Question
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I believe the following uncertainty principle to be true in the regime v<<c :
ΔgΔp≥|⟨˙U⟩+⟨p⟩⟨g⟩|
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]iℏU
Now we can again differentiate to get acceleration ˆa:
ˆa(t)=ˆU(t)†[[ˆH,ˆx],ˆx]iℏˆU(t)=ˆU(t)†(ˆH2ˆx+ˆxˆH2−2ˆHˆxˆH)ˆU(t)
Going back to the Schrodinger picture:
ˆa=[[ˆH,ˆx],ˆx]iℏˆU(t)=(ˆH2ˆx+ˆxˆH2−2ˆHˆxˆH)
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+xˆT2−2ˆTˆxˆT
We also know [ˆV,ˆx]=0 as potential is a function of position. Thus, we can simplify acceleration as:
ˆa=ˆVˆTˆx+ˆxˆTˆV−ˆVˆxˆT−ˆTˆxˆV
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⟩+⟨p⟩⟨g⟩|
By Schwatz inequality |⟨˙U⟩+⟨p⟩⟨g⟩| is 0 only when U is constant.