# Is there any astrophysical situation where this idea is testable?

+ 0 like - 1 dislike
81 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.

recategorized Jul 13

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)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.