Quantum Entanglement information transmission idea to knock down....

In summary: at least not without some sort of comparison value being sent between the remote observer and the sender beforehand.
  • #36
PeroK said:
That feels more like Alice in Wonderland, following Bob down a rabbit hole!
不不不
 
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  • #37
LozD said:
Now, here I'm introducing another degree of freedom (my aka "property") of the particle: velocity. And that's entangled between the two particles too ...

View attachment 295634
Again I've shown the local value boxes greyed out to signify the value is not really there but rather in the shared place above, attendant arrows pointing to the shared place too. (Note I've put a slight purple tinge on this greyed-ness for Velocity to distinguish it from Spin, which I'll use later).

So, we now have two entangled particles, One and Two, with their Spin and Velocity degrees of freedom entangled.

Next, let's say the Velocity can be some value between 0 and 1. When I prepare the particles, I set the probability like this: |v> = 99% |11> + 1% |00> . (This is LozD using ket-speak for the first time so beware more confusion by me, feel free to critique me, and hat-tip @ Nugatory!) In plain English I'm saying "I'm preparing the velocity degree of freedom of the entangled particles to 99% probability it is 1, 1% it is 0.

We introduce Alice and Bob and give them each a particle; Alice has One and Bob has Two, and we separate them. Here they are with the prepared pair of particles ...

View attachment 295636

We want Bob to communicate a 1 or 0 in the following manner ...

At midnight, if Bob wants to send a 0 he must not measure anything on his particle. Not Spin, not Velocity, nada. On the other hand, if he wants to send a 1 then he will examine the Spin. This protocol is pre-arranged with Alice and Bob.

At one minute past midnight, Alice will measure Velocity. If Bob didn't measure anything at midnight, then Alice's measuring of Velocity will have the probability |v> = 99% |11> + 1% |00> . She can interpret this as a zero being been sent by Bob. On the other hand, if Bob did measure Spin he will have collapsed the entanglement, then Alice's measuring of Velocity will have the probability |v> = 50% |11> + 50% |00> . She can interpret this as a one being been sent by Bob. Hence we have information communication between the two observers at a distance, and this will be happening instantaneously (or however many thousands of times faster than the speed of light someone posted that disentanglement happens at). And no subluminal back channel is required to send the information for comparison.

Now it might be said (someone did on this thread) that the entanglement collapse will only happen for the Spin, not the Velocity. So the spin will remain |v> = 99% |11> + 1% |00> as it was originally prepared and not revert to |v> = 50% |11> + 50% |00> .

Well, this is where the "degree-of-freedom to degree-of-freedom" entanglement comes in. If in addition to the particles being entangled, the Spin and Velocity are also entangled, then Bob's entanglement collapse of Spin will cascade to Velocity. See this diagram ...

View attachment 295640
The entanglement is shown by greying with a butterscotch tinge the two degrees of freedom, Spin and Velocity. (Perhaps I should have shown the entanglement in the shared place in black to be consistent with the notion that this is where the real values are and not on the particles?). But the entanglement collapse path if Bob transmits a 1 is: Bob measures Particle Two's Spin, that collapses the entanglement in the notional shared values Particle One & Two (right hand side butterscotch arrow), which in turn collapses the entanglement between Spin and Velocity in the shared notional particle (curved lines on either side of the shared notional particle One & Two), which in turn collapses the Spin and Velocity in the Particle One (left hand side butterscotch arrow), which means that when Alice measures Particle One's Velocity at a minute past midnight it will be |v> = 50% |11> + 50% |00> - Bob sent a 1. And if Bob doesn't measure Spin on Particle Two at midnight none of that cascading collapse will happen, and Alice's Velocity probability will be |v> = 99% |11> + 1% |00> - Bob sent a 0.

Is that clear now?
Damn, there's a confusing typo.

Where I said ...

Now it might be said (someone did on this thread) that the entanglement collapse will only happen for the Spin, not the Velocity. So the spin will remain |v> = 99% |11> + 1% |00> as it was originally prepared and not revert to |v> = 50% |11> + 50% |00> .

I maent ...

Now it might be said (someone did on this thread) that the entanglement collapse will only happen for the Spin, not the Velocity. So the velocity will remain |v> = 99% |11> + 1% |00> as it was originally prepared and not revert to |v> = 50% |11> + 50% |00> .
 
  • #38
Honestly, I think you would be better focusing on the basics of QM for now. You can't really learn QM very well by trying to do FTL messaging by entanglement.
 
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  • #39
LozD said:
... as it was originally prepared ...

Any quantum property that is entangled cannot also be in a "prepared" state. To the extent you know what the value of an observable is, it is NOT entangled. Any measurement on a particle's observable will "collapse" the entanglement for that particle (and that observable). It will also affect the conjugate observables (which will then take on undefined values that will read as being random if subsequently measured).

It is possible to measure spin without affecting velocity (momentum), because spin and velocity are not conjugate.
 
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  • #40
PeroK said:
Honestly, I think you would be better focusing on the basics of QM for now. You can't really learn QM very well by trying to do FTL messaging by entanglement.
I might well do, but it's a big ask. Keeping up with my day job tech and dev is tough enough.

Thanks for your input PeroK, and all others who chipped in.

Did y'all read about the tardigrade? How does an entire critter become entangled with a qubit? Bet the quantum maths and physicals around that gets pretty hectic?! ...

https://www.livescience.com/tardigrade-quantum-entangled-experiment

I think I'll quit while I'm ahead - OK, while I'm behind but not as far as I will be if I carry on!

Cheers!
 
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  • #41
LozD said:
Is that clear now?
Yes, but it's also clear that it is based on a misunderstanding of how QM works.

To get past this point you will have to develop the mathematical background needed to follow the presentation in https://en.wikipedia.org/wiki/No-communication_theorem (which unfortunately is one of the less good wikipedia articles - I'm not aware of any better readily accessible proof).

Without digging into the math, the theorem says that the quantum mechanical calculation will always show that no matter how cleverly we set up our experiment, the probabilities on one side of an entanglement are not affected by anything done or not done at the other side. Thus, your idea can only work if the quantum mechanical calculation is wrong.

This thread has drifted to the wrong side of the forum rules about personal speculation and challenges to mainstream physics so it is closed. As with all thread closures it can be reopened if someone has something more to add - suggestions for better explanations of the no-communication theorem would be particularly welcome.
 
  • #42
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