Klein-Gordon equation coupled to scalar curvature

Question

Consider the Klein-Gordon equation of the form

$$\square_g \psi - m^2 \psi - \xi R \psi \enspace = \enspace 0 \quad .$$

This equation describes the relativistic propagation of a scalar field with mass $m$ on a curved spacetime with metric $g$. My question now concerns the interpretation of the term which couples the field to the scalar curvature $R$. Since the d'Alembertian operator $\square_g$ already includes the properties of the curved space on which the wave propagates, why is there any need to couple the equation to the scalar curvature? The d'Alembertian operator should already "cover" the influence of the curvature on the wave, shouldn't it?

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E D I T :

Thank you for your answers so far. GHOSTER's answer of "allowing all possible terms" makes of course sense in how to fundamentally justify the occurence of this term in the Klein-Gordon equations most general form. What still confuses me - and this question may only arise due to a lack of basic knowledge in this sector - is the question: what makes the scalar curvature term "possible" in contrast to whatever other term, say e.g. the Kretschmann scalar $R^{abcd} \, R_{abcd}$? In other words, what condition does $R$ satisfy such that it is considered "possible"? Some condition must be satisfied otherwise all scalar fields on the spacetime would be "possible" and therefore by the principle "allow all possible terms" be added to the equation, wouldn't they?

Answer 1

You can put $\xi=0$ and you'll get a minimally-coupled massive scalar in curved spacetime. Nothing prevents you from doing that. The reason you usually see the $\xi R \psi$ term in the equations of motion is because it is that, beyond minimal coupling, this is the simplest way in which you can couple the scalar to gravity (it comes from a quadratic in $\psi$ term in the action $\sim \xi R \psi^2$). You are free to add higher dimensional/derivative operators to the action to get a more general (effective field) theory, like $\frac{1}{\Lambda^2}\xi \psi^2 R^2$ or $\frac{1}{\Lambda^8} \xi \psi^2 R^5$ or $\frac{1}{\Lambda^4} \xi \psi^2 R \square R$ or $\frac{1}{\Lambda^2} \xi \psi^2 R^{\alpha \beta \gamma \beta} R_{\alpha \beta \gamma \beta}$, etc. Note the high energy mass scale $\Lambda$ which must accompany these terms with appropriate powers so that the action remains dimensionless. The point is that these higher order terms are not relevant at low energies ($< \Lambda$). Moreover, these negative dimensional couplings make the theory non-renormalizable in general (which is not necessarily a problem, though). See my answer here: https://physics.stackexchange.com/a/467869/133418

Answer 2

You ask whether there is any need to introduce this non-minimal coupling, and the answer depends what you mean by 'need'. If you mean, is there some reason it must be there, then I'd say no. You can look at this both from a mathematical consistency point of view.

There is no theoretical necessity, because as you said, the field is already coupled to geometry through $\nabla$ and the action is covariant. There is motivation in various different contexts (e.g., from string theory and theories of quantum gravity or EFT considerations), but some of these arguments are speculative and they don't constitute any real reason that non-minimal coupling is incorrect. If you want to read more, there's a tonne of literature on minimal coupling in different areas (especially inflation).

And, as far as I'm aware, there's no physical necessity from observational evidence. Depending on the specific context and type of data you're looking it, it may or may not be favoured over the standard minimally coupled models. But there is not hard evidence to indicate it's presence in nature. For a bit more about observational constraints, here's an example https://arxiv.org/abs/1511.08736.