Background
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I recently read a blog post on measuring gravity waves of objects in superposition:http://backreaction.blogspot.in/2016/03/researchers-propose-experiment-to.html
And did some calculations regarding the same.
Calculations
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We go into the heisenberg picture to define velocity:
$$ \hat v = \frac{dU^\dagger x U}{dt} = U^\dagger\frac{[H,x]}{- i \hbar}U$$
where $\hat v$ is the velocity operator $U^\dagger$ 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:
$$ [\hat a , \hat x ]=0$$
Implications
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Now by the equivalence principle: "acceleration of the object is indistinguishable from gravity." Hence, I argue if the quantum version of the Ricci curvature tensor is measured then the position operator will also collapse and the superposition will exist no more.
Questions
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Is this line of reasoning correct and also existent in the literature? What do other theories of quantum gravity predict for the gravitational waves for objects in superposition?
Note: I used a similar line of reasoning here (for something else). However, this is a different question.
https://www.physicsoverflow.org/38402/there-any-astrophysical-situation-where-this-idea-testable