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Is there any astrophysical situation where this idea is testable?

+ 0 like - 1 dislike
77 views

Question
---
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
---
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
---
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.
      

asked Feb 9 in Phenomenology by Asaint (20 points) [ revision history ]
recategorized Jul 13 by dimension10

what are $U$, $g$, $p$,$x$?

$U$ is the unitary operator. 
$g$ is the $g$ field
$p$ is the momentum
$x$ is the position

which unitary operator? What is the $g$ field? 

$g$ is the gravitational field https://en.wikipedia.org/wiki/Gravitational_field

I'm using the standard definition of the unitary operator $U=e^{i H t}$ where $H$ is the Hamiltonian.
 

Usually one writes this $U(t)$, not $U$. - The first equality in the formula for $\hat a$ is incorrect. - When invoking the equivalence principle, you need to work with general relativity, where the gravitational field is something different from your $g$.

not graduate + level. Users with 500+ reputation may vote here to get it closed.

Ummm ...  http://physics.stackexchange.com/questions/67046/can-one-define-an-acceleration-operator-in-quantum-mechanics ... It seems correct to me ... Also the equivalence principle also holds in General relativity and Newtonian mechanics ... Im trying to construct a theory unifying quantum mechanics and gravity rather than gravity and QFT. The physical motivation can be understood by imagining oneself in a lift and trying to differeniate between acceleration and gravity (I believe one will be unable to do so even in the quantum mechanical regime)   

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