Entanglement swapping, monogamy, and realism

[kurt101]

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?

 

[PeterDonis]

I think the entanglement swapping experiment is compatible with realistic interpretations that respect cause and effect

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.

 

[PeterDonis]

My first question is how could you practically even test this?

By preparing two qubits in a maximally entangled state and preparing a bunch of other qubits in whatever state you like, and verifying, by a large number of repeated trials with identical preparations, that the two qubits you prepared in the maximally entangled state show the appropriate correlations with each other, and no correlations with any of the other qubits.

 

[PeterDonis]

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

Perhaps you can, but that is not the argument others were making in the previous threads on this topic. They were arguing that 1 & 4 were entangled the whole time.

 

[Demystifier]

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)

Realistic interpretations are compatible with cause and effect, but not with locality. This means that causal effects in realistic interpretations can relate events which are spatially separated. The reference to particles that no longer exist is rather naive, because in theories (typically QFT's) in which particles may be destroyed there is no destruction of fundamental degrees of freedom, meaning that something remains even when the particles are destroyed. For example, if a photon is absorbed by an atom, then the atom takes information that was carried by the photon.

 

[PeterDonis]

Where I think @DrChinese and the paper disagree is the case where the BSM test is done after 1 & 4 being measured.

While the paper does say that the BSM test does not cause the entanglement of 1 & 4 in this case, it offers no alternative explanation of what does cause the entanglement of 1 & 4.

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

But how do 1 & 4 become entangled? If it isn't the BSM measurement (which you say it isn't in this case), what is it? You offer no alternative explanation, any more than the paper does (see above).

 

[PeterDonis]

Realistic interpretations are compatible with cause and effect, but not with locality.

What they're not compatible with is not just "locality" but time ordering. In the case under discussion, a BSM measurement, on the realist interpretation @DrChinese has been using, is the reason for photons 1 & 4 being entangled, even if the BSM measurement event is in the future light cone of the 1 & 4 measurement events.

 

[PeterDonis]

in theories (typically QFT's) in which particles may be destroyed there is no destruction of fundamental degrees of freedom, meaning that something remains even when the particles are destroyed.

This is an interesting possibility to explore as far as interpretations are concerned, but I am not aware of any significant exploration of it in the literature. Do you know if anyone has tried to construct a realist interpretation along these lines? It wouldn't make the time ordering issue I described in post #7 any less weird, but it might at least offer a different viewpoint on this type of scenario.

 

[Demystifier]

What they're not compatible with is not just "locality" but time ordering. In the case under discussion, a BSM measurement, on the realist interpretation @DrChinese has been using, is the reason for photons 1 & 4 being entangled, even if the BSM measurement event is in the future light cone of the 1 & 4 measurement events.

Nonlocal realistic theories typically keep time ordering but violate Lorentz covariance. This means that time ordering and causality is not formulated in terms of light cones.

 

[Demystifier]

This is an interesting possibility to explore as far as interpretations are concerned, but I am not aware of any significant exploration of it in the literature. Do you know if anyone has tried to construct a realist interpretation along these lines? It wouldn't make the time ordering issue I described in post #7 any less weird, but it might at least offer a different viewpoint on this type of scenario.

Of course, Bohmian QFT is well developed in the literature. Perhaps nobody analyzed this specific experiment with Bohmian QFT, but in principle it's easy once you understand the general principles of Bohmian QFT.

 

[PeterDonis]

Perhaps nobody analyzed this specific experiment with Bohmian QFT, but in principle it's easy once you understand the general principles of Bohmian QFT.

Could you give a brief outline of how such an analysis would go?

 

[Demystifier]

Could you give a brief outline of how such an analysis would go?

The most difficult part is to first outline how such analysis would go with standard QFT. If you can present this part by yourself, in a way that you understand well, then I can easily explain to you how your analysis with standard QFT modifies with Bohmian QFT.

 

[vanhees71]

Perhaps you can, but that is not the argument others were making in the previous threads on this topic. They were arguing that 1 & 4 were entangled the whole time.

This is at best misleading. In the entanglement-swapping experiment the initial state is
$$\hat{\rho}=\hat{\rho}_{12} \otimes \hat{\rho}_{34},$$
i.e., photons 1 and 4 are not entangled, no matter in which state the photon pairs (12) and (34) are, because the (reduced) state of the pair (14) factorizes
$$\hat{\rho}_{14}=\mathrm{Tr}_{23} \hat{\rho}=\hat{\rho}_1 \otimes \hat{\rho}_2.$$
If now $\hat{\rho}_{12}$ and $\hat{\rho}_{34}$ are Bell states, i.e., each of the pairs is maximally entangled you can make the pair (14) maximally entangled by projecting the pair (34) to a Bell state, i.e., you select a subensemble from the original ensemble, for which (14) is in a Bell state.

The projection of (34) to a Bell state is, however, not the cause for the entanglement of (14), which can be ensured, if the measurement events ("clicks of detectors") on photons 1 and 4 are spacelike separated from the projection measurement on (34). At least this is, as far as I know, the consensus in the quantum-optics community, and formally that's well justified due to the validity of the microcausality constraint of standard QED, which correctly predicts the outcome of this and all other Bell-test/EPR-type measurements.

The possibility for entanglement swapping is thus due to the entanglement of photon pairs (12) and (34) in the initial state. Of course, to which of the Bell states you project, is completely random. In the most simple case one only selects the polarization-singlet state, because this can be most simply realized by simply selecting those events, when photons 2 and 3 end up at different detectors behind the PBS. This happens completely randomly with a probability of 25%.

 

[Morbert]

This is at best misleading. In the entanglement-swapping experiment the initial state is
$$\hat{\rho}=\hat{\rho}_{12} \otimes \hat{\rho}_{34},$$
i.e., photons 1 and 4 are not entangled, no matter in which state the photon pairs (12) and (34) are, because the (reduced) state of the pair (14) factorizes
$$\hat{\rho}_{14}=\mathrm{Tr}_{23} \hat{\rho}=\hat{\rho}_1 \otimes \hat{\rho}_2.$$
If now $\hat{\rho}_{12}$ and $\hat{\rho}_{34}$ are Bell states, i.e., each of the pairs is maximally entangled you can make the pair (14) maximally entangled by projecting the pair (34) to a Bell state, i.e., you select a subensemble from the original ensemble, for which (14) is in a Bell state.

The projection of (34) to a Bell state is, however, not the cause for the entanglement of (14), which can be ensured, if the measurement events ("clicks of detectors") on photons 1 and 4 are spacelike separated from the projection measurement on (34). At least this is, as far as I know, the consensus in the quantum-optics community, and formally that's well justified due to the validity of the microcausality constraint of standard QED, which correctly predicts the outcome of this and all other Bell-test/EPR-type measurements.

The possibility for entanglement swapping is thus due to the entanglement of photon pairs (12) and (34) in the initial state. Of course, to which of the Bell states you project, is completely random. In the most simple case one only selects the polarization-singlet state, because this can be most simply realized by simply selecting those events, when photons 2 and 3 end up at different detectors behind the PBS. This happens completely randomly with a probability of 25%.

I take it there are some typos here? Do you mean

This is at best misleading. In the entanglement-swapping experiment the initial state is
$$\hat{\rho}=\hat{\rho}_{12} \otimes \hat{\rho}_{34},$$
i.e., photons 1 and 4 are not entangled, no matter in which state the photon pairs (12) and (34) are, because the (reduced) state of the pair (14) factorizes
$$\hat{\rho}_{14}=\mathrm{Tr}_{23} \hat{\rho}=\hat{\rho}_1 \otimes \hat{\rho}_4.$$
If now $\hat{\rho}_{12}$ and $\hat{\rho}_{34}$ are Bell states, i.e., each of the pairs is maximally entangled you can make the pair (14) maximally entangled by projecting the pair (23) to a Bell state, i.e., you select a subensemble from the original ensemble, for which (14) is in a Bell state.

The projection of (23) to a Bell state is, however, not the cause for the entanglement of (14), which can be ensured, if the measurement events ("clicks of detectors") on photons 1 and 4 are spacelike separated from the projection measurement on (23). At least this is, as far as I know, the consensus in the quantum-optics community, and formally that's well justified due to the validity of the microcausality constraint of standard QED, which correctly predicts the outcome of this and all other Bell-test/EPR-type measurements.

 

[Morbert]

Consider the statement
"The subensemble selected by an outcome of the measurement on pair (23) will show entanglement in the pair (14). Given this same subensemble, if the physicist had, for these specific runs, decided not to perform any measurement on (23), there would be no correlation in (14) in these runs"

This is presumably the sense in which people discuss a measurement on (23) doing something to (14) even with spacelike separations. But QM on its own (i.e. without some additional interpretation-dependent commitment) doesn't let us conclude this. It doesn't let us evaluate the above statement as true.

 

[Demystifier]

The projection of (34) to a Bell state is, however, not the cause for the entanglement of (14)

I would like to reformulate the question: can projection ever cause entanglement? I will argue that it can, in a simpler context that does not involve entanglement swapping, but something simpler.

Let initial state be
$$|\Psi\rangle=|\psi\rangle|\psi\rangle$$
where $|\psi\rangle$ is a 1-particle state
$$|\psi\rangle=|L\rangle+|R\rangle .$$
Here $|L\rangle$ and $|R\rangle$ are some orthogonal 1-particle states, and for simplicity I don't care about normalization factors. Clearly, in the initial state $|\Psi\rangle$ there is no entanglement at all. However, $|\Psi\rangle$ can be written as
$$|\Psi\rangle=|{\rm Hardy}\rangle + |R\rangle|R\rangle \;\;\;\;\; (1)$$
where
$$|{\rm Hardy}\rangle=|L\rangle|L\rangle+|L\rangle|R\rangle+|R\rangle|L\rangle$$
is the Hardy state. The Hardy state is entangled. Hence, if we project $|\Psi\rangle$ onto the first term in (1), we get entangled state by projection. Since there was no entanglement at all in the initial state, it seems clear that entanglement was created by projection.

 

[martinbn]

I would like to reformulate the question: can projection ever cause entanglement? I will argue that it can, in a simpler context that does not involve entanglement swapping, but something simpler.

Let initial state be
$$|\Psi\rangle=|\psi\rangle|\psi\rangle$$
where $|\psi\rangle$ is a 1-particle state
$$|\psi\rangle=|L\rangle+|R\rangle .$$
Here $|L\rangle$ and $|R\rangle$ are some orthogonal 1-particle states, and for simplicity I don't care about normalization factors. Clearly, in the initial state $|\Psi\rangle$ there is no entanglement at all. However, $|\Psi\rangle$ can be written as
$$|\Psi\rangle=|{\rm Hardy}\rangle + |R\rangle|R\rangle \;\;\;\;\; (1)$$
where
$$|{\rm Hardy}\rangle=|L\rangle|L\rangle+|L\rangle|R\rangle+|R\rangle|L\rangle$$
is the Hardy state. The Hardy state is entangled. Hence, if we project $|\Psi\rangle$ onto the first term in (1), we get entangled state by projection. Since there was no entanglement at all in the initial state, it seems clear that entanglement was created by projection.

But the projection will not result in that state 100% of the trials. So the entanglement will be of a subensemble. Your example is, in this respect, not different than the swapping.

 

[lodbrok]

Consider the statement
"The subensemble selected by an outcome of the measurement on pair (23) will show entanglement in the pair (14). Given this same subensemble, if the physicist had, for these specific runs, decided not to perform any measurement on (23), there would be no correlation in (14) in these runs"

Statements like these are why there continues to be confusion about this issue. How do you make a subensemble of one pair? This makes no sense.
Until the participants in these discussions learn to use language correctly, it is best to shut up and just calculate.

 

[Demystifier]

But the projection will not result in that state 100% of the trials. So the entanglement will be of a subensemble. Your example is, in this respect, not different than the swapping.

Indeed, it was my intention to construct something very similar to swapping, but simpler. In this way one can better understand what about swapping is essential, and what is not.

 

[Demystifier]

So the entanglement will be of a subensemble.

While this is true, I don't see how is this relevant. It does not diminish the strangeness of entanglement.

To demonstrate this, let me consider another example. Consider a machine that first flips a coin, and then, depending on the result of coin flipping, creates a quantum state. If the coin turns out heads, the machine creates an entangled state. If the coin turns out tails, the machine creates a non-entangled state. Here again the entanglement is a property of a subensemble, but it does not make entanglement any less strange.

 

[mattt]

Statements like these are why there continues to be confusion about this issue. How do you make a subensemble of one pair? This makes no sense.
Until the participants in these discussions learn to use language correctly, it is best to shut up and just ...

What do you mean by one pair?

You need a sufficiently large number of runs of every pair to test anything.

I think what Morbert wrote there is crystal clear.

 

[martinbn]

While this is true, I don't see how is this relevant. It does not diminish the strangeness of entanglement.

To demonstrate this, let me consider another example. Consider a machine that first flips a coin, and then, depending on the result of coin flipping, creates a quantum state. If the coin turns out heads, the machine creates an entangled state. If the coin turns out tails, the machine creates a non-entangled state. Here again the entanglement is a property of a subensemble, but it does not make entanglement any less strange.

Yes, but one point in the older threads was that swapping doesn't bring anything new in terms of strangeness. The strangeness is already there in entangled pairs.

 

[vanhees71]

But the projection will not result in that state 100% of the trials. So the entanglement will be of a subensemble. Your example is, in this respect, not different than the swapping.

You can never have 100% success for swapping to a specific Bell state. That cannot be, because the full ensemble is unentangled by preparation, and indeed it's not different from this example.

 

[Morbert]

Statements like these are why there continues to be confusion about this issue. How do you make a subensemble of one pair? This makes no sense.

The pairings mentioned, like (23) or (14), are pairings that exist in every experimental run. You construct a subensemble by doing many runs.

 

[vanhees71]

You cannot do any such experiments with realizing just one pair. It's about probabilities, which can be tested only on ensembles. I'm a proponent of the ensemble interpretation, i.e., for me a quantum state describes the stastistical properties for the outcome of measurements on ensembles of equally prepared systems.

 

[PeterDonis]

This is at best misleading.

As a description of the actual math, which you present, it might be. But it's not misleading as a description of what other posters (not you) said in those previous threads. They explicitly claimed that 1 & 4 were entangled in the original prepared state (based on them being able to always pick a "subensemble" that reflected this).

 

[PeterDonis]

you can make the pair (14) maximally entangled by projecting the pair (34) to a Bell state

Not (34), (23). That is the pair that the BSM is done on.

 

[Nullstein]

As a description of the actual math, which you present, it might be. But it's not misleading as a description of what other posters (not you) said in those previous threads. They explicitly claimed that 1 & 4 were entangled in the original prepared state (based on them being able to always pick a "subensemble" that reflected this).

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.

 

[lodbrok]

What do you mean by one pair?

You need a sufficiently large number of runs of every pair to test anything.

I think what Morbert wrote there is crystal clear.

My issue wasn't with Morbert; it was with the statement he quoted. The statement he quoted was, "The subensemble selected by an outcome of the measurement on pair (23) will show entanglement in the pair (14)." Pair (23) is precisely two particles, and pair (14) is exactly two particles. So where does the subensemble come into it?

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."
The fact that many now consider these statements equivalent is the problem I was highlighting. Just re-read this thread and appreciate the near-universal misuse of "a pair," "the pair," "pair (12)", and "pair (23)" by all sides of the argument. 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. This impression is furthered even by those who actively argue against this view.

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.

 

[lodbrok]

The pairings mentioned, like (23) or (14), are pairings that exist in every experimental run. You construct a subensemble by doing many runs.

Correct, but those pairings in each experimental iteration are not the same as what is implied in a statement like the following from this thread:

you can make the pair (14) maximally entangled by projecting the pair (34) to a Bell state

The sad part is that the person who made this statement does not believe that what happens to two of the four particles in a particular iteration of the experiment changes anything about the other two.
So we keep going around in circles, talking past each other because we don't bother to be precise with our language.

 

[mattt]

My issue wasn't with Morbert; it was with the statement he quoted. The statement he quoted was, "The subensemble selected by an outcome of the measurement on pair (23) will show entanglement in the pair (14)." Pair (23) is precisely two particles, and pair (14) is exactly two particles. So where does the subensemble come into it?

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."
The fact that many now consider these statements equivalent is the problem I was highlighting. Just re-read this thread and appreciate the near-universal misuse of "a pair," "the pair," "pair (12)", and "pair (23)" by all sides of the argument. 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. This impression is furthered even by those who actively argue against this view.

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.

Well, yes, it's kind of an "economical way of writing" (not dangerous if everybody knows what we are talking about), after all I think everybody here knows that you need an ensemble of identically prepared particle-pairs (in this case), thousands of them indeed, the to be able to statistically test anything with a given desired statistical significance.

 

[Demystifier]

Yes, but one point in the older threads was that swapping doesn't bring anything new in terms of strangeness. The strangeness is already there in entangled pairs.

OK, but in my example in #16, do you agree that the strangeness is not already there before projection, and that the projection creates strangeness?

 

[mattt]

Correct, but those pairings in each experimental iteration are not the same as what is implied in a statement like the following from this thread:
"you can make the pair (14) maximally entangled by projecting the pair (34) to a Bell state"

I see nothing problematic about that statement (if you understand what it means).

It just means that, out of the millions of (1,4) pairs created, the approximately 25% of them that are the "companions" respectively of exactly the (2,3) pairs that ended up in the BSM state, satisfy (this (1,4) subensemble) satisfy the statistics that we call "a Maximally Entangled State", that's what Quantum Mechanics says, and that is what experimenters can test, up to technological limitations.

 

[PeterDonis]

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.

Thank you for the clarification.

Now let's look at particles 1 & 2. Their "before" state is a maximally entangled state. It was prepared that way. What is their "after" state? By your reasoning, it should be the same, just as the state of 1 & 4 "after" is the same as "before"--which would mean that the BSM done on particles 2 & 3 has no physical effect on anything at all. Is that your position?

 

[Nullstein]

Now let's look at particles 1 & 2. Their "before" state is a maximally entangled state. It was prepared that way. What is their "after" state? By your reasoning, it should be the same, just as the state of 1 & 4 "after" is the same as "before"--which would mean that the BSM done on particles 2 & 3 has no physical effect on anything at all. Is that your position?

No, that's not my position. 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. The physical effect of the BSM at 2&3 is to decrease the entanglement between 1&2 and 3&4 and to increase the entanglement between 2&3.