- #1
al onestone
- 61
- 0
Here is an instant communication protocol for Alice and Bob. Alice and Bob are given one each of a pair of momentum entangled photons (or particles).
The two particles are prepared with entangled momentum states and the momentum distribution for both has uncertainty Δpi. The position distribution has the complementary uncertainty Δxi .
Alice modifies her particle by simply choosing to measure the position with accuracy Δxf where Δxf << Δxi. Because the position of Alice’s particle is now known with the much greater accuracy of Δxf, the accuracy of our knowledge of the position of Bob’s particle must also be Δxf. Since Bob’s particle now has positional uncertainty of Δxf, then Bob’s particle must now have a momentum uncertainty which is complementary to it, Δpf = Δxf/ħ >> Δpi. This being the case, all Bob has to do is prepare his particle at a greater distance from the source than Alice’s, by passing it through a band pass filter that only allows the momentum range Δpi , which should have the result that Bob’s particle will pass the filter with much less likelyhood due to the increased momentum distribution of Bob’s particle which was due to the measurement that Alice performed. Remember, if Alice doesn’t measure her particle then Bob’s particle should pass the filter with 100% reliability. (Rather than a bandpass filter you could use its inverse which would have opposite results). So Alice’s performance of a measurement on her particle should have measurable results on Bob’s particle by the above reasoning, results which are instantly projected from a distance. By modulating between the extremes(measurement and non-measurement) of Alice’s preparation we can use this thought experiment as an instant communication protocol.
Just to let you know, this is approximately my 100th instant communication protocol thought experiment. Usually I’ve gotten my reasoning wrong somewhere. So tell me, where have I gone wrong this time?
The two particles are prepared with entangled momentum states and the momentum distribution for both has uncertainty Δpi. The position distribution has the complementary uncertainty Δxi .
Alice modifies her particle by simply choosing to measure the position with accuracy Δxf where Δxf << Δxi. Because the position of Alice’s particle is now known with the much greater accuracy of Δxf, the accuracy of our knowledge of the position of Bob’s particle must also be Δxf. Since Bob’s particle now has positional uncertainty of Δxf, then Bob’s particle must now have a momentum uncertainty which is complementary to it, Δpf = Δxf/ħ >> Δpi. This being the case, all Bob has to do is prepare his particle at a greater distance from the source than Alice’s, by passing it through a band pass filter that only allows the momentum range Δpi , which should have the result that Bob’s particle will pass the filter with much less likelyhood due to the increased momentum distribution of Bob’s particle which was due to the measurement that Alice performed. Remember, if Alice doesn’t measure her particle then Bob’s particle should pass the filter with 100% reliability. (Rather than a bandpass filter you could use its inverse which would have opposite results). So Alice’s performance of a measurement on her particle should have measurable results on Bob’s particle by the above reasoning, results which are instantly projected from a distance. By modulating between the extremes(measurement and non-measurement) of Alice’s preparation we can use this thought experiment as an instant communication protocol.
Just to let you know, this is approximately my 100th instant communication protocol thought experiment. Usually I’ve gotten my reasoning wrong somewhere. So tell me, where have I gone wrong this time?