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

[lugita15]

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.

What does it mean for two photon pairs to be identically prepared? And whatever your definition of identically prepared is, do you consider two photon pairs, each of which is polarization-entangled, to be identically prepared? At least quantum mechanics views them as having the same spin part for their wavefunctions.

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?

What I was envisioning is an experiment in which, for every photon pair, the experimenter just randomly decides on some pair of angles to measure at. So some pairs he measures at (θ3,θ4), and some pairs he measures at (θ6,θ7), and maybe for other pairs he measures all kinds of different angle combinations. So then, for any given angle pair (θ3,θ4), you can ask, "What percentage of photon pairs measured at (θ3,θ4) had f(θ1,θ2) equal to (1,-1) or (-1,1)?" And I was saying that the answer to this question stays the same even if you replace (θ3,θ4) with (θ6,θ7). Of course if you replace (θ3,θ4) with (θ6,θ7) in that expression, then you're changing what photon pairs you're talking about.

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.

I'm sorry for any confusion. I'm still agreeing with what you said in post #310. Now that we're agreed on one point, I'm trying to see whether we can get agreement on another point.

 

[billschnieder]

What does it mean for two photon pairs to be identically prepared?

It means in every aspect relevant for the outcome you are interested in, the two systems are identical. In other words, if the important property of the system is a hidden property λ, and a source is producing a set of photons with such hidden properties, then the probability distribution ρ(λ) in system "p" is identical to the probability distribution ρ(λ) in system "q".

And whatever your definition of identically prepared is, do you consider two photon pairs, each of which is polarization-entangled, to be identically prepared?

It depends. If what you are measuring depends on anything else which is different between the two pairs, then they are not. But if whatever you are measuring depends ONLY on the fact that they are polarization entangled, then you can say they are identically prepared. In other words, the definition of "identically prepared" cannot be separated from the experiment you are actually performing.

For two pairs of polarization entangled photons, let us say for example that in each pair the polarization vectors are perfectly opposite each other in direction. If in the experiment your outcome is also dependent on the relative angle between the polarizer and the direction of the polarization vector, then the original direction of the polarization vector is also important. And if this initial direction is not the same between the two pairs, then you can't say they are identically prepared for this experiment and you will get different results after measurement. However, if your source produces two sets of photons in which the original direction of the polarization vector is generated randomly in a probability distribution which is identical between the first set and the second, then you will get the same results for the experiment and you can say the two sets of photons are identically prepared for the experiment. Only in this case can you conclude that the results in set "p" are identical to the results in set "q".

If this is not clear, what I mean is: if polarization entanglement governs the relationship between the directions of the polarization vectors of the two photons in a pair, but the experiment measures the relationship between the direction of the polarization vector of each photon and the vector of the polarizer, two polarization entangled pairs of photons will not be considered identically prepared unless they have the same relative angle wrt to the polarizer . Only then will you expect to have the exact same result from measuring both pairs.

If this is what you meant, then I agree that the relative frequencies will be the same in both "p" and "q".

 

[billschnieder]

What I was envisioning is an experiment in which, for every photon pair, the experimenter just randomly decides on some pair of angles to measure at. So some pairs he measures at (θ3,θ4), and some pairs he measures at (θ6,θ7), and maybe for other pairs he measures all kinds of different angle combinations. So then, for any given angle pair (θ3,θ4), you can ask, "What percentage of photon pairs measured at (θ3,θ4) had f(θ1,θ2) equal to (1,-1) or (-1,1)?" And I was saying that the answer to this question stays the same even if you replace (θ3,θ4) with (θ6,θ7). Of course if you replace (θ3,θ4) with (θ6,θ7) in that expression, then you're changing what photon pairs you're talking about.

As explained in my previous post, provided the source is producing the photons with such a uniform distribution of the relevant parameters that the probability distribution of those parameters in the set actually measured in (θ3,θ4) is the same as the probability distribution in the set actually measured at (θ6,θ7), then I agree that the answer will be the same for both sets. Technically one may say in order for the correlations to be the same, the source must be operating as a stationary process with respect to the relevant hidden parameters.

 

[DrChinese]

What I was envisioning is an experiment in which, for every photon pair, the experimenter just randomly decides on some pair of angles to measure at.

I have always held that there is no purpose to changing angle settings randomly EXCEPT to demonstrate that the detectors could not be exchanging signals which would affect the outcome. This is easily ruled out for local realistic theories by experiments such as Weihs et al. Hopefully Bill is not arguing that such rapport is occurring between the detectors per their respective settings.

So to my thinking, the point is that the experimenter selects the desired 2 settings and let's the process run. And the experimenter 's selections are independent of what is occurring at the source. Since Bill thinks the outcomes from the source are predetermined, the question is what would you expect? And here is where Bill's challenges fall short. We know there is perfect correlation (as he expects) for P(a,a). Ditto for P(b,b) and P(c,c). He accepts this despite these are drawn from different sets (his terminology). But he rejects the same notion for P(a,b), P(a,c) and P(b,c).

Further: he has tried to construct LR data sets that mimic the QM results IFF the experimenter selects special angle pairs that have such property. But many don't. So for example, it is possible to construct LR data sets where P(a,b)=.25 and P(a,c)=.25 [agreeing with QM] but P(b,c) is necessarily .50 in this case [which does not agree with the QM expectation].

So I guess my point is that he is not going to accept your reasoning even if the experimenter were to choose a pair at a time, as you reasonably suggest.

 

[DrChinese]

So Bill, here is DrC challenge #2.

I have a laser pump (oriented at 45 degrees) to a single Type I crystal oriented H. The down converted output stream will be definite VV (let's call them LeftV and RightV). I run each V to a beam splitter set at 45 degrees with detectors at each output. Because each V has a 50-50 chance of being transmitter/refracted, the coincidence rate is 50%.

I have a laser pump (oriented at 45 degrees) to a single Type I crystal oriented V. The down converted output stream will be definite HH (let's call them LeftH and RightH). I run each H to a beam splitter set at 45 degrees with detectors at each output. Because each H has a 50-50 chance of being transmitter/refracted, the coincidence rate is 50%.

In neither of the above scenarios is the coincidence rate higher than 50%.

Yet if I combine the LeftV and LeftH streams, and then combine the RightV and RightH streams such that the source pump cannot be distinguished, they will be entangled. Using a common detector series for both, the correlation will now be 100%. How, prey tell, is this possible in your scenario when everything was predetermined from the get-go... and nothing matched more than 50%?

 

[billschnieder]

And here is where Bill's challenges fall short. We know there is perfect correlation (as he expects) for P(a,a). Ditto for P(b,b) and P(c,c). He accepts this despite these are drawn from different sets (his terminology). But he rejects the same notion for P(a,b), P(a,c) and P(b,c).

I don't think you fully understand the argument yet. In your challenge #2, you are asking me to provide a LR model for the experimental setup you described which I do not believe will advance the discussion. As I explained in post #311, the argument here does not claim that a LR model is possible (although I'm not necessarily admitting that one is impossible).

The main issue here is whether correlations from three sets "x","y","z" are equivalent to those in a single set "w". Lugita is in the process of demonstrating why they must be equal. If you can please clarify how your challenge helps us to establish that they are equal then it will be very beneficial for the discussion.

 

[billschnieder]

So Bill, here is DrC challenge #2.

I have a laser pump (oriented at 45 degrees) to a single Type I crystal oriented H. The down converted output stream will be definite VV (let's call them LeftV and RightV). I run each V to a beam splitter set at 45 degrees with detectors at each output. Because each V has a 50-50 chance of being transmitter/refracted, the coincidence rate is 50%.

I have a laser pump (oriented at 45 degrees) to a single Type I crystal oriented V. The down converted output stream will be definite HH (let's call them LeftH and RightH). I run each H to a beam splitter set at 45 degrees with detectors at each output. Because each H has a 50-50 chance of being transmitter/refracted, the coincidence rate is 50%.

In neither of the above scenarios is the coincidence rate higher than 50%.

Yet if I combine the LeftV and LeftH streams, and then combine the RightV and RightH streams such that the source pump cannot be distinguished, they will be entangled. Using a common detector series for both, the correlation will now be 100%. How, prey tell, is this possible in your scenario when everything was predetermined from the get-go... and nothing matched more than 50%?

That is an interesting challenge nonetheless so I'll be happy if you can clarify it a bit. From what I understand you have.

1) V-filtered non-entangled pairs : 50% coincidence.
2) H-filtered non-entagled pairs : 50% coincidence.
3) Unfiltered entangled pairs : 100 % coincidence.

Is this accurate? Or are the pairs in (1) and (2) also entangled.

 

[DrChinese]

That is an interesting challenge nonetheless so I'll be happy if you can clarify it a bit. From what I understand you have.

1) V-filtered non-entangled pairs : 50% coincidence.
2) H-filtered non-entagled pairs : 50% coincidence.
3) Unfiltered entangled pairs : 100 % coincidence.

Is this accurate? Or are the pairs in (1) and (2) also entangled.

You have it correct. The 1/2 pairs are not polarization entangled, but they are entangled as to momentum. The 3) group is the combination of streams like 1) and 2) - and I realize you in a sense consider these different sets.

 

[DrChinese]

I don't think you fully understand the argument yet. In your challenge #2, you are asking me to provide a LR model for the experimental setup you described which I do not believe will advance the discussion. As I explained in post #311, the argument here does not claim that a LR model is possible (although I'm not necessarily admitting that one is impossible).

The main issue here is whether correlations from three sets "x","y","z" are equivalent to those in a single set "w". Lugita is in the process of demonstrating why they must be equal. If you can please clarify how your challenge helps us to establish that they are equal then it will be very beneficial for the discussion.

What I am saying is that we are talking about a series of attributes of an entangled stream of pairs. This stream has correlated properties can be described in some variety of ways, such as P(a,a) as well as P(a,b). P(a,a) is fully consistent with a local HV hypothesis, as envisioned by EPR. You don't have any problem with P(a,a), P(b,b) etc but you have a problem with P(a,b) and P(a,c) etc.

 

[lugita15]

As explained in my previous post, provided the source is producing the photons with such a uniform distribution of the relevant parameters that the probability distribution of those parameters in the set actually measured in (θ3,θ4) is the same as the probability distribution in the set actually measured at (θ6,θ7), then I agree that the answer will be the same for both sets.

OK, then the question becomes, do you believe that the sources used in Bell tests, i.e. sources that produce type-I spontaneous parametric down-conversion, obey this criterion of the probability distribution of parameters in different sets being the same?

Let me also ask you this: do you at least agree that, if the photon pairs in both p and q are measured at (θ3,θ4), then the relative frequency of getting (1,-1) or (-1,1) for f(θ3,θ4) is the same for p and q?

 

[billschnieder]

What I am saying is that we are talking about a series of attributes of an entangled stream of pairs. This stream has correlated properties can be described in some variety of ways, such as P(a,a) as well as P(a,b). P(a,a) is fully consistent with a local HV hypothesis, as envisioned by EPR. You don't have any problem with P(a,a), P(b,b) etc but you have a problem with P(a,b) and P(a,c) etc.

I don't get what this has to do with the argument I'm making. Your last sentence there does not correctly characterize the argument. In the most simplistic terms, I'm saying it is reasonable to expect correlations between properties in a single set of entangled pairs, but it is unreasonable to expect the same correlations between properties from one set of entangled pairs and properties from a completely different set of entangled pairs even if the two sets are indistinguishable. So I do not understand why you would say I have a problem with P(a,b) and P(a,c). I don't. I have a problem with taking 3 correlations from 3 different independent sets which have no relationship with each other and claiming that the correlations are the same with 3 correlations from within a single set. This is the issue.

 

[billschnieder]

OK, then the question becomes, do you believe that the sources used in Bell tests, i.e. sources that produce type-I spontaneous parametric down-conversion, obey this criterion of the probability distribution of parameters in different sets being the same?

Yes, for a large enough number of photon pairs from such sources, I have no reason believe they are not, based on what I know. The fact that repeated measurements on such sets produce results which match QM is convincing evidence that the sets are identically prepared.

Let me also ask you this: do you at least agree that, if the photon pairs in both p and q are measured at (θ3,θ4), then the relative frequency of getting (1,-1) or (-1,1) for f(θ3,θ4) is the same for p and q?

I thought I already agreed to this point. Yes, for a large enough number photons in separate identically prepared sets of photons "p" and "q", f(θ3,θ4) should give the same relative frequencies.

 

[billschnieder]

You have it correct. The 1/2 pairs are not polarization entangled, but they are entangled as to momentum. The 3) group is the combination of streams like 1) and 2) - and I realize you in a sense consider these different sets.

I am baffled as to why you would even be comparing the first two streams with the third. Are you suggesting that stream 3 is simply a linear combination of the first two?

 

[lugita15]

Yes, for a large enough number of photon pairs from such sources, I have no reason believe they are not based on what I know. The fact that repeated measurements on such sets produce results which match QM is convincing evidence that the sets are identically prepared.

I'm glad (although again surprised) to hear that we're agreed on that. In future discussion, let's assume that we're dealing with sufficiently large sets of photons, so that we don't need to worry about whether they're identically prepared.

So now you've agreed that if the photon pairs in p are measured at (θ3,θ4), and the photon pairs in q are measured at (θ6,θ7), then the relative frequency of getting (1,-1) or (-1,1) for f(θ1,θ2) is the same for p and q.

In particular, if the photon pairs in p are measured at (-30,0), and the photon pairs in q are measured at (0,30), then the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for p and q. And similarly, if the photon pairs in r are measured at (-30,30), then the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for q and r. Therefore, the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for p, q, and r. Do you agree with that?

 

[billschnieder]

I'm glad (although again surprised) to hear that we're agreed on that.

I'm not surprised

In future discussion, let's assume that we're dealing with sufficiently large sets of photons, so that we don't need to worry about whether they're identically prepared.

Fine with me. I was just trying to be very clear so there was no chance for misunderstanding.

So now you've agreed that if the photon pairs in p are measured at (θ3,θ4), and the photon pairs in q are measured at (θ6,θ7), then the relative frequency of getting (1,-1) or (-1,1) for f(θ1,θ2) is the same for p and q.

Agreed.

In particular, if the photon pairs in p are measured at (-30,0), and the photon pairs in q are measured at (0,30), then the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for p and q. And similarly, if the photon pairs in r are measured at (-30,30), then the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for q and r. Therefore, the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for p, q, and r. Do you agree with that?

Agreed.

 

[lugita15]

Agreed.

Great! Here's my next question. If the relative frequency of something is the same for three different sets, then it's also the same for their union. So if we let s be the union of p, q, and r (the sets from my previous post), then the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for p and s. Do you agree with that?

 

[billschnieder]

Great! Here's my next question. If the relative frequency of something is the same for three different sets, then it's also the same for their union. So if we let s be the union of p, q, and r (the sets from my previous post), then the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for p and s. Do you agree with that?

Yes. if p, q, and r are identically prepared, then of course the union of (p + q + r) is also identically prepared to any of the component sets p, q and r.

 

[lugita15]

Yes. if p, q, and r are identically prepared, then of course the union of (p + q + r) is also identically prepared to any of the component sets p, q and r.

OK, so we're now agreed on the following 3 statements:
1. The relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for p and s
2. The relative frequency of getting (1,-1) or (-1,1) for f(0,30) is the same for q and s
3. The relative frequency of getting (1,-1) or (-1,1) for f(-30,30) is the same for r and s

So now my next question is, do you agree that
the relative frequency of getting (1,-1) or (-1,1) for f(-30,30) in s is less than or equal to the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) in s, plus the relative frequency of getting (1,-1) or (-1,1) for f(0,30) in s?

 

[DevilsAvocado]

[Bill, I’ll get back to your replies, but first the “Holy Grail” that could finally solve the elongated “Bill-Bell debates”! ;]

I'm glad (although again surprised) to hear that we're agreed on that. In future discussion, let's assume that we're dealing with sufficiently large sets of photons, so that we don't need to worry about whether they're identically prepared.

Fine with me. I was just trying to be very clear so there was no chance for misunderstanding.

And I’m extremely glad! This is what I’ve been shouting about for a couple of weeks now! ()

So now my next question is, do you agree that
the relative frequency of getting (1,-1) or (-1,1) for f(-30,30) in s is less than or equal to the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) in s, plus the relative frequency of getting (1,-1) or (-1,1) for f(0,30) in s?

WTF!? <-- oral cavity state vectors

I can’t believe we FINALLY gotten to the “Holy Grail”!?

Beautiful Lugita! Let’s just add Bell’s inequality for complete clarity:

N(+30°, -30°) ≤ N(+30°, 0°) + N(0°, -30°)​

Yay!

 

[billschnieder]

OK, so we're now agreed on the following 3 statements:
1. The relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for p and s
2. The relative frequency of getting (1,-1) or (-1,1) for f(0,30) is the same for q and s
3. The relative frequency of getting (1,-1) or (-1,1) for f(-30,30) is the same for r and s

I agree to each of those statements considered independently, but I do not agree to all of them considered together. In other words I would not agree if the s in 1, 2 and 3 were the exact same set, rather than 3 bigger arbitrarily larger identically prepared sets. In other words I do not agree that the operations for measuring f(-30,0), and f(-30,30) on the same set as f(0,30) on the exact same set commute.

 

[DrChinese]

I am baffled as to why you would even be comparing the first two streams with the third. Are you suggesting that stream 3 is simply a linear combination of the first two?

It (3) is a physical combination of 2 different streams. It is polarization entangled. Yet, according to your world view, all of the member pairs are predetermined with known polarization. Their statistics can never result in perfect correlation. So that does not match experiment, which has them giving entangled state statistics.

So you should assert that the result of 3) is a linear combination of 1) and 2), yes. There will be approximately twice as many photons detected and matched. You should assert there is no physically entangled state, since it is all predetermined. After all, each stream's pairs originates in a different crystal after undergoing down conversion. In terms of a photon's time, they arrive far apart.

 

[billschnieder]

It (3) is a physical combination of 2 different streams. It is polarization entangled.

I don't understand how you take 2 different streams, each not polarization entangled and then simply combine them linearly and get a polarization entangled stream.

Yet, according to your world view, all of the member pairs are predetermined with known polarization.

I do not believe you understand my world view yet, as I've explained already multiple times. I'm talking apples and you are talking oranges.

So you should assert that the result of 3) is a linear combination of 1) and 2), yes.

I've made no such assertion, I'm still trying to understand what you mean by 1, 2 and 3 and how linear combination of 1 and 2 results in 3.

 

[billschnieder]

So now my next question is, do you agree that
the relative frequency of getting (1,-1) or (-1,1) for f(-30,30) in s is less than or equal to the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) in s, plus the relative frequency of getting (1,-1) or (-1,1) for f(0,30) in s?

If the set "s" is the exact same set of photons in all three measurements, then I agree that the relative frequency of getting (1,-1) or (-1,1) for f(-30,30) in s is less than or equal to the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) in s, plus the relative frequency of getting (1,-1) or (-1,1) for f(0,30) in s.

Do you now see where the disagreement is? (after considering my answer to this question and the one answered in my last response to you) If you do see the point of disagreement but do not understand what I mean, I can explain further, but only after it is clear to you on exactly which point we disagree.

 

[Nugatory]

I agree to each of those statements considered independently, but I do not agree to all of them considered together.

Lugita15, that was a very good try and I'm impressed that you got as close as you did... But it looks as if once again Lucy has snatched the football away at the last moment.

 

[DrChinese]

Lugita15, that was a very good try and I'm impressed that you got as close as you did... But it looks as if once again Lucy has snatched the football away at the last moment.

We all know we are never going to change Bill's* mind. Every experimental form of entanglement is essentially a refutation of every local realistic viewpoint. If you simply deny this, as Bill does, well... here we are.


*As he is not likely to change mine.

 

[DrChinese]

I don't understand how you take 2 different streams, each not polarization entangled and then simply combine them linearly and get a polarization entangled stream.

In your world, you can't. In the real world, you get Type I polarization entangled pairs.

http://arxiv.org/abs/quant-ph/0205171

See Fig. 2. Note that here they use a single pump laser. But you could accomplish the same thing with 2 phase locked pumps too if you combine the outputs so the source is indeterminate.

 

[Nugatory]

I agree to each of those statements considered independently, but I do not agree to all of them considered together. In other words I would not agree if the s in 1, 2 and 3 were the exact same set, rather than 3 bigger arbitrarily larger identically prepared sets. In other words I do not agree that the operations for measuring f(-30,0), and f(-30,30) on the same set as f(0,30) on the exact same set commute.

So f I understand you properly, you're challenging the promiscuous transfer of measurement results from one set to another set in which the same measurements have not been made?

I mentioned much further up in the thread that there is an assumption embedded in the construction of the Bell experiments (but not in the theorem itself), and that this assumption is essential to accepting the experiments as falsification of the inequality. That assumption is fair sampling, which along with counterfactual definiteness allows us to transfer the results from measuring properties in one set to another set in which we do not measure the those properties.

If a property is counterfactually definite and I am measuring it within one statistically representative subset of a larger population, I can conclude that my measurements of that property are applicable to (by fair sampling) any other statistically representative subset of that population, even if I don't measure the property on that subset (by counterfactual definiteness).

 

[stevendaryl]

We all know we are never going to change Bill's* mind. Every experimental form of entanglement is essentially a refutation of every local realistic viewpoint. If you simply deny this, as Bill does, well... here we are.


*As he is not likely to change mine.

I like the attempt to get past disagreements about the meaningfulness or validity of abstract arguments by making it into an Amazing-Randy style wager:

The Bell side makes a bet, that there is no way to simulate EPR-style correlations without nonlocal communication, using a combination of deterministic devices plus random number generators.

The challenge is to design a "pair generator" that will produce a sequence of pairs of "secret messages", together with a box, "Alice's detector" and "Bob's detector" that will receive a secret message, together with a real-number input from Alice or Bob, and will output either +1 or -1.

The challenge proceeds as follows: We pick a number of rounds, say 100. Each round proceeds as follows: On round number $n$,

1. The pair generator creates a pair of secret messages $m_{A,n}$ and $m_{B,n}$, and sends $m_{A,n}$ to Alice's detector, and $m_{B,n}$to Bob's detector.
2. Alice rolls a 6-sided die. If the result is 1 or 2, she picks $\alpha_n =$ 0°. If the result is 3 or 4, she picks $\alpha_n =$ 120°. If the result is 5 or 6, she picks $\alpha_n =$ 240°. She inputs the value into her detector.
3. The detector produces an output, either $A_n =$+1 or -1, which is only seen by Alice. She records her choice of $\alpha_n$ and the output $A_n$ from the detector.
4. Bob similarly chooses $\beta_n$ from the set { 0°, 120°, 240° }.
5. Bob's detector produces an output, either +1 or -1, which is only seen by Bob. He records his choice of $\beta_n$ and the output, $B_n$ from his detector.

After many rounds, Alice and Bob each have a list of pairs. They put their lists together to compute joint probabilities as follows:

$P(\alpha, \beta, A, B) = \dfrac{N_{\alpha, \beta, A, B}}{N_{\alpha,\beta}}$

where $N_{\alpha, \beta, A, B}$ is the number of rounds in which Alice chose $\alpha_n = \alpha$, and Bob chose $\beta_n = \beta$, and $A_n = A$ and $B_n = B$, and where $N_{\alpha, \beta}$ is the number of rounds in which Alice chose $\alpha_n = \alpha$ and Bob chose $\beta_n = \beta$.

The bet is that there is no way to design the "pair generator" and the "detectors" so that the simulated joint probability distribution $P(\alpha, \beta, A, B)$ agrees with the quantum spin-1/2 EPR predictions:

$P(\alpha, \beta, +1, +1) = P(\alpha, \beta, -1, -1) =\frac{1}{2} sin^2(\frac{1}{2} (\beta - \alpha))$

$P(\alpha, \beta, +1, -1) = P(\alpha, \beta, -1, +1) =\frac{1}{2} cos^2(\frac{1}{2} (\beta - \alpha))$

Where the "local" comes in is the assumption that Bob's detector is not allowed to use Alice's input, and vice-versa, and that the pair generator is not allowed to use either input. If you violate these locality restrictions, it's easy to get the QM results.

 

[lugita15]

I agree to each of those statements considered independently, but I do not agree to all of them considered together. In other words I would not agree if the s in 1, 2 and 3 were the exact same set, rather than 3 bigger arbitrarily larger identically prepared sets. I

I definitely intend s to be the same set in all three statements. After all, s is just the union of p, q, and r. So you're saying that if I use the same set s, then my 3 statements aren't all true. Well, if they're not all true, then at least one of them is wrong, say the first one. So are you saying it's possible for the relative frequency of something to be the same for three different sets, but not to be the same for their union?

 

[billschnieder]

I definitely intend s to be the same set in all three statements.

I'm saying a relationship between 3 properties within a single set p, is the same relationship between 3 properties within another single q, and the same relationship between 3 properties within another single set r, and the same relationship between 3 properties within a bigger set s=(p+q+r). But I'm also saying that same relationship does not necessarily apply between one property from "p" and a second property from "q" and a third property from "r", even if p, q and r are each components of "s".

For example, Let us assume that we have a pair of perfectly correlated photons hidden properties which predetermine outcomes at three angles a,b,c for each photon and each of the outcomes can be +1 or -1.We can calculate the sum of the paired product of outcomes ab and ac as follows that

ab + ac = a(b + c) = ab(1+bc)
since b = 1/b, and taking the absolute value of both sides of this equation we get

|ab + ac| = |ab (1+bc) | ≤ 1 + bc or |ab + ac| - bc ≤ 1, a Bell inequality.

Clearly, we have derived this as a condition which applies to 3 outcomes a,b,c from a single pair of photons, without any other assumption than that the 3 outcomes exist. In fact, we might have started with an abstract assumption that we have an arbitrary set of 3 variables a,b,c each with values +1 or -1, without any physical meaning assigned to the set, and still obtain the inequalities. Furthermore, if we have a very large number of such photons, because individually they obey the inequality, collectively, set of photons also obeys the inequality
|<ab> + <ac>| - <bc> ≤ 1, another Bell inequality, or if you prefer |C(a,b) + C(a,c)| - C(b,c) ≤ 1, Bell's original form.

Do you agree?

 

[billschnieder]

Continuing ...

The terms C(a,b), C(a,c) and C(b,c) in the above inequality are each from the exact same set of photons, not three different sets. And the reason why it MUST be so, is because in order to derive the inequality, we started with two terms "ab + ac", and we factored out "a(b+c)", then we factored out b=1/b again to get "ab(1+bc)" and in this way we "created" the "bc" term by stitching together part from "ab" and part from "ac".

So what is the problem? You may ask.

Suppose instead we have 3 different pairs of photons each with predetermined outcomes at 3 angles, say (a1, b1, c1) for pair 1, (a2,b2,c2) for pair 2 and (a3, b3, c3) for pair 3. Obviously, each of those pairs will independently satisfy the inequality since we could simply say (a=a1,b=b1,c=c1) etc. Similarly, we could have 3 sets, each with a large number of photon pairs and each one will independently satisfy the inequalities.

However if we calculate the "ab" term from the first photon pair, the "ac" term from the second photon pair and "bc" from the third pair (ie, a1b1, a2c2, b3c3), is it reasonable to expect the inequality |ab + ac| - bc ≤ 1 to be obeyed?

Similarly if we calculate the "ab" correlation from the first set of photons, the "ac" correlation from the second set of photons and the "bc" correlation from the third set of photons (ie a1b1, a2c2, b3c3), is it reasonable to expect the inequality |C(a,b) + C(a,c)| - C(b,c) ≤ 1 to be obeyed?

I say the answer to the above two questions is NO. Let us see why, by verifying the maximum value of left hand side |a1b1 + a2c2| - b3c3
Clearly, the LHS is maximum when b3c3 = -1, a1b1 = a2c2 = ±1
Therefore the correct inequality for three different pairs is
|a1b1 + a2c2| - b3c3 ≤ 3 NOT |a1b1 + a2c2| - b3c3 ≤ 1

And, the correct inequality for 3 different sets is
|<a1b1> + <a2c2>| - <b3c3> ≤ 3 NOT |<a1b1> + <a2c2>| - <b3c3> ≤ 1

In other words, the relationship we derived from properties within a single pair of photons does not apply between properties from 3 different pairs of photons.
Do you agree?

 

[billschnieder]

Some may wonder why we got a maximum of 3 for different sets and 1 for the same set. I thought I should clarify again. For three sets we have |a1b1 + a2c2| - b3c3 , with the maximum of 3 obtained when b3c3 = -1, a1b1 = a2c2 = +1.

This can happen if b3 = -1, c3 = 1, a1 = 1, b1 = 1, a2 = 1, c2 = 1 which is possible within three different sets since b3 is allowed to be different from b1. However in a single set, it is impossible for the "b" used to calculate "bc" term to be different from the "b" used to calculate the "ab" term.

 

[Nugatory]

But I'm also saying that same relationship does not necessarily apply between one property from "p" and a second property from "q" and a third property from "r", even if p, q and r are each components of "s".

Of course it does not necessarily apply. But it does apply if we make two additional assumptions, namely counterfactual definiteness and fair sampling within statistically representative sets. Which of these assumptions do you reject?

 

[rlduncan]

Bill, your #346 and #347 posts are excellent summaries of your argument. They tie in nicely with your post #270 which reveal that Bell-Type inequalities cannot apply to both scenarios.

 

[lugita15]

I'm saying a relationship between 3 properties within a single set p, is the same relationship between 3 properties within another single q, and the same relationship between 3 properties within another single set r, and the same relationship between 3 properties within a bigger set s=(p+q+r). But I'm also saying that same relationship does not necessarily apply between one property from "p" and a second property from "q" and a third property from "r", even if p, q and r are each components of "s".

I'm not sure what properties you're talking about. Do you mean relative frequencies?

Anyway, let me just ask you this. For each of the four sets p, q, r, and s, we can calculate the following relative frequencies:
1. The relative frequency of getting (1,-1) or (-1,1) for f(-30,0)
2. The relative frequency of getting (1,-1) or (-1,1) for f(0,30)
3. The relative frequency of getting (1,-1) or (-1,1) for f(-30,30)

Now you've already agreed that all of these relative frequencies are the same for p, q, and r. But you've disagreed that all of these relative frequencies are the same for p, q, r, and s. Well then, you must believe that at least one of these relative frequencies can different for s than for p, q, and r. In other words, you must believe that the relative frequency of something can be different for the union of three sets even though it's the same for all three sets.

So first of all, do you in fact believe that? If so, I think I can prove you wrong with a very simple argument.