Understanding bell's theorem: why hidden variables imply a linear relationship?

[lugita15]

Are you a real person!? Have you read anything I've written in this thread, at all. Do you understand any of it?
I have already shown you that they must not always be equal. I have used several examples, with quotes from Bell himself to explain to you that they are not equal in the two scenarios. Yet you ask me again if I agree that they are always equal.

As far as I can tell the only response you've given to the argument involving the two questions (which I just restated in my previous post) was in post #262, where you give the example of Bernoulli's urn. But as I said, Bernoulli's urn is an illustration of something involving probabilities, and my two questions do not deal with probability at all.

The two Bell quotes above are also irrelevant to this point: one talks about how the sum of two measurements is not the measurement of the sum, and the other quote talks about probability.

If anything else you've said was a response to my two questions, then I must have missed it, so please tell me if there are any other examples or arguments you have against my two questions having the same answer.

 

[DevilsAvocado]

OK, then that's what we need to focus on. Again, I still want to know whether you agree or disagree with the answers to my two questions are always the same. Because if they are always the same, then I can show that P(A|x) must always equal P(A|w).

Lugita, you aren’t getting anywhere with this. Check out my post in #274 and Bill’s reply in #275 and you’ll see that Bill can’t keep two balls in the air at once. He – for real – thinks that:

• QM is wrong.
• QM measurement outcome is “definite” and “only appears random”.
• QM measurement of a single entangled photon is not 100% random 50/50 +1/-1.
• Bell's theorem is not about statistical correlations, but probabilities for a single photon.
• The cos^2(a - b) rule is not about the relative angle between Alice & Bob.
• Instead we can apply cos^2(θ) for any single photon/angle outcome.
• We can use Bayesian probability theory to investigate single photon properties, before measurement.
• At lot of measurements (hundreds/thousands) is not needed to get the (statistical) correlations.
• Experimental outcome/statistics of EPR-Bell test experiments is wrong.
• The definite property of LHV is not obligatory.
• Bell's inequality is not an upper limit for Local Realism only.
• To test QM against Bell's inequality, QM needs to posses the same basic properties as Local Realism, otherwise the test is groundless, i.e. Local Realism and QM has to be on the same footing, to be rigorously tested against Bell's inequality.

And AFAIK, Bill seems to think that QM and Local Realism is really the same thing, except for some “mysticism buffoons” [i.e. the rest of the world] who got it all wrong...

This discussion will take you nowhere.

 

[billschnieder]

Because if they are always the same, then I can show that P(A|x) must always equal P(A|w)..

Are you for real? You want to argue that two probabilities are always the same, yet you do not want me to mention probabilities when explaining to you that they are not always equal.

 

[billschnieder]

But as I said, Bernoulli's urn is an illustration of something involving probabilities, and my two questions do not deal with probability at all.

Contrary to your claims, they in fact do because you are using them to claim that two probabilities are the same. Isn't that what this is all about. I don't care about your two statements, I go directly to your claim that the probabilities are equal in the two scenarios and I show you that they are not. Besides, I have given you examples that have nothing to do with probability. For example:

T_b is just a fixed number like 5 in the physical situation in which you are only putting on socks. T_b is just a fixed number like 8, in the physical situation in which we are putting on shoes and then socks. T_b in in the first physical situation, is not the same value as T_b in the second physical situation, even though in both cases T_b is still the time it takes to put on socks.

The two Bell quotes above are also irrelevant to this point

That is because you do not understand the argument. They are very relevant. You are making the same tacit assumption Bell refers to in his second quote, that a measurement being made in one scenario will always give the same result in a different scenario, you naively assume that probabilities from scenario "xyz" can be used in scenario "w". You are making the same error, that Bell talks about in the first quote, because you assume that the sum of two noncommuting observables P(A|w) + P(B|w) can be obtained by combining trivially the results of separate observations on the two terms P(A|x) + P(B|y)

 

[billschnieder]

Consider a photon before it is measured. By counterfactual definiteness, for every angle setting θ1, there exists a definite, predetermined answer to the question "What result would you get if the photon were measured at angle θ1?". Now if the experimenter can freely choose his measurement decision, then the predetermined answer to this question cannot depend on what he is going to do, so it cannot depend on the angle θ2 he is going to measure at, whether θ1=θ2 or θ1≠θ2. Do you agree or disagree with that?

Ok for the last time, please pay attention carefully.
1) You do not have a single angle, you have two angles (a,b) for the photon pair, where your θ1 = (a-b). And you have two angles (a,c) where your θ2 = (a-c).
2) You are not considering each result independtly, you are considering them jointly in the same expression, in fact adding them together "answer_at(a,b) + answer_at(a,c)" ...
3) You are claiming that in the expression you have which includes both answers, the answer at (a,c) can not depend on the answer at (a,b).

However, NOTE: we are talking about the same photon pair measured at (a,b) and counterfactually measured at (a,c). Let us call the photons p1 and p2 where p1 is sent to Bob and p2 is sent to Alice. In the factual measurement at (a,b), Bob measures p1 at angle a, and Alice measures p2 at angle b. Now in the counterfactual measurement, Bob is not doing anything differently that what he already did. ONLY Alice's measurement would have been different because she is now doing the measurement at angle "c" instead of "b" at which she measured previously.

Therefore, contrary to what you naively thought, the result you would have obtained counterfactually at (a,c) does depend on the result you obtained for (a,b) for the simple reason that the same photon can not counterfactually give you something different when measured at "a", than what was already obtained at "a". In other words, Bob's result at "a" for the (a,b) measurement, constraints Bob's result at "a" for the counterfactual measurement (a,c). Get it?

Let us assume that the outcomes for each photon can be either + or -. Let us now replace the angles in the expression "answer_at(a,b) + answer_at(a,c)" with the corresponding answers for the counterfactual situation. If the measurement at (a,b) gave the answer (+,-), we now have "answers(+,-) + answers(+,c)". Do you notice that because we already have the answer for the first term, we therefore already have part of the answer for the second term, since we are talking about the same photon? Do you realize that although the result (a,c) is counterfactual to (a,b), it is in fact only Alice's part of the measurement that is counterfactual? Do you realize that to be consistent, Bob's part of the answer can not change from the factual?

Now what about three different photon pairs where the (a,b) measurement is performed on (p1, p2) and the (a,c) measurement is performed on a different pair of photons (p3, p4). In this case, we do not have any counterfactual measurements. Bob's result for p3 is not constrained by Bob's result for p1, because they are different photons which are free to have different results.

In other words, in scenario "xyz", there are no logical relationships between results from the sets of photons "x", "y" and "z" because they are three different sets. However in the scenario "w", we are talking about the same set, and therefore there are logical relationships between measurements on that set.

 

[billschnieder]

Lugita, you aren’t getting anywhere with this. Check out my post in #274 and Bill’s reply in #275 and you’ll see that Bill can’t keep two balls in the air at once. He – for real – thinks that:
...

Obviously, you are wrong and completely off base.

 

[BenjaminTR]

[...]

Therefore, contrary to what you naively thought, the result you would have obtained counterfactually at (a,c) does depend on the result you obtained for (a,b) for the simple reason that the same photon can not counterfactually give you something different when measured at "a", than what was already obtained at "a". In other words, Bob's result at "a" for the (a,b) measurement, constraints Bob's result at "a" for the counterfactual measurement (a,c).

[...]

If I understand, the independence lugita15 is talking about is independence between what properties a particle has when it becomes entangled and what measurement we decide to make. Since the LHV theory is that properties of the particles determine measurement outcomes, we can use measurement outcomes as proxy for properties.
Let A+ be the proposition that the particle would measured + in orientation A, and x be the event that we actually measure in orientation A. Let B+ be the proposition that particle would be measured + in orientation B and y be the event that we actually measure orientation B. Let w be the event that we measure the particle at all. Independence says that P(A+|x) = P(A+|y) = P(A+|w). Same for B+.
In the quoted passage, you are talking about P(B+|A+ & x). No one assumes B+ is independent of A+ & x, so your argument here does not undermine lugita15's or Herbert's proofs.

 

[billschnieder]

If I understand, ...
and x be the event that we actually measure in orientation A. ...
and y be the event that we actually measure orientation B.
Let w be the event that we measure the particle at all.

...

Huh? You have completely misunderstood the argument. Please go back and re-read the thread from page 15, carefully in order to understand what we are discussing.

Le us represent the sets of photons as one set "w" for scenario (c), and three different sets ("x", "y","z") for scenario (b). The the probabilities for (c) are more accurately represented as P(A|w), P(B|w), P(C|w). ...

However the probabilities for scenario (b) are more accurately represented as P(A|x), P(B|y), P(C|z).

 

[lugita15]

1) You do not have a single angle, you have two angles (a,b) for the photon pair, where your θ1 = (a-b). And you have two angles (a,c) where your θ2 = (a-c).

I was talking about individual photons, and then I was going to move to photon pairs. But if you want to start with photon pairs that's fine.

3) You are claiming that in the expression you have which includes both answers, the answer at (a,c) can not depend on the answer at (a,b).

What I'm claiming is that before measurement, the answer you would get at (a,c) is entirely predetermined, and it can't depend on whether you're going to measure at (a,b) or (a,c).

However, NOTE: we are talking about the same photon pair measured at (a,b) and counterfactually measured at (a,c). Let us call the photons p1 and p2 where p1 is sent to Bob and p2 is sent to Alice. In the factual measurement at (a,b), Bob measures p1 at angle a, and Alice measures p2 at angle b. Now in the counterfactual measurement, Bob is not doing anything differently that what he already did. ONLY Alice's measurement would have been different because she is now doing the measurement at angle "c" instead of "b" at which she measured previously.

I agree with this.

Therefore, contrary to what you naively thought, the result you would have obtained counterfactually at (a,c) does depend on the result you obtained for (a,b) for the simple reason that the same photon can not counterfactually give you something different when measured at "a", than what was already obtained at "a". In other words, Bob's result at "a" for the (a,b) measurement, constraints Bob's result at "a" for the counterfactual measurement (a,c). Get it?

The thing is, for any given photon pair, both what you would get at (a,b) and what you would get at (a,c) is entirely predetermined beforehand, before you make the measurement at (a,b). Do you agree or disagree with that with that?

The point is, that the predetermined answer to the question "What would you get at (a,c)" can't depend on what you're going to measure later on, because (by assumption) you have free will and can choose to measure anything you want.

Do you realize that although the result (a,c) is counterfactual to (a,b), it is in fact only Alice's part of the measurement that is counterfactual? Do you realize that to be consistent, Bob's part of the answer can not change from the factual?

Yes, this is all fine.

In other words, in scenario "xyz", there are no logical relationships between results from the sets of photons "x", "y" and "z" because they are three different sets. However in the scenario "w", we are talking about the same set, and therefore there are logical relationships between measurements on that set.

The point is, for any given photon pair, what you would get at (-30,0) is entirely independent of whether you're actually going to measure at (-30,0) or (0,30), because at the time that the answer to the question "What would you get at (-30,0)?" is determined, the experimenter hasn't yet made up his mind about whether he'll measure at (-30,0) or (0,30).

So the question of what you would get at (-30,0) is independent of whether the photon pair is going to be in set x or set y, and thus the percentage of pairs in set x for which you would get R1 at (-30,0) is the same as the percentage of pairs in set y for which you would get R1 at (-30,0). Do you agree or disagree with that?

 

[billschnieder]

What I'm claiming is that before measurement, the answer you would get at (a,c) is entirely predetermined

I agree with this. I also agree that the outcome of any measurement is predetermined no matter the angle.

, and it can't depend on whether you're going to measure at (a,b) or (a,c).

I don't think you fully understand the implication of this claim. It is not correct to say the outcome at (a,b) does not depend on the angle (a,b) or the outcome at (a,c) does not depend on the angle (a,c), since those settings are part of the set-up which produces the outcome. Probably what you are trying to say is that the outcome of the single photon measured at angle "a" in the (a,b) pair must be exactly the same as the outcome of the same photon measured at angle "a" in the (a,c) pair. In other words, the outcome for that particular single photon should not change from (a,b) to (a,c) just because it's sibling was measured at "c" in (a,c) instead of "b" in (a,b). So Alice's choice of setting should not influence Bobs outcome for the same setting "a" and the exact same photon. If this is what you mean, then I agree. But if you mean that the outcome at (a,b) can not depend on the angle pair (a,b) then I disagree with that.

The thing is, for any given photon pair, both what you would get at (a,b) and what you would get at (a,c) is entirely predetermined beforehand, before you make the measurement at (a,b). Do you agree or disagree with that with that?

I agree.

The point is, that the predetermined answer to the question "What would you get at (a,c)" can't depend on what you're going to measure later on, because (by assumption) you have free will and can choose to measure anything you want.

No. I disagree with that. Just because it is predetermined to produce the result at (a,c) doesn't mean it can produce the result without (a,c). Without the setting (a,c) you will not get the result at (a,c) so how can you say the result obtained at (a,c) does not depend on what you actually choose (a,c)? What is predetermined is the condition that "if you set the device to (a,c), you will obtain such and such result". It doesn't make much sense to say the result you obtain at (a,c) does not depend on the setting, which is required to obtain the result! Free will has nothing to do with it. You are free to choose a different setting, you will get a different result. But whenever you freely choose (a,c) you will get the result predetermined for (a,c). Probably what you are trying to say here is that the result for the single photon measured at angle "a", should not depend on whether you measured the pair at (a,c) or you measured (a,b). Again, that means Bob's result for the same photon measured at the same angle should not depend on what angle Alice chose to measure it's sibling photon at. If this is what you mean, I agree.

The point is, for any given photon pair, what you would get at (-30,0) is entirely independent of whether you're actually going to measure at (-30,0) or (0,30), because at the time that the answer to the question "What would you get at (-30,0)?" is determined, the experimenter hasn't yet made up his mind about whether he'll measure at (-30,0) or (0,30).

See explanation above. The results predetermined for a given angle pair can not be independent of the angle pair for which it is predetermined.

 

[billschnieder]

So the question of what you would get at (-30,0) is independent of whether the photon pair is going to be in set x or set y, and thus the percentage of pairs in set x for which you would get R1 at (-30,0) is the same as the percentage of pairs in set y for which you would get R1 at (-30,0). Do you agree or disagree with that?

No! The same photon can not belong to two different sets. If a specific photon is predetermined to give a result at -30, then obviously the result when that specific photon and it's sibling are measured at (-30, 0) respectively CANNOT CONTRADICT the result when that same photon and it's sibling were instead measured at (-30, 30). In other words, the same photon can not be predetermined to have different results at the same angle, however, two different photons can be predetermined to have opposite results at the same angle. Do you agree with this?

Remember I'm explaining why the results from three different sets "x", "y", "z" are not the same as the results from one single set "w".

Scenario "xyz"
A_x: results from measuring photon set x at angles (a,b)
B_y: results from measuring photon set y at angles (a,c)
C_z: results from measuring photon set z at angles (c,b)

Scenario "w"
A_w: results from measuring photon set x at angles (a,b)
B_w: results from measuring photon set y at angles (a,c)
C_w: results from measuring photon set z at angles (c,b)

Your claim is that the results from Scenario "xyz" are the same as the results from scenario "w", ie, A_x = A_w, B_y = B_w and C_y = C_w.

For scenario "w", the results at angle "a" when measured at the angle pair (a,b) must be the same as the results for angle "a" when measured at angle pair (a,c) and the results for angle "b" when measured at angle pair (a,b) must be exactly the same as the results for angle "b" when measured at angle pair (c,b) etc.. for "c" too. This is the case because we are dealing with the same set of photons "w". There is therefore a cyclic logical relationship between the results at (a,b), (a,c) and (c,b).

However, for scenario "xyz", since each result is obtained from a different set of photons, the results at "a" in the (a,b) pair can be different from the results for "a" in the (a,c) pair and the same for "b", and "c". There is therefore no cyclic logical relationship between the results at (a,b), (a,c) and (c,b) in this scenario.

Note. In all of this explanation, the results for a specific photon measured angle "a" does not depend on the angle at which it's sibling was measured at, and all the results are predetermined.

Another way to look at this is to say that results drawn from a single set "w" have additional constraints between them that results drawn from three different sets "x", "y", "z" do not have. These are the constraints that Boole worked on 150 years ago. So knowing the result (a,b) from "w" give's you insight into the properties of "w" which might be important for the results (a,c) and (c,b). However knowing (a,b) from the set "x", gives you no insight into the properties of "y" which are important for the results (a,c) or the results (c,b) from another set z. Because the sets are all different.

This is why I disagree with your claim is that the results from Scenario "xyz" are the same as the results from scenario "w", ie, A_x = A_w, B_y = B_w and C_y = C_w.

 

[DevilsAvocado]

In other words, the same photon can not be predetermined to have different results at the same angle,

Pretty sweet, you’ve summarized Bell’s theorem in a single line.

These are the constraints that Boole worked on 150 years ago.

And many will draw the logical conclusion that Boole was probably not completely familiar with the life of QM photons (that wasn’t ‘invented’ until 1926), hence Boole’s work will almost certainly say very little (if anything?) about the properties of unmeasured QM photons.

This of course, is just a personal speculation.

 

[billschnieder]

Pretty sweet, you’ve summarized Bell’s theorem in a single line.

As usual, you say much about stuff you know little about. This is not Bell's theorem.

And many will draw the logical conclusion that Boole was probably not completely familiar with the life of QM photons (that wasn’t ‘invented’ until 1926)

What you do not know is that Bell's inequalities were originally discovered by Boole. Boole found that one of the constraints that properties from a given set must obey to be logically consistent was Bell's inequalities. Bell re-discovered the same rules without knowing that Boole had already discovered them a century earlier. One thing Boole also found out, was that properties from a single set can never violate the inequalities but properties from different sets can.

Remember this article I mentioned earlier:

George Boole's ‘Conditions of Possible Experience’ and the Quantum Puzzle
ITAMAR PITOWSKY, Br J Philos Sci (1994) 45 (1): 95-125. doi: 10.1093/bjps/45.1.95

One thing should be clear at the outset: none of Boole's conditions of possible experience can ever be violated when all the relative frequencies involved have been measured in a single sample. The reason is that such a violation entails a logical contradiction. For example, suppose that we sample at random a hundred balls from an urn. Suppose, moreover that 60 of the balls sampled are red, 75 are wooden and 32 are both red and wooden. We have p1=0.6, p2= 0.75, p12 = 0.32. But then p1+p2-p12 > 1. This clearly represents a logical impossibility, for there must be a ball in the sample (in fact three balls) which is 'red', is 'wooden', but not 'red and wooden'; absurd.
Similar logical absurdities can be derived if we assume a violation of any of the relevant conditions, no matter how complex they appear to be. This is the reason for the title 'conditions of possible experience'. In case we deal with relative frequencies in a single sample, a violation of any of the relevant Boole's conditions is a logical impossibility.
But sometimes, for various reasons, we may choose or be forced to measure the relative frequencies of (logically connected) events, in several distinct samples. In this case a violation of Boole's conditions may occur.

Bell's inequality is simply one of "Boole's conditions of possible experience".

 

[billschnieder]

The point is, that the predetermined answer to the question "What would you get at (a,c)" can't depend on what you're going to measure later on, because (by assumption) you have free will and can choose to measure anything you want.

It occurs to me that when you use the word "depend", you may be assuming causal dependence. You are thinking that since what happens later can not change what happens earlier, it means what happens earlier cannot "depend" on what happens later. But this is just wrong, dependence does not have to mean causation, that is why I gave you the Bernouli urn example. Two events can be causally independent and yet logically dependent.

 

[DevilsAvocado]

This is not Bell's theorem.

I didn’t expect you to recognize it.

Bell's inequalities were originally discovered by Boole.

There’s only one “little” glitch, that you constantly refuse to discuss.

We can use classical logic to calculate the properties of unmeasured Local Hidden Variables that acts the classical way, but I sure hope you are not saying that Boole – 150 years ago – developed a “quantum logic” that makes it possible to deal with superposition, entanglement, uncertainty principle, etc??

Besides this erroneous approach – you’re arguing that a single QM photon/pair could give you the correct probabilities/correlations, when we all know that a single QM outcome is always 100% random, no matter what angle. We need hundreds of measured entangled pairs to get the correlations, which you refuse to discuss or admit.

The stochastic nature of QM is fundamental – and absolutely not only “appears random” – to quote your own catastrophic lack of knowledge.

Hence, Boole works excellent for LHV in every instance – before & after measurement – and it even works for a single “LHV photon” outcome, because they must be DEFINITE/PREDETERMINED. However, Boole can’t say anything about QM photons before measurement, and a single QM measurement outcome is always 100% random.

This is what I mean when I say that you can’t keep two balls in the air at once. You refuse to see there’s a BIG difference between unmeasured “LHV photons” and unmeasured QM photons, and not even dear old Boole can help you get them on same footing.

The only thing you can do is to deal with measurement STATISTICS for hundreds/thousands of QM photons, and there goes your “personal theory” down the drain, and this is why you react the way you do.

I’m not expecting any other reaction this time.

 

[billschnieder]

There’s only one “little” glitch, that you constantly refuse to discuss.

We can use classical logic to calculate the properties of unmeasured Local Hidden Variables that acts the classical way, but I sure hope you are not saying that Boole – 150 years ago – developed a “quantum logic” that makes it possible to deal with superposition, entanglement, uncertainty principle, etc??

Besides this erroneous approach – you’re arguing that a single QM photon/pair could give you the correct probabilities/correlations, when we all know that a single QM outcome is always 100% random, no matter what angle. We need hundreds of measured entangled pairs to get the correlations, which you refuse to discuss or admit.

The stochastic nature of QM is fundamental – and absolutely not only “appears random” – to quote your own catastrophic lack of knowledge.

Hence, Boole works excellent for LHV in every instance – before & after measurement – and it even works for a single “LHV photon” outcome, because they must be DEFINITE/PREDETERMINED. However, Boole can’t say anything about QM photons before measurement, and a single QM measurement outcome is always 100% random.

This is what I mean when I say that you can’t keep two balls in the air at once. You refuse to see there’s a BIG difference between unmeasured “LHV photons” and unmeasured QM photons, and not even dear old Boole can help you get them on same footing.

The only thing you can do is to deal with measurement STATISTICS for hundreds/thousands of QM photons, and there goes your “personal theory” down the drain, and this is why you react the way you do.

I’m not expecting any other reaction this time.

There is nothing coherent here to respond to. You probably think you are making a lot of sense but somebody needs to tell you that this is nonsense. It is difficult to discuss with somebody who does not even understand the basics of the topic let alone thinking rationally about any topic.

 

[billschnieder]

Scenario "w"
A_w: results from measuring photon set x at angles (a,b)
B_w: results from measuring photon set y at angles (a,c)
C_w: results from measuring photon set z at angles (c,b)

What I meant to say here is:
A_w: results from measuring photon set w at angles (a,b)
B_w: results from measuring photon set w at angles (a,c)
C_w: results from measuring photon set w at angles (c,b)

 

[DevilsAvocado]

Lugita15, ask Buffalo* Bill exactly what set x and set w represents in terms of number of photons and exactly where in the process – before/after measurement – he perform his magical pet theory.


(and don’t forget to wear raincoat against the flood of personal insults)


*droppings

 

[billschnieder]

Lugita15, ask Buffalo* Bill exactly what set x and set w represents in terms of number of photons and exactly where in the process – before/after measurement – he perform his magical pet theory.

In Bell-test experiments, to measure the probability of mismatches at angle pair (a,b), the Alice sets her polarizer at angle "a", Bob sets his to angle "b", the source is turned on and hundreds of thousands of photons or even millions of photons go through. Everytime a photon goes through Alice's device she records the detector which fired, either a +1 or a -1 for D+ or D- respectively. At the end of the "run", Alice has a long list consisting of +'s and -'s in random order and Bob has a corresponding list. The lists are brought together and you have something that looks like:

ab
-+
++
+-
--
... etc
for as long as there were photon pairs produced, where each row corresponds to results from one single photon pair. It is this list that is used to calculate the C(a,b), or probability of mismatches, or any other statistic you may want to calculate for the angle pair (a,b) in the Bell test experiment. Let us call the set of photons which produced this list "x".

Each photon can only be measured once, being destroyed in the process. Therefore to measure at the angle pair (a,c), a different set of photons has to be used, in a similar manner as above. Let us call this set "y". And similarly, yet a different set "z" is needed to measure at the angle pair (b,c).

Each of the sets "x", "y", "z", is a different set of photons. In other words, no photon belongs to more than one set. No photon is common between those sets because no photon was measured more than once.

An alternative method of doing the experiment involves random switching of the angles by Alice and Bob. However, in the end, to calculate statistics for (a,b), they select only those photons measured when Alice had her device at angle "a" and Bob had his at angle "b". it is this set that we called "x", and the other two are "y" and "z".

Therefore, it should be obvious that for any Bell test experiment the correlations, C(a,b), C(b,c), C(a,c) are each measured on a distinct set of photons. If you prefer, P(A|x), P(B|y), P(C|z) are each measured on a different set of photons. In other words, we have 3 different sets of photons (x, y, z). This is what is referred to in the above discussion as scenario "xyz" and it refers to the way experimental correlations are measured.

The QM predictions are similarly made for separate sets of photons, and not surprisingly, the QM predictions match the experimental results.

 

[DrChinese]

...The QM predictions are similarly made for separate sets of photons, and not surprisingly, the QM predictions match the experimental results.

That is ALL that a Bell test is. Because Bell says that LR theories cannot match QM results, ergo at least one of QM or LR theories are wrong. It isn't QM, as experiment shows.

What I'm claiming is that before measurement, the answer you would get at (a,c) is entirely predetermined, and it can't depend on whether you're going to measure at (a,b) or (a,c).

Glad to see you finally admit that LR requires that for any photon, the results of a measurement must be predetermined. Now all you have left to do is draw the Bell conclusion, and we will let you join the rest of us.

 

[DevilsAvocado]

The QM predictions are similarly made for separate sets of photons, and not surprisingly, the QM predictions match the experimental results.

Thank you Bill, for the first time I can say that I agree 100%, and it feels like a ‘relief’... and as DrC I’m also very glad we finally gotten this far.

Now, if you could only explain – What’s the problem?

Should we be grumpy that LHV runs into problems with its definite nature and CFD?

Should we dismiss the outcome of EPR-Bell test experiments and say – No! This is not happening, because LHV can’t keep up with the pace!

Should we reject the fact that when only chosen two aligned/counter-aligned angles the compatibility between QM & LHV is 100%, and everything works like a dream?

Should we prohibit Bell to draw the logical conclusion that LHV gets dysfunctional when tested against more than aligned/counter-aligned angles?

How could one not see that this is the whole point of Bell’s theorem; to clearly point out the breakdown of LHV when tested against three angles?

How could one not understand that it’s exactly this ingenious move that finally solved the 20+ year long Bohr–Einstein debate?

And if refuting everything above – How on Earth could we ever test LHV in any other way?? Or should we draw the bizarre conclusion that there is a Local Reality indeed, but we are not allowed to test it thoroughly?

Many in this forum have tried very hard, but no one understands your main objection??

 

[billschnieder]

Thank you Bill, for the first time I can say that I agree 100%, and it feels like a ‘relief’... and as DrC I’m also very glad we finally gotten this far.

Now, if you could only explain – What’s the problem?

Continuing from my last post. Now that we all understand and agree how Bell test experiments are performed, and what experiments QM predictions are made for, let us now examine how Bell inequalities are derived.

Unlike in the experiments and QM in which we have 3 different sets "x", "y", "z", we start with a single set "w" and we ask

* What would we get if we measure set "w" at (a,b)
* What would we get if we measure set "w" at (a,c)
* What would we get if we measure set "w" at (b,c)

Note we are asking the questions for the exact same set of photons "w" not three different sets "x", "y", "z". In other words, we assume that each pair of photons in this set has a definite outcome at a given angle. In other words we are assuming that the outcomes for the set "w" look like:

abc
+++
++-
+-+
-++
+--
-+-
--+
...
etc for as many as there are photons in the set. It is this single list that is used to calculate all three correlations or probabilities used in Bell's inequality.

To calculate the (a,b) correlation/probability, we take the first column and the second column of this same list, since we are dealing with a single set of photons "w". To calculate the (a,c) correlation/probability, we take the first column and the third column and to calculate the (b,c) term we take the second and the third column of the exact same set "w". Notice all three correlations C(a,b), C(a,c), and C(b,c) are calculated from the exact same set. These are the correlations in Bell's inequality.

The issue Lugita and I have been discussing has been:

1) Lugita claims that the three correlations from three different sets "x", "y", "z" as described in my previous post from experiments and QM are the same and equal to the three correlations from a single set "w" from which Bell's inequalities are derived.

2) I claim that the three correlations from three different sets "x", "y", "z" from experiments and QM are different from the three correlations from a single set "w" from which Bell's inequalities are derived.

 

[billschnieder]

Should we be grumpy that LHV runs into problems with its definite nature and CFD?

The reason CFD comes in is because since we are dealing with a single set of photon "w" in deriving the inequality, but we can only measure one of the angle pairs from this set of photons, the other two hypothetical measurements will be counterfactual. In other words, Bell's inequalities is a relationship between three correlations, each of which which could possibly be measured, but only one of which can in fact be measured.

 

[billschnieder]

Many in this forum have tried very hard, but no one understands your main objection??

1) Do you understand my previous two posts? Do you agree that Bell's inequalities are derived for a single set of photons "w", and Bell tests experiments and QM produce correlations for three different sets of photons "x", "y", "z"? If you understand and agree with this, then you are half way to understand the problem.

2) Now if you understand (1), then you have Bell's inequalities such as P(C|w) <= P(A|w) + P(B|w) derived from a single set "w", and probabilities P(A|x), P(B|y), P(C|z) from QM and experiments obtained from three different sets. In order to claim that the probabilities from QM and experiments violate the inequalities (aka Bell's theorem), you must be also making the extra assumption that P(A|x) = P(A|w), P(B|y) = P(B|w), P(C|z) = P(C|w). This is in fact what lugita has been claiming. Without this assumption you can not even formulate Bell's theorem. If you understand and agree with this, then you are 75% of the way to understand the problem.

3) If you understand (1) and (2) then the only question we have left is this. Is it reasonable to make the assumption that P(A|x) = P(A|w), P(B|y) = P(B|w), P(C|z) = P(C|w)? In otherwords, can this assumption by itself be responsible for the violation without anything spooky going on? Another way of asking the same question: Are the correlations from QM and experiments legitimate terms to be used for comparison with Bell's inequality? The implication being that if the correlations from QM and experiments are not legimate terms for doing the comparison, then the assumption P(A|x) = P(A|w), P(B|y) = P(B|w), P(C|z) = P(C|w) is unreasonable, and we do not have Bell's theorem. In yet other words, if those correlations from three different sets are not the same as those from a single set, then violation of the inequality based on a single set, simply tells us that the correlations are not from a single set, which should be obvious already.

If you understand and agree with all of this, then you are 90% of the way there and all I have left is to convince you that three different sets can violate the inequality but a single set can never violate the inequality, and therefore the assumption P(A|x) = P(A|w), P(B|y) = P(B|w), P(C|z) = P(C|w) is not only unreasonable, it is in fact responsible for the violation.

 

[billschnieder]

That is ALL that a Bell test is. Because Bell says that LR theories cannot match QM results, ergo at least one of QM or LR theories are wrong. It isn't QM, as experiment shows.

But Bell draws this conclusion by using QM results predicted for three different sets of photons in an inequality derived from a single set of photons. Which is an implicit assumption that those correlations from three sets should be the same as those from a single set. It is this assumption that is at the core of the issue, not the ones used to derive the inequality.

Glad to see you finally admit that LR requires that for any photon, the results of a measurement must be predetermined.

I have never ever questioned this.

 

[DrChinese]

But Bell draws this conclusion by using QM results predicted for three different sets of photons ...

No, it doesn't. The coincidence rate QM predicts for anyone set of photons is cos^2(a-b). See, just one term there. So for example, QM predicts .25 for settings 120 degrees apart. A local realistic theory might predict instead .33* for the same setup. But that would be inconsistent with actual experiments, which yield very close to .25.

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

There are currently no local realistic theory candidates which make the same predictions as QM. Although there are a few odd papers out there that claim to be local realistic models mimicking QM, so far none have held up.
*.33 would be the linear expected value for a LRT. As per the title of this thread.

 

[billschnieder]

No, it doesn't. The coincidence rate QM predicts for anyone set of photons is cos^2(a-b).

But this is where you are mistaken. The prediction QM makes for (a,b) is for one set of photons. The prediction QM makes for (b,c) ie cos^2(b-c) is for a completely different experiment (aka, different set of photons), and also cos^2(a-c) another completely different set of photons.

See, just one term there.

It looks like one term. But you need three terms to compare with Bell's inequality, where does QM get the other two, except by also predicting them? The question QM is answering is not:

If we have a single set of photons, what would we observe if we measure at angle (a,b), and at the angle (a,c) and at angle (b,c).

But rather it is.

If we measure a set of photons at a pair of angles (i,j) what would we observe.

You think you can simply substitute i,j with (a,b,c) to get the answers for (a,b), (a,c) and (b,c) and those answers together will be equivalent to the answer to the question "If we have a single set of photons, what would we observe if we measure at angle (a,b), and at the angle (a,c) and at angle (b,c)" but by doing that you making the same mistake Bell was talking about when he said:

The essential assumption can be criticized as follows. At first sight the required additivity of expectation values seems very reasonable, and it is rather the non-additivity of allowed values (eigenvalues) which requires explanation. Of course the explanation is well known: A measurement of a sum of noncommuting observables cannot be made by combining trivially the results of separate observations on the two terms.

...

The danger in fact was not in the explicit but in the implicit assumptions. It was tacitly assumed that measurement of an observable must yield the same value independently of what other measurements may be made simultaneously. Thus as well as P(Φ3) say, one might measure either P(Φ2) or P(ψ2), here Φ2 and ψ2 are orthogonal to Φ3 but not to one another. These different possibilities require different experimental arrangements; there is no a priori reason to believe that the results for P(Φ3) should be the same.

 

[DrChinese]

It looks like one term. But you need three terms to compare with Bell's inequality, ...

The usual Bell inequality has more than 1 term, sure. Who cares?* What we care about is that Bell says no local realistic (hidden variable) theory can match the predictions of QM. The inequality is simply a convenience for seeing which scenarios don't match. Because some do! Here is one, for example:

P(a,a)=1 for both QM and some LR theories.

Similarly, P(b,b)=1 for both QM and some LRTs. And likewise P(c,c)=1.

So P(a,a)+P(b,b)+P(c,c)=3. A shocker I am sure - no disagreement between QM and some LRTs here. Now tell me, you deny this relationship as well on the grounds that it is 3 different sets of photons? What a dopey argument. It is obvious you are on a different page than everyone else, why don't you admit it and move on?


*To prove it, take the DrChinese challenge. As you have always failed that test, as we already have found out. And guess what - you don't even need Bell for that one! You only need to try to match QM.

 

[lugita15]

I don't think you fully understand the implication of this claim. It is not correct to say the outcome at (a,b) does not depend on the angle (a,b) or the outcome at (a,c) does not depend on the angle (a,c), since those settings are part of the set-up which produces the outcome.

I think we're having a misunderstanding here. You agreed that before a given photon pair is measured, answer to the question "What result would you get if you measured this photon pair at (θ1,θ2)?" is predetermined for all possible polarizer settings θ1 and θ2, right? Let us denote this predetermined answer by f(θ1,θ2). The domain of f is ℝ² and its codomain (or range) is {1,-1}×{1,-1}. So for instance, if photon 1 would go through and photon 2 would not go through if the polarizers are set at -30 degrees and 0 degrees respectively, then we denote that by f(-30,0)=(1,-1).

Now when you say "It is not correct to say the outcome at (a,b) does not depend on the angle (a,b)", I assume what you mean by that is that in order to find out what f(a,b) is, you need to know what a and b are. If that's what you meant, I fully agree. For any angles θ1 and θ2, then value of f(θ1,θ2) is certainly dependent on the values of θ1 and θ2.

But what I'm saying is that the function f does not depend on the angle settings θ3 and θ4 that you actually turn the polarizers to. In other words, for all possible angle settings θ1 and θ2, the value of f(θ1.θ2) does not depend on whether θ1=θ3 or θ2=θ4. That's because for each photon pair, the function f is entirely predetermined before the photon pair even "knows" what angles it's actually going to be measured at. Do you agree or disagree with that?

Probably what you are trying to say is that the outcome of the single photon measured at angle "a" in the (a,b) pair must be exactly the same as the outcome of the same photon measured at angle "a" in the (a,c) pair. In other words, the outcome for that particular single photon should not change from (a,b) to (a,c) just because it's sibling was measured at "c" in (a,c) instead of "b" in (a,b). So Alice's choice of setting should not influence Bobs outcome for the same setting "a" and the exact same photon. If this is what you mean, then I agree.

No, this isn't what I was trying to say, although I do agree with it.

But if you mean that the outcome at (a,b) can not depend on the angle pair (a,b) then I disagree with that.

What I mean is that the outcome you would get at (a,b) depends on what (a,b) is, but it doesn't depend on whether you're actually going to measure at (a,b) or (a,c). See above.

No. I disagree with that. Just because it is predetermined to produce the result at (a,c) doesn't mean it can produce the result without (a,c). Without the setting (a,c) you will not get the result at (a,c) so how can you say the result obtained at (a,c) does not depend on what you actually choose (a,c)? What is predetermined is the condition that "if you set the device to (a,c), you will obtain such and such result". It doesn't make much sense to say the result you obtain at (a,c) does not depend on the setting, which is required to obtain the result!

Again, I think we're having a misunderstanding. See what I said above.

Free will has nothing to do with it. You are free to choose a different setting, you will get a different result. But whenever you freely choose (a,c) you will get the result predetermined for (a,c).

I agree with this.

Probably what you are trying to say here is that the result for the single photon measured at angle "a", should not depend on whether you measured the pair at (a,c) or you measured (a,b). Again, that means Bob's result for the same photon measured at the same angle should not depend on what angle Alice chose to measure it's sibling photon at. If this is what you mean, I agree.

Again, that's not what I meant, but I do agree with it.

See explanation above. The results predetermined for a given angle pair can not be independent of the angle pair for which it is predetermined.

Yes, the predetermined result for a given angle pair does depend on the angle pair for which it is predetermined, but it does NOT depend on the actually measured angle pair.

 

[billschnieder]

I think we're having a misunderstanding here. You agreed that before a given photon pair is measured, answer to the question "What result would you get if you measured this photon pair at (θ1,θ2)?" is predetermined for all possible polarizer settings θ1 and θ2, right? Let us denote this predetermined answer by f(θ1,θ2). The domain of f is ℝ² and its codomain (or range) is {1,-1}×{1,-1}. So for instance, if photon 1 would go through and photon 2 would not go through if the polarizers are set at -30 degrees and 0 degrees respectively, then we denote that by f(-30,0)=(1,-1).

I agree.

Now when you say "It is not correct to say the outcome at (a,b) does not depend on the angle (a,b)", I assume what you mean by that is that in order to find out what f(a,b) is, you need to know what a and b are. If that's what you meant, I fully agree. For any angles θ1 and θ2, then value of f(θ1,θ2) is certainly dependent on the values of θ1 and θ2.

Yes.

But what I'm saying is that the function f does not depend on the angle settings θ3 and θ4 that you actually turn the polarizers to. In other words, for all possible angle settings θ1 and θ2, the value of f(θ1.θ2) does not depend on whether θ1=θ3 or θ2=θ4. That's because for each photon pair, the function f is entirely predetermined before the photon pair even "knows" what angles it's actually going to be measured at. Do you agree or disagree with that?

I agree, whatever hidden mechanism is producing the outcome for the given photon pair (your function f), is the same hidden mechanism for the exact same photon pair irrespective of what angles are actually chosen.

What I mean is that the outcome you would get at (a,b) depends on what (a,b) is, but it doesn't depend on whether you're actually going to measure at (a,b) or (a,c). See above.

I agree that what the photon pair would produce at (a,b) is the same and won't be different whether that photon pair was measured at (a,b) or at (a,c). If this is what you mean? I agree.

 

[billschnieder]

The usual Bell inequality has more than 1 term, sure. Who cares?* What we care about is that Bell says no local realistic (hidden variable) theory can match the predictions of QM.

If you care about what Bell's theorem says, then you also care (or should care) about the logic of how Bell arrived at that conclusion, which was by generating 3 correlations from QM to compare with 3 correlations from the inequalities.

And you also care (or should care) about the logic that is used to claim that 3 correlations from experiments violate the inequality.

If your mind is made up about those two aspects and you already believe that the logic is iron-clad and beyond discussion, then this discussion will not interest you. I do not question the inequalities, they are valid. I do not question the QM predictions, they are valid. I do not question the experimental results, they are valid. I do not claim to have a LR model that matches QM so asking me to provide one does not advance the discussion. What I question is the suggestion that the correlations from QM and experiments are equivalent correlations to those in the inequalities and this goes to the heart of the logic used to obtain Bell's theorem or demonstrate experimental violation.

At least, if you already believe that the correlations from QM and experiments are equivalent to those in the inequalities, you should consider the question: Would Bell's theorem still follow in case they were not equivalent? If you can at least understand my argument and agree that Bell's theorem will not be a valid conclusion unless those correlations from QM and experiments are equivalent to those in the inequalities, then the only question left would be for you to convince me (if you are interested) that they are equivalent, and for me to convince you that they are not. Of course we won't have to agree on anything but the issues will be clearly laid out.

 

[lugita15]

I agree, whatever hidden mechanism is producing the outcome for the given photon pair (your function f), is the same hidden mechanism for the exact same photon pair irrespective of what angles are actually chosen.


I agree that what the photon pair would produce at (a,b) is the same and won't be different whether that photon pair was measured at (a,b) or at (a,c). If this is what you mean? I agree.

Yes, this is exactly what I meant. I am astounded that we agree on this point, because I was expecting this to be the crux of our disagreement.

Well, let me ask you a follow-up question now: do you agree or disagree that the value of the following fraction is independent of θ3 and θ4?

(Number of photon pairs, actually measured at (θ3,θ4), for which f(θ1,θ2) is equal to (1,-1) or (-1,1))/(Number of photon pairs which were actually measured at (θ3,θ4))

 

[billschnieder]

Well, let me ask you a follow-up question now: do you agree or disagree that the value of the following fraction is independent of θ3 and θ4?

(Number of photon pairs, actually measured at (θ3,θ4), for which f(θ1,θ2) is equal to (1,-1) or (-1,1))/(Number of photon pairs which were actually measured at (θ3,θ4))

That question is not sufficiently clear to me. Let me try to rephrase it and you can tell me if I understood it correctly or not.

You are saying we take a set of photon pairs "p" and we measure at the angles (θ3,θ4) and then we calculate the relative frequency for getting the result of either (+, -) or (-, +). Which is calculated as
$\frac{ N_{(+-)} + N_{(-+)} }{N_p}$
That is the total number of photon pairs which gave (+1, -1) plus the total number which gave (-1, +1) divided by the total number of photon pairs in the set "p".

If by not depending on (θ3,θ4), you mean the result we would obtain for the set of photon pairs "p" at angles (θ3,θ4), would not be different whether we actually measured at angles (θ3,θ4) or at angles (θ6,θ7), then I agree with this.

 

[lugita15]

You are saying we take a set of photon pairs "p" and we measure at the angles (θ3,θ4) and then we calculate the relative frequency for getting the result of either (+, -) or (-, +). Which is calculated as
$\frac{ N_{(+-)} + N_{(-+)} }{N_p}$
That is the total number of photon pairs which gave (+1, -1) plus the total number which gave (-1, +1) divided by the total number of photon pairs in the set "p".

Yes, that's what I mean.

If by not depending on (θ3,θ4), you mean the result we would obtain for the set of photon pairs "p" at angles (θ3,θ4), would not be different whether we actually measured at angles (θ3,θ4) or at angles (θ6,θ7), then I agree with this.

No, that's not what I mean. What I mean is, we take one set of photon pairs p measured at (θ3,θ4), and we take another set of photon pairs q measured at (θ6,θ7). And then I'm claiming that the relative frequency of getting (1,-1) or (-1,1) for f(θ1,θ2) is the same for p and q. Do you agree or disagree with that?

 

[billschnieder]

If by not depending on (θ3,θ4), you mean the result we would obtain for the set of photon pairs "p" at angles (θ3,θ4), would not be different whether we actually measured at angles (θ3,θ4) or at angles (θ6,θ7), then I agree with this.

No, that's not what I mean. What I mean is, we take one set of photon pairs p measured at (θ3,θ4), and we take another set of photon pairs q measured at (θ6,θ7). And then I'm claiming that the relative frequency of getting (1,-1) or (-1,1) for f(θ1,θ2) is the same for p and q. Do you agree or disagree with that?

No, I do not agree with this. Since you are now talking about two different sets of photons, the two relative frequencies can only be the same if the two sets of photons were identically prepared. So the answer is, yes they can be the same (if identically prepared) but they are not necessarily the same.

Furthermore, I do not understand what this has to do with what you asked earlier that

do you agree or disagree that the value of the following fraction is independent of θ3 and θ4?

And earlier when I answered your question about "depends", you said:

I agree that what the photon pair would produce at (a,b) is the same and won't be different whether that photon pair was measured at (a,b) or at (a,c). If this is what you mean? I agree.

Yes, this is exactly what I meant. I am astounded that we agree on this point, because I was expecting this to be the crux of our disagreement.

But then now you seem to be changing what you mean by "depends", because what I said now about a single set is almost word for word what I said earlier about a single pair, and you agreed then.