A How Many Hidden Variables Would It Take to Model Photon Polarization?

[DrChinese]

1. Exactly, and that's why I suggest to open another thread if you are interested in discussing it.

2. One can discuss this only in terms of math, if one doesn't want to run into the problem of frequently misunderstanding each other. Let me repeat the arguments from the other thread. We discuss an experiment like the one described by Jennewein et al in

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

In the subensemble due to the projection measurement on 2&3 you have also 1&4 in a Bell state. That's a fact from the above calculation as well as the experiment. That's after all what entanglement swapping means. It's a sophisticated version of "teleportation".

3. Of course also your so-called "monogamy theorem" is fulfilled since obviously in the subensemble 1&2 as well as 3&4 are now completely uncorrelated. That's why it's called "swapping".

... In teleportation the original entanglement is then completely destroyed for the so prepared subensemble. I guess that's what you call "monogamy".

Of course photons 2&3 interact with the beam splitter and finally the detectors to enable you to select the subensemble by projection.

1. I never brought up GHZ, you did.

2. Fully agreed. This is just another recap of the experiment. I will point out your statement: "...due to the projection measurement on 2&3 you have also 1&4 in a Bell state." I couldn't say it better myself.

3. The "so-called" MoE insures exactly as you say: 1 & 2 are no longer entangled, and 3 & 4 are no longer entangled, even though they previously were.

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

We started with 2 biphotons. We end up with a single biphoton (1 & 4). It was created by the action of a projection measurement upon (2 & 3), exactly as you say.

However, both 1 and 4 are distant from each other when that happens (since they don't need to be in a common light cone). And the projection measurement is distant as well (far from 1, and far from 4). We are fully agreed to this point.

So the only thing you need to do is make the leap to the most obvious element here: accept the action at a distance which causes the creation of the (1 & 4) biphoton. How you are able to hand-wave this straightforward deduction away is a mystery. The BSM event creates a new and distant quantum system that did not previously exist. From the abstract of the reference: "Demonstrating quantum nonlocality for photons that never interacted... the nonlocality is confirmed by observing a violation of Bell's inequality by 4.5 standard deviations." That's the math we're talking about. In the words of the experimentalists we're talking about.

We walked up to the door together, we knocked on the door together, but only one of us is willing to step inside and say hello to quantum nonlocality. Come on in, this home is very comfortable inside.

 

[vanhees71]

1. I never brought up GHZ, you did.

You brought up a discussion about entanglement of more than 2 photons.

2. Fully agreed. This is just another recap of the experiment. I will point out your statement: "...due to the projection measurement on 2&3 you have also 1&4 in a Bell state." I couldn't say it better myself.

3. The "so-called" MoE insures exactly as you say: 1 & 2 are no longer entangled, and 3 & 4 are no longer entangled, even though they previously were.

So finally we agree about the mathematical facts in agreement with the experiments we discuss ;-).

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

We started with 2 biphotons. We end up with a single biphoton (1 & 4). It was created by the action of a projection measurement upon (2 & 3), exactly as you say.

However, both 1 and 4 are distant from each other when that happens (since they don't need to be in a common light cone). And the projection measurement is distant as well (far from 1, and far from 4). We are fully agreed to this point.

So the only thing you need to do is make the leap to the most obvious element here: accept the action at a distance which causes the creation of the (1 & 4) biphoton. How you are able to hand-wave this straightforward deduction away is a mystery. The BSM event creates a new and distant quantum system that did not previously exist. From the abstract of the reference: "Demonstrating quantum nonlocality for photons that never interacted... the nonlocality is confirmed by observing a violation of Bell's inequality by 4.5 standard deviations." That's the math we're talking about. In the words of the experimentalists we're talking about.

There is no action at a distance. This claim simply contradicts the properaties of standard relativistic QFT, for which microcausality is enforced by construction for the very reason to avoid any action at a distance or faster-than-light signal propagation. The BSM doesn't create anything new, it just selects a subensemble, and it's a local measurement on photons 2&3. The correlations due to the entanglement of 1&4 in this subensemble are due to the fact that in the initial state 1&2 were entangled as well as 3&4 (but not (2&3) and (1&4)). The selection of one of the 4 Bell states for (2&3) implies that also (1&4) is in (in this case the same) Bell state for this subensemble. Which of the 4 Bell states in each measurement is selected, of course, is random with a probability 1/4 for either outcome.

We walked up to the door together, we knocked on the door together, but only one of us is willing to step inside and say hello to quantum nonlocality. Come on in, this home is very comfortable inside.

There are "nonlocal correlations", i.e., strong correlations between far distantly observed parts of a quantum system but no violations of Einstein causality, because in relativistic QFT this is excluded by construction, i.e., only consider such QFTs, for which the microcausality constraint for all local observables is fulfilled. You should adapt your view to the mathematical facts. Any interpretation of a theory must respect the mathematical properties of this theory!

 

[Demystifier]

How Many Hidden Variables Would It Take to Model Photon Polarization?

I claim that this number is larger than or equal to $x$, where $x$ is the number of angels that can dance on the head of a pin. (SICR )

 

[martinbn]

How Many Hidden Variables Would It Take to Model Photon Polarization?

I claim that this number is larger than or equal to $x$, where $x$ is the number of angels that can dance on the head of a pin. (SICR )

Are angels bosons or fermions?

 

[vanhees71]

anyons...

 

[Fra]

Whenever one tries to "count" things in theoryspace one often get lost fast due to the curse of the real numbers.

I would agree with that, probably no two are alike (I think you are saying the continuum would be infinite). But I don't think that can be proven by direct experimental means

I would prefer to ask say, how far can an agent count? Which is similar to ask what we can experimentally distinguish. Which is hardly anywhere close to uncountably many.

This raises an interesting thought that at some point of complexity when you either scale the system complexity up, or agent cpmplexity down..an agent may loose count of things and decople in its actions because its input is saturated or overloaded. Its like when information is thrown at you above your level or in a code you cant dechipher. It is likely perceived as noise.

So i see the "agent stance" functioning as a natural regularization, but one with a more clear physical motivation rather than replacing the real number with an arbitrary and anbigous limit process.

/Fredrik

 

[Demystifier]

Are angels bosons or fermions?

Angels are ghosts, and ghosts are fermions with zero spin.

 

[Demystifier]

anyons...

So how can they live in 3 dimensions?

 

[vanhees71]

The head of the pin is a 2D surface ;-).

 

[Leureka]

Aren't we forgetting the measurement apparatus in all of this discussion?

Malus' law is a non-linear law. Non-linear effects often arise when we ignore contributions external to the system we are analyzing.

What if the hidden variable is not "inside" the particle, but only in the whole system particle + detector? I.e. the outcome of a measurement depends on both the particle's state AND the detector's state.

Notably, we don't have a theory that represents microscopically the detector, so we're left with a description of microscopic particle + macroscopic detector which is quantum mechanics.

 

[DrChinese]

Aren't we forgetting the measurement apparatus in all of this discussion?

Malus' law is a non-linear law. Non-linear effects often arise when we ignore contributions external to the system we are analyzing.

What if the hidden variable is not "inside" the particle, but only in the whole system particle + detector? I.e. the outcome of a measurement depends on both the particle's state AND the detector's state.

Notably, we don't have a theory that represents microscopically the detector, so we're left with a description of microscopic particle + macroscopic detector which is quantum mechanics.

The detector's state can be ruled out as a factor in measurements. Neglecting experimental imperfection: there is demonstrably no contribution from the apparatus itself, and that is a theoretical prediction of QM.

a. For example, suppose you have a beam splitter with detectors at the 2 output ports H and V. If you send a stream of H polarized photons to it, they will all be detected as H. No contribution from the detectors whatsoever.

b. Or suppose you have a stream of entangled photons (each in a superposition of spin). Run it through a beam splitter with detectors at the 2 output ports H and V. You can predict the outcome with certainty every time, demonstrating absolutely no contribution whatsoever from any part of the measurement apparatus itself.

 

[Leureka]

The detector's state can be ruled out as a factor in measurements. Neglecting experimental imperfection: there is demonstrably no contribution from the apparatus itself, and that is a theoretical prediction of QM.

a. For example, suppose you have a beam splitter with detectors at the 2 output ports H and V. If you send a stream of H polarized photons to it, they will all be detected as H. No contribution from the detectors whatsoever.

A stream of H polarized photons is very different from a stream of polarized photons in other directions, there is no "collapse" problem and I don't think you even need QM to figure out the outcome. It's exactly like saying "I measured the photon to be in state H, so I perform another measurement in H and I get the same result". You can't discard the detector (i.e. the polarizing filter + actual detector system) contribution for other angles, for the simple fact that rotations don't commute (detecting polarization with angle A and then with angle B is different than first angle B and then angle A).

Also, what do you mean by "predict with certainty" in

b. Or suppose you have a stream of entangled photons (each in a superposition of spin). Run it through a beam splitter with detectors at the 2 output ports H and V. You can predict the outcome with certainty every time, demonstrating absolutely no contribution whatsoever from any part of the measurement apparatus itself.

Are you talking about the statistical outcomes? Yes, that's entirely in the realm of prediction of QM. But it seemed to me hidden variables are used to try explaining the outcome of single measurements, not of an ensemble. And again, for each single measurement I don't see how you can exclude the contribution from detectors. The impossibility of separating quantum phenomena and their measurement is the core of the measurement problem, so I don't think I understand your points.On a slightly unrelated note, here's a wild thought: how can we experimentally "prove" the existence of photons when they are not being detected? Every single detection method involves an energy exchange from the EM field to a matter field, like the electron around an atom. It feels more logical to me it's the quantized nature of these interactions (due to confinement in potentials) that makes us "see" a photon. In turn, there would be no reason to describe the EM field as these particle-like entities during the journey to the detectors, hence the statistical outcome of a photon detection following a Malus' law - like probability distribution would be entirely the consequence of how EM waves interact with polarizers, which is a classical result. The real mystery would then be how do extended entities like EM waves "collapse" to point-like in detectors, but again it sounds more a problem with our description of the photon-detector interaction rather than the photon itself.

I'm genuinely asking for an opinion here.

 

[DrChinese]

1. A stream of H polarized photons is very different from a stream of polarized photons in other directions, there is no "collapse" problem and I don't think you even need QM to figure out the outcome. It's exactly like saying "I measured the photon to be in state H, so I perform another measurement in H and I get the same result". You can't discard the detector (i.e. the polarizing filter + actual detector system) contribution for other angles...

2, Also, what do you mean by "predict with certainty" in...

Are you talking about the statistical outcomes? Yes, that's entirely in the realm of prediction of QM. But it seemed to me hidden variables are used to try explaining the outcome of single measurements, not of an ensemble.

3. On a slightly unrelated note, here's a wild thought: how can we experimentally "prove" the existence of photons when they are not being detected? Every single detection method involves an energy exchange from the EM field to a matter field, like the electron around an atom. It feels more logical to me it's the quantized nature of these interactions (due to confinement in potentials) that makes us "see" a photon. In turn, there would be no reason to describe the EM field as these particle-like entities during the journey to the detectors, hence the statistical outcome of a photon detection following a Malus' law - like probability distribution would be entirely the consequence of how EM waves interact with polarizers, which is a classical result. The real mystery would then be how do extended entities like EM waves "collapse" to point-like in detectors, but again it sounds more a problem with our description of the photon-detector interaction rather than the photon itself.

I'm genuinely asking for an opinion here.

1. Well, clearly the detector is not accounting for any change from the expected H>. So that is the opposite of your hypothesis. You are just saying that when you measure the same thing over and over, you make the contribution of the detection system to be zero. (But not zero at other times.)

2. No, I'm not talking about statistical outcomes. I can predict the outcome exactly every time for each individual trial. Alice measures one of an entangled pair first (call it A), and sends me (I'm Bob in this example) her result. I receive it before I measure my photon (B). If the measurement apparatus was a variable in any individual measurement of B, then I would not get it right for each individual trial (due to perfect correlation).

3. This may not be the answer to your question ("prove the existence of photons when they are not being detected"), but it comes as close as it gets without going into the philosophical side of things.

http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf
"...while well-known phenomena such as the photoelectric effect and Compton scattering strongly suggest the existence of photons, they are not definitive proof of their existence. Here we present an experiment, suitable for an undergraduate laboratory, that unequivocally demonstrates the quantum nature of light. Spontaneously downconverted light is incident on a beamsplitter and the outputs are monitored with single-photon counting detectors."

Or, on the philosophical side, how do you know there is anything... when it is not being detected?

 

[Leureka]

1. Well, clearly the detector is not accounting for any change from the expected H>. So that is the opposite of your hypothesis. You are just saying that when you measure the same thing over and over, you make the contribution of the detection system to be zero. (But not zero at other times.)

The difference is that you don't need to change the system to measure H again. In the context of polarizers, any other angle necessarily rotates polarized light and doing so changes it, while this does not happen if the angle is correctly aligned. It's the nature of that change that would hide the detector's variable.

No, I'm not talking about statistical outcomes. I can predict the outcome exactly every time for each individual trial. Alice measures one of an entangled pair first (call it A), and sends me (I'm Bob in this example) her result. I receive it before I measure my photon (B). If the measurement apparatus was a variable in any individual measurement of B, then I would not get it right for each individual trial (due to perfect correlation).

You would not get it right everytime if A was at an angle with B. Let's say the photon is polarized with angle A, so it passes test A 100% of the time. Polarizer B is at a 45° angle to the first, so it will pass with probability cos(45) = 0.5, so 50% and you'll be wrong 50% of the time. This is independent on whether you received a message from Alice or not. Even if you didn't know the polarization angle beforehand the distribution of outcomes would still be consistent with an unknown, pre-determined polarization of the photon. This is the reason why I find it weird polarization experiments are used as a test of Bell's theorem.

Let me clarify: suppose you have a photon whose polarization is pre-determined but you don't know it before hand. You send it through one polarizer, which is NOT aligned with the photons hidden polarization. Here, the photon does not make any choices: whether it passes or not depends on the particular state of the detector at that moment, because the polarizer HAS to act on the photon to make its polarization "compatible". The probability of it passing is determined by both the photon's polarization and the decision of the polarizer: if the angle between the two is 22.5°, there are 85 states of the polarizer out of 100 where the result will be "rotate the photon and let it pass", and 15 states where the photon is simply absorbed.

 

[DrChinese]

1. You would not get it right everytime if A was at an angle with B.

Let's say the photon is polarized with angle A, so it passes test A 100% of the time. Polarizer B is at a 45° angle to the first, so it will pass with probability cos(45) = 0.5, so 50% and you'll be wrong 50% of the time. This is independent on whether you received a message from Alice or not.

2. Even if you didn't know the polarization angle beforehand the distribution of outcomes would still be consistent with an unknown, pre-determined polarization of the photon. This is the reason why I find it weird polarization experiments are used as a test of Bell's theorem.

1. Your hypothesis is simply wrong if both are measured at the same angle. You say the outcome is influenced by the detection apparatus, when obviously it isn't. If it was, there would not be perfect agreement - because both A and B would not be reported correctly.

It is also just as completely wrong at almost all angles. You just cannot see that on each individual trial. But the statistics would be flat out wrong - except when the angles are 45 degrees apart. The match rates would be off from the QM prediction of cos^2(A-B).2. The point of Bell is that the outcomes would NOT be "consistent with an unknown, pre-determined polarization of the photon" as you claim. I am surprised you would be posting in this sub-forum without knowing and understanding the Bell result more clearly, as you need this to understand why all interpretations need to address Bell as a central issue.

Bell shows that the assumption of pre-determined but unknown polarizations for particles fails when A≠B and Alice and Bob's choice of angle settings are not known to each other in advance. You can't even hand select an ensemble of pairs that will match the predictions of QM for entangled photons. You are welcome to take the "DrChinese Challenge" if you don't know why this is true. If you pick certain settings, you can't even HAND PICK results that match QM. You only need about 8 or so examples to see the impossibility. Once you hit that wall, you quickly see why Bell rules out ALL local hidden variable theories. The only way to "win" the challenge is to "cheat". That is, you hand pick knowing what you plan to measure in advance. But that defeats the purpose of having the outcomes predetermined at ALL angles simultaneously (but unknown).

Consider Type I entangled photon pairs - they are always at the same polarization for identical angle settings. The DrChinese Challenge is where you hand pick the results on both sides (Alice and Bob) for the angle settings A=0, B=120, C=240 degrees. Then I pick which pair of angles (A&B, B&C, or A&C) Alice and Bob actually measure. The challenge is to produce a dataset (ensemble) that will match the statistical predictions of QM. The QM prediction is 25% match rate when the Alice and Bob settings are different. But your hand picked ensemble can't get closer than 33% rate for all 3 combinations (A&B, B&C, or A&C) - which of course is entirely incorrect. You might get it for one, or maybe even two, but you can't for all three. Of course, a single counterexample invalidates any hypothesis.

Don't believe me, try it yourself. Or check out one of my web pages that explains:
Bell's Theorem with Easy Math

 

[Leureka]

1. Your hypothesis is simply wrong if both are measured at the same angle. You say the outcome is influenced by the detection apparatus, when obviously it isn't. If it was, there would not be perfect agreement - because both A and B would not be reported correctly.

It is also just as completely wrong at almost all angles. You just cannot see that on each individual trial. But the statistics would be flat out wrong - except when the angles are 45 degrees apart. The match rates would be off from the QM prediction of cos^2(A-B).2. The point of Bell is that the outcomes would NOT be "consistent with an unknown, pre-determined polarization of the photon" as you claim. I am surprised you would be posting in this sub-forum without knowing and understanding the Bell result more clearly, as you need this to understand why all interpretations need to address Bell as a central issue.

Bell shows that the assumption of pre-determined but unknown polarizations for particles fails when A≠B and Alice and Bob's choice of angle settings are not known to each other in advance. You can't even hand select an ensemble of pairs that will match the predictions of QM for entangled photons. You are welcome to take the "DrChinese Challenge" if you don't know why this is true. If you pick certain settings, you can't even HAND PICK results that match QM. You only need about 8 or so examples to see the impossibility. Once you hit that wall, you quickly see why Bell rules out ALL local hidden variable theories. The only way to "win" the challenge is to "cheat". That is, you hand pick knowing what you plan to measure in advance. But that defeats the purpose of having the outcomes predetermined at ALL angles simultaneously (but unknown).

Consider Type I entangled photon pairs - they are always at the same polarization for identical angle settings. The DrChinese Challenge is where you hand pick the results on both sides (Alice and Bob) for the angle settings A=0, B=120, C=240 degrees. Then I pick which pair of angles (A&B, B&C, or A&C) Alice and Bob actually measure. The challenge is to produce a dataset (ensemble) that will match the statistical predictions of QM. The QM prediction is 25% match rate when the Alice and Bob settings are different. But your hand picked ensemble can't get closer than 33% rate for all 3 combinations (A&B, B&C, or A&C) - which of course is entirely incorrect. You might get it for one, or maybe even two, but you can't for all three. Of course, a single counterexample invalidates any hypothesis.

Don't believe me, try it yourself. Or check out one of my web pages that explains:
Bell's Theorem with Easy Math

It's not that I don't believe you, I'm actually thankful for your replies. This is a public forum, if I already understood everything I wouldn't bother posting at all.
I do understand now the issue with the perfect correlation, having two detectors at the same angle always produces the same result, which wouldn't be the case for a predetermined random polarization. It wasn't clear from your first answer.

You know, I still have the gut feeling that we must be missing something anyway. It feels like if we forgo local realism we're actually giving up on making sense of the world. I guess nature doesn't HAVE to make sense to us, I just find it difficult to believe the moon is not there when we don't look at it.