# What does the non-commuting nature of the translation and dilation generators mean for the scaling dimension of a field?

I am reading about CFTs from the book by Di Francesco, Mathieu and Senechal and in page 98 was introduced to the conformal group and the algebra of the generators. In particular, we have the dilation/scaling generator given by $$D = -i x^{\mu}\partial_{\mu}$$ and the translation generator $$P_{\mu} = -i \partial_{\mu}$$. These two generators (as an example) do not commute, $$[D,P_{\mu}] = iP_{\mu}\tag{4.19}.$$

Now, if I apply a dilation on a scalar field $$\phi$$, then I have:

$$\phi \rightarrow D\phi = \lambda^{\Delta}\phi ,$$ where $$\Delta$$ is the scaling dimension of the field. However, if I now apply a dilation on the translated field, i.e. on $$D\left(P_{\mu}\phi\right)$$, the transformation law is no longer a power law with exponent $$\Delta$$.

What does this mean for the field, and the scaling dimension since my understanding was that the scaling dimension is intrinsic/specific to a field.

• The expressions for $D$ and $P_\mu$ given are the infinitesimal generators, but the transformation on $\phi$ seems to be a finite one. Jun 6, 2022 at 19:37
• @MengCheng, I could just take $\lambda = e^{\rho}$ and then expand it, so the infinitesimal transformation for $\phi$ would be $\phi \rightarrow \Delta \phi$ maybe up to a constant. My question is not about the specific kind of transformation under $D$ but rather what it means for translation to change the transformation under $D$ Jun 6, 2022 at 19:42
• Sure, I know how to go from finite to infinitesimal. But I think it's better not to mix symmetry transformation and their generator. For what's worth, $P_\mu \phi$ is a descendant operator, with scaling dimension $\Delta+1$. Jun 6, 2022 at 20:23
• Oh okay, I understand. Thanks! Jun 7, 2022 at 5:30