Would the state collapse when quantum entanglement is being observed?

[particleinmc]

TL;DR

What will be the result of the following example?

There are a pair of AB particle in entangled state: |fin>=|00>+|11>

(where the first digit is the state of particle A and the second digit is the state of B particle)



Lets prepare 100 pairs of AB particles.

Alice and Bob each take 100 A particles and 100 B particles respectively, and they are far apart.

If B has an observation method ObsB:

Let ObsB = ( 0 1 ; 1 0 )

That is, ObsB will have the following result for different state:

ObsB ( |0> ) => Observed as 50% +1, 50% -1

ObsB ( |1> ) => Observed as 50% +1, 50% -1

ObsB ( |0> + |1>) => Observed as 100% +1

If Alice observes their 100 particles at (t=1000), finding each A particle to be either |0> or |1>,

For Alice, each B particle is either |0> or |1> (for example, if Alice observes the first A particle as |0>, then Alice knows the first B particle is |0>,

But for Bob, will the state of Bob’s particles collapse to |0> or |1> as known by Alice? If Bob then uses ObsB to observe the 100 B particles:



Case 1:

If the state of Bob’s particles collapses to |0> or |1> as known by Alice, and Bob uses ObsB to observe, resulting in 50% +1 and 50% -1, then Bob can immediately know that Alice has already observed. This could be used for faster-than-light communication, for example, if Alice and Bob agree:

If Alice’s message is “yes”, Alice observes the 100 Alice particles before (t=1000) (Bob uses ObsB to observe the 100 B particles, resulting in 50% +1, 50% -1),

If the message is “no”, Alice does not observe before (t=1000) (Bob uses ObsB to observe the 100 B particles, resulting in 100% +1).



Q1) Can Alice use the decision of when to observe to achieve faster-than-light communication as described above?







Case 2:

For Bob, the B particles are still |0> + |1>, and if Bob uses ObsB to observe, the result is 100% +1.



Q2a) If Alice phone Bob and tells Bob that each particle is either |0> or |1>, will this cause the first B particle to collapse to |0> or |1>? If Bob then uses ObsB to observe, will the result be 50% +1, 50% -1?



Follow-up questions:

Q2b) If Alice is lying and has not actually observed, what will Bob’s observation result be using ObsB?

Q2c) If the first B particle is |0> for Alice, but |0> + |1> for Bob, is there a contradiction (i.e. does the B particle lack an objective state)?





(I may not sure whether I have correct understanding on Observation calculation, please correct me if I am wrong)

 

[DrChinese]

It is a bit difficult to understand what you are imagining. I can help you with a few things though...

I'm sure you have figured out that no FTL signaling is possible using any technique remotely like what you describe. It's not like scientists haven't explored this in literally thousands of papers. What I assume is: You want to know why it's not possible. Here are some details that will explain that:

a. All Bob ever sees is a series of random outcomes. The outcome of any observation on an entangled particle in Bob's possession is random, and the same for Alice. (In your notation, that would be a series of +1 and -1 values.) Obviously, there is not a lot of useful information to be found in a list of random values.

b. It doesn't matter when anyone observes anything. It's not like "Alice measures first" and then can "send a message to Bob". The order of observation between Alice and Bob has no known or observable effect on any statistical outcomes. Therefore there is nothing to distinguish "Alice sends Bob a message" as opposed to "Bob sends Alice a message". If you like, you can claim (assume) there is a difference between these based on order, there just isn't any actual evidence to support this case.

 

[particleinmc]

Thanks for reading and comment. I find out that I need to understand mixed state, Density matrix and etc. before understand No-communication theorem. But it seems it is difficult to understand for me.

 

[Xuedong He]

Hello Particleinmc,

I am interested in your topic and I want to have a discussion with you. But first I think I need to understand your message as better as possible.
Question 1.
What is the meaning of "Let ObsB = ( 0 1 ; 1 0 )"? Does it mean the observation result of Particle B is 0, 1, 1, 0?

That is, ObsB will have the following result for different state:

Question 2.
ObsB ( |0> ) => Observed as 50% +1, 50% -1, Xuedong: What is the meaning of 50% +1, 50% -1? Does it mean that the probability that a B particle is at "0" state is 50% ± 1?

Thank you,

Xuedong He

 

[PeterDonis]

will the state of Bob’s particles collapse

It depends on what you mean by "collapse" and what interpretation of QM you are using.

 

[PeterDonis]

Moderator's note: Thread moved to QM interpretations subforum since the OP question is interpretation dependent.