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particleinmc

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- TL;DR Summary
- 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)

(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)