Should kinetic energy be conformally invariant for a QFT in curved space?

Question

When coupling a scalar field to gravity, one sometimes introduces an additional term into the action: $S=\int d^4x \sqrt{-g}(L-\frac{1}{2}H_0 R\phi_0^2)$ where $R$ is the Ricci scalar, $L$ is the matter lagrangian, and $H_0$ is a parameter tuned to make the kinetic energy conformally invariant. This so-called "non-minimal" term is zero in flat space, but it's derivative with respect to the metric gives a modification to the canonical energy momentum tensor, defining the "Improved Energy Momentum Tensor".

In the paper https://doi.org/10.1103/PhysRevD.14.1965 (apologies for the paywall), Collins writes that

"One might say the minimal way to go from flat to curved space is not for the kinetic energy term to be $\frac{1}{2}(\partial\phi)^2$, but for it to be conformally invariant".

If a theory exhibits conformal invariance, great. But I don't understand why such a thing should be imposed, especially for a massive theory. Is there an intuitive reason the kinetic term in a lagrangian should be conformally invariant?

Answer 1

The massless scalar in flat space is classically conformal, so it is natural to do whatever is necessary to maintain that symmetry when lifting to a curved background.

The massive scalar is not conformal, and cannot be conformal.

Edit

The above is in some sense correct, but it requires clarification. A conformal transformation is a special case of a general coordinate transformation, and so any theory on a curved background is invariant under conformal transformations! When we say a theory is conformal, the nontrivial statement is that it is conformal on flat space. In order for a theory to be conformal on flat space, the theory on curved space must be both diffeomorphism invariant and Weyl invariant.

A Weyl transformation is not a coordinate transformation, it is a local rescaling of the metric $g'_{\mu\nu}(x)=\Omega^2(x)g_{\mu\nu}(x)$ (note that a conformal transformation is coordinate transformation which results in the metric transforming the same way). Now, in order for the theory on curved space to correspond to the one on flat space, the symmetries in the curved space need to correspond to the ones on flat space. So, the massless scalar which is classically conformal on flat space needs to have Weyl invariance on curved space.

Just so we have a concrete theory, the action is $$S=\frac{1}{2}\int d^dx\sqrt{g}\Big(g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi+\frac{d-2}{4(d-1)}R\phi^2-V(\phi)\Big)$$

The point is if you don't include the $R\phi^2$ term, and take the flat space limit, the theory actually won't be conformal.