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 $\xi = 0$)?

1 Answer 1


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.

  • 1
    $\begingroup$ Could someone argue that this form of coupling is an effective mass term of gravitational origin? I've seen papers where it is mentioned that this is a mass term, for example arxiv.org/abs/1707.03483 equation 2.13. $\endgroup$
    – Noone
    Jun 1 at 21:19
  • $\begingroup$ @Noone Perhaps, but I find it a little bit weird. It is a mass-like term in the sense it is quadratic, but it certainly does not behave energetically in the same way a mass term does and it has the odd behavior of depending on spacetime event (because it depends on the Ricci scalar, which is a function). I'd say you can call it a "mass term" on a case-by-case basis. It depends on the point you're trying to make $\endgroup$ Jun 2 at 6:12
  • $\begingroup$ I would like to reinforce @Noone 's argument since throughout the literature authors seem to imply that conformally coupled scalars are massive. For example iopscience.iop.org/article/10.1088/1475-7516/2021/08/003/pdf eq (2.9) - (2.11) there are two cases: massless and conformally coupled. Or another example in arxiv.org/pdf/2203.15330.pdf eq (75) where there is a case distinguishment between "massless" and "conformally coupled", implying that conformally coupled means massive. $\endgroup$ Jun 2 at 11:43
  • $\begingroup$ @Welcome_Green it depends on the particular context. A mass parameter introduces a preferred scale that breaks scale invariance, so it breaks conformal symmetry. However, the conformal coupling can be "mass-like", in the sense it might occur in the propagator just like a mass term would and so on. In some contexts, it can be seen as mass-like, but it is not a mass term. Some papers might use "conformally coupled vs massless" as a way to quickly distinguish between massless conformal coupling and massless minimal coupling, but there are quite some differences between a mass term and a + $\endgroup$ Jun 2 at 17:59
  • $\begingroup$ curvature coupling (as one can see in the stress tensor, for example) $\endgroup$ Jun 2 at 17:59

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