phi'(x') = \phi (x)$. This means that in $3+1$ the action of the massless scalar field is not scale invariant but in $1+1$ it is but the Lagrangian density goes the otherway. But from the point of view of beta-functions isn't it more consistent to call the $1+1$ theory as a CFT but not the one in $3+1$?

Isn't a massless scalar field theory in $3+1$ guaranteed to produce mass by RG flow whereas the $1+1$ theory will not?

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A classical field theory is said to be conformally invariant iff the action is left invariant under the action of the conformal transformations (as usual when we define a symmetry of a theory). The action for the free massless scalar field is invariant under scale transformations in any dimensions because the (classical) scaling dimension for the scalar field is taken to be $(d-2)/2$. In the interacting case (but still classical), it is enough for the couplings to be dimensionless to ensure scale invariance. So for exemple in four dimensions, you can have only $\phi^4$ interactions. In any cases, a mass term will always break scale invariance.
I didn't understand your comment about the action being scale invariant for a massless scalar in any dimension. That was the point of my question - under $x' = \lambda x$, $\phi'(x') = \lambda \phi(x)$, $\int d^4x (\partial _\mu \phi(x))^2$ is not scale invariant. The action in primed variables is $\lambda ^4$ times the unprimed one - it comes from the measure (though the Lagrangian density is covariant) – user6818 Mar 30 '13 at 2:45
The Lagrangian density has mass dimension $4$, that's easy to see: $[\partial_\mu] = [\phi] = 1$ (in 4D). So the action, including the measure, will be invariant. – Vibert Mar 30 '13 at 9:56
Actually there is a mistake in your formula, user6818. In four dimensions, $\phi'(\lambda x)=\lambda^{-1} \phi (x)$. – Bru Mar 30 '13 at 10:05
@Bru Ah! Here I guess the power of $\lambda$ is being determined by trying to ensure that as a $1-form$, $\phi'(x')dx' = \phi(x)dx$..right? In general for a primary operator of weight $(h,\bar{h})$ one would want $A(z',\bar{z}')(dz')^h (d\bar{z}')^{\bar{h}} = A(z,\bar{z})dz d\bar{z}$ So for every $d$ I guess one will have to first derive the values of $h$ and $\bar{h}$ such that one can satisfy $h + \bar{h} = mass\text{ }dimension$ and $h - \bar{h} = spin$..right? – user6818 Mar 30 '13 at 22:40