Question

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I believe the following uncertainty principle to be true in the regime $ v << c$ :

$$ \Delta g \Delta p \geq | \langle \dot U \rangle + \langle p \rangle \langle g \rangle|$$

Is there any astrophysical situation which can test this?

Background

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I recently had the following idea in the regime of quantum mechanics and gravity $ v << c$: we can the Heisenberg picture we define an acceleration operator and using Einstein's equivalence principle we state it must be equivalent to $g$ operator (quantum mechanical version of $g$ field) and thus we proceed to find an uncertainty principle.

Mathematical Details

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

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

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

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

Going back to the Schrodinger picture:

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

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:

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

Now from the equivalence principle we know that the effect of acceleration is indistinguishable from gravity. Using this fact:

$$ \hat g = \hat a $$

However, $\hat g$ does not commute with $\hat p$ (it commutes with position). Using the anti-commutator (rather than the commutator) we should observe:

$$ \Delta g \Delta p \geq | \langle \dot U \rangle + \langle p \rangle \langle g \rangle|$$

By Schwatz inequality $ | \langle \dot U \rangle + \langle p \rangle \langle g \rangle| $ is $0$ only when $U$ is constant.