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Imagine I am doing Einstein's famous thought experiment: I'm in a lift and I'm asked to do local experiments to determine if I am accelerating in a non-inertial reference frame or in some gravitational potential accelerating me.

Now, I get clever and measure the standard deviation of acceleration. Standard deviation is a property of a quantum mechanical operator. This is a legal move see:

Can one define an acceleration operator in quantum mechanics?

And check if it mimics the standard deviation in a non-inertial reference frame:

The 3 fictitious forces of the rotating frame in Quantum Mechanics

By the equivalence principle they must be the same as otherwise I could tell you if I was locally in a gravitational field or not. Can I claim the quantization of the gravitational potential must also be the same? Note, one can imagine an array of of possible cases for example of the potential being $mg \hat x$ (where $m$ is mass, $g$ is the gravitational field and $x$ is the position operator)

Qmechanic
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  • The gravitational potential is not covariant so it makes no sense to try and quantise it. In quantum gravity we quantise the metric i.e. the graviton is a quantum of curvature. – John Rennie Oct 31 '23 at 14:11
  • @JohnRennie we are thinking of it differently. I suspect the equivalence principle is optional for you? And no not all theories of quantum gravity quantize the metric. – More Anonymous Oct 31 '23 at 14:21
  • Also V= mgx is a valid potential of Newtonian gravity – More Anonymous Oct 31 '23 at 14:23

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