- #1

LozD

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- TL;DR Summary
- Many entangled particle pairs are prepared with property A (1 or 0) at 99% probability of 1, and property B (1 or 0) at 1% probability of 1. The pairs are separated with 2 observers, X and Y, each having a particle of each pair. They pre-arrange that at a certain time: Y will "send a 0" by NOT measuring anything on his particles or Y will "send a 1" by measuring property B on every one of his particles. X will measure property A on her particles: If 99% = 1, Y sent a 1; if 50%, Y sent a 0.

I work in IT and am a layman in the quantum world. I have obviously misunderstood something in my amateur reading of quantum, but if someone could explain my mistake in the above scenario it might be very insightful for me! Forgive me if the terminology is not correct - or if indeed lay folks' enquiries on this forum are inappropriate then please ignore me!

Here's the above again with a bit more detail and the concepts that lead me to assume the main steps ...

My reasoning around the "complementarity" and Heisenberg Uncertainty Principle pieces came from the narrative in the section entitled "Bohm's variant" in this link ...

https://en.m.wikipedia.org/wiki/EPR_paradox

... and from the narrative in this excellent layman's explanation of Quantum Entanglement ...

https://www.quantamagazine.org/entanglement-made-simple-20160428/

If anyone who knows what they're talking about has got a spare moment to put me straight, please fire away!

Here's the above again with a bit more detail and the concepts that lead me to assume the main steps ...

- I take a pair of particles with two properties (spin, velocity, whatever) A and B with values 1 or 0 and prepare them (entangle them): 99% A is 1, 1% B is 1.
- Send one of the particles with another observer to another place.
- Tell them that at a pre-arranged time they must either (a) not observe their particle if they wish to send a 0, or, (b) observe (measure) the B property if they wish to send a 1.
- At the pre-arranged time (or just after!) I then observe my particle's A value.
- If the sender *didn't* observe her particle then the A property of my particle will be 1 with the original prepared 99% probability, because I am the first observer of any property of the entangled pair. (Btw, thereafter, property B, if examined by me or the other observer, would only have a 50% chance, not 1%, of being 1, due to "complementarity" and the Heisenberg Uncertainty Principle, but this is not directly relevant here). So, per the pre-arranged plan, the remote observer has "sent a 0" if the probability of property A being 1 is 99%.
- If the sender *did* observe her particle's B property then the A property of my particle will be 1 with only a 50% probability, because the sender has observed property B and due to "complementarity" and the Heisenberg's Uncertainty Principle the probability of property A being 1 has collapsed to only 50%.
- By using multiple pairs of such entangled particles the probability difference of 99% vs 50% for 0 or 1 respectively will allow me to derive a 0 was sent if the number of 1's on A of my particles is near 99%, or a 1 was sent if the number of 1's is near 50%, with good accuracy.

My reasoning around the "complementarity" and Heisenberg Uncertainty Principle pieces came from the narrative in the section entitled "Bohm's variant" in this link ...

https://en.m.wikipedia.org/wiki/EPR_paradox

*Whatever axis their spins are measured along, they are always found to be opposite. In quantum mechanics, the x-spin and z-spin are "incompatible observables", meaning the Heisenberg uncertainty principle applies to alternating measurements of them: a quantum state cannot possesses a definite value for both of these variables. Suppose Alice measures the z-spin and obtains +z, so that the quantum state collapses into state I. Now, instead of measuring the z-spin as well, Bob measures the x-spin. According to quantum mechanics, when the system is in state I, Bob's x-spin measurement will have a 50% probability of producing +x and a 50% probability of -x. It is impossible to predict which outcome will appear until Bob actually performs the measurement. Therefore, Bob's positron will have a definite spin when measured along the same axis as Alice's electron, but when measured in the perpendicular axis its spin will be uniformly random. It seems as if information has propagated (faster than light) from Alice's apparatus to make Bob's positron assume a definite spin in the appropriate axis.*... and from the narrative in this excellent layman's explanation of Quantum Entanglement ...

https://www.quantamagazine.org/entanglement-made-simple-20160428/

*Upon deeper reflection, the paradox dissolves further. Indeed, let us consider again the state of the second system, given that the first has been measured to be red. If we choose to measure the second q-on’s color, we will surely get red. But as we discussed earlier, when introducing complementarity, if we choose to measure a q-on’s shape, when it is in the “red” state, we will have equal probability to find a square or a circle. Thus, far from introducing a paradox, the EPR outcome is logically forced. It is, in essence, simply a repackaging of complementarity.*If anyone who knows what they're talking about has got a spare moment to put me straight, please fire away!