Photons from separated sources can be entangled - after they were detected

[DrChinese]

Yes, I know the thread name sounds like something you might see on the cover of Discover (or sadly, Scientific American these days). Hey, I could have called it: Proof that the future changes the past! Because that is precisely what happens.

The subject relates to entanglement swapping. In typical entanglement swapping setups, two pairs of photons are polarization entangled (often labeled A & B and C & D). The pairs are created closely together in time from different (preferably) sources. The trick is to swap the entanglement so that A & D are polarization entangled, and they will violate a Bell Inequality.

See Figure 1 attached for a general diagram, and Figure 2 for a more detailed schematic. Both of these are from a paper realizing this experiment: http://arxiv.org/abs/quant-ph/0409093"

1. The reported results show violation of a Bell Inequality for photons (A & D) which each traveled over a 1 km fiber. These are post-selected based on results of a suitable Bell state measurement (BSM) on photons B & C. The BSM is what causes the previously otherwise independent photons A & D to become entangled. Note that in this particular experiment, the same pump laser is used to created both sets of photons pairs (A & B and C & D, each pair coming from different PDC crystals however). However, subsequent experiments have performed the same trick using 2 different lasers that have been pulse synchronized.

2. Here is where it gets interesting. According to QM, the path lengths of the various measuring devices can be set to any length in principle. And length implies time as well. The following can be made arbitrarily long (or short), for instance:

a. Location of Alice measuring the polarization of photon A.
b. Location of Bob measuring the polarization of photon D.
c. Location of Charlie performing the Bell state measurement (BSM) on photons B & C.
d. The distance separating the laser/PDC setup creating photon pairs A & B and the laser/PDC setup creating photon pairs C & D.

3. So here is what we do: we pick a. and b. to be short and c. to be long. Thus photons A and D have their polarization measured shortly after being created, and long before the BSM is performed on B & C. And we have d. be long as well so that A and D are created and destroyed (upon measurement) far before their light cones could have ever overlapped. See Figure 3.

Charlie is designated as the person doing the BSM, and it takes a joint measurement of the Bell state to cause A & D to become entangled via entanglment swapping. Charlie's BSM does not itself directly identify the actual polarization results of A and D, which Alice and Bob may select to be anything. For our example, let's assume Alice and Bob always set their observations to be at the same angle but something Charlie does not know. In that case, Alice and Bob will see perfectly correlated/anti-correlated results for any A & D pair that were entangled by Charlie's BSM.

Alice and Bob have already performed their measurements on A & D by the time B & C arrive at Charlie's BSM. Now, Charlie can decide whether or not to actually allow his BSM to entangle A & D. If he DOES perform the BSM, A & D WILL become entangled. They will then show either perfect correlation or perfect anti-correlation (depending on the actual Bell state measured, which occurs at random). Alternately, he can choose NOT to perform the BSM, and A & D will NOT become entangled.

And yet A & D don't exist any longer. They were detected and destroyed BEFORE Charlie decides whether or not to perform the BSM. A & D were never in each other's light cone, and were in fact created from separate and independent laser sources. And yet, subsequent analysis can show that A & D were entangled in the cases when Charlie decides to perform the BSM (a Bell Inequality will be violated). In the cases when Charlies does NOT perform the BSM, a Bell Inequality will NOT be violated.

Thus: Charlie's future decision appears to change the past by entangling photons which no longer exist and have never been in contact with a common light cone.

Not an easy trick to explain by any physical mechanism, and in fact I don't believe any interpretation (other than perhaps the retrocausal/time symmetric variations) would even be able to predict this situation. Such is predicted by workman-like application of our poor ol' abused friend: Standard Quantum Mechanics. So I guess the reports of its demise and uselessness might be premature.

 

[SpectraCat]

Yes, I know the thread name sounds like something you might see on the cover of Discover (or sadly, Scientific American these days). Hey, I could have called it: Proof that the future changes the past! Because that is precisely what happens.

The subject relates to entanglement swapping. In typical entanglement swapping setups, two pairs of photons are polarization entangled (often labeled A & B and C & D). The pairs are created closely together in time from different (preferably) sources. The trick is to swap the entanglement so that A & D are polarization entangled, and they will violate a Bell Inequality.

See Figure 1 attached for a general diagram, and Figure 2 for a more detailed schematic. Both of these are from a paper realizing this experiment: http://arxiv.org/abs/quant-ph/0409093"

1. The reported results show violation of a Bell Inequality for photons (A & D) which each traveled over a 1 km fiber. These are post-selected based on results of a suitable Bell state measurement (BSM) on photons B & C. The BSM is what causes the previously otherwise independent photons A & D to become entangled. Note that in this particular experiment, the same pump laser is used to created both sets of photons pairs (A & B and C & D, each pair coming from different PDC crystals however). However, subsequent experiments have performed the same trick using 2 different lasers that have been pulse synchronized.

2. Here is where it gets interesting. According to QM, the path lengths of the various measuring devices can be set to any length in principle. And length implies time as well. The following can be made arbitrarily long (or short), for instance:

a. Location of Alice measuring the polarization of photon A.
b. Location of Bob measuring the polarization of photon D.
c. Location of Charlie performing the Bell state measurement (BSM) on photons B & C.
d. The distance separating the laser/PDC setup creating photon pairs A & B and the laser/PDC setup creating photon pairs C & D.

3. So here is what we do: we pick a. and b. to be short and c. to be long. Thus photons A and D have their polarization measured shortly after being created, and long before the BSM is performed on B & C. And we have d. be long as well so that A and D are created and destroyed (upon measurement) far before their light cones could have ever overlapped.

Charlie is designated as the person doing the BSM, and it takes a joint measurement of the Bell state to cause A & D to become entangled via entanglment swapping. Charlie's BSM does not itself directly identify the actual polarization results of A and D, which Alice and Bob may select to be anything. For our example, let's assume Alice and Bob always set their observations to be at the same angle but something Charlie does not know. In that case, Alice and Bob will see perfectly correlated/anti-correlated results for any A & D pair that were entangled by Charlie's BSM.

Alice and Bob have already performed their measurements on A & D by the time B & C arrive at Charlie's BSM. Now, Charlie can decide whether or not to actually allow his BSM to entangle A & D. If he DOES perform the BSM, A & D WILL become entangled. They will then show either perfect correlation or perfect anti-correlation (depending on the actual Bell state measured, which occurs at random). Alternately, he can choose NOT to perform the BSM, and A & D will NOT become entangled.

And yet A & D don't exist any longer. They were detected and destroyed BEFORE Charlie decides whether or not to perform the BSM. A & D were never in each other's light cone, and were in fact created from separate and independent laser sources. And yet, subsequent analysis can show that A & D were entangled in the cases when Charlie decides to perform the BSM (a Bell Inequality will be violated). In the cases when Charlies does NOT perform the BSM, a Bell Inequality will NOT be violated.

Thus: Charlie's future decision appears to change the past by entangling photons which no longer exist and have never been in contact with a common light cone.

Not an easy trick to explain by any physical mechanism, and in fact I don't believe any interpretation (other than perhaps the retrocausal/time symmetric variations) would even be able to predict this situation. Such is predicted by workman-like application of our poor ol' abused friend: Standard Quantum Mechanics. So I guess the reports of its demise and uselessness might be premature.

Ok, I think I follow what you are saying, but just to be clear, you are just extrapolating here, right? No one has done the experiment yet where the BSM is made after Alice and Bob make their detections, right? Not that I see anything wrong with your conclusions ... I just want to know if the measurement has been made.

So, if this is correct, doesn't it constitute experimental proof that free-will is an illusion, at least where Q.M. is concerned? I say this because, if Alice and Bob are located next to each other in a lab, but isolated from Charlie's location, then they would be able to predict with 100% accuracy Charlie's future "decisions".

Now, what if Charlie were right there in the lab next to them (since all of the beamlines are optical fibers)? Alice and Bob could conduct their measurements, and tell him the results ... is he free to change his mind? Because if they detect entanglement between A&D, then Charlie must make the BSM measurement on B&C, right? It sure seems to me like he could "decide" that he isn't going to, just to hack off those know-it-all's, Alice and Bob. But that is impossible based on the laws of Q.M., assuming I follow your analysis correctly.

Now take it one level further ... what if Alice, Bob and Charlie are all the same person? Doesn't that mean that Charlie could determine with 100% certainty his own future actions? Isn't that a pretty good definition of not having free will?

So, I have thought about this for a while now, and I can't see my way through this apparent paradox. Can Charlie change his mind or can't he? What am I missing?

 

[DrChinese]

Ok, I think I follow what you are saying, but just to be clear, you are just extrapolating here, right? No one has done the experiment yet where the BSM is made after Alice and Bob make their detections, right? Not that I see anything wrong with your conclusions ... I just want to know if the measurement has been made.

So, if this is correct, doesn't it constitute experimental proof that free-will is an illusion, at least where Q.M. is concerned? I say this because, if Alice and Bob are located next to each other in a lab, but isolated from Charlie's location, then they would be able to predict with 100% accuracy Charlie's future "decisions".

Now, what if Charlie were right there in the lab next to them (since all of the beamlines are optical fibers)? Alice and Bob could conduct their measurements, and tell him the results ... is he free to change his mind? Because if they detect entanglement between A&D, then Charlie must make the BSM measurement on B&C, right? It sure seems to me like he could "decide" that he isn't going to, just to hack off those know-it-all's, Alice and Bob. But that is impossible based on the laws of Q.M., assuming I follow your analysis correctly.

Now take it one level further ... what if Alice, Bob and Charlie are all the same person? Doesn't that mean that Charlie could determine with 100% certainty his own future actions? Isn't that a pretty good definition of not having free will?

So, I have thought about this for a while now, and I can't see my way through this apparent paradox. Can Charlie change his mind or can't he? What am I missing?

You are correct that as far as I know, the variation I describe has not been performed. And by the way, I added another diagram to the earlier 2 to show the setup I describe.

The trick is as follows: the Bell State Measurement (BSM) can send the entangled state into one of two states: one in which Alice and Bob are correlated (HH or VV) and one in which they are anti-correlated (HV or VH). THIS OCCURS RANDOMLY. At the time Charlie performs the BSM, the entanglement occurs and one of those 2 states occurs. So the pattern at Alice and Bob does not look like anything other than a random string of values anyway. If Charlies SKIPS the BSM, then Alice and Bob don't see entanglement... and they similarly see nothing other than a random string of values. So even if you put Alice and Bob's results together, nothing obvious ever changes when Charlie does - or does not - perform the BSM to cause the entanglement. No FTL signalling is possible (no surprise there).

(Now let's suppose when Charlie skips performing the BSM, he instead simply checks B & C separately to see if they are correlated or anti-correlated at some angle he chooses independently of Alice/Bob. But now A and D are not polarization entangled.)

When Charlie hands over his results - i.e. tells Alice and Bob which pairs are entangled and which should be correlated vs. anti-correlated (he can see that information), then magically it turns out all along that Alice and Bob's photons A & D were entangled as Charlie indicates. Presto magico! Charlie's decision changes the past, as the pairs that were not entangled do not show any such pattern (after considering whether they are correlated/anti-correlated per his alternate measurement).

Charlie's joint measurement of the B & C Bell State using the BSM cause A & D to become polarization entangled (and he finds out if they are correlated or anti-correlated), while his alternate separate measurements of the B & C polarizations does not cause them to be entangled.

 

[Frame Dragger]

Nature is so... weird. =o
I checked your diagrams... is there any reason that this experiment cannot be performed? I should say, is there any way to make this experiment useful in this application? I'm a fan of the TI and the notion that time and space independant particles might become entangled is... did I mention WEIRD?!

 

[SpectraCat]

You are correct that as far as I know, the variation I describe has not been performed. And by the way, I added another diagram to the earlier 2 to show the setup I describe.

The trick is as follows: the Bell State Measurement (BSM) can send the entangled state into one of two states: one in which Alice and Bob are correlated (HH or VV) and one in which they are anti-correlated (HV or VH). THIS OCCURS RANDOMLY. At the time Charlie performs the BSM, the entanglement occurs and one of those 2 states occurs. So the pattern at Alice and Bob does not look like anything other than a random string of values anyway. If Charlies SKIPS the BSM, then Alice and Bob don't see entanglement... and they similarly see nothing other than a random string of values. So even if you put Alice and Bob's results together, nothing obvious ever changes when Charlie does - or does not - perform the BSM to cause the entanglement. No FTL signalling is possible (no surprise there).

(Now let's suppose when Charlie skips performing the BSM, he instead simply checks B & C separately to see if they are correlated or anti-correlated at some angle he chooses independently of Alice/Bob. But now A and D are not polarization entangled.)

When Charlie hands over his results - i.e. tells Alice and Bob which pairs are entangled and which should be correlated vs. anti-correlated (he can see that information), then magically it turns out all along that Alice and Bob's photons A & D were entangled as Charlie indicates. Presto magico! Charlie's decision changes the past, as the pairs that were not entangled do not show any such pattern (after considering whether they are correlated/anti-correlated per his alternate measurement).

Charlie's joint measurement of the B & C Bell State using the BSM cause A & D to become polarization entangled (and he finds out if they are correlated or anti-correlated), while his alternate separate measurements of the B & C polarizations does not cause them to be entangled.

Ok, thanks ... I missed the randomness aspect of the correlated (HH + VV) and anti-correlated (HV + VH) entanglements. So Alice and Bob still can't predict Charlie's future actions ... that is good. Otherwise that experiment probably would have "torn open the fabric of spacetime", or something similiarly dire.

I guess I should have realized that Eberhard's theorem must forbid the scenarios I originally described, since it would have meant FTL information transmission. I don't think I'll ever get used to (or tired of) the weirdness of QM.

 

[SpectraCat]

Nature is so... weird. =o
I checked your diagrams... is there any reason that this experiment cannot be performed? I should say, is there any way to make this experiment useful in this application? I'm a fan of the TI and the notion that time and space independant particles might become entangled is... did I mention WEIRD?!

I guess the experiment really already has been done, as I think Dr. Chinese was implying in his first post. Once it has been shown that there is a Bell's inequality violation when A&D are detected "after" B&C are entangled, it must also happen when they are detected (and destroyed) "before" B&C are entangled. The choices of the relative lengths of the delay lines just "seems" important to us (because of our observer's bias I guess), but the quantum particles don't care at all about it. They are either entangled or not ... that's all.

I can't even decide if I think this has ramifications for free will. I oscillate wildly back and forth when I try to think about it ... perhaps it is just a philosophical distinction anyway.

 

[Frame Dragger]

I guess the experiment really already has been done, as I think Dr. Chinese was implying in his first post. Once it has been shown that there is a Bell's inequality violation when A&D are detected "after" B&C are entangled, it must also happen when they are detected (and destroyed) "before" B&C are entangled. The choices of the relative lengths of the delay lines just "seems" important to us (because of our observer's bias I guess), but the quantum particles don't care at all about it. They are either entangled or not ... that's all.

I can't even decide if I think this has ramifications for free will. I oscillate wildly back and forth when I try to think about it ... perhaps it is just a philosophical distinction anyway.

See?! WEIRD!

 

[Demystifier]

Not an easy trick to explain by any physical mechanism, and in fact I don't believe any interpretation (other than perhaps the retrocausal/time symmetric variations) would even be able to predict this situation. Such is predicted by workman-like application of our poor ol' abused friend: Standard Quantum Mechanics.

I don't understand what you are saying here. If it is predicted by standard QM, then it is also predicted by standard INTERPRETATION of QM. Yet, the standard interpretation of QM does not involve retrocausal/time symmetry. Then why do you say that no interpretation (without retrocausal/time symmetry) can predict it?

 

[Demystifier]

So Alice and Bob still can't predict Charlie's future actions ... that is good.

But then does it really make sense to say that Charlie's action influences the past? If it does, in what sense?

 

[Dmitry67]

wow... wow...
so... how is it handled in different interpretations?
(TI has no problems with it, I know)

 

[Demystifier]

You are correct that as far as I know, the variation I describe has not been performed.

Has it at least been proposed in the literature? If yes, then why don't you give us a link to the appropriate reference? If no, i.e., if it is your original idea (and this is certainly a very important idea if there is no an error in it), then why don't you write a paper before sharing your idea with us? Perhaps you are not certain that your conclusions are correct?

 

[Demystifier]

wow... wow...
so... how is it handled in different interpretations?
(TI has no problems with it, I know)

I am convinced that, if it can be explained by the "standard" interpretation, then it can also be explained by the many world and Bohmian interpretations too. However, I must admit that I still don't fully understand the whole idea in ANY view of QM, which is why I ask various questions in the posts above.

 

[Demystifier]

I don't think I'll ever get used to (or tired of) the weirdness of QM.

I don't think I'll ever get used to the weirdness of various inconsistent personal INTERPRETATIONS of QM. (Consistent interpretations such as Copenhagen, many world, Bohmian, etc are excluded.) On the other hand, when one of the consistent interpretations that exist on the market is chosen, then QM is not weird at all. It looks weird only if you have not yet picked one of the interpretations.

 

[Demystifier]

Alice and Bob have already performed their measurements on A & D by the time B & C arrive at Charlie's BSM. Now, Charlie can decide whether or not to actually allow his BSM to entangle A & D.

I think I have found the mistake in your reasoning. I think the above is wrong if standard QM is correct. If Alice and Bob have performed their measurements, then these measurements have destroyed the coherence (i.e., caused the decoherence) of the total wave function describing the joint system consisting of A, D, B , and C. Consequently, contrary to your claim, the Charlie's decision will NOT affect the entanglement between A & D. Charlie can do whatever he wants, but his choice will not be correlated with the presence or absence of the entanglement between A and D.

Nice try, but in my opinion you forgot to take into account the role of decoherence.

 

[Dmitry67]

Demystifier, it can not be about who is doing the measurement first - because otherwise (knowing that the 'arm' is quite long) different relativistic observers would disagree on the result.

So what you are saying is that it is irrelevant at all what Charlie is doing?

 

[Frame Dragger]

I am convinced that, if it can be explained by the "standard" interpretation, then it can also be explained by the many world and Bohmian interpretations too. However, I must admit that I still don't fully understand the whole idea in ANY view of QM, which is why I ask various questions in the posts above.

That is definitely why Dmitry mentions the Transactional Interpretation. http://en.wikipedia.org/wiki/Transactional_interpretation

Even there it's mentioned that decoherence in that context is atemporal, so A and D should still be entangled in that scenario. As for the 'standard' against which dBB is meant to be compared, I don't know. Then again, is this really any stranger than DCQE from a non-Bohmian perspective?

 

[Frame Dragger]

I don't think I'll ever get used to the weirdness of various inconsistent personal INTERPRETATIONS of QM. (Consistent interpretations such as Copenhagen, many world, Bohmian, etc are excluded.) On the other hand, when one of the consistent interpretations that exist on the market is chosen, then QM is not weird at all. It looks weird only if you have not yet picked one of the interpretations.

deBB is subject to EPR is it not? It must be if it matches the predictions of QM, and if so then in theory it's not a complete description either. Different approaches, one singular and evolving, and the other branching, branching, branching, then consolidating (deBB vs SQM) to form a new platform for progress.

Given that both are just approaches to the same problem, the main difference is... what? That EPR paradox never arises because of MWI? deBB is attractive in that it replaces a probabilistic cloud with trajectories, but that doesn't help with the inherent oddness of non-locality without the crutch of MWI, or Determinism.

 

[Demystifier]

Demystifier, it can not be about who is doing the measurement first - because otherwise (knowing that the 'arm' is quite long) different relativistic observers would disagree on the result.

The point is that the environment (that causes decoherence) defines a "preferred" Lorentz frame. The environment possesses a time arrow (a timelike vector pointing in the direction of entropy increase), which defines the "preferred" Lorentz frame. In practice, this is usually the laboratory frame.

For some details see also
http://xxx.lanl.gov/abs/gr-qc/0403121 [Found.Phys.Lett. 19 (2006) 259]

So what you are saying is that it is irrelevant at all what Charlie is doing?

Yes.

 

[Demystifier]

deBB is subject to EPR is it not? It must be if it matches the predictions of QM, and if so then in theory it's not a complete description either. Different approaches, one singular and evolving, and the other branching, branching, branching, then consolidating (deBB vs SQM) to form a new platform for progress.

Given that both are just approaches to the same problem, the main difference is... what? That EPR paradox never arises because of MWI? deBB is attractive in that it replaces a probabilistic cloud with trajectories, but that doesn't help with the inherent oddness of non-locality without the crutch of MWI, or Determinism.

These are interesting questions, but off-topic in my opinion.

 

[Dmitry67]

The point is that the environment (that causes decoherence) defines a "preferred" Lorentz frame. The environment possesses a time arrow (a timelike vector pointing in the direction of entropy increase), which defines the "preferred" Lorentz frame. In practice, this is usually the laboratory frame.

Yes, but you can put Alice, Bob, Charlie and the Emitter on the different moving spaceships not sharing the same frame. Just a sidenote

 

[SpectraCat]

I think I have found the mistake in your reasoning. I think the above is wrong if standard QM is correct. If Alice and Bob have performed their measurements, then these measurements have destroyed the coherence (i.e., caused the decoherence) of the total wave function describing the joint system consisting of A, D, B , and C. Consequently, contrary to your claim, the Charlie's decision will NOT affect the entanglement between A & D. Charlie can do whatever he wants, but his choice will not be correlated with the presence or absence of the entanglement between A and D.

Nice try, but in my opinion you forgot to take into account the role of decoherence.

Hmmm ... I was trying to come up with a similar argument to this before I posted my first response in this thread, (i.e. the measurement of A&D is a time-specific destruction of the coherent state) but I couldn't nail it down. I have noticed that there is usually a similar flaw in other arguments that claim that measurements in the future can affect measurements in the past, but I couldn't see it in this case. Of course now that you point it out, it seems obvious.

But wait, we have a testable hypothesis! I guess someone needs to call up those experimenters and suggest that they switch the lengths of their delay lines to do the appropriate Bell's inequality test.

 

[Demystifier]

That is definitely why Dmitry mentions the Transactional Interpretation. http://en.wikipedia.org/wiki/Transactional_interpretation

Even there it's mentioned that decoherence in that context is atemporal, so A and D should still be entangled in that scenario.

In "ordinary" interpretations (such as Copenhagen, many world, Bohm, ...) decoherence is temporal and, in my opinion, plays a mayor role for a correct treatment of the DrChinese proposal.

As for the 'standard' against which dBB is meant to be compared, I don't know. Then again, is this really any stranger than DCQE from a non-Bohmian perspective?

What is DCQE?

 

[Demystifier]

Yes, but you can put Alice, Bob, Charlie and the Emitter on the different moving spaceships not sharing the same frame. Just a sidenote

In that case the time arrow may point in slightly different directions at different places. This means that the global "preferred" frame will not be a Lorentz one, but a curved one. Everything else will be the same.

 

[Demystifier]

Hmmm ... I was trying to come up with a similar argument to this before I posted my first response in this thread, (i.e. the measurement of A&D is a time-specific destruction of the coherent state) but I couldn't nail it down. I have noticed that there is usually a similar flaw in other arguments that claim that measurements in the future can affect measurements in the past, but I couldn't see it in this case. Of course now that you point it out, it seems obvious.

But wait, we have a testable hypothesis! I guess someone needs to call up those experimenters and suggest that they switch the lengths of their delay lines to do the appropriate Bell's inequality test.

I completely agree.

 

[Frame Dragger]

In "ordinary" interpretations (such as Copenhagen, many world, Bohm, ...) decoherence is temporal and, in my opinion, plays a mayor role for a correct treatment of the DrChinese proposal.


What is DCQE?

Sorry, my bad. Delayed Choice Quantum Eraser. I'm normally quite a fine writer, but online I get lazy to the point of incoherence.

If my earier questions were off-topic (and they were) I'd still love to hear your view on them in a different thread or PM. You and Zenith are the two people who've really shown me how to research (and respect even if it isn't my preference) dBB.

As for the decoherence being atemporal as a means of maintaining coherence... to be honest it was the best thing I could think of. This isn't something I find easy to understand even in the oft puzzling world of QM.

To be fair, if they change the lengths and get the same result then this is seemingly paradoxical in both QM and dBB and various interpretations. I can't see how it wouldn't be; in fact it would be the ULTIMATE "Spukhafte Fernwirking"!

 

[DrChinese]

I think I have found the mistake in your reasoning. I think the above is wrong if standard QM is correct. If Alice and Bob have performed their measurements, then these measurements have destroyed the coherence (i.e., caused the decoherence) of the total wave function describing the joint system consisting of A, D, B , and C. Consequently, contrary to your claim, the Charlie's decision will NOT affect the entanglement between A & D. Charlie can do whatever he wants, but his choice will not be correlated with the presence or absence of the entanglement between A and D.

Nice try, but in my opinion you forgot to take into account the role of decoherence.

I think you have seen how difficult the interpretations issue really is. Because in my mind, there is no question of the result. But I also must preface this by saying, as far as I know the experiment has NOT been run yet where we have delayed choice by Charlie. Now, why do I say the result should be as I describe? Here is my reasoning. I will make 2 arguments in favor:

1. Because I follow SQM, I do not view the entangled particles A & B as individuals. They are completely described by their wavefunction, and that expands in spacetime to join the wavefunction of C & D. So I believe the wavefunction acts as if it is real until collapse. Collapse is context sensitive, and depends on the entire setup. Clearly, we already know that a wave function - once split - can in fact be re-assembled. After all, that is how quantum erasers work. This experiment is a form of delayed choice quantum eraser (DCQE). So timing of the context must be irrelevant: Alice, Bob and Charlie can make their measurements in any order and the outcome is going to be the same.

2. If there were the kind of decoherence you describe, then such could be exploited to locate an absolute time frame. After all, there is definitely entanglement when Charlie chooses BEFORE Alice and Bob make their measurements. So by progressively having Charlie wait just a bit longer, you could locate the decoherence point you hypothesize and voila! There is an absolute frame after all.

So what I am saying in essence is: I can predict the outcome of this experiment by using my interpretation, which is SQM. I know you will argue that BM always gives the same predictions as SQM, so this is your prediction too (we can put a * by that until you make a final decision). But I will say that if you believe particles follow realistic trajectories and have non-local interactions, that you will NOT be able to predict that particles can be entangled AFTER they are destroyed - or not, depending on Charlie's later choice.

Now of course there is a way out. You can also say the detections at A & D subselect for both of Charlie's possible choices. But that now makes the Bell State itself an element of reality (since it can predicted with 100% certainty) ! Something which makes perfect sense in SQM, but not so much in most other interpretations. Doesn't really make sense to me in LR (which was already ruled out but here is another stake), BM or MWI. But does make sense in the Time Symmetric group like RBW. And of course within poor ol' SQM.

 

[Demystifier]

If my earier questions were off-topic (and they were) I'd still love to hear your view on them in a different thread or PM.

If you ask the questions in another thread, I will be happy to answer them.

 

[zonde]

Thus: Charlie's future decision appears to change the past by entangling photons which no longer exist and have never been in contact with a common light cone.

Not an easy trick to explain by any physical mechanism, and in fact I don't believe any interpretation (other than perhaps the retrocausal/time symmetric variations) would even be able to predict this situation. Such is predicted by workman-like application of our poor ol' abused friend: Standard Quantum Mechanics. So I guess the reports of its demise and uselessness might be premature.

Without the information about Charlie's BSM there is no interference pattern between A&D.
When you consider Charlie's BSM you actually perform post selection based on this measurement and now you observe correlations between A/B&C/D.
So pure correlations between A&D does appear in any case like in delayed choice quantum eraser experiment double slit interference pattern appear only after applying post selection filter from entangled photon.

If I am not mistaken according to QM entanglement is swapped between two interacting photons so that joined measurement of B&C entangle C with A and B with D and not A with D.
So you can easily go astray if you assume physical versus statistical connection between entangled photons.

 

[Demystifier]

Collapse is context sensitive, and depends on the entire setup.

Not only true, but essential.

Clearly, we already know that a wave function - once split - can in fact be re-assembled. After all, that is how quantum erasers work.

True, but it has nothing to do with collapse. The collapse is irreversible.

This experiment is a form of delayed choice quantum eraser (DCQE).

Not really. The eraser does not involve decoherence, so it is reversible. Your measurements by Alice and Bob involve decoherence so they are irreversible.

If there were the kind of decoherence you describe, then such could be exploited to locate an absolute time frame.

It is not more "absolute" than, e.g., the frame with respect to which Earth is at rest.

So what I am saying in essence is: I can predict the outcome of this experiment by using my interpretation, which is SQM.

No it isn't. Modern SQM involves decoherence, while your interpretation does not seem to involve it.

I know you will argue that BM always gives the same predictions as SQM, so this is your prediction too (we can put a * by that until you make a final decision). But I will say that if you believe particles follow realistic trajectories and have non-local interactions, that you will NOT be able to predict that particles can be entangled AFTER they are destroyed - or not, depending on Charlie's later choice.

Note that in my analysis of your thought experiment I have not used the Bohmian interpretation at all.

But that now makes the Bell State itself an element of reality (since it can predicted with 100% certainty) ! Something which makes perfect sense in SQM, but not so much in most other interpretations. Doesn't really make sense to me in LR (which was already ruled out but here is another stake), BM or MWI.

By Bell state you mean a certain many particle wave function, right? Just to remind you that the wave function IS an element of reality in BM and MWI.

 

[zonde]

You can also say the detections at A & D subselect for both of Charlie's possible choices.

If you subselect from both Charlie's possible choices you still don't have correlations at A&D. You have correlations only when you make BSM at Charlie and detect simultaneously photons at both detectors. Check the paper you linked (fig.3) - there is not correlations for 3 photon coincidences. Correlations appear only for 4 photon coincidences.

 

[DrChinese]

If I am not mistaken according to QM entanglement is swapped between two interacting photons so that joined measurement of B&C entangle C with A and B with D and not A with D.
So you can easily go astray if you assume physical versus statistical connection between entangled photons.

A & D definitely get entangled. And in fact you could continue/extend the chain by adding a 3rd entangled pair, E & F. Then you could get A & F entangled (by performing the BSM on D & E) even though B & E never meet. And in fact the person who chooses to perform that experiment - or not - (let's call her Dee) could make the decision to do so before or after Charlie. Such a setup - where there is extension of the quantum protocol - is called a quantunm repeater. There is substantial work on this precise scenario - using entanglement repeaters - already in the literature. So that is strong confirmation - to me at least - that I have interpreted my proposed setup correctly.

http://arxiv.org/abs/0912.3871"

 

[DrChinese]

If you subselect from both Charlie's possible choices you still don't have correlations at A&D. You have correlations only when you make BSM at Charlie and detect simultaneously photons at both detectors. Check the paper you linked (fig.3) - there is not correlations for 3 photon coincidences. Correlations appear only for 4 photon coincidences.

4 fold coincidences allow you to see the sets where A & D are entangled. Sometimes there are photons at A & D but not both B & C, so the Bell State Measurement does not occur and there is no entanglement.

But Charlie can see several different Bell states. According to which detectors go off, the Bell state can imply correlated or anti-correlated entanglement of A & D. That occurs randomly.

 

[DrChinese]

Note that in my analysis of your thought experiment I have not used the Bohmian interpretation at all.

And yet: you don't see anything weird about your prediction that if Charlie performs the BSM after A & D are detected, then there is no entanglement. Because that prediction seems weird to me. To me it implies (and would confirm) the absolute existence of a uniform time frame. Something which as far as I know has never been observed.

And what about the scenario in which the order of measurements is Alice, Charlie, Bob? Is there entanglement then? I say order does not matter to outcome, period. And I say that is the prediction of SQM too.

 

[Demystifier]

And yet: you don't see anything weird about your prediction that if Charlie performs the BSM after A & D are detected, then there is no entanglement. Because that prediction seems weird to me.

But not to me.

To me it implies (and would confirm) the absolute existence of a uniform time frame. Something which as far as I know has never been observed.

The point is that this "absolute" time frame is not really absolute, but depends on the environment. For example, one time frame is defined by cosmic background radiation, another time frame by motion of Earth, etc. The physically relevant frame is the one which corresponds to the environment that causes physical decoherence.

And what about the scenario in which the order of measurements is Alice, Charlie, Bob? Is there entanglement then? I say order does not matter to outcome, period. And I say that is the prediction of SQM too.

I simply disagree. I say the order does matter the outcome, whenever there is an irreversible process such as decoherence involved.

Anyway, you are an experimentalist. Can you perform an experiment to see who is right?

 

[Dmitry67]

If order is important, and parts of the experimental setup are put on spaceships, then you can experimentally detect freferred frame for that experiment. And as ships can fly in different directions, that frame has nothing to do with the 'environment' or 'lab'. It is something really mysterious (not for dBB may be :) )

 

[SpectraCat]

And yet: you don't see anything weird about your prediction that if Charlie performs the BSM after A & D are detected, then there is no entanglement. Because that prediction seems weird to me. To me it implies (and would confirm) the absolute existence of a uniform time frame. Something which as far as I know has never been observed.

I am also fairly convinced by DeMystifier's arguments here, for the following reason.

When Alice and Bob measure A & D, then that destroys the entanglement between A&B, and between C & D. Thus, when Charlie performs his operation, he is working with photons that have already undergone decoherence, so there is nowhere to transfer the entanglement to. Sure, B & C become entangled if Charlie makes a BSM, but that no longer has any ramifications for A & D, because they are no longer entangled with B & C, respectively, when Charlie makes the BSM (or not).

So, based on this, I think there will be no violation of a Bell's inequality for A & D based on Charlie's choice of whether or not to conduct a BSM on B & C. But I really want to see the results of the experiment!

 

[Dmitry67]

decoherence can be used as explanation only if lab is small so Bob, Alice and Charlie constantly decohere with each other. In case of perfectly isolated spaceships it can not serve as explanation.

 

[SpectraCat]

decoherence can be used as explanation only if lab is small so Bob, Alice and Charlie constantly decohere with each other. In case of perfectly isolated spaceships it can not serve as explanation.

I can't see how that matters ... as was pointed out to me in the https://www.physicsforums.com/showthread.php?t=374888", the entire apparatus has to be considered as a single unit. So there is no way for the spaceships to be "perfectly isolated", because they have to be connected by the photon paths. AFAICS the relative locations of Alice, Bob, Charlie and the emitter are irrelevant to the phenomenon of decoherence.

 

[DrChinese]

I am also fairly convinced by DeMystifier's arguments here, for the following reason.

When Alice and Bob measure A & D, then that destroys the entanglement between A&B, and between C & D. Thus, when Charlie performs his operation, he is working with photons that have already undergone decoherence, so there is nowhere to transfer the entanglement to. Sure, B & C become entangled if Charlie makes a BSM, but that no longer has any ramifications for A & D, because they are no longer entangled with B & C, respectively, when Charlie makes the BSM (or not).

So, based on this, I think there will be no violation of a Bell's inequality for A & D based on Charlie's choice of whether or not to conduct a BSM on B & C. But I really want to see the results of the experiment!

OOOOO, fun...

We have a great question here, and now we have folks on different sides of the predictions. I honestly haven't seen the answer, but now I will need to look even deeper just to see... Does anyone else want to weigh in on an answer?

 

[Dmitry67]

They can be isolated ENOUGH. In order for spacehips to decohere they must exchange several photons. While you can make a small detector looking in precisely calculated direction, ready to detect a single photon. Difficult, but possible. As you know, it is possible to decohere - at least for short time - machroscopic systems on the lab table - an experiment with superconductive ring in the entangled state.

 

[zonde]

4 fold coincidences allow you to see the sets where A & D are entangled. Sometimes there are photons at A & D but not both B & C, so the Bell State Measurement does not occur and there is no entanglement.

Sometimes photons B & C take different paths out of BS and then there can be simultaneous detection at both detectors. But sometimes B & C take the same path out of BS and then there can be only one detection in principle. So you clearly make unfair postselection.

But Charlie can see several different Bell states. According to which detectors go off, the Bell state can imply correlated or anti-correlated entanglement of A & D. That occurs randomly.

Please explain the thing about different detectors and different correlations. How do you mean that?

 

[Demystifier]

If order is important, and parts of the experimental setup are put on spaceships, then you can experimentally detect freferred frame for that experiment. And as ships can fly in different directions, that frame has nothing to do with the 'environment' or 'lab'. It is something really mysterious (not for dBB may be :) )

You cannot determine the whole global preferred frame everywhere, but only 3 small pieces (for 3 local spaceships) of that frame. These pieces are determined by local labs. (I don't need dBB for that.)

 

[Dmitry67]

Demystifier, if ORDER of events in 3 spacially-separated locations affects the result, then different observers would not agree on the results. Let me say it more careful: i don't detect the frame, I don't care about it, I just show that nature is not Lorentz-invriant if you're right.

I can agree that we can be overlooking something and there won't be entanglement in any case, no matter what the order is, but the opposite, that order affects the result is absolutely impossible

 

[DrChinese]

I have a reference on the SQM interpretation. Will post it in a bit... everyone still has a chance to weigh in! Does decoherence mean the order of measurements matters?

 

[SpectraCat]

And what about the scenario in which the order of measurements is Alice, Charlie, Bob? Is there entanglement then? I say order does not matter to outcome, period. And I say that is the prediction of SQM too.

Based on my earlier arguments, I would say that, once Alice has made her measurements, B represents a free, unentangled photon with a definite polarization state, which we know because it must be opposite to whatever Alice measured. So, B is indistinguishable from any other random photon with a well-defined polarization that would become entangled with C when Charlie makes his measurement.

So, I would predict the results of that experiment should be the same as in the 3-photon case where Charlie and Bob are the only two measurements, B is a "free" photon and C & D are an entangled pair. I don't actually know what those results are ... I have seen 3-photon entanglement before (c.f. the Steinberg group's paper in Nature 429, p.161 [2004]), but never in the 2+1 case, i.e. from an entangled pair with a "free" photon.

 

[DrChinese]

http://arxiv.org/abs/quant-ph/0201134"

[Note: My Charlie is here labeled as Alice; and my photons A/B/C/D are labeled as 0/1/2/3. So the question to be answered below that compares to my question is: Can Alice delay her BSM measurement and still end up with photons 0 and 3 entangled?]

"A seemingly paradoxical situation arises — as suggested by Peres [4] — when Alice’s Bellstate analysis is delayed long after Bob’s measurements. This seems paradoxical, because
Alice’s measurement projects photons 0 and 3 into an entangled state after they have been measured. Nevertheless, quantum mechanics predicts the same correlations. Remarkably, Alice is even free to choose the kind of measurement she wants to perform on photons 1 and 2. Instead of a Bell-state measurement [BSM] she could also measure the polarizations of these photons individually. Thus depending on Alice’s later measurement, Bob’s earlier results either indicate that photons 0 and 3 were entangled or photons 0 and 1 and photons 2 and 3. This means that the physical interpretation of his results depends on Alice’s later decision.

"Such a delayed-choice experiment was performed by including two 10 m optical fiber delays for both outputs of the BSA. In this case photons 1 and 2 hit the detectors delayed by about 50 ns. As shown in Fig. 3, the observed fidelity of the entanglement of photon 0 and photon 3 matches the fidelity in the non-delayed case within experimental errors. Therefore, this result indicate that the time ordering of the detection events has no influence on the results and strengthens the argument of A. Peres [4]: this paradox does not arise if the correctness of quantum mechanics is firmly believed."

--------------------

So the experiment has been performed, and the ordering is not important. I think some beers are due me. Demystifier, don't think I won't come over there to collect. SpectraCat, not sure where you are but I'm in Texas.

 

[SpectraCat]

http://arxiv.org/abs/quant-ph/0201134"

[Note: My Charlie is here labeled as Alice; and my photons A/B/C/D are labeled as 0/1/2/3. So the question to be answered below that compares to my question is: Can Alice delay her BSM measurement and still end up with photons 0 and 3 entangled?]

"A seemingly paradoxical situation arises — as suggested by Peres [4] — when Alice’s Bellstate analysis is delayed long after Bob’s measurements. This seems paradoxical, because
Alice’s measurement projects photons 0 and 3 into an entangled state after they have been measured. Nevertheless, quantum mechanics predicts the same correlations. Remarkably, Alice is even free to choose the kind of measurement she wants to perform on photons 1 and 2. Instead of a Bell-state measurement [BSM] she could also measure the polarizations of these photons individually. Thus depending on Alice’s later measurement, Bob’s earlier results either indicate that photons 0 and 3 were entangled or photons 0 and 1 and photons 2 and 3. This means that the physical interpretation of his results depends on Alice’s later decision.

"Such a delayed-choice experiment was performed by including two 10 m optical fiber delays for both outputs of the BSA. In this case photons 1 and 2 hit the detectors delayed by about 50 ns. As shown in Fig. 3, the observed fidelity of the entanglement of photon 0 and photon 3 matches the fidelity in the non-delayed case within experimental errors. Therefore, this result indicate that the time ordering of the detection events has no influence on the results and strengthens the argument of A. Peres [4]: this paradox does not arise if the correctness of quantum mechanics is firmly believed."

--------------------

So the experiment has been performed, and the ordering is not important. I think some beers are due me. Demystifier, don't think I won't come over there to collect. SpectraCat, not sure where you are but I'm in Texas.

Ok, so I looked at the published version of that paper (PRL 88 [2002], art. 017903-1), and I don't think their case corresponds to the case we are discussing. I also don't think that all of the statements you quoted from their paper are correct. Their "delayed-choice" measurement involved adding 10 m extensions to the OUTPUTS of the BSA device. That delayed the detection of photons 1 and 2, but that is not the crucial event. The crucial event was the interaction of photons 1 & 2 in the BSA, which is defines the quantum teleportation event that we are discussing in this thread.

Our discussion here is debating what would happen if the extensions were added to the INPUTS of the BSA device described in the paper. Since the interaction of the photons in the BSA device constitutes a measurement in SQM (which is why I prefer the designmation BSM for this), for the purposes of our discussion, it doesn't matter how long one delays detection of photons 1 and 2 after the BSM.

So, I think my beer money is safe for now.

EDIT: I am currently digging through the 75 or so papers that cite that one, to see if any of them mention our case explicitly ... but it might take a while.

 

[DrChinese]

Ok, so I looked at the published version of that paper (PRL 88 [2002], art. 017903-1), and I don't think their case corresponds to the case we are discussing. I also don't think that all of the statements you quoted from their paper are correct. Their "delayed-choice" measurement involved adding 10 m extensions to the OUTPUTS of the BSA device. That delayed the detection of photons 1 and 2, but that is not the crucial event. The crucial event was the interaction of photons 1 & 2 in the BSA, which is defines the quantum teleportation event that we are discussing in this thread.

Our discussion here is debating what would happen if the extensions were added to the INPUTS of the BSA device described in the paper. Since the interaction of the photons in the BSA device constitutes a measurement in SQM (which is why I prefer the designmation BSM for this), for the purposes of our discussion, it doesn't matter how long one delays detection of photons 1 and 2 after the BSM.

So, I think my beer money is safe for now.

OK, now you are doubling the bet! :grin:

You see, it makes NO difference where in the BSM the extra 10m extensions are added as long as the delay occurs. Now, why do I make this claim? Because we already know that once a beam is split, it can be recombined to restore the quantum state of the original beam. (That is not easy to do, but it is feasible.) So the measurement cannot be the point at which the beamsplitter is encountered.

Ultimately, the measurement occurs at the point at which results can no longer be erased and the process of information gain is not reversible (at the quantum level). That is so even though it was the beamsplitter that was the key element in the measurement process. That might not be true for an electron, but it is true for a photon - it must be detected/detectable.

So line up them beers, I'm coming to collect! (Although I don't know where I need to head yet...)

 

[Frame Dragger]

OK, now you are doubling the bet! :grin:

You see, it makes NO difference where in the BSM the extra 10m extensions are added as long as the delay occurs. Now, why do I make this claim? Because we already know that once a beam is split, it can be recombined to restore the quantum state of the original beam. (That is not easy to do, but it is feasible.) So the measurement cannot be the point at which the beamsplitter is encountered.

Ultimately, the measurement occurs at the point at which results can no longer be erased and the process of information gain is not reversible (at the quantum level). That is so even though it was the beamsplitter that was the key element in the measurement process. That might not be true for an electron, but it is true for a photon - it must be detected/detectable.

So line up them beers, I'm coming to collect! (Although I don't know where I need to head yet...)

Can you elaborate on the point of just how that quantum state could reasonably be restored? I'm not disagreeing... I just don't understand.

 

[Demystifier]

Now I have found THE RIGHT solution of the problem,
completely different from my previous one.
It is so simple and obvious that, I am convinced,
everybody will accept it.
The solution consists of several conceptual steps.

0. Forget everything that I said in my previous posts
of this thread!

1. The standard delayed-choice experiment involves
two entangled particles. It cannot be used for FTL
transfer of information because the interference
is encoded in the COINCIDENCES between the entangled particles.
To observe the coincidences, one needs a CLASSICAL COMMUNICATION
between entangled systems, and classical communication
cannot be FTL.

2. Point 1. above is a special case of the general
property of QM: Without classical communication,
entanglement cannot be used to transmit information.
NOT EVEN SLOWER THAN LIGHT FORWARD IN TIME.

3. As we all know from everyday life,
by classical communication, information can be
transfered ONLY FORWARD IN TIME. This is related
to the second law of thermodynamics.

Now let us apply these facts to two variants of the
DrChinese setting.

4. Assume that Charlie does his choice BEFORE Alice and
Bob make their measurements. Can Alice and Bob observe
any consequences of this choice? Yes, but only if
Charlie sends a classical information to Alice and Bob.

5. Now consider a different situation.
Now assume that Charlie does his choice AFTER Alice and
Bob make their measurements. Can Alice and Bob observe
any consequences of this choice? They could if
Charlie could send a classical information to Alice and Bob.
However, Charlie cannot send classical information to the
past. Therefore, Alice and Bob cannot observe
any consequences of the Charlie's choice.

Q.E.D.