# Proof that free scalar field is conformally invariant

So, under conformal transformations $$x\mapsto x'\\ \phi\mapsto\phi'(x')=\Omega^{(2-D)/2}\phi(x),$$ where $$\eta_{\mu\nu}\frac{\partial x^\mu}{\partial x^{'\alpha}}\frac{\partial x^\nu}{\partial x^{'\beta}}=\Omega^{-2}(x)\eta_{\alpha\beta},$$ the action transforms like $$S\mapsto\int d^Dx\,\Omega^{D-2}\partial_\mu(\Omega^{(2-D)/2}\phi)\partial^\mu(\Omega^{(2-D)/2}\phi)$$ (A quick way to get to this equations is by instead considering the associated Weyl transformation as described in an answer in Simple conceptual question conformal field theory). It is then obvious that the action is invariant under scale transformations, i.e. when $$\Omega$$ is constant. However, why is it invariant when $$\Omega$$ is not constant? What does one do with the terms involving derivatives of $$\Omega$$?

• The only non-constant $\Omega$ for which this will still be invariant is the one corresponding to special conformal transformations. Jun 23 at 23:32
• So this theory is not Weyl invariant? In any case, let us assume that $\Omega$ is restricted by the equation relating the metrics in the different coordinate systems. This should be enough to proof the invariance of the action since that restriction is already enough to show that $\Omega$ comes from dilations or special conformal transformations (or compositions thereof). Jun 23 at 23:40
• I think this is because $\phi$ is not a primary field. Rather, $\partial\phi$ is a primary field, and the action is invariant under conformal transform. This is true in 2d. For general dimensions, I am not sure. Jun 25 at 12:28
• Oh yeah, for sure. So the general action is Weyl invariant (so the $\Omega$ doesn't have to come from a conformal transformation) if we scale $\partial\phi$ instead of $\phi$. I remember something similar happening in electrodynamics (maybe in a book by Wald?). Jun 25 at 14:47
• $\partial_\mu \phi$ is not primary in $d > 2$. Jun 28 at 16:10