Entanglement swapping, monogamy, and realism

[Nullstein]

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.

Even if you take a subensemble of the initial full ensemble, then DrChinese's state after the BSM would be a subensemble of that subensemble, so they still can't be compared.

 

[PeterDonis]

Even if you take a subensemble of the initial full ensemble, then DrChinese's state after the BSM would be a subensemble of that subensemble

Not if you take the subensemble of runs where the BSM gives an "event ready" signal. That is precisely the same subensemble that is used to ground the claim of entanglement swapping, since for that subensemble 1&4 are maximally entangled.

 

[PeterDonis]

Physical cause and effect relationships must be inferred from the evolution of the full ensembles.

In the evolution of the full ensemble, 2&3 are not entangled. Does that mean the BSM has no causal effect on anything?

 

[lodbrok]

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

The experimental results at (14) are available before the (23) measurement. Those results do not change after the (23) measurement. Therefore the (23) measurement does not affect the (14) results. The fact that the results are the same before and after the (23) measurement is direct experimental evidence that the (23) measurement has absolutely no effect on the (14) results. Are you really arguing the contrary?

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.

On the contrary, we have evidence from that experiment. The moment when the (14) measurement is complete and the results recorded but the (23) measurement is yet to be done. Those (14) results are exactly the same results as those after the (23) measurements are done. That is direct evidence that the results would not have changed if the (23) measurements were never done. That is direct evidence that the (23) measurement has no effect on the (14) results.

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.

No, this is not the assertion. We are not simply talking of a different experiment showing similar statistics. We are talking about the same data already recorded, for every single particle pair, and not one of those results changed as a result of the (23) measurement done later. Therefore the (23) measurement had no effect on the (14) results. This is an experimental fact.

 

[Simple question]

There is no entanglement between 1&4 before the BSM and there is no entanglement between 1&4 after the BSM.

Show me a Quantum mechanical process that allow you two pick a sub ensemble of entangle pair, in a full set of non-entangled pair, with %100 certitude.

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.

No let's not. Projection are not part of the preparation of state.

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.

You conclude your preparation state from your results ? How convenient ... and backward.

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

Nope, post selection happens only at BSM site.

 

[PeterDonis]

The experimental results at (14) are available before the (23) measurement.

So what? The QM math is clear that the time ordering of the measurements makes no difference.

The moment when the (14) measurement is complete and the results recorded but the (23) measurement is yet to be done. Those (14) results are exactly the same results as those after the (23) measurements are done. That is direct evidence that the results would not have changed if the (23) measurements were never done.

It is no such thing. The (23) measurements are made in both cases, and the QM math says there is no dependence on the time ordering of the measurements, and that prediction is borne out by the results. So that's all the results are telling you.

To get experimental evidence about what happens if a (23) measurement is not done, you need to actually run the experiment with the (23) measurement not being done.

We are talking about the same data already recorded, for every single particle pair, and not one of those results changed as a result of the (23) measurement done later.

No. The experiment with the (23) measurement done before the (14) measurement, and the experiment with the (23) measurement done after the (14) measurement, are two different experiments. QM predicts that the statistics are the same in both cases, but that doesn't mean you can just do one experiment and claim to have evidence about both--still less about an experiment that is never done at all. To test the QM prediction, you need to do the experiment both ways and compare the results. That is what was actually done.

It is of course impossible to do a single experiment in which the (23) measurement is both before and after the (14) measurement. So we are not talking about "the same data already recorded"--it's not as if we record the (14) measurement after a (23) measurement, and then magically move the same (23) measurement to occur after the same (14) measurement (or the reverse). We are only talking about comparing data from different experiments in which the time ordering of the measurements is different. And of course this does not justify any claims about an experiment that was never done.

 

[Simple question]

The experimental results at (14) are available before the (23) measurement.

... or after

Those results do not change after the (23) measurement.

Nobody said the results change

Therefore the (23) measurement does not affect the (14) results.

Quite the opposite. 2&3 measurement confirmed them. And this would be impossible if 1&4 measurement was just (classic) random. They are "Quantum random" with a different bound impossible to explain locally. And doubly impossible if 1&4 are not entangled at the start.

 

[Nullstein]

Not if you take the subensemble of runs where the BSM gives an "event ready" signal. That is precisely the same subensemble that is used to ground the claim of entanglement swapping, since for that subensemble 1&4 are maximally entangled.

I'm talking about an idealized experiment here. The event-ready scheme is an experimental complication in real world experiments in order to avoid measurement errors. The event ready signal is not produced by the BSM, but at the time of preparation. If the event ready signal occurs, we know that we can expect a properly prepared state. So the subensemble of event ready photons is taken as the ensemble that is being studied. This is fine, but this ensemble is not the subensemble we get from the BSM. The BSM at 2&3 still decomposes this event ready ensemble into the four subensembles that are studied in the idealized setting.

In the evolution of the full ensemble, 2&3 are not entangled. Does that mean the BSM has no causal effect on anything?

No, there are other subsystems $\rho_{i_1,\ldots,i_n}$ that are influenced by the BSM. The state $\rho_{1234,\text{after}}$ is definitely different from the state $\rho_{1234,\text{before}}$, so some subsystems are subject to change.

By the way, I'm not even saying that we cannot perform causal inference from subensembles. But there are some well understood principles that need to be taken care of. If correlations remain after controlling for common causes and after summing over common effects, then this is definitely a hint for causality. But in the case of entanglement swapping, certain correlations go away after summing over the common effects.

 

[Nullstein]

Show me a Quantum mechanical process that allow you two pick a sub ensemble of entangle pair, in a full set of non-entangled pair, with %100 certitude.

Picking a subensemble is not a quantum mechanical process. It's performed by the experimenter on his computer. But what does that have to do with the fact that the 1&4 subsystem is in the state $\frac 1 4\mathbb 1\otimes\mathbb 1$, both before and after the BSM?

No let's not. Projection are not part of the preparation of state.

This sentence is quite meaningless. It's a mathematical fact that the equality $\frac 1 4\mathbb 1\otimes\mathbb 1 = \sum_i \frac 1 4 P_i$ holds. So it is a mathematical fact that the full ensemble at 1&4 is equivalent to a mixed state of Bell states, weighted with the probabilities $\frac 1 4$.

 

[PeterDonis]

The event ready signal is not produced by the BSM, but at the time of preparation.

This is wrong. The "event ready" signal is generated by a combination of things: the 2 & 3 photons arriving at the BSM device within the same narrow time window, and the output of the BSM indicating the particular Bell state that the BSM is set up to distinguish. This happens at the BSM, not at the initial preparation. @DrChinese is more familiar than I am with the specific papers describing the experiments, and I'm sure can give specific references to the descriptions in those papers that match the above.

 

[StevieTNZ]

I hope people realise that the WolframAlpha file I was sent from one of the co-authors, that presents the calculations of photons 2 and 3 through the Mach-Zehnder interferometer (with various wave plates) in the Zeilinger et al (2012) experiment before a bell-state measurement occurs on them treats those photons as if they were already entangled in a particular bell-state before entering the interferometer. I found that rather strange.

Again, I point to my comments: https://www.physicsforums.com/threa...ing-monogamy-and-realism.1050017/post-6855670

 

[Morbert]

I hope people realise that the WolframAlpha file I was sent from one of the co-authors, that presents the calculations of photons 2 and 3 through the Mach-Zehnder interferometer (with various wave plates) in the Zeilinger et al (2012) experiment before a bell-state measurement occurs on them treats those photons as if they were already entangled in a particular bell-state before entering the interferometer. I found that rather strange.

This is ok to do if you are careful. The subensemble selected by the projector $\mathbb 1 \otimes P_i \otimes \mathbb 1$ is also selected by a history $C_i$ that projects to a bell state for (23) before the measurement occurs.

 

[DrChinese]

The "event ready" signal is generated by a combination of things: the 2 & 3 photons arriving at the BSM device within the same narrow time window, and the output of the BSM indicating the particular Bell state that the BSM is set up to detect. This happens at the BSM, not at the initial preparation. @DrChinese is more familiar than I am with the specific papers describing the experiments, and I'm sure can give specific references to the descriptions in those papers that match the above.

Most folks probably don't know some of the details in these experiments, and most of the time they don't matter. Everyone gets the main idea. But as @Nullstein shows us, the devil is sometimes in the details. So allow me to correct a few notions. What I am about to say below is the theoretical case of the perfect experiment, but swapping experiments are quite difficult to execute in practice so they won't work quite like I describe.

Imagine our same scenario, starting off with distant independent sources A and B generating entangled pairs (1 & 2) and (3 & 4) from Type I PDC (both H or both V). These pairs are produced spontaneously and asynchronously, no particular time of creation from either source. The initial quantum state is a Product State of 2 entangled pairs: $\hat{\rho}=\hat{\rho}_{12} \otimes \hat{\rho}_{34}.$ That of course is ONLY true when we have one pair from Source A and one pair from Source B that we choose to discuss as a candidate for a swap. We need to connect two pairs to have a set of 4 to make any sense of the pairs we start with.

What connects these pairs? Is it time of emission? No. Is it time of arrival of the 1 & 4 photons? No. What connects them is the time the 2 & 3 photons cross the Beam Splitter (BS) during the Bell State Measurement (BSM). They must cross within a narrow time window, let's call it 10 nanoseconds to have a number. And they must be indistinguishable as to whether the source was A or B once they pass the BS. So same wavelength, etc. and no other identifying characteristics (cannot be polarized). This is how the pairs from different sources interact, the creation time of the pairs itself is not a factor.

When they come out of the BS they are routed to 2 Polarizing Beam Splitters (PBS) and then to an array of 4 detectors. Some Bell States can be distinguished by this method, but unfortunately not all can. There will always be a percentage of cases that cannot be used for a Bell test on 1 & 4 pairs. However, that does NOT mean they aren't entangled. They are, we just cannot determine how.

For our purposes, let's assume that the 1 & 4 measurement systems (PBS plus 2 detectors for each) are positioned such that when we have a successful BSM, the 1 & 4 detectors go off at the same time (within our designated time window). We add fiber cable to make that work out, and we place them at the same spot. Additionally, we do the same with the BSM detector array. We use fiber to adjust the travel time, and route them to the same location as the 1 & 4 detectors. All 4 photons will arrive at location where all of the detectors are, and the photons will all arrive within the same time window. I am adding this little twist so you can see exactly what should be discussed when we talk about an ensemble or subset or subensemble. So what we expect, with a successful BSM, is that 4 detectors will click at almost precisely the same time. For the 1 & 4 photons, each will generate one click indicating their polarization relative to their respective PBS. The BSM detector array will register 2 clicks, one for the 2 photon and one for the 3 photon - but we won't know which is which. So 4 "simultaneous" clicks means we have a successful BSM in this setup.

Here is where the labeling of ensembles gets confusing. Most everyone assumes we are ignoring numerous sets of clicks. That's not really accurate. We often get a click on the photon 1 detectors and a single click on the BSM detector array, but nothing in the time window for the photon 4 detectors. And we often get a click on the photon 4 detectors and a single click on the BSM detector array, but nothing in the time window for the photon 1 detectors. Since there is no 1 click to match with a 4 click, there is no way to make any sense of the earlier statement $\hat{\rho}=\hat{\rho}_{12} \otimes \hat{\rho}_{34}.$ For me, only cases in which the 1 & 4 photons are detected within the time window should be considered. I would call that the full universe or data set. Note that this definition does not reference whether or not the BSM succeeded.

In theory: for every single case where the 1 & 4 photons are detected within the time window: they are entangled. We may not know which of the 4 Bell states they are in - and therefore we can't perform a Bell test on them - but they ARE entangled. The new state is $\hat{\rho}=\hat{\rho}_{14} \otimes \hat{\rho}_{23}.$ If the experimental setup is perfect, then every single time we have clicks for 1 and 4, we will also have 2 clicks at the BSM array. So once again, every 1 & 4 pair is entangled. Of course, that wouldn't be true if we messed up the BSM setup so that the source was distinguishable (we could do that if we wanted). But assuming we are good scientists, there is no subensemble yet.

The subensemble comes when we ask what kind of entanglement did we get? Each of the 4 Bell states is equally likely, but usually only 2 can be identified. So half the results are thrown out, but of course they were still entangled. So when you talk about the subensemble, it is something quite different than many here are picturing. To me, the subensemble is the group for which we know the specific Bell state. The full data set are the cases in which we have matched clicks for 1 & 4. And in the ideal case, they are all entangled and there are also clicks at the BSM array.

 

[berkeman]

[Mentor Note -- @lodbrok has been thread banned]

 

[kurt101]

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.

I am ok if others think it is strained. I am not ok if others claim it is impossible without a solid argument.

Perhaps you can

If that is the case, then the monogamy argument is not a problem for the cause and effect realistic interpretation that I have brought into this thread.

There are no "coincidental" violations of Bell Inequalities.

Maybe we are just using the term "coincidental" differently. See my explanation of how I am using "coincidental" below.

1. 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.

I can live with that explanation for now. Taking your definition, it sounds like this is not a blocker for the cause and effect interpretation I am using to explain the "coincidental" entanglement.

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).

I will give an explanation that I think will stay in the rules of this forum. It is not a proof, but it is enough for a discussion.

Here is what I mean by "coincidental" in the context of a realist cause and effect interpretation that I am using and in the case where the BSM test is done after the measurements of 1 & 4:

Measuring the polarization of photon 1 changes the polarization of photon 2. Measuring the polarization of photon 4 changes the polarization of photon 3. When 2 and 3 are identical in almost all properties including polarization, the BSM test will determine they are indistinguishable and this indicates that the measurements of 1 & 4 are maximally entangled.

That is how I would describe what is happening in a realistic interpretation of this case. I use the short hand term "coincidental" to describe it and differentiate it from the other case, but clearly there is more going on then just coincidence, but fundamentally coincidence is the driving factor that allows 1 & 4 to be entangled.

So in this realistic description, at no time is the monogamy argument violated. Certainly, the details matter on whether this can be a valid explanation, but these important details set aside, as far as I can tell, there is nothing that @DrChinese is arguing that prevents this from being an explanation of what is actually happening. It produces the same correlation as when the BSM test in the other order and does not violate monogamy.

 

[PeterDonis]

If that is the case,

I only said "perhaps". If you really want to make such an argument, you need to make it. I was only pointing out that it's not an argument anyone else has made in these discussions.

a realist cause and effect interpretation that I am using

Is it just your own, or is it one you have encountered in the literature? It should be the latter, in which case you should be able to provide a reference.

 

[DrChinese]

Here is what I mean by "coincidental" in the context of a realist cause and effect interpretation that I am using and in the case where the BSM test is done after the measurements of 1 & 4:

Measuring the polarization of photon 1 changes the polarization of photon 2. Measuring the polarization of photon 4 changes the polarization of photon 3. When 2 and 3 are identical in almost all properties including polarization, the BSM test will determine they are indistinguishable and this indicates that the measurements of 1 & 4 are maximally entangled.

That is how I would describe what is happening in a realistic interpretation of this case. I use the short hand term "coincidental" to describe it and differentiate it from the other case, but clearly there is more going on then just coincidence, but fundamentally coincidence is the driving factor that allows 1 & 4 to be entangled.

So in this realistic description, at no time is the monogamy argument violated. Certainly, the details matter on whether this can be a valid explanation, but these important details set aside, as far as I can tell, there is nothing that @DrChinese is arguing that prevents this from being an explanation of what is actually happening. It produces the same correlation as when the BSM test in the other order and does not violate monogamy.

You have mixed some good points and some factual errors. "At no time is the monogamy argument violated" is good. "Measuring the polarization of photon 1 changes the polarization of photon 2" is fine as long as you accept it could as easily be the other way around (since ordering makes no difference). The important point is that the swap (BSM) is an essential action.

The polarization of 2 & 3 is one of the things that does not need to be identical (they might or might not). That is tested to determine the Bell state that 1 & 4 are cast into. But going into the beam splitter, the polarization must be unknown for both so that they can be indistinguishable, as you say.

Yours is not what I would call a realistic interpretation. "Fundamentally coincidence is the driving factor that allows 1 & 4 to be entangled" doesn't make sense, because the BSM is the driving factor. Each and every BSM results in 1 & 4 entanglement, and that does not occur otherwise.

 

[kurt101]

Is it just your own, or is it one you have encountered in the literature? It should be the latter, in which case you should be able to provide a reference.

I am using the general realist position that John Bell and Einstein use when they refer to spooky action at a distance. I don't go into any details as to what that action is beyond that the polarization is changed. That is why I figured it would be safe to use. But to @DrChinese point, I am fine with just calling it some action that influences their respective polarizations versus assigning a particular direction to the action.

"Measuring the polarization of photon 1 changes the polarization of photon 2" is fine as long as you accept it could as easily be the other way around (since ordering makes no difference). The important point is that the swap (BSM) is an essential action.

Yes, I can agree that it can be the other way around or some mix. And I can agree that the BSM is essential, but in a realist cause and effect interpretation that I am using it is just the test that reveals the correlation, not what causes the correlation.

Yours is not what I would call a realistic interpretation. "Fundamentally coincidence is the driving factor that allows 1 & 4 to be entangled" doesn't make sense, because the BSM is the driving factor. Each and every BSM results in 1 & 4 entanglement, and that does not occur otherwise.

In the realist cause and effect interpretation that I am using, the BSM test is the test that reveals the correlation, not what causes the correlation. All the actions that happened prior to the BSM test being performed is what caused the correlation between 1 & 4. And if you chose not to do the BSM test, the correlation between 1 & 4 is still there, you just don't have a BSM test to select it from all of your measurement data.

 

[Demystifier]

I would like to say something about how projection can create entanglement. In the example in #16, the required projector is
$$\Pi=1\otimes 1-\pi_R\otimes\pi_R$$
where $\pi_R=|R\rangle\langle R|$. The projector $\Pi$ is not separable, i.e. cannot be written in the form $\pi_1\otimes\pi_2$. In particular, it is not equal to the separable projector
$$(1-\pi_R)\otimes(1-\pi_R).$$
The separable projector above would project the state in #16 to the non-entangled state $|L\rangle \otimes |L\rangle$.

So the moral is, a projection can create entanglement when the projector is not separable. I'm sure it can be formulated precisely and proved rigorously as a general theorem, but it looks so intuitive that I don't really feel a need for a precision and proof.

 

[vanhees71]

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:$$\hat{\rho}'=\hat{\rho}_{23} \otimes \hat{\rho}_{14}.$$

One has to be careful. The point here is that, at least within the minimal statistical interpretation, the complete ensemble is always described by state (a), i.e., no matter what you measure on the pair (23), if considering coincidence measurements on all four-photon preparations, the statistical outcome of these measurements is predicted by (a).

(b) describes the subensemble, selected by projecting (23) to one and only one of the four Bell states. For each of these four possible subensembles the pairs (14) and (23) are described by the state (b) and thus are entangled.

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.

Of course to measure the entanglement of (14) one has to do a coincidence experiment, i.e., one has to be sure to take the pair (23) doing the projection measurement to a Bell state with them associated with the photons 1 and 4 prepared in state (a), but this has been obviously done in these experiments.

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.

Sure, you need to select the photons according to the result of the measurement on (23).

In all this it doesn't matter in which temporal order the measurements on (23) and on photons 1 and 4 are done. They can even be space-like separated, and you can post-select the subensemble(s), provided you have the accurate measurement protocols for all these coincidence measurements.

Concerning the causal question, I think due to micorcausality and within the minimal statistical interpretation entanglement swapping demonstrates no non-local causality-violating effects of one local measurement (measurement on the pair (23)) and the others (measurements on photons 1 and 4) but the correlations on the selected subensemble described by state (b) are already due to the correlations implies by the preparation of the full ensemble in state (a).

At least this is a consistent interpretation of the local-QFT formalism which doesn't contradict its mathematical defining properties (in this context most importantly microcausality), and this resolves the EPR paradox, and what has to be given up is the assumption of "reality", i.e., the assumption that all observables take predetermined values and the probabilities were thus only due to ignorance of these values due to the ignorance about the values of some "hidden variables". In this sense what has to be given up is the assumption of separability, i.e., precisely what Einstein considered the most serious quibble he had with QT (according to his Dialectica paper of 1948, which is the far more clearly formulated source about Einstein's point of view than the EPR paper, which Einstein himself didn't like too much).

 

[vanhees71]

In the evolution of the full ensemble, 2&3 are not entangled. Does that mean the BSM has no causal effect on anything?

The "causal effect" is "local", i.e., due to the "local interactions" of these two photons with the polarizing beam splitter and the detectors. Any possible causal effect on the outcome of measurements on photons 1 and/or 4 can only be in the future lightcone of the registration of photons 2&3 in this BSM, i.e., if you do the local measurements on photons 1 and 4 spacelike separated from this BSM there cannot be any causal influence of the BSM on photons 1 and 4 before measured, provided you accept microcausality, which seem to be the case in all scientific papers on Bell-test measurements, particularly when it comes to the discussion about the various loopholes.

 

[martinbn]

So the moral is, a projection can create entanglement when the projector is not separable. I'm sure it can be formulated precisely and proved rigorously as a general theorem, but it looks so intuitive that I don't really feel a need for a precision and proof.

If the projector is a product, then it cannot create entanglement. That is clear. It isnt obvious that not being a product is enough to create entanglement.

 

[Morbert]

Yours is not what I would call a realistic interpretation. "Fundamentally coincidence is the driving factor that allows 1 & 4 to be entangled" doesn't make sense, because the BSM is the driving factor. Each and every BSM results in 1 & 4 entanglement, and that does not occur otherwise

Consider a pair of spacelike separated entangled particles in the state $|\uparrow_A\uparrow_B\rangle + |\downarrow_A\downarrow_B\rangle$. You have particle $A$ and I have $B$. You measure $A$ in the $\{\uparrow, \downarrow\}$ basis and obtain the outcome $\uparrow$, so you know that if I measure $B$, I will obtain the outcome $\uparrow$. My questions to you:

i) If, for the same experimental run (i.e. not merely a new/different run) you had instead not performed a measurement, would I still have been guaranteed to obtain the outcome $\uparrow$?

ii) Is your answer to i) interpretantion-dependent, or enforced by the formalism of QM? Or can QM even concern such questions that are not strictly resolvable by experiment?

 

[vanhees71]

If I'm also allowed to answer:

ad i) No, because all you know for the full ensemble is that your particle is ideally unpolarized. You simply get in 50% of the cases up and in 50% of the cases down (in the usual statistical sense).

ad ii) It may be interpretation dependent. At the end all that counts, however, is what's observed in the lab, and in all such experiments all that's observed are the statistical properties predicted by Q(F)T, as measured on sufficiently large ensembles (for a given level of statistical significance), and that's what the Q(F)T formalism predicts in the minimal statistical interpretation, maybe extended by the POVM formalism when weak measurements are performed or to describe the reality of non-ideal preparation and detection equipment.

As far as the physics is concerned, the minimal statistical interpretation is all that's needed. You don't need additional metaphysical balast, including "state collapse", "many worlds", "de Broglie-Bohm trajectories", and all other kinds of esoterics.

 

[Demystifier]

It isnt obvious that not being a product is enough to create entanglement.

It's not enough.

 

[martinbn]

It's not enough.

So, what you hoped for is not true then?

 

[Morbert]

If I'm also allowed to answer:

ad i) No, because all you know for the full ensemble is that your particle is ideally unpolarized. You simply get in 50% of the cases up and in 50% of the cases down (in the usual statistical sense).

What I'm trying to tease out here is the significance of "counterfactual knowledge". All you know for a given experimental run is that your particle is ideally unpolarized. But for this very specific run, you have some knowledge (your observed outcome) that you can premise some counterfactual reasoning on.

We do this in classical scenarios all the time: You are thinking of going for a walk but decide to stay inside. It starts raining. You reason counterfactually that if you had gone for a walk, you would have gotten wet, even though at the beginning of the scenario you didn't know if it would rain.

ad ii) It may be interpretation dependent. At the end all that counts, however, is what's observed in the lab, and in all such experiments all that's observed are the statistical properties predicted by Q(F)T, as measured on sufficiently large ensembles (for a given level of statistical significance), and that's what the Q(F)T formalism predicts in the minimal statistical interpretation, maybe extended by the POVM formalism when weak measurements are performed or to describe the reality of non-ideal preparation and detection equipment.

As far as the physics is concerned, the minimal statistical interpretation is all that's needed. You don't need additional metaphysical balast, including "state collapse", "many worlds", "de Broglie-Bohm trajectories", and all other kinds of esoterics.

The rest of your answer gets closer to what I think is the point of disagreement between you and Dr. Chinese. I think the statistical interpretation discards these kinds of counterfactual scenarios, so that there is no unique answer to i). The significance of your outcome "up" is only the selection of a subensemble of experimental runs, and not some ontological assertion about how the spacelike separated particle would have behaved, contingent on your action. I think this discarding of counterfactual questions as meaningless is how the statistical interpretation stays compatible with locality.

I predict that Dr. Chinese would disagree with the minimalist statistical position, and say that the behaviour of the distant particle (not merely your knowledge of the particle) *is* contingent on whether or not you performed a measurement and observed "up"

[edit] - I'm considering this simple entangled pair because I think it is analogous to the BSM experiment, insofar as there is disagreement about whether the relation between 1 and 4 is contingent on a BSM of 2 and 3

 

[Demystifier]

So, what you hoped for is not true then?

It's true, note that I said "can". Like in "real numbers can have infinite number of non-repeating digits after decimal point".

 

[vanhees71]

What I'm trying to tease out here is the significance of "counterfactual knowledge". All you know for a given experimental run is that your particle is ideally unpolarized. But for this very specific run, you have some knowledge (your observed outcome) that you can premise some counterfactual reasoning on.

There is no "counterfactual knowledge" when you just give the state (in terms of a statistical operator). It describes, how your particle is prepared in each realization of the experiment, defining an "ensemble of equallyt prepared particles". There is no more knowledge than that, and if it's prepared as one particle in a pair of maximally entangled particles, it's perfectly unpolarized.

We do this in classical scenarios all the time: You are thinking of going for a walk but decide to stay inside. It starts raining. You reason counterfactually that if you had gone for a walk, you would have gotten wet, even though at the beginning of the scenario you didn't know if it would rain.

The rest of your answer gets closer to what I think is the point of disagreement between you and Dr. Chinese. I think the statistical interpretation discards these kinds of counterfactual scenarios, so that there is no unique answer to i). The significance of your outcome "up" is only the selection of a subensemble of experimental runs, and not some ontological assertion about how the spacelike separated particle would have behaved, contingent on your action. I think this discarding of counterfactual questions as meaningless is how the statistical interpretation stays compatible with locality.

Physics is about what's observed, and QT with the minimal statistical interpretation used to connect the mathematical formalism with observable facts just tells you about the statistical outcome when measuring what can be measured.

If you measure the spin of a particle in an direction $\vec{n}$, you cannot say anything what would have happened when instead measureing the spin in another direction $\vec{n}'$, which is not collinear to $\vec{n}$. It's simply physically impossible to measure two spin components in different directions. So it's just an empty speculation about something that's not realizable in Nature and thus not part of physics. Indeed this is the merit of the minimal statistical interpretation: It just says what's the meaning of the formalism for real-world observations and not some philosophical (aka esoteric) speculation.

I predict that Dr. Chinese would disagree with the minimalist statistical position, and say that the behaviour of the distant particle (not merely your knowledge of the particle) *is* contingent on whether or not you performed a measurement and observed "up"

[edit] - I'm considering this simple entangled pair because I think it is analogous to the BSM experiment, insofar as there is disagreement about whether the relation between 1 and 4 is contingent on a BSM of 2 and 3

 

[Morbert]

If you measure the spin of a particle in an direction $\vec{n}$, you cannot say anything what would have happened when instead measureing the spin in another direction $\vec{n}'$, which is not collinear to $\vec{n}$.

As an aside, I think with the minimal statistical interpretation we have an even stronger restriction. If you measure the spin of a particle in a direction $\vec{n}$, you cannot say anything about what would have happened when instead measuring the spin in another direction, collinear or otherwise, even the same direction. Counterfactual statements are thoroughly off limits, and we can only concern ourselves with future measurements on the particle or, in statistical language, subensembles of sequences of measurements, selected by the projectors $P_{\uparrow_\vec{n}}, P_{\downarrow_\vec{n}}$

 

[PeterDonis]

I am using the general realist position that John Bell and Einstein use when they refer to spooky action at a distance.

I'm not sure either Bell or Einstein made the same claims as you are making. That is why you should give a specific reference.

in a realist cause and effect interpretation that I am using it is just the test that reveals the correlation, not what causes the correlation.

If you think either Einstein or Bell made this claim, please give a specific reference. On its face, what you describe here and in the rest of your post looks like a local hidden variable interpretation of the sort that is ruled out by Bell's Theorem.

 

[PeterDonis]

Any possible causal effect on the outcome of measurements on photons 1 and/or 4 can only be in the future lightcone of the registration of photons 2&3 in this BSM, i.e., if you do the local measurements on photons 1 and 4 spacelike separated from this BSM there cannot be any causal influence of the BSM on photons 1 and 4 before measured, provided you accept microcausality

As I commented earlier (in response to someone else, I believe), this leads to what seems like a highly implausible conclusion: that the BSM has a causal effect on photons 1 & 4 only for the case where the BSM is in the past light cone of the 1 & 4 measurements. But both experimental results and the QM math predict the same outcomes regardless of the spacetime relationship of the measurements--whether the BSM is in the past light cone, spacelike separated, or in the future light cone of the 1 & 4 measurements. The simplest conclusion from all this is that, whatever is going on, it's the same for all three of those cases, so if whatever is going on can't be a "causal effect" by whatever definition of "causal effect" you are using, this would be the case when the BSM is in the past light cone of the 1 & 4 measurements, as well as when it is not.

 

[vanhees71]

[EDIT: This is not well formulated, as pointed out in #104; One should never discuss QT in words, only in formulae. See the clarification in #105]

There is no causal effect of the BSM on photons 2&3 on the outcome of measurments on photons 1&4 when the BSM and these measurements are space-like separated. This is a mathematical fact, if you use microcausal QFT, but that's not an implausible conclusion, if you accept that the observed entanglement of 1&4 in the ensemble selected through the BSM on 2&3 is due to the correlations described by the entanglement of the photon pairs 1&2 and 3&4 in the initial state, and that's indeed all what the QFT formalism tells you. So there's no "causation" but "correlation", and there's no implausibility or contradiction whatsoever.

Indeed, in this measurement it doesn't matter, whether the BSM on 2&3 and the measurements on 1&4 are space-like or time-like separated. The only point is that if you make sure that the measurements are space-like separated that there cannot be a causal effect of the BSM on the measurement outcome on 1&4, and this argument you find regularly in experiments with entangled states, and indeed this argument is consistent with the microcausality condition of QFT.

E.g., it was a big breakthrough when Zeilinger et al realized the "teleportation protocol" where this spacelike separation was ensured to exclude some (hidden) causal influences, i.e., excluding the "locality loophole". I think this was in a paper in 1997 or so. In fact it was also realized in Aspects seminal measurements in the early 1980ies. If needed, I can try to find the references.

 

[PeterDonis]

if you accept that the observed entanglement of 1&4 in the ensemble selected through the BSM on 2&3 is due to the correlations described by the entanglement of the photon pairs 1&2 and 3&4 in the initial state, and that's indeed all what the QFT formalism tells you.

I don't understand. In the initial state there is no correlation between the 1&2 and 3&4 pairs. So how can there be any correlations between 2&3 or 1&4 in the initial state?

 

[vanhees71]

1&2 are in a Bell state as well as 3&4, i.e., each of these pairs is maximally entangled. There are no correlations between 2&3 or 1&4 in the initial state. I haven't claimed anything like that.

Now if you project to a Bell state of 2&3, in the so prepared subensemble 2&3 are fully entangled, because they are prepared to be in a Bell state. Due to the entanglement of 1&2 and 3&4 this implies that for this subensemble also 1&4 are in a Bell state.

It's a easy, although some elaborate calculation, as described by Jennewein et al in

https://arxiv.org/abs/quant-ph/0201134

It's more easily seen in the 2nd-quantization notation. The four Bell states of a photon pair with momentum labels $j$ and $k$ are created from the Vakuum by
$$\hat{\Psi}_{jk}^{\dagger \pm} \rangle=\frac{1}{\sqrt{2}}[\hat{a}^{\dagger}(\vec{p}_j,H) \hat{a}^{\dagger}(\vec{p}_k,V) \pm \hat{a}^{\dagger}(\vec{p}_j,V) \hat{a}^{\dagger}(\vec{p}_k,H)],$$
$$\hat{\Phi}_{jk}^{\dagger \pm} \rangle=\frac{1}{\sqrt{2}}[\hat{a}^{\dagger}(\vec{p}_j,H) \hat{a}^{\dagger}(\vec{p}_k,H) \pm \hat{a}^{\dagger}(\vec{p}_j,V) \hat{a}^{\dagger}(\vec{p}_k,V)].$$
Then the initial four-photon state can then be written in two forms,
$$|\Psi_{1234} \rangle = \hat{\Psi}_{12}^{\dagger -} \hat{\Psi}_{34}^{\dagger-} |\Omega \rangle,$$
but this can as well be written as
$$|\Psi_{1234} \rangle=\frac{1}{2} (\hat{\Psi}_{23}^{\dagger +} \hat{\Psi}_{14}^{\dagger +} - \hat{\Psi}_{23}^{\dagger -} \hat{\Psi}_{14}^{\dagger -}-\hat{\Phi}_{23}^{\dagger +} \hat{\Phi}_{14}^{\dagger +} + \hat{\Phi}_{23}^{\dagger -} \hat{\Phi}_{14}^{\dagger -})|\Omega \rangle.$$
The former notation shows that photon pairs (12) and (34) are each in the polarization-singlet Bell state but (14) and (23), i.e, are uncorrelated.

The latter notation shows that if you project pair (23) to either of the four Bell state the pair (14) must be found in the same Bell state. In Pan et al's work, which we discuss here, (23) has been projected to the polarization-singlet state, and it has been demonstrated that then also the pair (14) is then in the same polarization-singlet state. This happens with probability 1/4.