Entangled particle decoherence question

[brajesh]

TL;DR Summary

When an entangled particle is observed, does the other one need to be observed too?

Suppose there are two entangled particles A and B, separated by a few miles or light years.

If the spin for particle A is observed, then from my understanding, A will experience quantum decoherence.
And now we can be sure that B will have the opposite spin since B is entangled with A.

Assuming I got my provious assumptions right, my question is, does particle B also experience decoherence when A is observed?
Or does B need to be observed first and experience decoherence before it can be confirmed that it has the opposite spin?

 

[StevieTNZ]

Summary:: When an entangled particle is observed, does the other one need to be observed too?

Suppose there are two entangled particles A and B, separated by a few miles or light years.

If the spin for particle A is observed, then from my understanding, A will experience quantum decoherence.
And now we can be sure that B will have the opposite spin since B is entangled with A.

Assuming I got my provious assumptions right, my question is, does particle B also experience decoherence when A is observed?
Or does B need to be observed first and experience decoherence before it can be confirmed that it has the opposite spin?

What happens when system A is measured by an apparatus is entanglement with the measurement device. No collapse of the wave function, at least if you follow the fundamental Schrodinger equation, occurs (even in the presence of decoherence). From then it'd simply be entanglement between the apparatus that measured system A and system B.

 

[PeterDonis]

does particle B also experience decoherence when A is observed?

No. Particle B only experiences decoherence when it itself interacts with something like a measuring device.

 

[anuttarasammyak]

Summary:: When an entangled particle is observed, does the other one need to be observed too?

f the spin for particle A is observed, then from my understanding, A will experience quantum decoherence.
And now we can be sure that B will have the opposite spin since B is entangled with A.

Say A observes up spin. Decoherence for A takes place then. A judges that B's spin is down. A sends this result to B by light speed signal (slower signal admitted.). B gets it at time T.

Case 1: B observes spin down at t<T. B judges that A's spin is up. He confirms it later at time T by news from B. Decoherence for B takes place at time t
Case 2: After A got news from B at time T, he observes spin down at t<T and confirm the entangled result.
Decoherence for B takes place at time T
Case 3: A got news from B at time T but does no observation. He is sure that spin is down with no need of observation.
Decoherence for B takes place at time T

From this I understand that decoherence is not universal but observer specific.
B's decoherence takes place at earlier time of the two: B's observation and B's receipt of A's observation result.
symmetrically
A's decoherence takes place at earlier time of the two: A's observation and A's receipt of B's observation result.
In other words one's decoherence timing is determined by when he gets the knowledge by his own observation or news from his partner.

[EDIT1]
WRONG : news from B, After A got news from B at time T
CORRECTION: news from A, After B got news from A at time T
Sorry for confusion. Thanks @PeterDonis for pointing it out in #5.

[EDIT2]addition:
For other general observer C, decoherence of electron spin for him, C's decoherence in short, takes place when he first get the knowledge. E.g. by observing A's or B's observation outside the window or by live broadcasting, watching their interview in evening TV news, reading a mail sent from (one of ) them, and taught by mom who read the newspaper.

 

[PeterDonis]

Say A observes up spin. Decoherence for A takes place then. A judges that B's spin is down. A sends this result to B by light speed signal (slower signal admitted.). B gets it at time T.

Case 1: B observes spin down at t<T. B judges that A's spin is up. He confirms it later at time T by news from B.

You mean news from A: A sent the news of his observation of spin up to B, it arrived at B at time T.

Decoherence for B takes place at time t

For this case, yes. But see notes at the end of this post.

Case 2: After A got news from B at time T

A didn't get news from B at time T. B received the news from A at time T. That's what your description above says.

he observes spin down at t<T and confirm the entangled result.

B can't go back in time and make his own measurement at t < T after he receives a message from A at time T. Either B already made a measurement at t < T (which is Case 1 above), or he didn't (which is I assume what this Case 2 is supposed to be).

Decoherence for B takes place at time T

For the case where B makes no measurement prior to time T, yes, B will decohere because of becoming entangled with the information arriving from A. But note that B's *particle* itself will *not* decohere, since B has not measured it. (We're assuming the particles are kept isolated until they are measured.) Also see further notes at the end of this post.

Case 3: A got news from B at time T but does no observation. He is sure that spin is down with no need of observation.

You keep confusing A with B. B gets the news from A at time T. There is no Case 3 for B: either he makes a measurement on his own particle before he receives the news from A, or he doesn't. Those are Cases 1 and 2 above, and they exhaust the possibilities.

From this I understand that decoherence is not universal but observer specific.

Decoherence is a local process: it happens when a quantum system gets entangled with another system that has a large number of degrees of freedom that can't be individually tracked, like a measuring device, or even the environment in general. The entanglement interactions involved are local.

B's decoherence takes place at earlier times of the two: B's observation and B's receipt of A's observation result.

Yes, this is basically true. But see notes at the end of this post.

symmetrically
A's decoherence takes place at earlier times of the two: A's observation and A's receipt of B's observation result.

Yes, this is basically true. But see notes at the end of this post.

In other words one's decoherence timing is determined when he got the knowledge by his own observation or news from his partner.

It's not "getting the knowledge" that causes the decoherence, it's becoming entangled with a system that has a large number of degrees of freedom that can't be individually tracked. Since B himself (the observer--you used A and B to refer to the observers, not the particles themselves) is such a system, he is basically continually decohering himself. The "decoherence" specifically related to the measurement results happens as soon as he becomes entangled with any degree of freedom that is itself entangled with one of those results--either his own particle and measuring device, if he makes his own measurement first, or the signal arriving from A that is entangled with A and A's particle and measuring device, if B receives the signal from A first.