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  Is there any astrophysical situation where this idea is testable?

+ 0 like - 1 dislike
1407 views

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

ΔgΔp|˙U+pg|

 

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 ˆv:

ˆv(t)=dU(t)xU(t)dt=U[H,x]iU

Now we can again differentiate to get acceleration ˆa:

ˆa(t)=ˆU(t)[[ˆH,ˆx],ˆx]iˆU(t)=ˆU(t)(ˆH2ˆx+ˆxˆH22ˆHˆxˆH)ˆU(t)

 

Going back to the Schrodinger picture:

ˆa=[[ˆH,ˆx],ˆx]iˆU(t)=(ˆH2ˆx+ˆxˆH22ˆHˆxˆH)

We can simplify the calculation by splitting the Hamiltonian into potential  ˆV and kinetic energy ˆT: ˆH=ˆT+ˆV

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

ˆ0=ˆT2x+xˆT22ˆTˆxˆT

We also know [ˆV,ˆx]=0 as potential is a function of position. Thus, we can simplify acceleration as:

ˆa=ˆVˆTˆx+ˆxˆTˆVˆVˆxˆTˆTˆxˆV

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

ˆg=ˆa

However, ˆg does not commute with ˆp (it commutes with position). Using the anti-commutator (rather than the commutator) we should observe:

ΔgΔp|˙U+pg|

 

By Schwatz inequality |˙U+pg| is 0 only when U is constant.
      

Closed as per community consensus as the post is not graduate-level
asked Feb 9, 2017 in Closed Questions by Asaint (90 points) [ revision history ]
recategorized Nov 26, 2017 by Dilaton

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=eiHt where H is the Hamiltonian.
 

Usually one writes this U(t), not U. - The first equality in the formula for ˆa 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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