1
$\begingroup$

It is my understanding that you can have two entangled particles, A and B, and measure their spins from two different angles, X and Y, such that: If you measure A using one angle (e.g. X), there's about an 85% chance B will have the opposite spin. If you measure A using a different angle (Y), there's about an 85% chance that B will have the same spin.

And, after the measurement of A, A and B are no longer entangled.

From numerous questions and answers on this site and others, it is not clear to me how multi-particle entanglement works.

Suppose I have three entangled particles, A, B, and C. If I measure A using angle X, what can we say about the likelihood of spins for B and C? If I measure A using angle Y, what can we say about the likelihood of spins for B and C?

Also, after the measurement of A, I assume none of A, B, or C are entangled with the others. In particular, B and C are no longer entangled. Is that correct?

Edit 1:

Adding background and a follow up question:

This question came about because I reasoned that either FTL communication is possible (ultimately anyway, even if not yet figured out) or I misunderstood the behavior of entangled particles. I wanted to start by better understanding entangled particles. I've done a lot of reading. After asking the question, I'm still left wondering whether A) FTL communication is possible and physicists just haven't considered beyond basic scenarios, or B) nobody active on this forum understands entanglement well enough to explain, or C) my question is so poor that it deserves no response.

As can be seen in my comments, a follow up question for GHZ 3 particle entanglement abruptly ended the conversation.

Please help me understand this. What is wrong with the proposal?

Suppose you had 1000 GHZ triplets (1000 As, 1000 Bs, and 1000 Cs each in their own A-B-C entangled groups). We put the As in one location, and their B-C counterparts in another location. We measure the As.

A portion of the entangled groups are "lost", meaning they are no longer entangled and the measurements of their B-C partners will be random. However, the entangled ones will yield B-C pairs that are 00 or 11.

Couldn't we figure out a way to minimize the number that are lost resulting in a statistically significant percent of the B-C pairs as 00 or 11, thereby providing a way for FTL communication?

An answer of "no, FTL communication is not possible" does not help explain what is wrong with the proposal. Explaining how FTL is not possible with one triplet does not help either. An answer of, "you understood the meaning of lost correctly, but 99.9% are lost and always will be, we can not improve upon this because ..." would help. An answer of, "measuring A does not make it more likely that B-Cs will be 00 or 11, and this is because the real meaning of GHZ 3 particle entanglement is ..." would help.

Thanks.

Edit 2: I found a paper by Raymond Jensen (http://www.academia.edu/5364451/On_using_Greenberger-Horne-Zeilinger_three-particle_states_for_superluminal_communication) on using GHZ for FTL communication. It doesn't even use many ABCs like my proposal did to account for some being lost. After finding the paper I did another search expecting to find others refuting Raymond's idea. After all, some people are so insistent that you can't communicate using entangled particles. However, I can't find anything refuting it. This is puzzling.

Side note: I realize that one could argue this not being "FTL", reasoning that the particles are actually not separated by a distance, kind of like a worm hole. Either way, it's intriguing.

$\endgroup$
  • 1
    $\begingroup$ if A & B share a Bell state - and all the measurements are at computational basis $|\pm z\rangle$ you will get perfect (100%) correlations. For entangled states of more than 2 particles check GHZ (which wont be entangled after measurement of a single qubit) and W state (which will be partially entangled after measurement of a single qubit) en.wikipedia.org/wiki/Bell_state en.wikipedia.org/wiki/… en.wikipedia.org/wiki/W_state $\endgroup$ – Alexander Jul 17 '15 at 18:23
  • $\begingroup$ @Alexander In the GHZ explanation it says "if we were to measure one of the subsystems, in such a way that the measurement distinguishes between the states 0 and 1, we will leave behind either |00\rangle or |11\rangle which are unentangled pure states." I think the answer at physics.stackexchange.com/a/115035/84813 refers to this. I don't understand how that doesn't provide a way to communicate via 3 particle entanglement. Any help? $\endgroup$ – user3141592 Jul 17 '15 at 21:36
  • $\begingroup$ Are we able to chose to measure "in such a way that the measurement distinguishes between the states 0 and 1", and thereby make two particles match? $\endgroup$ – user3141592 Jul 17 '15 at 21:45
  • $\begingroup$ The meaning of that paragraph is that there is a difference between losing a qubit and measuring one - if you lose it you can't identify the remaining state as a pure state, but if you measure it - GHZ collapses to a pure state (00 or 11, depends on what the measurement yielded). In both cases the remaining state, pure or mixed, is NOT entangled. $\endgroup$ – Alexander Jul 17 '15 at 22:44
  • $\begingroup$ You can't communicate by measuring entangled shared states because every measurement yields a random product, while encrypted messages have a very organized sequence of 0's and 1's. (There exist even cooler tricks you can do with entangled states but you never can use it to signal a message.) What you can do with this is share a perfectly random sequence of numbers and use it as a one time pad for encryption. It's called "Quantum key distribution". en.wikipedia.org/wiki/One-time_pad en.wikipedia.org/wiki/Quantum_key_distribution $\endgroup$ – Alexander Jul 17 '15 at 22:59
2
$\begingroup$

I'm still not certain I understand things correctly, but I'll answer my own question with my current understanding.

There are different types of multi-particle quantum entangled states; GHZ is one of them. The proposed scenario with GHZ would NOT work. It mistakenly assumed that after one particle is measured, the other two would measure the opposite (agreeing with each other but opposite the first measured). However, all 3 are the same. Therefore, if you read B and C they will always agree regardless of whether or not A was read.

$\endgroup$
  • $\begingroup$ yes ! the signal to communicate would be 'no more entangled' and there is not such signal $\endgroup$ – user46925 Dec 21 '15 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.