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What I have learned after studying QFT on curved spacetime is that we replace the Minkowski metric, used in flat spacetime QFT, with the general metric and partial derivatives goes to covariant derivatives and also take other machinery of GR but the main point: we [formalism used in Birrell or Parker book and I'm not dwelling into modified theory of gravity to find the spacetime metric] always take $g_{\mu\nu}$ to be given by classical Einstein field equation. I have the following doubt regarding this setup:

  1. Why not take $g_{\mu\nu}$ to be a quantum operator/field as well? I know that quantum gravity when taking the Hilbert-Einstein action is nonrenormalizable but nonrenormalizable theory are still predictive. To be a bit clear suppose I do QED in curved spacetime and do quantum gravity coupled to electrodynamics both of them have a scale up to which they'll be predictive so does the first one wins in general case (larger scale of prediction) or is it related to computational complexity involved?
Níckolas Alves
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aitfel
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  • I still don't understand the question. There is plenty of work on quantising gravity. Are you asking why QFT on (classical) curved spacetime is useful? – fqq Jul 06 '20 at 11:39
  • @fqq yes! Why is it more useful than canonically quantized gravity? – aitfel Jul 06 '20 at 11:42
  • who says it is more useful? – fqq Jul 06 '20 at 11:44
  • @fqq when I said useful I meant it's pros and cons when compared to canonical quantized gravity? That would have been most natural interpretation of my statement. – aitfel Jul 06 '20 at 11:54
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    "QFT in curved spacetime" is, by definition, the field of research in which gravity is kept classical and everything else is quantized. You can't ask why. If instead you're asking - why do we not just quantize gravity as we do other fields in QFT and get on with it, that's a totally different question which has nothing to do with "QFT in curved spacetimes". You should rephrase. – Prahar Jul 06 '20 at 12:58
  • @fqq I'm extremely sorry. I only noticed (after thinking about Prahar's comment) that my question is very badly phrased what I had in mind was [this][1]. I'm not asking you to look at the question I just wanted to apologize for acting unjustified clearly when I was at mistake m(_ _)m. [1]: https://physics.stackexchange.com/q/567713/ – aitfel Jul 23 '20 at 17:43

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I'll address the question by means of an analogy with Electrodynamics.

We know how to quantize Electrodynamics pretty well. So well that nowadays it is a common subject in many graduate courses. However, if you take a look at some books in non-relativistic Quantum Mechanics, you'll see that many books (I can't check now, but I think Sakurai is an example) still deal with the interaction of atoms with classical electromagnetic fields. Using time-dependent perturbation theory, they study how an atom undergoes transitions in the presence of external electromagnetic fields. All of that analysis is often carried out assuming the background electromagnetic field to be classical.

In this situation, why do we use a classical electromagnetic field? Because it is way easier than doing the computations with Quantum Electrodynamics, and it still allows us to obtain quite interesting results. A similar idea holds for QFTCS: even though we do have a considerably reliable theory of Quantum Gravity at low energies (namely, General Relativity itself), there's many interesting consequences can be derived in the simpler case in which gravity is taken to be classical. It is simpler, but still extremely rich.

While this last paragraph is certainly dubious, I add for historical significance in the future that so far we still do not have experimental evidence of the quantization of gravity (although I've seen arguments in favor of it, I've seen physicists arguing that gravity might just be classical). Hence, another possible answer would be that QFTCS is interesting due to the possibility of gravity not being quantized at all, but from what I've seen I do not believe this to be the opinion of the majority of the community. In other words, I think most physicists working in QFTCS do it because it is an interesting formalism that leads to deep consequences, even though they don't see it as a final theory.

Níckolas Alves
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