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)!

  • $\begingroup$ 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. $\endgroup$ – Ryan Thorngren Dec 8 '18 at 14:35
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    $\begingroup$ @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. $\endgroup$ – octonion Dec 8 '18 at 16:02

To be brief, no you integrate over all field configurations. Field configurations are not operators they are ordinary functions that are summed over in the path integral. Primary fields are operators. They appear in correlation functions which involve an expectation value over all field configurations. The conformal invariance happens at the level of operators and correlation functions.

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  • $\begingroup$ Another naive question about the language. Usually, the path integral formulation of QFT does not use "operators", so when you say primary fields are operators you mean the operator formulation of QFT? While in the path integral language, primary fields are special (classical) fields that transform in a special way, and then they appear in correlation functions that involve an expectation value over all field configurations. Of course correlation can also be formulated using operator language. Is this right? $\endgroup$ – Wenzhe Dec 8 '18 at 17:12
  • $\begingroup$ @Wenzhe, when you find a correlation function of the primary fields in the path integral language, you integrate a product of field configurations at different spacetime points weighted by the action in an exponential factor. These field configurations do not themselves transform in that special way, even though they are associated to primary fields. It is only the expectation value over all field configuration that has that transformation property. That's why I'm making the distinction between 'operators' (things that live in correlation functions) and individual field configurations. $\endgroup$ – octonion Dec 9 '18 at 1:48
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    $\begingroup$ A very simple example: scalar field theory is invariant under translations. The actual individual field configurations you integrate over are messy arbitrary things, but the correlation functions are indeed translation invariant. $\endgroup$ – octonion Dec 9 '18 at 1:52
  • $\begingroup$ 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? $\endgroup$ – Wenzhe Dec 15 '18 at 4:52
  • $\begingroup$ @Wenzhe, Good question, why do you think it obviously satisfies the definition? Since the correlation function is a log, under dilations it will shift by a constant term. You might want to ask this as a new question. $\endgroup$ – octonion Dec 19 '18 at 10:00

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