Quantum entanglement, wave-function collapse

Tara Sunray

Main Question or Discussion Point

I have 2 questions i would really like to find an answer to: if we have a quantum pair, we can determine whether 2 quanta are entangled or not assuming that we have access to the information of both of them. so if we have a quantum "trio" instead of a quantum pair, does the information about 2 out of 3 quanta suffice to determine whether those two quanta are entangled? if not, why not. the only entanglement i am interested in is the one of the 2 quanta that are tested. the 3rd one is not important for question one. the second question is: if one out of those 3 entangled particles wave functions collapses, this should break the entanglement of all 3 particles and collapse their wave-functions, right?

this is a rephrased version of my original question, in hopes that there would be less confusion.

original question was:

It is my understanding that you can entangle multiple quanta, so that you don't get "quantum pairs" but for example "quantum-trios" etc. I base this assumption on the following study: https://www.nature.com/articles/ncomms13251 Further I believe to know that once the wave function of an entangled particle is collapsed, the entanglement is "broken". I know that there are several resctrictions for faster than light communication with a quantum pair/quantum pairs. I do base following thought experiment on the assumption that one does need only access to the information about 2 out of three quanta to determine whether an entanglement is existent or not.

If we assume a entanglement of atleast 3 quanta(trio) and person (A) has 1 quantum, person (B) has two quanta and person (A) collapses the wave-function, (B) should be able to determine that his 2 quanta are no longer entangled, and faster than light communication would be possible. I don't think it is that simple, but I lack the knowledge. What am I missing?
 

Answers and Replies

33,579
9,311
so if we have a quantum "trio" instead of a quantum pair, does the information about 2 out of 3 quanta suffice to determine whether those two quanta are entangled?
It depends on what exactly is entangled and what you measure.
And keep in mind that you can't do that on an event-by-event basis, but only with many samples. Unentangled particles can give the same measurement results entangled particles will give - they just don't give that every time.
f one out of those 3 entangled particles wave functions collapses, this should break the entanglement of all 3 particles and collapse their wave-functions, right?
Again, it depends on the initial state and what you measure.
Further I believe to know that once the wave function of an entangled particle is collapsed, the entanglement is "broken".
Not necessarily completely. It depends on what exactly you do.
know that there are several resctrictions for faster than light communication with a quantum pair/quantum pairs.
There is just one restriction: You cannot do it.

Whatever you do to the third particle doesn't influence what you measure with the other two particles. Or, more general: Whatever you do to some particles somewhere doesn't change what you measure with the other particles elsewhere. Only the correlation between these measurements in different places is special, and to see this you have to compare the results (e. g. via transmitting them in classical ways, slower than light).
 
Tara Sunray
It depends on what exactly is entangled and what you measure.
And keep in mind that you can't do that on an event-by-event basis, but only with many samples. Unentangled particles can give the same measurement results entangled particles will give - they just don't give that every time.Again, it depends on the initial state and what you measure.Not necessarily completely. It depends on what exactly you do.There is just one restriction: You cannot do it.

Whatever you do to the third particle doesn't influence what you measure with the other two particles. Or, more general: Whatever you do to some particles somewhere doesn't change what you measure with the other particles elsewhere. Only the correlation between these measurements in different places is special, and to see this you have to compare the results (e. g. via transmitting them in classical ways, slower than light).

i know about the no information theorem. my thought experiment is not affected by it.
ok, lets say we have an array of entangled photon-trios for each bit of information we want to send(per bit an array of lets say 100 trios). now we can use stochastic to determine entanglement. we collapse the wave-function by measuring the spin. if the entanglement is: 1_2, 2_3 and 1_3 for each trio (numbers represent photons), and person (A) has photons 1 and (B) has photons 2+3. (B) can now measure the spin of his 2's and 3's, and if his statistics tell him that they have the expected spins he knows the entanglement was intact. or not? what do i still not get?
i mean 2 and 3 are entangled too, why would that not work.
 
Last edited by a moderator:
Leo1233783
i know about the no information theorem. my thought experiment is not affected by it.
ok, lets say we have an array of entangled photon-trios for each bit of information we want to send(per bit an array of lets say 100 trios). now we can use stochastic to determine entanglement. we collapse the wave-function by measuring the spin. if the entanglement is: 1_2, 2_3 and 1_3 for each trio (numbers represent photons), and person (A) has photons 1 and (B) has photons 2+3. (B) can now measure the spin of his 2's and 3's, and if his statistics tell him that they have the expected spins he knows the entanglement was intact. or not? what do i still not get?
i mean 2 and 3 are entangled too, why would that not work.
a short answer: to know if the particles are still entangled or not, the protocol needs that A and B and C ( or in some conditions, 2 of them ) meet. Thus, there is not such kind of detectable signal on/off before the meeting. Continue to investigate this way, add D , E ... after a good understanding of the protocol. No hope of TFL but interesting ...
 
Tara Sunray
" Thus, there is not such kind of detectable signal on/off before the meeting." - that is not necessary for my thought experiment. if you for example agree on periodically checking arrays for entanglement and discard every array after analysis, status quo for each array is entanglement. if person (A) collapses the wavefunctions of an array of 1's, it can be used like morsecode, because (B) has arrays with 2 out of 3 particles, and since (B) has two particles(well, an array of them), they can interfere with each other and they can be examined. (B) can now determine for each array when entanglement between his 2 particles is broken.
 
Leo1233783
How B or C can know when the 1st measurement was or not done by A ?
ps : don't imagine it is possible at the particle level. You must use samples.
 
Tara Sunray
first off, there is no person (C).
secondly, if (A) collapses the wave-function of his particle, the entanglement of the two particles that (B) has and his (A's) particle is broken. (B) has two particles. he can test them for entanglement. I don't see why people have such a hard time wrapping their head about such a simple idea. this has to be the most misunderstood question i've ever asked.
 
Leo1233783
Sorry, yes, B owns 2. Always the same question : how could he know that his measure was done before A ? Measuring his 1st photon, he will produce the same effect on the second photon. ( in this context, photon means a sample, a class of objects having a distribution probability of states )
 
Tara Sunray
because if (A) collapses the wave-function, the entanglement of all 3 particles is broken. (B) will detect no correlation between his two quanta when testing them, and knows the entanglement is broken.
if (A) does not collapse the wave-function, the entanglement is not broken and (B) can detect an entanglement when he is testing his 2 quanta for it.

edit: when i talk of (A)'s particle or (B)'s 2 particles i always mean an array of particles as said in the beginning
 
Last edited by a moderator:
1,458
366
You would have a GHZ state, so even if one particle has been measured, you can't suddenly say particles B and C are now entangled by themselves.
 
Last edited:
Strilanc
Science Advisor
586
210
If we assume a entanglement of atleast 3 quanta(trio) and person (A) has 1 quantum, person (B) has two quanta and person (A) collapses the wave-function, (B) should be able to determine that his 2 quanta are no longer entangled
If you don't understand quantum mechanics, the idea "B can check for entanglement, A can break entanglement, therefore B can talk to A" seems reasonable enough. But the two instances of "entanglement" in that sentence are not referring to the same thing. One refers to entanglement within B's local system, the other refers to entanglement between B and A. Those are not the same. This simple ambiguity is what kills the initially-reasonable-sounding idea.

There is never an observable difference at B based on what A does. Any "entanglement test" will either require B to condition on A's result or require that A's result didn't matter in the first place. For any specific mechanism you give, one or both of those problems will kill it.

This kind of mistake is symptomatic of reasoning with oversimplified pop-science explanations, instead of just doing the math. The problem is immediately obvious if you actually write down what operations the two people apply and simulate a few runs. (If you're not able to do that kind of thing, you should label the thread "Beginner" instead of "Intermediate". That's especially true if you don't know how to write down or pick the operations.)
 
Tara Sunray
If you don't understand quantum mechanics, the idea "B can check for entanglement, A can break entanglement, therefore B can talk to A" seems reasonable enough. But the two instances of "entanglement" in that sentence are not referring to the same thing. One refers to entanglement within B's local system, the other refers to entanglement between B and A. Those are not the same. This simple ambiguity is what kills the initially-reasonable-sounding idea.

There is never an observable difference at B based on what A does. Any "entanglement test" will either require B to condition on A's result or require that A's result didn't matter in the first place. For any specific mechanism you give, one or both of those problems will kill it.

This kind of mistake is symptomatic of reasoning with oversimplified pop-science explanations, instead of just doing the math. The problem is immediately obvious if you actually write down what operations the two people apply and simulate a few runs. (If you're not able to do that kind of thing, you should label the thread "Beginner" instead of "Intermediate". That's especially true if you don't know how to write down or pick the operations.)

that actually kinda helped me to understand where i am wrong. as said in the beginning, i knew/assumed i was wrong and that the solution couldn't be that simple, but i didn't know why. seems like i really have to look into the math though if i really want to get a grasp on it. and no, i don't know how to do the math. i never bothered so far to look into it - seems like a huge mistake now.
 
Last edited by a moderator:
Simon Phoenix
Science Advisor
Gold Member
291
224
so if we have a quantum "trio" instead of a quantum pair, does the information about 2 out of 3 quanta suffice to determine whether those two quanta are entangled?
The short answer to this question is yes. You can test for entanglement on pairs of particles (you'll need an ensemble of particle pairs all prepared in the same state) to see if there is entanglement between them. What does it matter about some 3rd particle?

However, there are some quite deep issues here that I'm not certain you've got to grips with
(a) 3 particle entanglement is quite a different beast to 2 particle entanglement
(b) if you have 3 particles in a pure state then the state of particles 2 and 3 alone is described by a mixed state - the notion of entanglement for bi-partite mixed states requires you to know what a separable density operator is
(c) if you have a 3 particle entangled state then measurement of one does not necessarily 'break' any entanglement there might be between the other 2. An example here is the 3 particle GHZ state - measure one thing for particle 1 and you can destroy entanglement between all 3 particles - measure another thing for particle 1 and you can end up with particles 2 and 3 being perfectly entangled. Also in the GHZ state there is no entanglement between particles 2 and 3 alone - that is, if you had an ensemble of 3 particle GHZ states and just did entanglement tests on particles 2 and 3 you would find no evidence of non-classical correlation (entanglement).
(d) GHZ states are in some sense the maximally entangled states for 3 qubits - but as mentioned above there is no bipartite entanglement.

Not sure these comments help - as Strilanc says you really need to get into the ring and go a few rounds with the math.
 
Tara Sunray
The short answer to this question is yes. You can test for entanglement on pairs of particles (you'll need an ensemble of particle pairs all prepared in the same state) to see if there is entanglement between them. What does it matter about some 3rd particle?

However, there are some quite deep issues here that I'm not certain you've got to grips with
(a) 3 particle entanglement is quite a different beast to 2 particle entanglement
(b) if you have 3 particles in a pure state then the state of particles 2 and 3 alone is described by a mixed state - the notion of entanglement for bi-partite mixed states requires you to know what a separable density operator is
(c) if you have a 3 particle entangled state then measurement of one does not necessarily 'break' any entanglement there might be between the other 2. An example here is the 3 particle GHZ state - measure one thing for particle 1 and you can destroy entanglement between all 3 particles - measure another thing for particle 1 and you can end up with particles 2 and 3 being perfectly entangled. Also in the GHZ state there is no entanglement between particles 2 and 3 alone - that is, if you had an ensemble of 3 particle GHZ states and just did entanglement tests on particles 2 and 3 you would find no evidence of non-classical correlation (entanglement).
(d) GHZ states are in some sense the maximally entangled states for 3 qubits - but as mentioned above there is no bipartite entanglement.

Not sure these comments help - as Strilanc says you really need to get into the ring and go a few rounds with the math.

thanks a lot, i very much appreciate the answer. it does help, but i think you are both right, i will have to learn the math to fully wrap my head around it.
 

Related Threads for: Quantum entanglement, wave-function collapse

Replies
1
Views
764
Replies
103
Views
82K
  • Last Post
2
Replies
25
Views
6K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
2K
Replies
21
Views
5K
Replies
8
Views
4K
Top