# Using GHz multi-particle entanglement for FTL communication

#### r20

Although there are numerous questions about entanglement and FTL communication, I can't find anything directly related to this. I proposed a method for FTL communication via GHZ multi-particle entanglement at http://physics.stackexchange.com/questions/194566/how-does-measurement-affect-multi-particle-entangled-particles

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.​

After a few weeks I still haven't received any response so I'm trying this community. Any help?

Thanks.

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#### DEvens

Education Advisor
Gold Member
Entanglement does not allow you to send a message because you cannot predict what the observed value will be on the "send" end in order to send a particular value to the "receive" end.

Say you have agreed with your correspondent to set up an entanglement apparatus. Suppose it is even many light years long.Say you have agreed that you will observe particles from some central source. So there are always particles arriving at your observation point that are paired with particles arriving at your correspondent's observation point. And suppose for ease of reference that the two of you are at rest relative to each other, and agree on what "simultaneous" means for purposes of sending messages.

For specificity, suppose the entanglement means that if you see an "up" he has to see a "down." And if you see "down" he has to see "up." And you have agreed that when he sees "up" it is 1. And suppose, for specificity, that it is equally likely to be 1 and 0.

So you want to send a 1. So you set in to look at the particles and look for a 0. But you have some random chance to see 1's and the corresponding chance to see 0's. You can't decide if you will observe a 1 or a 0. You get what you get. He gets the other, but you can't pick which you see. So if you get the 0 you want and stop, he can't tell that you did so. You might have been intending to send a 0, so are still waiting at your dial looking for a 1.

In other words, the fact that you are busy observing is not apparent to the other guy. He sees exactly what he would have seen if you were not observing. It is only when you gather your results with his and compare them (after suitably synchronizing) that you see that your 1's correspond to his 0's, and your 0's to his 1's.

So you can't sit down and try to send 1100 because your correspondent will see the same results no matter what you want to send. Nothing you can do to your observation equipment will change the statistics he sees in any way. And the fact that he knows you got 1's for each of his 0's does not transmit the 1100 you wanted him to get.

#### r20

Entanglement does not allow you to send a message because you cannot predict what the observed value will be on the "send" end in order to send a particular value to the "receive" end.

Say you have agreed with your correspondent to set up an entanglement apparatus. Suppose it is even many light years long.Say you have agreed that you will observe particles from some central source. So there are always particles arriving at your observation point that are paired with particles arriving at your correspondent's observation point. And suppose for ease of reference that the two of you are at rest relative to each other, and agree on what "simultaneous" means for purposes of sending messages.

For specificity, suppose the entanglement means that if you see an "up" he has to see a "down." And if you see "down" he has to see "up." And you have agreed that when he sees "up" it is 1. And suppose, for specificity, that it is equally likely to be 1 and 0.

So you want to send a 1. So you set in to look at the particles and look for a 0. But you have some random chance to see 1's and the corresponding chance to see 0's. You can't decide if you will observe a 1 or a 0. You get what you get. He gets the other, but you can't pick which you see. So if you get the 0 you want and stop, he can't tell that you did so. You might have been intending to send a 0, so are still waiting at your dial looking for a 1.

In other words, the fact that you are busy observing is not apparent to the other guy. He sees exactly what he would have seen if you were not observing. It is only when you gather your results with his and compare them (after suitably synchronizing) that you see that your 1's correspond to his 0's, and your 0's to his 1's.

So you can't sit down and try to send 1100 because your correspondent will see the same results no matter what you want to send. Nothing you can do to your observation equipment will change the statistics he sees in any way. And the fact that he knows you got 1's for each of his 0's does not transmit the 1100 you wanted him to get.
I know there are limitations with 2 particle entanglement. My proposal is different. Please re-read it.

#### kith

Science Advisor
I proposed a method for FTL communication via GHZ multi-particle entanglement
I don't see this. Let's make it explicit: how would you transmit the message "110"?

Also, I don't get what you mean by "a portion of the entangled groups are lost". What is the physics behind this supposed to be? And if such an effect is present in real experiments, does it really matter to the thought experiment of measuring 3 idealized qubits?

#### mfb

Mentor
I know there are limitations with 2 particle entanglement. My proposal is different. Please re-read it.
The limitations are independent of the number of particles. Are you aware of the No-cloning theorem? Your description of the setup is unclear, but I think it would need a violation of this theorem. Or something else - but I don't know how the setup is supposed to work at all.

Please keep in mind that the forum rules do not allow the discussion of personal theories. "Why is this not working?" is somewhere at the edge.

#### StevieTNZ

I know there are limitations with 2 particle entanglement. My proposal is different. Please re-read it.
The main point that should be taken from DEvens post is:
Entanglement does not allow you to send a message because you cannot predict what the observed value will be on the "send" end in order to send a particular value to the "receive" end.

#### Nugatory

Mentor
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
You will get the exact same correlations between the B's and the C's when they're measured whether the A's have been measured or not.

There's no way of knowing whether the correlated B and C results are what they are because a measurement at A has already determined the result for all three particles, or whether the first measurement on one of B/C has determined the state for all three particles including the result at A when and if that measurement is made.

#### DrChinese

Science Advisor
Gold Member
Although there are numerous questions about entanglement and FTL communication, I can't find anything directly related to this. I proposed a method for FTL communication via GHZ multi-particle entanglement at http://physics.stackexchange.com/questions/194566/how-does-measurement-affect-multi-particle-entangled-particles

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.​

After a few weeks I still haven't received any response so I'm trying this community. Any help?

Thanks.
Welcome to PhysicsForums, r20!

This is an oversimplification of the GHZ protocol over several levels.

1. How do you think you are going to measure some of the As and not others? Eventually the As will end up somewhere. The ordering of measurements (A first, B first, etc) does not matter.
2. The Bs and Cs will show a random sequence that would not be limited to 00 and 11 pairs.

In other words: the pattern will only appear when A, B and C are brought together.

#### DrChinese

Science Advisor
Gold Member
Or looking at it another way: there is nothing preferred about A relative to B or C.

#### r20

Thanks everyone for responding. Sorry, I suddenly had to go out of town.

You will get the exact same correlations between the B's and the C's when they're measured whether the A's have been measured or not.

There's no way of knowing whether the correlated B and C results are what they are because a measurement at A has already determined the result for all three particles, or whether the first measurement on one of B/C has determined the state for all three particles including the result at A when and if that measurement is made.
Thanks, Nugatory. That helped a lot. I misunderstood a few things. One of which was that I was under the impression that after one of the 3 particles is measured, the other two would be opposite (agreeing with each other but opposite to the first measured). If this were true, you could use it to communicate. I won't bother explaining how, because after reading your reply and reading more information on the subject I see this is not the case; all 3 will be the same as the first measured.

I still have another question. I came across a paper [reference removed] that talks about "destroying path information of the photon (erasure)", and using this to differentiate from when A is measured and B and C agree with eachother. Is this rubbish?
Can you do something to A so that when B and C are measured they will not necessarily be the same?

And another question. Since measuring A breaks the entanglement, is there a test you could do with B and C to see if they are still entangled (and thereby know whether A was measured)?

Thanks.

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#### DrChinese

Science Advisor
Gold Member
And another question. Since measuring A breaks the entanglement, is there a test you could do with B and C to see if they are still entangled (and thereby know whether A was measured).
To repeat: time ordering is irrelevant. There is no observational difference in which you measure first. B and C will always present the same statistics. This is still the same question as your original actually.

I would politely request that you remove the reference to Jensen's paper as it does not qualify for discussion here under PF guidelines.

#### Nugatory

Mentor
And another question. Since measuring A breaks the entanglement, is there a test you could do with B and C to see if they are still entangled (and thereby know whether A was measured)?
No. The only way to detect entanglement is for all of the observers to get together after the fact and compare their observations, see if they have the right correlations.

One consequence of this is that you can never determine whether the members of a single set of particles were entangled. Instead, you always have to look at a large number of measurements and see if any correlations emerge when you compare notes afterwards. Consider the simpler two-particle case: I have a source that emits pairs of particles. If the particles are spin-entangled, then measurements of their spin in some given direction will always give opposite results: one up and down. How would I determine whether the particles are in fact spin-entangled?

Naturally I'm going to set up two detectors aligned on the same axis, one on each side of the source. Any time that both trigger at about the same time, I'll know that I've observed a pair, and I can compare the results after the fact and see if one is up and one is down. But a single observation won't do the job; there's a fifty-fifty chance of getting one up and one down even if the two particles are not entangled. Instead, I have to observe a large number of pairs - if every pair I see has opposite spins, then I can say that the source is emitting entangled pairs.
(Actually, it's even harder than this, because there are many sources of experimental error and stray/bogus readings from my detectors cluttering up the measurements).

#### r20

I would politely request that you remove the reference to Jensen's paper as it does not qualify for discussion here under PF guidelines.
I have removed the reference.

Science Advisor
Gold Member

#### r20

All,
I don't have any other questions on this.
Thank you for the information.

#### mfb

Mentor
One consequence of this is that you can never determine whether the members of a single set of particles were entangled. Instead, you always have to look at a large number of measurements and see if any correlations emerge when you compare notes afterwards. Consider the simpler two-particle case: I have a source that emits pairs of particles. If the particles are spin-entangled, then measurements of their spin in some given direction will always give opposite results: one up and down. How would I determine whether the particles are in fact spin-entangled?

Naturally I'm going to set up two detectors aligned on the same axis, one on each side of the source. Any time that both trigger at about the same time, I'll know that I've observed a pair, and I can compare the results after the fact and see if one is up and one is down. But a single observation won't do the job; there's a fifty-fifty chance of getting one up and one down even if the two particles are not entangled. Instead, I have to observe a large number of pairs - if every pair I see has opposite spins, then I can say that the source is emitting entangled pairs.
(Actually, it's even harder than this, because there are many sources of experimental error and stray/bogus readings from my detectors cluttering up the measurements).
You have to do more - you have to repeat this for different polarization axes. Your source could emit unentangled photons in opposite polarity along some axis - if this axis is aligned with your detector you don't see a difference between entangled pairs and unentangled ones.

#### Nugatory

Mentor
You have to do more - you have to repeat this for different polarization axes. Your source could emit unentangled photons in opposite polarity along some axis - if this axis is aligned with your detector you don't see a difference between entangled pairs and unentangled ones.
Right - thanks.

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