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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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1 Answer 1

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.

For the quantum aspects, it is good to keep in mind that generically, a classical interacting conformal theory will not remains conformal after quantization (in particular, this is true also in two dimensions, answering one of your questions). Typically, to define the quantum theory one need to introduce a UV-cutoff, and this will generically trigger some non-trivial RG flow, breaking scale invariance. I think the answer to the previous question Defining a CFT using beta-functions contains good elements on this discussion, so I shall refrain to reproduce it here.

I hope this help !

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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
@user6818: I think there is some confusion in your messages. A mass term will automatically break conformal invariance, as it introduce a mass scale in the theory. Moreover, the very notion of RG flow is intrinsically quantum-mechanical, and thus it is meaningless to talk about RG flows in a classical theory. Finally, a free quantum theory never runs. These three facts should be usefull to solve some puzzles you raised. – Bru Mar 31 '13 at 11:37

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