Entanglement swapping, monogamy, and realism

[DrChinese]

TL;DR Summary: I have some questions about entanglement swapping, realistic interpretations, and the monogamy entanglement argument that @DrChinese has been making that I want to ask about in an I-level thread.

For this discussion I would like to continue referring to the entangled pairs 1,2 and 3,4 as we have in previous discussions. For background on this discussion refer to post-selection: pre-existing correlations or action at a distance where @DrChinese describes the entanglement swapping experiments and his monogamy argument.

I think the entanglement swapping experiment is compatible with realistic interpretations that respect cause and effect such as argued in this paper http://philsci-archive.pitt.edu/9427/1/Delayed_Choice.pdf (Go to chapter 4 for the argument on the entanglement swapping experiment)

This paper is not disputing entanglement between 1 & 4 in the case where the BSM test is done prior to 1 & 4 being measured. I think it is accurate to say the paper and @DrChinese are in agreement here. And in the context of a cause and effect realistic interpretation, I don't see how anyone could argue for a coincidental explanation for this case, because you can always change the way you measure 1, 4 after knowing the result of the BSM test and so you have a cause and effect problem if you are suggesting a coincidental explanation for this case.

Where I think @DrChinese and the paper disagree is the case where the BSM test is done after 1 & 4 being measured. In a realistic interpretation with cause and effect, the cause can't happen after the effect. So while nobody disputes that 1 & 4 have an entangled correlation, the realistic explanation for the entanglement of 1 & 4 is one that requires coincidence between 1 & 4 versus causation; in other words 1 & 4 happened to have just the right initial states where their measurements resulted in an entangled correlation and a change to 2 & 3 so that the BSM test showed maximum entanglement.

It is my understanding that @DrChinese argument using monogamy of entanglement is in dispute with this second case where the BSM test is done after 1 & 4 being measured. I have read many definitions and gone through the proof that @DrChinese provided and I don't see how this second case where the BSM test is done after 1 & 4 is being measured is in violation of monogamy.

Taking one of the definitions that @DrChinese has previously referred to from https://www.quantiki.org/wiki/monogamy-entanglement
''Monogamy ''' is one of the most fundamental properties of entanglement and can, in its extremal form, be expressed as follows: If two qubits A and B are maximally quantumly correlated they cannot be correlated at all with a third qubit C.

My first question is how could you practically even test this? If monogamy is true in a realistic sense, then by definition it is untestable, since trying to measure a pair of entangled photons changes the entangled pair and as far as we know they would no longer be entangled after the measurement.

The other question, is in the case where the BSM test is done after 1 & 4 being measured, I can argue that 1 only became entangled with 4 after 1 was measured and after 4 was measured which means at no time was 1 entangled with both 2 & 4 and at no time was 4 entangled with both 3 & 1. So even if I did take the most strict view of entanglement of monogamy, it still is not violated in this case.

So please help me understand:
1) Is @DrChinese and his monogamy of entanglement argument in dispute with this view of cause and effect realism?
2) Assuming we are truly in disagreement, how does his monogamy of entanglement argument refute cause and effect realism in the case where the BSM test is done last?

I think @PeterDonis, @Morbert , @Demystifier et al have addressed these questions one way or another. But I will recap:

1. There is no experimental difference when the BSM on 2 & 3 (which is a required event) is performed before OR after the Bell test is performed on 1 & 4. Either way, 1 & 4 entanglement is confirmed by the violation of a Bell Inequality.
2. Accordingly, from the reference, the following is simply incorrect: "Without delayed choice, this has physical significance, because each (A, D) pair actually is in such a state after the (B, C) measurement. But if the (A, D) measurements precede the (B, C) measurement, the (A, D) pair never is in any
   of these states." The 1 & 4 (Egg's A & D) pairs are entangled if and only if a successful BSM occurs, regardless of when they occur. There are no "coincidental" violations of Bell Inequalities.
3. Monogamy of Entanglement applies regardless of when the BSM occurs. The timing defies the usual cause and effect relationship/ordering. You cannot reasonably say "The swap occurred at time X". You can only make the statement that "1 was initially entangled to 2, and was later entangled to 4." But at no time was 1 maximally entangled with both 2 and 4, as that defies MoE.
4. For that to happen means that there was a change in the quantum state. We started with biphoton 1 & 2 and ended with biphoton 1 & 4. A biphoton has temporospatial extent which can exceed the usual Einsteinian limitations.
5. If you are asking about "local realistic" interpretations, clearly those are long excluded due to Bell anyway. Extending the discussion to swapping with independent sources is simply going to fall further flat. If you want to be "realistic", you need to explain why the Bell State Measurement - which according to your position is NOT responsible for the ultimate entanglement of 1 & 4 (being just to identify 1 & 4 pairs with certain characteristics) - requires indistinguishability between 2 & 3. Because you can easily identify those same characteristics WITHOUT them being indistinguishable - and then 1 & 4 are not entangled! If the BSM is not an actual event responsible for the swap, why would they need to be indistinguishable? After all, you can still apply the same "filter" to identify the Bell State. Same 2 clicks.

 

[PeterDonis]

The state of 1&2 after the BSM does change, it becomes a mixed state, because 2 is now entangled with 3 due to the local action at 2&3. Before the BSM, it was a pure state. It means that some of the initial monogamous entanglement between 1&2 "spilled over" to the 2&3 system.

No, not "some" of the monogamous entanglement between 1 & 2 -- all of it, because the state of 2 & 3 after the BSM is maximally entangled. So the state of 1 & 2 after the BSM must be a product state. (And similarly for the state of 3 & 4.)

But even that isn't a complete description, because what happens to 2 & 3 is not the same on every run. There are four possible Bell states of 2 & 3 that can be produced by the BSM, each with equal probability, but only one of them leads to an "event ready" signal that is then used to post-select the runs for which 1 & 4 end up maximally entangled. And even the four Bell states taken all together only get produced on a subset of runs, because particles 2 & 3 don't always arrive at the Bell state measurement device in a narrow enough time window for a BSM to happen at all. So actually, there are four different possible maximally entangled 2 & 3 states, each with its own subset of runs, plus another subset of runs (in practice this is by far the largest one) where nothing happens to 2 & 3 at all.

Which means that, in order to justify the logic in what I quoted from you above (with the correction I gave in the first paragraph of mine above), you have to pick out a particular subensemble of the full ensemble, according to the result of the BSM. And this is exactly the same logic that @DrChinese is using to pick out the subensemble in which 1 & 4 are maximally entangled. So either you agree with @DrChinese, or your own position is inconsistent--you allow the picking of subensembles to have a physical effect when it suits you, but not when it doesn't.

 

[akvadrako]

What people were trying to explain was that selecting a subensemble of that state does not have any physical effect on system 1&4, so the fact that one can pick an entangled subensemble does not imply that the system 1&4 is entangled. Neither cherry picking nor picking by conditioning on a measurement at 2&3 has any physical effect on the 1&4 system. Selecting a subensemble is something that is done long after the experiment has been performed and the measurement results have become manifest and been recorded. The entanglement in the subensembles is spurious and infering causal relationships from them is an instance of Berkson's fallacy, as explained here: https://arxiv.org/abs/1606.04523

I would agree that nothing physical happens to 1&4, which is clear in the post-selection case, especially if one assumes that effects don't travel back in time.

But when you select the subensemble using the swapping procedure before 1&4 are measured, you can use that subensemble to violate Bell inequalities. So it must be entangled; I would say by definition.

 

[Nullstein]

No, not "some" of the monogamous entanglement between 1 & 2 -- all of it, because the state of 2 & 3 after the BSM is maximally entangled. So the state of 1 & 2 after the BSM must be a product state.

After the BSM, the state at 2&3 is given by a mixture of 4 fully entangled states.

But even that isn't a complete description, because what happens to 2 & 3 is not the same on every run. There are four possible Bell states of 2 & 3 that can be produced by the BSM, each with equal probability, and even that only happens on a subset of runs, because particles 2 & 3 don't always arrive at the Bell state measurement device in a narrow enough time window for a BSM to happen at all. So actually, there are four different possible maximally entangled 2 & 3 states, each with its own subset of runs, plus another subset of runs (in practice this is by far the largest one) where nothing happens to 2 & 3 at all.

This is why the full ensemble is in a mixture. This mixture is what is relevant for the physical considerations.

Which means that, in order to justify the logic in what I quoted from you above, you have to pick out a particular subensemble of the full ensemble, according to the result of the BSM. And this is exactly the same logic that @DrChinese is using to pick out the subensemble in which 1 & 4 are maximally entangled. So either you agree with @DrChinese, or your own position is inconsistent--you allow the picking of subensembles to have an effect when it suits you, but not when it doesn't.

No, I'm never picking subensembles. I'm only talking about the full ensemble and calculate states of subsystems.

Before the BSM, the full ensemble is given by
$$\rho_{1234,\text{before}}=\rho_{12}\otimes\rho_{34}$$
After the BSM, the full ensemble is given by
$$\rho_{1234,\text{after}}=\sum_i (\mathbb 1_1\otimes P_i\otimes\mathbb 1_4)\rho_{1234,\text{before}}(\mathbb 1_1\otimes P_i\otimes\mathbb 1_4)$$

The states of the subsystems are given by
$$\rho_{ij,\text{after}}=\mathrm{tr}_{kl}(\rho_{1234,\text{after}})$$
where $(i,j,k,l)$ is a permutation of $(1,2,3,4)$. It's just a mathematical exercise to compute them. I agree with whatever the result of that computation is.

 

[Nullstein]

I would agree that nothing physical happens to 1&4, which is clear in the post-selection case, especially if one assumes that effects don't travel back in time.

But when you select the subensemble using the swapping procedure before 1&4 are measured, you can use that subensemble to violate Bell inequalities. So it must be entangled; I would say by definition.

I agree that the subensembles are entangled. I'm just saying that no conclusion about causal relationships can be inferred from the subensembles unless one can exclude Berkson's fallacy, which we can't in the case of entanglement swapping.

 

[DrChinese]

The more precise statement would have been: "The measurement on the ensemble of particle pairs (23) allows us to select a subensemble of particle pairs (14) which show entanglement."
...

This gives an utterly false impression that during each iteration involving four particles, the measurement performed on two particles changes something about the other two. ...

The only thing we know that factually happens in these experiments is that for each iteration of 4 particles, the measurement result from two of them is used later to decide whether or not the other two belong to the desired sub-ensemble.

This is inaccurate. A BSM absolutely changes the state of remote 1 & 4. That's the essence of our disagreement.

Each and every time there are 2 clicks at the BSM signaling the arrival of photons 2 & 3 within the specified time window, then the corresponding 1 & 4 will be cast into an maximally entangled state (forming a different biphoton). Which of the 4 possible Bell states are identified varies from experiment to experiment, but there is always entanglement - as long as the event does not allow 2 to be distinguished from 3.

And that is another fly in the ointment to this whole debate. If all you are doing is "selecting" whether it is psi+ or psi- entanglement, why require indistinguishability? You can plainly see all of the same characteristics (DoFs) - even if they are distinguishable. Transmit or reflect at the beam splitter, H or V at the PBS, identical wavelengths, and within the time window. That's it. So why do the same selection parameters fail when there is distinguishability? Answer: because a successful BSM IS an event, and it remotely changes the quantum state of both 1 and 4. Without the successful BSM, there is no entanglement between any 1 & 4 pairs.

 

[lodbrok]

This is inaccurate. A BSM absolutely changes the state of remote 1 & 4. That's the essence of our disagreement.

There's no experimental evidence for this in any experiment done to date. You are not disagreeing with me. You are disagreeing with the experimental record.

 

[Nullstein]

This is inaccurate. A BSM absolutely changes the state of remote 1 & 4. That's the essence of our disagreement.

Nope. The state of the system 1&4 is given by $\rho_{14,\text{before}}=\rho_{14,\text{after}}=\frac 1 4 \mathbb 1\otimes\mathbb 1$. It is not entangled.

And that is another fly in the ointment to this whole debate. If all you are doing is "selecting" whether it is psi+ or psi- entanglement, why require indistinguishability?

Nobody requires indistinguishability. We just write down the prepared state, compute its time evolution according to the axioms of QM and compute the statistical distributions that arise from the resulting state. All of this is interpretation-independent math.

 

[vanhees71]

No, that's exactly the opposite of what people said in the previous threads. The system 1&4 is always non-entangled, both before and after the BSM at 2&3. The state of the 1&4 system is given by:
$$\rho_{14,\text{before}}=\rho_{14,\text{after}}=\frac 1 4\mathbb 1\otimes\mathbb 1$$
This state is a product state, so it is not entangled.

What people were trying to explain was that selecting a subensemble of that state does not have any physical effect on system 1&4, so the fact that one can pick an entangled subensemble does not imply that the system 1&4 is entangled. Neither cherry picking nor picking by conditioning on a measurement at 2&3 has any physical effect on the 1&4 system. Selecting a subensemble is something that is done long after the experiment has been performed and the measurement results have become manifest and been recorded. The entanglement in the subensembles is spurious and infering causal relationships from them is an instance of Berkson's fallacy, as explained here: https://arxiv.org/abs/1606.04523

The fact that $\rho_{14,\text{before}}=\rho_{14,\text{after}}=\frac 1 4\mathbb 1\otimes\mathbb 1$ also proves that the argument of DrChinese based on monogamy of entanglement is erroneous. Since the system is not entangled, it is in particular not monogamously entangled, so the very premise of DrChinese's argument is violated.

In the subensemble chosen by projecting (23) to one of the four Bell states (14) are in a Bell state too, i.e., they are entangled. That's why this is called enranglement swapping!

 

[Nullstein]

In the subensemble chosen by projecting (23) to one of the four Bell states (14) are in a Bell state too, i.e., they are entangled. That's why this is called enranglement swapping!

I agree. The subensembles are entangled, but the correlations that arise from this entanglement are spurious, i.e. not causal. Inferring a causal relationship from the correlations is a statistical fallacy called Berkson's paradox, as explained in the Spekkens paper I cited above. So it is illegal to conclude any non-local cause and effect relationship from the appearance of correlations in the subensembles.

 

[Simple question]

What people were trying to explain was that selecting a subensemble of that state does not have any physical effect on system 1&4,

...and failed. At preparation there is no entanglement (I think you agree with this), at measurement there is some. I think most physicist consider entanglement "physical".

so the fact that one can pick an entangled subensemble does not imply that the system 1&4 is entangled.

That sub ensemble cannot exist because 0% of the 1 & 4 pair are entangled. Furthermore 100% of 1 & 4 are fully entangled with another partner (2 & 3 repetitively). Basic QM.

Neither cherry picking nor picking by conditioning on a measurement at 2&3 has any physical effect on the 1&4 system.

Yet one sub ensemble is picked, while it "could not exist". Most people call this a change.

Selecting a subensemble is something that is done long after the experiment has been performed and the measurement results have become manifest and been recorded.

That's incorrect. The sub ensemble is selected/picked at BSM site. Whatever the place and time.
Furthermore it can ONLY be selected there. This is called a swapping.

The entanglement in the subensembles is spurious and infering causal relationships from them is an instance of Berkson's fallacy, as explained here: https://arxiv.org/abs/1606.04523

Why not calling it lucky entanglement than ? What are the chances that anyone, anytime, can pickup that (or those ?) sub-ensemble ? Show me the math.

The fact that $\rho_{14,\text{before}}=\rho_{14,\text{after}}=\frac 1 4\mathbb 1\otimes\mathbb 1$ also proves that the argument of DrChinese based on monogamy of entanglement is erroneous. Since the system is not entangled, it is in particular not monogamously entangled, so the very premise of DrChinese's argument is violated.

The system is completely entangled, contrary to your claim. It is prepared as such. Just not 1&4, but 1&2 and 3&4.

Can we cut to the chase ? Are you claiming that monogamy of entanglement is wrong ? Are you claiming that swapping is a fluke ? You really need reference here.

So it is illegal to conclude any non-local cause and effect relationship from the appearance of correlations in the subensembles.

Well, it will be fun to see the Quantum police arrest the people using those "appearance" (not change of course) to secure your communication.

 

[DrChinese]

1. ... especially if one assumes that effects don't travel back in time.

2. But when you select the subensemble using the swapping procedure before 1&4 are measured, you can use that subensemble to violate Bell inequalities. So it must be entangled; I would say by definition.

1. How do you not see that this is circular reasoning? Swapping is proof (by counterexample) that the quantum mechanical world does NOT obey Einsteinian causality.

2. There is no difference in outcomes - or description of what is occurring - regardless of ordering. That is an experimental fact. There are NO 1 & 4 pairs that are entangled - NO SUBSET or SUBENSEMBLE - unless a BSM occurs.

You must realize that you are saying that there were SOME such entangled 1 & 4 pairs - even if the experimenter insured no BSM occurred. But that actually means that every entangled photon - everywhere, created anytime in all of history - was potentially also entangled with those same 1 & 4 pairs. Because what otherwise connects them? The 1 & 4 photons are created from independent distant sources at completely different times! That same requirement (which is no requirement at all) means any photon could be similarly entangled with either 1 or 4 - if you say that there existed entanglement between 1 & 4 for some subensemble prior to the swap. (Keep in mind that 1 & 4 need not have ever even co-existed for a swap to succeed. And they still be swapped either before or after their Bell test.)

But no, that's not what happens. The swap is a necessary event to the results. It is not "merely" selection.

 

[Nullstein]

...and failed. At preparation there is no entanglement (I think you agree with this), at measurement there is some. I think most physicist consider entanglement "physical".

No I don't agree with this. Entanglement between which systems? There is entanglement between 1&2 and 3&4 before the BSM. There is no entanglement between 1&4 before the BSM and there is no entanglement between 1&4 after the BSM.

That sub ensemble cannot exist because 0% of the 1 & 4 pair are entangled. Furthermore 100% of 1 & 4 are fully entangled with another partner (2 & 3 repetitively). Basic QM.

False. The 1&4 system is in the state $\frac 1 4\mathbb 1\otimes\mathbb 1$. Let $P_i$ be projectors onto the Bell basis. The completeness relation says that $\sum_i P_i=\mathbb 1\otimes\mathbb 1$. So we conclude that $\rho_{14,\text{before}}=\sum_i \frac 1 4 P_i$, i.e. a statistical mixture of 4 entangled states.

That's incorrect. The sub ensemble is selected/picked at BSM site. Whatever the place and time.
Furthermore it can ONLY be selected there. This is called a swapping.

Nope, the post-selection is done once the full data is collected.

Show me the math.

See above.

The system is completely entangled, contrary to your claim. It is prepared as such. Just not 1&4, but 1&2 and 3&4.

$\frac 1 4\mathbb 1\otimes\mathbb 1$ is a product state, so not entangled.

Can we cut to the chase ? Are you claiming that monogamy of entanglement is wrong ? Are you claiming that swapping is a fluke ? You really need reference here.

Nope, I'm claiming none of it.

 

[DrChinese]

1. The state of the system 1&4 is given by $\rho_{14,\text{before}}=\rho_{14,\text{after}}=\frac 1 4 \mathbb 1\otimes\mathbb 1$. It is not entangled.

2. Nobody requires indistinguishability. We just write down the prepared state, compute its time evolution according to the axioms of QM and compute the statistical distributions that arise from the resulting state. All of this is interpretation-independent math.

1. You have really gone off the deep end. After a swap, 1 & 4 are maximally entangled and are not in a Product state. Particles in a Product state do not violate Bell inequalities, and if you don't follow this point, you need to go back a step or two.

2. "Indistinguishability" is an absolute requirement for a swap. There are many ways to distinguish the source for the 2 clicks at the BSM. All of them must be eliminated for the swap to succeed. For example: you need to place a filter from each source to insure the wavelengths of the 2 & 3 photons are the same - otherwise that would distinguish them. They need to arrive close together - otherwise that would distinguish them. And if you placed a polarizer too early in the path, that could distinguish them.

So this requirement - indistinguishability - has no counterpart in the selection criteria being thrown around. Because every other element of the selection criteria can be met and there will be no swap. This requirement is a strictly quantum thing, and the same requirement exists when entangled pairs are created via PDC using BBo crystals. If you know the source of each photon, they won't be entangled. The output cones must be overlapped to hide the source.

 

[PeterDonis]

There's no experimental evidence for this in any experiment done to date.

You cannot assert this as a fact, since this is precisely what the discussion is about: whether, and under what interpretations, the entanglement swapping experiments provide such evidence.

 

[Nullstein]

1. You have really gone off the deep end. After a swap, 1 & 4 are maximally entangled and are not in a Product state. Particles in a Product state do not violate Bell inequalities, and if you don't follow this point, you need to go back a step or two.

I have presented the math and I doubt that anyone other than you (not even PeterDonis) disagrees with the fact the full ensemble, after the BSM, is given by $\frac 1 4\mathbb 1\otimes\mathbb 1$. You are in disagreement with the basic math of QM.

It is correct that, after the BSM, the full data of 1&4 will not violate a Bell inequality in a real experiment. Only the post-selected subensembles will violate it.

 

[PeterDonis]

You are in disagreement with the basic math of QM.

No, we are in disagreement about what the meaning of particular subensembles of the full ensemble is. You have agreed that the 2 & 3 full ensemble is a mixture of entangled subensembles. The 1 & 4 full ensemble is similarly a mixture of entangled subensembles--the same subensembles in both cases, comprising the same subsets of the experimental runs. You are saying that, nevertheless, the physical meaning of the subensembles is different: the 1 & 4 subensembles being entangled has no physical meaning, while the 2 & 3 subensembles being entangled does. @DrChinese and I disagree: we think the entanglement of the subensembles has physical meaning in both cases.

I don't think this disagreement is going to be resolved by further discussion; at this point everyone has stated their positions many, many times.

 

[Nullstein]

You are saying that, nevertheless, the physical meaning of the subensembles is different: the 1 & 4 subensembles being entangled has no physical meaning, while the 2 & 3 subensembles being entangled does.

No, I'm not saying that. A subensemble is entangled if and only if it is not a product state. The 1&4 subensembles are entangled, I said this many times. The actual question we are discussing is whether one can draw causal conclusions from that fact. And because of Berkson's fallacy, this is illegal in this case, as explained in the Spekkens paper I cited earlier. So even though some 1&4 subensembles are entangled, we can not conclude that the BSM had a causal effect on the 1&4 system.

But DrChineses misunderstanding is one step earlier. Apparently he believes that the full ensemble of 1&4 is entangled, not only the subensembles. In fact, from what he has written so far, I suspect that he doesn't even recognize the difference between a full ensemble and a subensemble.

 

[PeterDonis]

The 1&4 subensembles are entangled, I said this many times. The actual question we are discussing is whether one can draw causal conclusions from that fact.

That is just another way of saying that we disagree on the physical meaning of the entangled subensembles. You are saying we can't draw causal conclusions from the 1 & 4 subensembles being entangled. We are saying we can. But you can't be saying that we can't draw causal conclusions from the 2 & 3 subensembles being entangled--unless you want to maintain that the BSM on 2 & 3 does not cause 2 & 3 to be entangled, and you don't seem to be claming that. So you are treating the two entangled subensembles (1 & 4 vs. 2 & 3) differently. We are not.

 

[DrChinese]

1. I have presented the math and I doubt that anyone other than you (not even PeterDonis) disagrees with the fact the full ensemble, after the BSM, is given by $\frac 1 4\mathbb 1\otimes\mathbb 1$. You are in disagreement with the basic math of QM.

2. It is correct that, after the BSM, the full data of 1&4 will not violate a Bell inequality in a real experiment. Only the post-selected subensembles will violate it.

1. No! We (@vanhees71, @PeterDonis and many others) have previously established as follows (in another thread you were in):

a. That the initial quantum state is a Product State of 2 entangled pairs: $$\hat{\rho}=\hat{\rho}_{12} \otimes \hat{\rho}_{34}.$$

b. The final quantum state is a Product State of 2 entangled pairs:$$\hat{\rho}'=\hat{\rho}_{23} \otimes \hat{\rho}_{14}.$$

c. And it should be obvious that: $$\hat{\rho}_{12} \otimes \hat{\rho}_{34} ≠ \hat{\rho}_{23} \otimes \hat{\rho}_{14}.$$

2. The full set of 1 & 4 will be entangled if and only if there is a successful BSM. But the issue here is that most of the 1 and 4 photons cannot be placed together as pairs. That is because they are created at random times by fully independent sources. So they are not otherwise synced or forced to pair up. Occasionally the sources fire in such sequence that the 2 & 3 photons arrive within a narrow time window and a swap can occur. The partner 1 & 4 photons can be identified in these cases, and all of them will be entangled. The others are ignored precisely because each cannot be matched to anything.

In principle, there are no 1 & 4 pairs detected that are not entangled (again assuming the BSM apparatus is running). 1 & 4 pairs are identified by having a specific difference in their arrival times. There will be many more lone 1s, and many more lone 4s. You don't actually need the BSM to identify them. Again, this is in principle. In practice, the issue is that you cannot tell which Bell state they arrive in. You need that to run the Bell test.

 

[lodbrok]

You cannot assert this as a fact, since this is precisely what the discussion is about: whether, and under what interpretations, the entanglement swapping experiments provide such evidence.

Then please point to the experimental evidence that the measurement of a specific particle pair (23) absolutely changes the physical situation of the other particle pair (14) in a given run of the experiment.

The delayed-choice swapping experiment shows that the measurement results for (14) are recorded well in advance of the (23) measurement. And those do not change after the (23) measurement is performed. This is evidence contradicting the suggestion that the (23) measurement changes the physical situation at (14).

This is not an interpretation issue as you explained previously:

The obvious issue with any such interpretation is that the actually observed experimental behavior (which agrees with the QM predictions) is the same regardless of the time ordering of the measurements. Any interpretation that attempts to preserve "cause and effect" therefore has to argue that, even though the actual behavior is exactly the same, which measurement is the "cause" and which is the "effect" depends on the order in which you do the measurements. Such an interpretation seems strained, to say the least.

 

[Nullstein]

That is just another way of saying that we disagree on the physical meaning of the entangled subensembles. You are saying we can't draw causal conclusions from the 1 & 4 subensembles being entangled. We are saying we can.

Well, but then you are subject to Berkson's fallacy and I haven't seen you or DrChinese address this argument so far.

But you can't be saying that we can't draw causal conclusions from the 2 & 3 subensembles being entangled--unless you want to maintain that the BSM on 2 & 3 does not cause 2 & 3 to be entangled, and you don't seem to be claming that. So you are treating the two entangled subensembles (1 & 4 vs. 2 & 3) differently. We are not.

No, I'm not treating anything differently. The BSM causes the evolution $\rho_{1234,\text{before}}\rightarrow\rho_{1234,\text{after}}$. It thus also causes the evolution $\rho_{ij,\text{before}}\rightarrow\rho_{ij,\text{after}}$. Some of these subsystems evolve, others don't. The 1&4 subsystem doesn't evolve, because $\rho_{14,\text{before}}=\rho_{14,\text{after}}$. I haven't done the calculation for the 2&3 subsystem, but I will agree to whatever the result of that calculation is.

 

[PeterDonis]

Then please point to the experimental evidence that the measurement of a specific particle pair (23) absolutely changes the physical situation of the other particle pair (14) in a given run of the experiment.

I already answered that. You can say you disagree with the interpretation @DrChinese and I are adopting (and instead agree with an interpretation like the one @Nullstein has been defending). But you can't just stick your fingers in your ears and ignore the whole discussion. If you continue to do that, you will receive a warning and be banned from further posting in this thread.

 

[StevieTNZ]

Then please point to the experimental evidence that the measurement of a specific particle pair (23) absolutely changes the physical situation of the other particle pair (14) in a given run of the experiment.

The delayed-choice swapping experiment shows that the measurement results for (14) are recorded well in advance of the (23) measurement. And those do not change after the (23) measurement is performed. This is evidence contradicting the suggestion that the (23) measurement changes the physical situation at (14).

This is not an interpretation issue as you explained previously:

I would say systems 1 and 4 are still in a superposition state when 'measured' - because the system entangles with the apparatus - no 'definite' state (ie polarization) arises, as per the linearity of quantum mechanics.

 

[PeterDonis]

This is not an interpretation issue as you explained previously:

What you quoted from me only says that, whatever interpretation we adopt, it should be the same regardless of the time ordering of the 2&3 vs. 1&4 measurements, since both the observed results and the QM math are the same in both cases. But that, in itself, does not tell us which intepretation to adopt. Both the interpretation @DrChinese and I are using, and the intepretation @Nullstein is using, are independent of the time ordering of the measurements. So pointing out that the interpretation should be independent of the time ordering of the measurements does not give any way of choosing between those two positions. Which means it doesn't give any argument either way about whether or not the BSM on 2&3 has an actual physical effect on 1&4.

 

[PeterDonis]

I'm not treating anything differently.

In the particular math you go on to quote, you're not--but only because you're ignoring the result of the BSM and just looking at the full ensemble. But you also agree that that full ensemble, in the case of 2&3, is a mixture of entangled subensembles. Are you then claming that, once we know that in a particular subensemble, 2&3 are in a particular maximally entangled state (the one indicated by the result of the BSM in that subensemble), that doesn't tell us that the BSM had an actual physical effect on 2&3?

 

[PeterDonis]

then you are subject to Berkson's fallacy

So you say. But you have not proven it.

I haven't seen you or DrChinese address this argument so far.

It's not up to us to disprove that Berkson's fallacy is in fact the correct explanation for the entangled 1&4 subensembles. It's up to you to prove it if you are going to assert it. Saying that it is possible (because of general heuristics about correlations not necessarily showing causation) is not the same as proving it in this particular case. In all of the other cases you cited in previous threads, Berkson's fallacy was proven by looking at additional relevant data that ruled out actual causal effects. You need to do that for the case under discussion if you want to assert that that is the correct explanation.

 

[lodbrok]

What you quoted from me only says that, whatever interpretation we adopt, it should be the same regardless of the time ordering of the 2&3 vs. 1&4 measurements, since both the observed results and the QM math are the same in both cases. But that, in itself, does not tell us which intepretation to adopt. Both the interpretation @DrChinese and I are using, and the intepretation @Nullstein is using, are independent of the time ordering of the measurements. So pointing out that the interpretation should be independent of the time ordering of the measurements does not give any way of choosing between those two positions. Which means it doesn't give any argument either way about whether or not the BSM on 2&3 has an actual physical effect on 1&4.

The experimental fact is that the already recorded (14) results never change. It's not simply an issue of time ordering. The (14) results will be the same even if the (23) measurement is never done. If the results do not change, there is no physical effect. It doesn't matter what interpretation you use. This is the experimental fact I was referring to and it has nothing to do with interpretation.

 

[akvadrako]

Well, but then you are subject to Berkson's fallacy and I haven't seen you or DrChinese address this argument so far.

Are you saying that Berkson's fallacy is (potentially) applicable to correlations resulting from all examples of entanglement and hence that one can not draw conclusions about causality from entanglement in general?

Or just the entangled 1&4 sub-ensemble in this case? That would be odd as entanglement should be a fungible resource.

 

[Nullstein]

1. No! We (@vanhees71, @PeterDonis and many others) have previously established as follows (in another thread you were in):

You don't understand what has been established, because you don't understand the difference between the full ensemble and the subensembles. I will try to explain it once more.

b. The final quantum state is a Product State of 2 entangled pairs:$$\hat{\rho}'=\hat{\rho}_{23} \otimes \hat{\rho}_{14}.$$

That depends on whether we talking about the full ensemble or the subensembles. What you have written here is the state of a subensemble after the BSM. The state of the full ensemble after the BSM is not of that form.

c. And it should be obvious that: $$\hat{\rho}_{12} \otimes \hat{\rho}_{34} ≠ \hat{\rho}_{23} \otimes \hat{\rho}_{14}.$$

You are comparing a full ensemble (before) with a subensemble (after). This makes no sense at all.

 

[PeterDonis]

The experimental fact is that the already recorded (14) results never change.

That's true of every experimental result. It tells you nothing special about any particular experiment.

The (14) results will be the same even if the (23) measurement is never done.

Here you are making assertions without evidence. Since the (23) measurement was done in the actual experiment, you cannot use that experiment as evidence about what would happen if the (23) measurement were not done.

You can assert that the statistics of the full ensemble of (14) measurements, taken in isolation, would be the same whether the (23) measurement was done or not, as a prediction of QM. But that prediction is only about statistics; it says nothing about the actual results of individual runs.

 

[PeterDonis]

You are comparing a full ensemble (before) with a subensemble (after).

No, he isn't. By construction, the "before" state of any subensemble of runs is the same as that of the full ensemble, because the preparation process is identical for every single run.

 

[Nullstein]

In the particular math you go on to quote, you're not--but only because you're ignoring the result of the BSM and just looking at the full ensemble. But you also agree that that full ensemble, in the case of 2&3, is a mixture of entangled subensembles. Are you then claming that, once we know that in a particular subensemble, 2&3 are in a particular maximally entangled state (the one indicated by the result of the BSM in that subensemble), that doesn't tell us that the BSM had an actual physical effect on 2&3?

The BSM is responsible for the transitions $\rho_{i_1,\ldots,i_n,\text{before}}\rightarrow\rho_{i_1,\ldots,i_n,\text{after}}$. The selection of subensembles is not a physical process. It happens long after the measurement results have become manifest and is performed by the experimenter on his computer when he analyzes the recorded data. Physical cause and effect relationships must be inferred from the evolution of the full ensembles.

 

[Nullstein]

So you say. But you have not proven it.

Well, I have provided a citation to a paper by Spekkens, where it is proved very concretely for the swapping situation. (I found that recently, so you may not have seen it yet.) In any case, the argument is that the measurement result at 2&3 is influenced by the preparation of 2 and 3, so it is a common effect and thus Berkson's effect applies.

It's not up to us to disprove that Berkson's fallacy is in fact the correct explanation for the entangled 1&4 subensembles.

If you are confronted with a proof, it is up to you to point out an error in the proof though.

You need to do that for the case under discussion if you want to assert that that is the correct explanation.

I'm not saying it is the correct explanation. I'm saying it is one viable explanation and thus a knowledge-based interpretation is viable. An interpretation with non-local cause and effect relationships is of course viable too, but not required.

 

[Nullstein]

Are you saying that Berkson's fallacy is (potentially) applicable to correlations resulting from all examples of entanglement and hence that one can not draw conclusions about causality from entanglement in general?

No, it is only applicable if conclusions are to be drawn from the correlations in subensembles. Entanglement produced by, e.g., parametric down-conversion occurs in the full ensemble and is genuinely mysterious.