What is the physical significance of Bell's math?

[edguy99]

... Then, since all classical situations known to me do just that -- they satisfy (X) -- Bell's theorem is a valid limit on all such classical situations. Full stop.

I disagree. Normally, hidden variables give a photon a specific orientation. This clearly does not work, as the normalized Jones Vector polarization vector is not the absolute orientation, but only the "best guess". Consider a classical model where the axis of a photon "wobbles" left and right as in this picture with of photons at various angles and various amount of "Wobble".

This model does not satisfy (X) as the split photons will only measure the same for sure when measured on their basis vector.

 

[Strilanc]

This is faulty. When you split photons and shoot them in opposite directions, you only measure them the same when they are measured on their basis vectors (vertical or horizontal). If measured off their basis vectors they do not always measure the same. Bell experiments pre-select only the photons that match, hence the statistics do not appear to make sense. They toss out unmatched measurements as noise.

No, singlet states anti-correlate along all directions. Even directions that aren't your basis vectors. This isn't some kind of trick of analysis in the experiment, the prediction comes directly from the math.

The singlet state doesn't even have a distinguished basis. For example, $\frac{1}{\sqrt 2}|01\rangle - \frac{1}{\sqrt 2}|10\rangle$ is exactly equal to $\frac{1}{\sqrt 2}|+-\rangle - \frac{1}{\sqrt 2}|-+\rangle$, where $|+\rangle = \frac{1}{\sqrt 2}|0\rangle + \frac{1}{\sqrt 2}|1\rangle$ and $|-\rangle = \frac{1}{\sqrt 2}|0\rangle - \frac{1}{\sqrt 2}|1\rangle$.

Tossing out possibly relevant measurements as noise sounds like leaving the detection loophole open. And some experiments do that. But there have also been experiments that specifically focus on closing the detection loophole. In the last couple years, there have even been experiments that close both the detection loophole and the signalling loophole at the same time.

 

[edguy99]

... Tossing out possibly relevant measurements as noise sounds like leaving the detection loophole open. And some experiments do that. But there have also been experiments that specifically focus on closing the detection loophole. In the last couple years, there have even been experiments that close both the detection loophole and the signalling loophole at the same time.

This is the paper they are quoting: https://arxiv.org/abs/1508.05949. They are using only "Event Ready" signals, where the two orientations match, despite generating each setup the same way. If they used every signal, they would find some did not match. They are pre-selecting the events.

 

[Zafa Pi]

The assumption that conflicts with reality:
As an answer to the question: . What are the assumptions (hypotheses) of Bell's Theorem that lead to a conflict with reality? I take it that you're making a tautology joke.This is faulty. When you split photons and shoot them in opposite directions, you only measure them the same when they are measured on their basis vectors (vertical or horizontal). If measured off their basis vectors they do not always measure the same. Bell experiments pre-select only the photons that match, hence the statistics do not appear to make sense. They toss out unmatched measurements as noise.

I agree with Strilanc's post #37. In a recent visit to Zeilinger's optics lab in Vienna the pros said the noise was now negligible. I take their word for it since I'm a mere mathematician, and don't know a photon from a fauxton.

 

[N88]

I disagree. Normally, hidden variables give a photon a specific orientation. This clearly does not work, as the normalized Jones Vector polarization vector is not the absolute orientation, but only the "best guess". Consider a classical model where the axis of a photon "wobbles" left and right as in this picture with of photons at various angles and various amount of "Wobble".

This model does not satisfy (X) as the split photons will only measure the same for sure when measured on their basis vector.

Thanks for this; however: Given your model, which is hardly classical, I'd like to see the calculation that breaches Bell's theorem (BT).

Note, there other systems that do no breach BT: Models that are poorly correlated do not breach BT. And, without the wobble, your model does not breach BT. So, as I see it: to breach BT, you need to show that the wobble improves the correlation.

 

[edguy99]

Thanks for this; however: Given your model, which is hardly classical, I'd like to see the calculation that breaches Bell's theorem (BT).

Note, there other systems that do no breach BT: Models that are poorly correlated do not breach BT. And, without the wobble, your model does not breach BT. So, as I see it: to breach BT, you need to show that the wobble improves the correlation.

The "wobble" model does not match the premise of BT. Ie. A photon that is prepared vertical, then split and measured by Bob and Alice at 45 degrees will not always measure the same. The wobble does significantly improve correlation. When Bob and Alice are off in measurement by 30 degrees, the wobble model will have them measure more matches then you would expect. This is because we are throwing out all the mismatches as "noise" prior to counting. When you use this kind of model in an experiment like this, it produces perfect correlation.

 

[stevendaryl]

I question this use of "nonlocal".

I've never met Bob, but I understand that he's on the other side of an idealised EPRB experiment. So, as I sit chatting with Alice, we discuss several things that we can predict with certainty. Examples include: (i) Certainly Bob's result, if he chooses the same detector setting as Alice. (ii) If Bob's setting is at $\theta$ wrt hers, then the probability they will both get the same result is certainly $sin^2(\frac{\theta}{2})$.

Should such information be described as nonlocal?

Well, it's a matter of definition. I'm defining nonlocal information as information about the state of an system extended in space that does not "factor" into knowledge about smaller parts of the system. It's part of the intuition behind Bell's definition of a local realistic theory that nonlocal information of this type is the result of lack of complete information about the local state. If I take a pair of shoes, and put one shoe into one box, another into another box, mix the boxes up and send one box to Alice and another box to Bob, then my information about the system is nonlocal: Either Alice got the left shoe and Bob got the right shoe, or vice-versa. But if I knew in perfect detail the action of "mixing up the boxes", I would know precisely which shoe Alice will get and which shoe Bob will get. So nonlocal information can be replaced by more detailed local information.

I consider this a very important concept, and I also consider the word "local" to be appropriate, since "local" means "having to do with neighborhoods".

 

[edguy99]

I agree with Strilanc's post #37. In a recent visit to Zeilinger's optics lab in Vienna the pros said the noise was now negligible. I take their word for it since I'm a mere mathematician, and don't know a photon from a fauxton.

The noise level may be very low, but it is the definition of entanglement that has the problem. Consider the mathematics of these two models:

Model1: Standard HVT used to model a photon in a Bell (γ − λ is the difference between the photon and detectors angles):

• if |γ − λ| ≤ π/4 then vertical
• if |γ − λ| > 3π/4 then vertical
• horizontal otherwise.

Where color represents the probability of a photon getting through the filter

Model2: Non-Bell style HVT model of photon:

• Chance of vertical measurement = (cos((γ − λ)*2)+1)/2
• Chance of horizontal measurement = (cos((γ − λ + π/2)*2)+1)/2

Where shading represents probability of a photon getting through the filter.

If we apply this mathematics to Experimental loophole-free violation of a Bell inequality using entangled electron spins, where two electrons in different places are prepared in the same state (ie. both are up). These electrons can be measured at different angles to see if they are up or down from that angle. A test is done to see if they measure the same. If they do, they are considered entangled.

Consider the experimental setup. If you measure both electrons from the vertical, they will always measure the same. If you measure them at a different angle from vertical, some will measure up, some will not. In model1, both electrons measure the same all the time, in model2, both electrons will not measure the same all the time. This experiment can be modeled with model2 (and will match it exactly) since all mis-matches are not considered entangled and are not counted.

Model2 makes a lot more sense to me. An electron (or photon) has an axis of spin called the normalized Jones vector. All model2 is doing is assuming the electron spin axis is precessing. This makes the measurement somewhat random when read from a different angle compared to the angle it was created at.

 

[Jilang]

Edguy,I could see how this might lead to the probabilities being higher than you might expect classically, but not how they could be lower.

 

[Zafa Pi]

Is time relevant here? Aren't the relevant orientations: Alice's in A(a,λ_i) under the event that tests the i-th of the twinned-particles and Bob's in B(b,λ'_i) under the event that tests the other twin; i = 1, 2, …, N?

I agree with you except for a technical matter. A(a,λi) and B(b,λ'i) are values of +1 or -1. The orientations are determined by the vectors a and b.

 

[Zafa Pi]

The noise level may be very low, but it is the definition of entanglement that has the problem. Consider the mathematics of these two models:

I am unable to follow your posts, perhaps due to my lack education. Many experimentalists have claimed to have violated Bell Inequalities for versions such as CHSH, GHZ, Hardy, Herbert using entangled photons, and discarding nothing of significance, making a zillion trials and averaging.
Are you saying they are reporting fake news? It is quite fashionable these days.

 

[Zafa Pi]

I would say that QM is nonlocal, in Bell's sense. A local theory in Bell's sense has a notion of "state" such that the state of the entire universe can "factor" into states for each little neighborhood. In QM, there is no notion of the state of a small region, there is only a notion of state for the entire universe as a whole.

In the paper cited in post #1, Bell says, "It is the requirement of locality, or more precisely that the result of a measurement on one system be unaffected by operations on a distant system with which it has interacted in the past". I find this simpler and more comprehensible. Just what would transpire if they were far apart, preformed their measurements at near the same time, and there were no influences that went faster than light.

 

[edguy99]

I am unable to follow your posts, perhaps due to my lack education.

Sorry if I was unclear in the posts and the problem is certainly not your education level. The model1 is taken from Dietrich Dehlinger and M. W. Mitchell “Entangled photons, nonlocality and Bell inequalities in the undergraduate laboratory”. In that paper, they compare the coincidences to the difference in measurement angles of Bob and Alice. This chart illustrates what their experiment got, compared to what a specific "Hidden Variable Theory" (model1) would have got.

From the paper: "Our HVT is very simple, and yet it agrees pretty well with quantum mechanics.". You can see if you follow the white dots on the Calculated chart, at 30 degrees, there is not enough matches and at 60 degrees, there are too many matches. It is probably easier to follow the paper then my postings.

Many experimentalists have claimed to have violated Bell Inequalities for versions such as CHSH, GHZ, Hardy, Herbert using entangled photons, and discarding nothing of significance, making a zillion trials and averaging.

Cant really comment on zillions, but if you have a specific example to post, it would be fun to have a look at it.

 

[edguy99]

Edguy,I could see how this might lead to the probabilities being higher than you might expect classically, but not how they could be lower.

The assumption would be that at small angles (closer to 0 degrees), you have lots of photons with at least a small amount of "wobble" so you get more then expected number of matches, just like QM. At large angles (closer to 90 degrees), you have very few photons with that amount of "wobble" so you get less then expected number of matches, just like QM.

 

[Zafa Pi]

Cant really comment on zillions, but if you have a specific example to post, it would be fun to have a look at it.

This has plenty of experimental errors, but can still disprove the Bell Equality (in this case). That is all that's necessary.
https://vcq.quantum.at/fileadmin/Publications/2002-12.pdf

You can go on line and find many with good accuracy.

 

[stevendaryl]

In the paper cited in post #1, Bell says, "It is the requirement of locality, or more precisely that the result of a measurement on one system be unaffected by operations on a distant system with which it has interacted in the past". I find this simpler and more comprehensible. Just what would transpire if they were far apart, preformed their measurements at near the same time, and there were no influences that went faster than light.

Yes, that is more intuitive, but what does it mean to say that one system affects another system? The description of the facts of EPR are ambiguous. Does Alice's measurement affect Bob's measurement? It's hard to say. On the one hand, there is no way for Alice to send FTL messages to Bob. On the other hand, in the "collapse" interpretation of QM, it is Alice's measurement that causes a collapse, and that in turn does affect what Bob measures. So it's hard to say. But we can definitely say that such a collapse is nonlocal.

 

[Jilang]

Yes, that is more intuitive, but what does it mean to say that one system affects another system? The description of the facts of EPR are ambiguous. Does Alice's measurement affect Bob's measurement? It's hard to say. On the one hand, there is no way for Alice to send FTL messages to Bob. On the other hand, in the "collapse" interpretation of QM, it is Alice's measurement that causes a collapse, and that in turn does affect what Bob measures. So it's hard to say. But we can definitely say that such a collapse is nonlocal.

It certainly causes a collapse for Alice. For Bob I am not convinced.

 

[stevendaryl]

It certainly causes a collapse for Alice. For Bob I am not convinced.

The point is that the most straightforward way of describing the situation after Alice's measurement and before Bob's measurement is nonlocal. There might be a completely local description of the situation, but the collapse interpretation isn't one.

I guess I would say that local/nonlocal is a matter of how something is modeled, rather than something intrinsic to the situation.

 

[Zafa Pi]

Yes, that is more intuitive, but what does it mean to say that one system affects another system? The description of the facts of EPR are ambiguous. Does Alice's measurement affect Bob's measurement? It's hard to say. On the one hand, there is no way for Alice to send FTL messages to Bob. On the other hand, in the "collapse" interpretation of QM, it is Alice's measurement that causes a collapse, and that in turn does affect what Bob measures. So it's hard to say. But we can definitely say that such a collapse is nonlocal.

I agree that what one system affecting another is somewhat ambiguous (as what constitutes a region), but that is Bell's statement not mine. I've previously posted the below which I feel is not ambiguous.

#1 The physical set up for Bell’s Theorem:
Alice and Bob are 2 light minutes apart, and Eve is half way between them. Alice has a fair coin (see probability appendix) and a device. Her device has 2 buttons labeled h and t, and a port to receive a signal from Eve. The device also has a screen that will display “Eve’s signal received” when a signal from Eve is received. It will also display either +1 or -1 if one of the buttons is pushed. Bob has the same equipment and shows the same values, though the internal workings of his device may be different.

#2 The following experiment is performed:
Eve simultaneously sends a light signal to each of Alice and Bob. When Alice’s device indicates Eve’s signal has been received she flips her coin. If it comes up heads she pushes button h, otherwise button t, and then notes what the screen displays. What Alice does takes less than 30 seconds. The same goes for Bob.

#3 Notation & assumption:
If Alice flipped a head and pushed button h, we let Ah be the value her screen would show. So Ah = 1 or -1 and is the result of some objective physical process. Similarly we let At be the value if she had flipped a tail. We let Bh and Bt be the analogous values for Bob. The values Ah etc. may come from a random process. P = probability. We assume no influence can go faster than light, called locality.

Bell’s Theorem: Let Ah, At, Bh, and Bt take on values of ±1. If Ah•Bh = 1, then we have
Bell’s Inequality: P(At•Bt = -1) ≤ P(At•Bh = -1) + P(Ah•Bt = -1).

Proof: P(At•Bt = -1) = P(At•Bt•Ah•Bh = -1) = P(At•Bh•Bt•Ah = -1) = P({At•Bh = -1 and Bt•Ah = 1} or {At•Bh = 1 and Bt•Ah = -1}) =
P(At•Bh = -1 and Bt•Ah =1) + P(At•Bh = 1 and Bt•Ah = -1) ≤ P(At•Bh = -1) + P(Ah•Bt = -1) QED

Now lab tests show Bell's Inequality in the theorem can be violated, yet satisfying #1 & #2.
Of course no bit of reality can refute a math theorem so I ask you what hypotheses of the theorem are violated?

 

[Jilang]

The assumption would be that at small angles (closer to 0 degrees), you have lots of photons with at least a small amount of "wobble" so you get more then expected number of matches, just like QM. At large angles (closer to 90 degrees), you have very few photons with that amount of "wobble" so you get less then expected number of matches, just like QM.

Sorry I don't see it. If there was no wobble wouldn't you get the expected number of matches?

 

[stevendaryl]

Now lab tests show Bell's Inequality in the theorem can be violated, yet satisfying #1 & #2.
Of course no bit of reality can refute a math theorem so I ask you what hypotheses of the theorem are violated?

I'm a little confused by your scenario. Are you trying to describe the actual EPR experiment, abstractly, or a classical experiment that is similar in flavor?

 

[N88]

I agree with you except for a technical matter. A(a,λi) and B(b,λ'i) are values of +1 or -1. The orientations are determined by the vectors a and b.

Thanks. My expression is a short-form colloquialism. Expanding on my note, to be clear: What is relevant is the orientation of Alice's detector during the event A(a,λ_i); where A(a,λ_i) denotes (in short) the interaction between a and λ_i. That orientation is a. The other relevant orientation is Bob's detector-setting b in A(b,λ_i). HTH.

 

[N88]

I agree that what one system affecting another is somewhat ambiguous (as what constitutes a region), but that is Bell's statement not mine. I've previously posted the below which I feel is not ambiguous.

#1 The physical set up for Bell’s Theorem:
Alice and Bob are 2 light minutes apart, and Eve is half way between them. Alice has a fair coin (see probability appendix) and a device. Her device has 2 buttons labeled h and t, and a port to receive a signal from Eve. The device also has a screen that will display “Eve’s signal received” when a signal from Eve is received. It will also display either +1 or -1 if one of the buttons is pushed. Bob has the same equipment and shows the same values, though the internal workings of his device may be different.

#2 The following experiment is performed:
Eve simultaneously sends a light signal to each of Alice and Bob. When Alice’s device indicates Eve’s signal has been received she flips her coin. If it comes up heads she pushes button h, otherwise button t, and then notes what the screen displays. What Alice does takes less than 30 seconds. The same goes for Bob.

#3 Notation & assumption:
If Alice flipped a head and pushed button h, we let Ah be the value her screen would show. So Ah = 1 or -1 and is the result of some objective physical process. Similarly we let At be the value if she had flipped a tail. We let Bh and Bt be the analogous values for Bob. The values Ah etc. may come from a random process. P = probability. We assume no influence can go faster than light, called locality.

Bell’s Theorem: Let Ah, At, Bh, and Bt take on values of ±1. If Ah•Bh = 1, then we have
Bell’s Inequality: P(At•Bt = -1) ≤ P(At•Bh = -1) + P(Ah•Bt = -1).

Proof: P(At•Bt = -1) = P(At•Bt•Ah•Bh = -1) = P(At•Bh•Bt•Ah = -1) = P({At•Bh = -1 and Bt•Ah = 1} or {At•Bh = 1 and Bt•Ah = -1}) =
P(At•Bh = -1 and Bt•Ah =1) + P(At•Bh = 1 and Bt•Ah = -1) ≤ P(At•Bh = -1) + P(Ah•Bt = -1) QED

Now lab tests show Bell's Inequality in the theorem can be violated, yet satisfying #1 & #2.
Of course no bit of reality can refute a math theorem so I ask you what hypotheses of the theorem are violated?

Dear Zapa Pi: as previously discussed, this example might make for some interesting discussion. It will certainly allow me to make several points that have not yet been made.

So, please: lay-out the equations in a regular math format and number EVERY separate equation and inequality.

Also: Since no one has answered to your satisfaction thus far, I suggest it's time for your own explanation. Mine (which you did not understand) is sealed (as you know) in my private note to you.

PS: Or are you asking the question because you are mystified and do not have an answer?

 

[Zafa Pi]

I'm a little confused by your scenario. Are you trying to describe the actual EPR experiment, abstractly, or a classical experiment that is similar in flavor?

I gave what I consider a coherent definition of locality. The experiment could either be classical or quantum and could include EPR.
You might prefer: nielsen and chuang p.111

 

[Zafa Pi]

Dear Zapa Pi: as previously discussed, this example might make for some interesting discussion. It will certainly allow me to make several points that have not yet been made.

So, please: lay-out the equations in a regular math format and number EVERY separate equation and inequality.

You do it.

Also: Since no one has answered to your satisfaction thus far, I suggest it's time for your own explanation. Mine (which you did not understand) is sealed (as you know) in my private note to you.

The answer is similar to the one I gave in post #2. In each trial of the experiment only two values are revealed by measurements, e.g. At and Bh. yet in Bell's Theorem we find four values all appearing in the same expression. What allows this is the assumption of CFD, without which we cannot consider an expression like At•Bh•Bt•Ah.
Thus CFD is incompatible with certain aspects of nature. Bohr said something like, "Unmeasured entities have no value." (You can get them for free)
CFD, hidden variables, realism, and determinism all come to the same thing in this context.

So if Alice flipped a head and found Ah = 1, while Bob flipped a tail and got -1 for Bt, Would Bob have got 1 if he flipped heads instead? (recall Ah = Bh)
Yes follows from realism, but how would one go about testing whether it's true. It's in general impossible, we need to preform a new experiment in which case maybe Ah = Bh = -1.

That is why in a previous thread I said:
Let us suppose that:
1) Alice and Bob are isolated from one another, so that no communication or influence can pass between them and neither knows what the other is doing.
2) If Alice and Bob both perform experiment X they will get the same result.
3) Alice performs experiment X and gets value 0, while Bob performs experiment Y and gets 1.
Then
4) If Bob had performed X instead of Y would he have necessarily gotten 0?

Classical physics says yes and quantum physics says no. This is a simple proposition that distinguishes classical and quantum requiring no knowledge of physics at all!

 

[edguy99]

Sorry I don't see it. If there was no wobble wouldn't you get the expected number of matches?

Using expected was unclear. To clarify, with no wobble you get the "calculated" graph in post #48, ie. it does not match the reality of QM. With a wobble, you match the "experimental", ie. it matches the reality of QM.

 

[N88]

I gave what I consider a coherent definition of locality. The experiment could either be classical or quantum and could include EPR.
You might prefer: nielsen and chuang p.111

? Is p.111 meant to be included in the linked material (which appears to end at p.20)?

 

[Zafa Pi]

? Is p.111 meant to be included in the linked material (which appears to end at p.20)?

Indeed. That's why I didn't link directly to the pdf. It is just a different Bell Inequality, CHSH, but I really like the they write. Their book is great for me since it requires minimal physics.

If you like the "Jabberwocky" you may also like http://www.askamathematician.com/2009/12/q-howwhy-are-quantum-mechanics-and-relativity-incompatible/
then scroll down till you get to the posts of Stephen Tuck. It's like what a lot of physics sounds like to me.

 

[edguy99]

This has plenty of experimental errors, but can still disprove the Bell Equality (in this case). That is all that's necessary.
https://vcq.quantum.at/fileadmin/Publications/2002-12.pdf

You can go on line and find many with good accuracy.

Page 231: "A correlation circuit extracts only those events where all four detectors registered a photon
within a small time window of a few ns."

This condition will exclude all mismatched entangled pairs and can be modeled by a photon with a "wobble". ie. if you model the setup with a photon with a wobble, mis-matches will not be counted and you end up with the same statistics as QM.

 

[N88]

Indeed. That's why I didn't link directly to the pdf. It is just a different Bell Inequality, CHSH, but I really like the they write. Their book is great for me since it requires minimal physics.

If you like the "Jabberwocky" you may also like http://www.askamathematician.com/2009/12/q-howwhy-are-quantum-mechanics-and-relativity-incompatible/
then scroll down till you get to the posts of Stephen Tuck. It's like what a lot of physics sounds like to me.

Let me try it another way. Please: Do you mean that the material your refer to is in pp.111-117?

 

[Zafa Pi]

Page 231: "A correlation circuit extracts only those events where all four detectors registered a photon
within a small time window of a few ns."

This condition will exclude all mismatched entangled pairs and can be modeled by a photon with a "wobble". ie. if you model the setup with a photon with a wobble, mis-matches will not be counted and you end up with the same statistics as QM.

I think what they are doing is making sure that the photon that was sent was the the one received. It's not about noise. It's like I'm testing the weight of frogs and I exclude a fish that got into my sample. If you think I'm wrong (highly possible here) can you talk to me like I've been talking here. I don't follow what your wobble is about.

 

[Zafa Pi]

Let me try it another way. Please: Do you mean that the material your refer to is in pp.111-117?

Yes.

 

[N88]

I think what they are doing is making sure that the photon that was sent was the the one received. It's not about noise. It's like I'm testing the weight of frogs and I exclude a fish that got into my sample. If you think I'm wrong (highly possible here) can you talk to me like I've been talking here. I don't follow what your wobble is about.

Zafa Pi and edguy99: It's my impression that the photons (in the cited example) do NOT wobble. Rather: the detectors (in the cited example) are such [sic] that they collect widely-differing input-polarizations and direct them to a single output. I have yet to see where this idea breaches Bell's theorem.

 

[edguy99]

I think what they are doing is making sure that the photon that was sent was the the one received. It's not about noise. It's like I'm testing the weight of frogs and I exclude a fish that got into my sample. If you think I'm wrong (highly possible here) can you talk to me like I've been talking here. I don't follow what your wobble is about.

Absolutely I agree on the reason they do it. But.. by doing this they are assuming the the photon obeys the Bell assumption that entangled photons will match no matter the angle they are measured at. Assume that the polarization is represented by the normalized Jones vector.

By introducing a "wobble", we can make some sense out of the QM fact that these polarization are only the "best guess" (amplitude of probability if you like) of what you will measure the polarization as. QM tells us that if you create a vertical photon, you will alway detect a vertical photon if measured vertical. But if you create a vertical photon and measure it an a different angle (say 45 degrees), the result is probabilistic, as if there was a wobble in the orientation. The point is that a photon with a wobble, will not match the starting conditions of a Bell Test since Bob and Alice are not guaranteed to get matches when photons are measured off of their basis vectors. But a photon with a wobble will match QM if we do not use the entangled pairs that do not match at weird angles as many experiments do.

 

[Zafa Pi]

Zafa Pi and edguy99: It's my impression that the photons (in the cited example) do NOT wobble. Rather: the detectors (in the cited example) are such [sic] that they collect widely-differing input-polarizations and direct them to a single output. I have yet to see where this idea breaches Bell's theorem.

The Bell Theorem in this case is the GHZ Theorem, and the purpose of the paper is to refute the equality in the theorem in the lab as does QM in theory (which I do understand).
If you believe they failed talk it over with edguy99. it is really out of my realm.