# Primary field in CFT and path integral

I should feel ashamed to ask such a naive question, but anyway let me start with the $$\phi^4$$ theory in the Minkowski spacetime, which has a Lagrangian of the form $$\frac{1}{2}(\partial\phi)^2-\frac{1}{4!}g\,\phi^4$$ One say that it is scale invariant if under the transformation $$x^\mu \rightarrow \lambda x^\mu$$, the field $$\phi$$ transforms as $$\phi(x)\rightarrow \frac{1}{\lambda}\phi(x)$$.

So when we consider QFT, we take the path integral of this Lagrangian over all the configuration of the field $$\phi(x)$$. But if we are interested in scale invariance of this theory, in path integral formulation, do we only integrate over the configurations of $$\phi$$ which transforms in the way $$\phi(x)\rightarrow \frac{1}{\lambda}\phi(x)$$?

Similarly in QFT, a primary field transforms in a very specific way (like the rules of the tensor transformation). When we consider the corresponding quantum theory, in the path integral, do we only integrate over the field configurations of primary fields? Instead of integrating over all the fields which might not be primary (of the same type)!

• Note that you want to action to be scale invariant. You want to choose the transformation rule of $\phi$ so that the kinetic term is scale invariant. So in $D$ spacetime dimensions it's $\phi \mapsto \lambda^{1-D/2}\phi$. This happens for $D = 4$ when $\phi \mapsto \lambda^{-1} \phi$ as you say, but your logic sounds strange. – Ryan Thorngren Dec 8 '18 at 14:35
• @Ryan Thorngren, It sounds strange because he is asking whether the field configurations you integrate over in the path integral themselves have the property that $\phi(x)=\lambda^{-1}\phi(\lambda x)$. That's a fine question, but it's not true, and it's due to a misconception between fields and operators. – octonion Dec 8 '18 at 16:02

• Sorry that I need to bother you again. From a Youtube video talking about CFT, the lecturer says that in the free (scalar) field theory defined on $\mathbb{C}$ ($\mathbb{R}^2$), where $\mathscr{L}=(\partial \phi)^2$ up to a constant, after quantization the operator valued field $\phi(x)$ is not primary, as the two point correlation function is $\log$ something. But $\phi(x)$ clearly satisfies the condition in the definition of primary field with weights $h=\bar{h}=0$, is this just a convention, or do I misunderstand something? – Wenzhe Dec 15 '18 at 4:52