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

In summary: EPR concept were true, we would need to encode a hidden variable for each degree angle in order to get results that looked like perfect correlations between particles. And we would need to encode a hidden variable for every 2 degrees.
  • #36
Whenever one tries to "count" things in theoryspace one often get lost fast due to the curse of the real numbers.
DrChinese said:
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
 
Physics news on Phys.org
  • #37
martinbn said:
Are angels bosons or fermions?
Angels are ghosts, and ghosts are fermions with zero spin.
 
  • #38
vanhees71 said:
anyons...
So how can they live in 3 dimensions?
 
  • Like
Likes vanhees71
  • #39
The head of the pin is a 2D surface ;-).
 
  • Haha
  • Like
Likes DrChinese, gentzen and Demystifier
  • #40
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.
 
  • #41
Leureka said:
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.
 
  • Like
Likes vanhees71
  • #42
DrChinese said:
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.
 
  • #43
Leureka said:
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? :smile:
 
  • Like
Likes vanhees71
  • #44
DrChinese said:
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.

DrChinese said:
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.
 
  • #45
Leureka said:
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
 
  • #46
DrChinese said:
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.
 
  • Like
Likes DrChinese

Similar threads

Replies
50
Views
5K
Replies
333
Views
14K
Replies
96
Views
6K
Replies
3
Views
3K
Replies
2
Views
2K
Replies
31
Views
2K
Replies
1
Views
1K
Back
Top