# Is a conformally coupled scalar always massive?

Maybe this is trivial, but the action of a conformally coupled scalar is $$S = \frac{1}{2} \int d^Dx \sqrt{g} (g^{ab} \partial_a \phi \partial_b \phi + \xi R \phi^2),$$ where $$\xi = (D-1)/4D$$. Does this imply that...

1. ... a theory of conformally coupled scalars is always massive (i.e., there exists by definition no massless, conformally coupled scalar theory) since there is always a mass term $$\propto \phi^2$$?
2. ... minimally coupled theories are always massless (since minimally coupled means $$\xi = 0$$)?

That isn't really a mass term, but rather a curvature-coupling term, due to that being the Ricci scalar. The difference is that $$R$$ is not constant, but rather a function of spacetime with a non-trivial metric dependence. This means, for example, that that term will lead to different contributions in a stress tensor than a mass term would. If you want to see a bit of the details, I wrote this pdf a while ago with the computations of the stress tensor for a scalar field. Notice that the curvature coupling $$\xi$$ and the potential $$V(\phi)$$ enter the expression in very different ways.
The correct conclusion is actually that conformal theories are always massless. Notice that conformal symmetry implies scale symmetry, which would not be possible if you had a mass term dictating a preferred energy scale (which implies a preferred length scale and etc). You could, of course, add a mass term to a conformally coupled theory, but you would spoil conformal symmetry. I don't see any general argument to be made concerning minimally coupled theories: they can be massless, they can be massive, and you won't get conformal symmetry in any of these cases$$^*$$.
$$^*$$: since a conformal coupling leads to a different stress tensor, this is also subtle in Minkowski spacetime. Even though $$R=0$$ in Minkowski spacetime, a conformally coupled field and a minimally coupled field have different stress tensors even in Minkowski spacetime.