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I am thinking that they aren't strictly compatible. I have the following logical argument for this:

  1. The unitary evolution postulate says that the state of a system is given by a time-depending state vector $\psi (t)$ which evolves unitarily.

  2. For the physical theory to have predictability, the parameter $t$ in $\psi (t)$ must be proportional to the time measured on the experimenter's clock.

  3. According to general relativity, the number of ticks measured on the experimenter's clock is proportional to the proper time along their worldline.

  4. The proper time along the worldline is defined in terms of an integral involving the metric tensor.

  5. The metric tensor is coupled to the stress-energy tensor.

  6. According to quantum field theory, the stress-energy tensor is not a classical entity and has quantum-uncertainties.

  7. This means that the proper time along the experimenter's worldline is strictly not well-defined in a fully quantum theory. This kills the unitary evolution postulate because you can't have unitary evolution when its parameter isn't well defined.

The above proves that the parameter of unitary evolution is well-defined only in semi-classical contexts where the proper time of the experimenter's worldline is well-defined. This semi-classical theory cannot be a fundamental theory because it's making quantum mechanics reliant on classical mechanics.

This also means that a fully quantum theory gravity cannot obey the usual postulates of quantum theories.

What, if anything, is wrong with this argument?

Peter Mortensen
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Ryder Rude
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  • None of what you are describing there is quantum gravity. It's classical gravity. The concepts you are trying to use don't make any sense in quantum mechanics, not even in flat spacetime. – FlatterMann Jun 19 '23 at 10:44
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    No one knows since we don't know "the" quantum theory of gravity. In string theory, unitary evolution is preserved. However, one resolution of the black hole information loss paradox is that quantum gravity does not respect unitary evolution. Many people have strong feelings about this topic but to date there is no single correct answer. – Andrew Jun 19 '23 at 13:51

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This doesn't argue that time-evolution is not unitary in quantum gravity. It argues that time-evolution is not defined. Congrats, you just landed on the problem of time.

ɪdɪət strəʊlə
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  • Is it standard belief among physicists that Quantum Gravity won't obey the usual quantum postulates? – Ryder Rude Jun 19 '23 at 10:44
  • @RyderRude There are no world lines in "the usual quantum postulates". – FlatterMann Jun 19 '23 at 11:03
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    @RyderRude well, it is a standard belief that something has to be tweaked a little bit. – ɪdɪət strəʊlə Jun 19 '23 at 12:35
  • @FlatterMann of course there are. So long as you sum over them in the path integral: https://ncatlab.org/nlab/show/worldline+formalism – ɪdɪət strəʊlə Jun 19 '23 at 12:35
  • @ɪdɪətstrəʊlə Path integrals aren't integrating over worldlines. In most calculations they aren't even integrating over positions but momenta. They are for sure not integrating over positions and momenta of classical objects but about the complex exponential of the classical action. In other words, they are quantization procedures, not descriptions of actual physics. – FlatterMann Jun 19 '23 at 17:13
  • @FlatterMann Check out section 2.1 of Israël's string theory notes (don't worry, despite being string theory notes this is section is just QM/QFT). There a worldline path integral for the free relativistic particle is set up and shown to be equivalent to the traditional formulation of QFT. – ɪdɪət strəʊlə Jun 19 '23 at 18:13
  • @ɪdɪətstrəʊlə There is, physically, no such thing as "a free particle" in QM because there are no particles at all. That's just a poorly chosen phrase from the early days of quantum mechanics. The only way to obtain a "worldline" in QM is to allow for continuous weak interactions of the quantum system with a matter background. That is what we do in "particle detectors". Such a physical process does, obviously, break relativity because the detector matter is a preferred physical coordinate system. One can not build a relativistic theory on an effect that is thoroughly non-relativistic. – FlatterMann Jun 19 '23 at 18:18
  • Let us continue this discussion in chat. – ɪdɪət strəʊlə Jun 19 '23 at 18:19