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Is there any program which attempts at unifying quantum mechanics and gravity rather than unifying quantum field theory and gravity?

Motivation

We use the heisenberg picture to define velocity $\hat v$:

$$ \hat v = \frac{dU^\dagger x U}{dt} = U^\dagger\frac{[H,x]}{- i \hbar}U$$

where $U^\dagger$ is the unitary operator and $H$ is the Hamiltonian.

Now we can again differentiate to get acceleration $\hat a$:

$$ \hat a = \hat U^\dagger\frac{[[\hat H, \hat x], \hat x]}{-i \hbar} \hat U = \hat U^\dagger\frac{(\hat H^2 \hat x + \hat x \hat H^2 - 2\hat H \hat x \hat H)}{\hbar^2} \hat U $$

We can simplify the calculation by splitting the Hamiltonian into potential $ \hat V $ and kinetic energy $ \hat T $: $\hat H = \hat T + \hat V$

By noticing (one can also calculate this) that the acceleration of an object in a constant potential is $0$:

$$ \hat 0 = \hat T^2 x + x \hat T^2 - 2 \hat T \hat x \hat T $$

We also know $ [\hat V, \hat x] = 0 $ as potential is a function of position. Thus, we can simplify acceleration as:

$$ \hat a = \hat V \hat T \hat x + \hat x \hat T \hat V - \hat V \hat x \hat T - \hat T \hat x \hat V $$

Note this acceleration operator also commutes with position:

$$ [\hat a , \hat x ]=0$$

Now from the equivalence principle we know that the effect of acceleration is indistinguishable from gravity. Hence, even the quantum mechanical version of the Riemann curvature tensor must also commute with position. Note, this argument does not work for high energies (QFT) as acceleration does not make sense in QFT.

drewdles
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    (didnt downvote) note that QFT is just an example of QM and viceversa, so the question is kind of meaningless. – AccidentalFourierTransform May 24 '17 at 16:12
  • @AccidentalFourierTransform I have edited the question ... Hopefully now it makes more sense where I'm coming from. – drewdles May 24 '17 at 16:19
  • Hi Anant, I have the feeling that you have asked this question several times already. If you are not satisfied with the answers you got, you should offer a bounty, not ask the same question again. Thanks. – AccidentalFourierTransform May 24 '17 at 16:29
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    @AccidentalFourierTransform I asked something different once before ... It was a similar argument but more to do with quantization with momentum ... It also got closed ... I have now refined my arguments and posted a new question to find out if anyone has made this argument in the past besides me ... I hope this is acceptable – drewdles May 24 '17 at 16:32
  • Why $\hat v$ is not defined as $d\hat x /dt$ ? or $\hat p/m$ –  May 24 '17 at 16:35
  • It is: $ \hat v = \frac{d \hat x(t)}{d t} = d(\frac{U^\dagger x U)}{dt} $ – drewdles May 24 '17 at 16:36
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    Due to Malament's theorem it's not really possible to have quantum mechanics with point particles in a relativistic theory. – Slereah May 24 '17 at 17:12
  • Related: https://physics.stackexchange.com/q/387/2451 – Qmechanic May 25 '17 at 14:32
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    "Now from the equivalence principle we know that the effect of acceleration is indistinguishable from gravity" --> The equivalence principle guarantees that acceleration is locally indistinguishable from gravity. How do you define a commutator between, say, the Ricci scalar and the position $\hat{x}$, keeping in mind that the Ricci scalar is a function of position? For example, the Ricci scalar involves derivative of the metric wrt spatial coordinates; how would you define differentiation with respect to an operator? – Andrew Feb 13 '21 at 18:18
  • What is $U$ in your first expression? – Lucas Baldo Jun 17 '21 at 01:06

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General relativity is a field theory, so any quantization of it is an example of QFT rather than non-relativistic QM. However, Newtonian gravity's quantization requires no QFT-level treatment; you just use a $1/r$ potential, as with the quantum hydrogenlike atom's electromagnetism. You can even use a $V=mgh$ approximation. There have been studies of cold neutrons falling in Earth's gravity, e.g. https://www.researchgate.net/profile/T_Jenke/publication/223269839_QuBounce_the_dynamics_of_ultra-cold_neutrons_falling_in_the_gravity_potential_of_the_Earth/links/57dbd38e08aeea1959355e6e.pdf & https://arxiv.org/pdf/1103.1015 .

J.G.
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  • This answer completely ignores my motivation though ... Can you also address that? Because it seems to me I can argue that unifying QM with the equivalence principle requires a change in standard QM. – drewdles May 24 '17 at 16:28
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    @AnantSaxena Newton's gravity is just a force, whereas the equivalence principle requires a non-trivial geometry. If you concede Lorentz invariance we can't work with QM alone, so what you're hoping for would require some kind of space-only equivalence principle, which Einstein abandoned even in an entirely non-quantum model. – J.G. May 24 '17 at 16:39
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    Would my argument work if I used the Dirac equation instead and used the same definition of acceleration? (I only know acceleration does not make sense in the Klien Gordon equation). – drewdles May 24 '17 at 16:50