Can Alice and Bob play FTL ping pong?

[andrewkirk]

Alice entangles a pair of particles and sends one of the pair - particle B - towards Bob, keeping the other one - particle A - trapped within some sort of containment device. Alice is on Earth and Bob is several light minutes away, say on Mars. Alice and Bob each have two apparatus for measuring spin, one (device X) measures in direction \vec{x} and the other - device Y - measures spin in direction \vec{y} which is perpendicular to \vec{x}. These directions are specified relative to the CMBR reference frame, so that Alice's and Bob's X measuring apparatuses always remain aligned with one another, as do their Y apparatuses.

When the other particle reaches Bob he traps it within a containment device.

Say Alice and Bob have worked out at what time (T) particle B will arrive at Bob and have arranged to take it in turns, at two second intervals, measuring the spin of their particle. So Bob measures the spin at T+1, Alice at T+2, Bob at T+3, Alice at T+4 and so on. Alice and Bob decide independently of one another which of their two apparatuses to use for each measurement. So Alice may use X, X, Y, X and Y at times T+2, T+4, T+6, T+8 and T+10, and Bob will likely follow a different pattern.

I have two questions:

1. Do the particles remain entangled throughout this process? If not, what causes them to become disentangled?

2. If so, does that then mean that whenever one of them measures using a different device from that which their partner used immediately prior (eg if Alice uses her X device at time T+k when Bob has just used his Y device at T+k-1), they instantly change the state of their partner's particle, and hence the probability distribution for their partner's next measurement?*

thank you

* My understanding is that, if Bob measures using Y at T+k-1 and gets an 'up' result and then measures using Y again at T+k+1 then:
- if Alice measured using device Y at T+k, Bob has a 100% chance of getting an 'up' result at T+k+1
- if Alice measured using device X at T+k, Bob has a 50% chance of getting an 'up' result at T+k+1

 

[mfb]

Say Alice and Bob have worked out at what time (T) particle B will arrive at Bob and have arranged to take it in turns, at two second intervals, measuring the spin of their particle.

The first measurement (=answer to question 1) destroys entanglement, you cannot "re-use" the particles.

and hence the probability distribution for their partner's next measurement?*

The measurement does not change the probability distribution as seen by the other one. Bob cannot know that Alice performed a measurement, unless Alice sends a light signal (or a rocket or whatever) to Bob. Only if they combine their measurement results, they see a correlation.

 

[Nugatory]

1. Do the particles remain entangled throughout this process?

No. Once you observe either particle at either end, the entanglement is broken; you're allowed one measurement after which you have two independent particles not one wave function describing two particles.

If not, what causes them to become disentangled?

Somewhat against my better judgment, I'll try describing it in terms of Schrodinger's cat. Before we open the box, we have a cat and a radioactive atom. They're entangled; there will be a perfect correlation between future measurements that find the atom decayed and future measurements that find the cat dead, and vice versa. But once I make an observation of either, the entanglement disappears. I may have observed a dead cat and a decayed atom, or a live cat and an undecayed atom, but no subsequent measurement of the cat will tell us anything more about the state of the atom and vice versa.

 

[andrewkirk]

Thank you for the replies Nugatory and mfb. I have two supplementary questions.

1. My understanding from your replies is that if Bob measures the momentum on particle B (without either of Alice or Bob having done any prior measurements), that will instantly change the state of particle A, but if Alice then measures say the position of particle A, it will have no effect on the state of particle B because the entanglement was broken when Bob did the first measurement. Is that correct?

2. Is this rule of 'one measurement and then the entanglement is broken' a consequence of the QM postulates? If so, could you please explain how it follows from them as I can't yet see it.

Thanks again.

 

[mfb]

1. My understanding from your replies is that if Bob measures the momentum on particle B (without either of Alice or Bob having done any prior measurements), that will instantly change the state of particle A, but if Alice then measures say the position of particle A, it will have no effect on the state of particle B because the entanglement was broken when Bob did the first measurement. Is that correct?

That is a possible way to describe it.
Actually, the order of the measurements does not matter, you get the same with the opposite order.

2. Is this rule of 'one measurement and then the entanglement is broken' a consequence of the QM postulates? If so, could you please explain how it follows from them as I can't yet see it.

The measurements breaks the superposition of multiple possible measurement results. It has to, otherwise it is not a measurement of the entangled property. Either you (can) know the state (-> no superposition, nothing which could be entangled) or you (cannot) know it (-> superposition possible, entanglement possible).

 

[andrewkirk]

I infer from mfb's reply that it is not possible to reliably generate a sequence of entangled pairs, all of whose states are eigenkets of the spin operator along a pre-defined axis. That is, the process of entangling means that we do not know the direction of spin prior to measurement. Is that correct?

Secondly, does it follow that passing one particle of an entangled pair through a polariser breaks the entanglement, by forcing the particle's state into an eigenket of a particular spin operator? Or am I misunderstanding what polarisation does here?

thank you

 

[audioloop]

Thank you for the replies Nugatory and mfb. I have two supplementary questions.

1. My understanding from your replies is that if Bob measures the momentum on particle B (without either of Alice or Bob having done any prior measurements), that will instantly change the state of particle A, but if Alice then measures say the position of particle A, it will have no effect on the state of particle B because the entanglement was broken when Bob did the first measurement. Is that correct?

2. Is this rule of 'one measurement and then the entanglement is broken' a consequence of the QM postulates? If so, could you please explain how it follows from them as I can't yet see it.

Thanks again.

but you can swap the particles in different entanglements.

 

[andrewkirk]

but you can swap the particles in different entanglements.

Can you explain what that means please?

 

[StevieTNZ]

There is the question, however, that the measurement apparatus does not cause collapse, thus does not break entanglement - it only leads to the system + apparatus entangled, and no definite result arises.

 

[audioloop]

Can you explain what that means please?

inter-exchange of photons of pairs entangled.


http://iopscience.iop.org/1367-2630/11/3/033008

 

[andrewkirk]

I infer from mfb's reply that it is not possible to reliably generate a sequence of entangled pairs, all of whose states are eigenkets of the spin operator along a pre-defined axis. That is, the process of entangling means that we do not know the direction of spin prior to measurement. Is that correct?

Does anybody know the answer to this question?

 

[mfb]

It depends on the property that is entangled. If it is spin: sure. It is the point of entanglement that the state is not determined in advance.

 

[andrewkirk]

Thank you mfb. Focusing on the first part of your answer, what would be a property that could be entangled and allow the experimenter to know in advance the values of the entangled property?

 

[mfb]

Nothing.
"You can know the result in advance" and "entangled" exclude each other by the definition of entanglement.