Scholarpedia article on Bell's Theorem

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stevendaryl

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If such a dataset is impossible then what dataset is being used to compare experiments to the inequalities, or are you now claiming that the experiments do not produce datasets?
I have no idea what you are talking about. What it boils down to is that there is a joint probability distribution for Alice and Bob: [itex]P(\alpha, \beta) = \frac{1}{2} sin^2(\frac{1}{2} (\beta - \alpha))[/itex]. This joint probability distribution gives rise to a particular correlation between Bob's measurements and Alice's measurements. This correlation can be tested experimentally, and the prediction is born out. So experiment confirms the predictions of quantum mechanics.

What Bell showed is that you can't simulate the joint probability distribution [itex]P(\alpha, \beta)[/itex] by a "factored" distribution of the form

[itex]\int d\lambda P_A(\alpha, \lambda) P_B(\beta, \lambda) P_L(\lambda)[/itex]

Bell's inequality gives a bound on the greatest correlation that can be simulated by "factored" probabilities of this form.

The dataset you are asking for is NOT what is measured in experiments. We already know ahead of time that there is no such dataset, so there's no point in testing that. What is measured in experimental is the correlations between Alice's and Bob's measurements.
 
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I have no idea what you are talking about.
I didn't think so. See posts #424 and #492 in this thread.
What it boils down to is that there is a joint probability distribution for Alice and Bob: [itex]P(\alpha, \beta) = \frac{1}{2} sin^2(\frac{1}{2} (\beta - \alpha))[/itex]. This joint probability distribution gives rise to a particular correlation between Bob's measurements and Alice's measurements. This correlation can be tested experimentally, and the prediction is born out. So experiment confirms the predictions of quantum mechanics.

What Bell showed is that you can't simulate the joint probability distribution [itex]P(\alpha, \beta)[/itex] by a "factored" distribution of the form

[itex]\int d\lambda P_A(\alpha, \lambda) P_B(\beta, \lambda) P_L(\lambda)[/itex]

Bell's inequality gives a bound on the greatest correlation that can be simulated by "factored" probabilities of this form.
I didn't want to get into this but you've made the error repeatedly. You do realize that the expression
[itex]\int d\lambda P_A(\alpha, \lambda) P_B(\beta, \lambda) P_L(\lambda)[/itex]

is not a conditional probability statement but a statement for the expectation value of the paired product of outcomes at A and B, don't you. Where [itex]P_A(\alpha, \lambda)[/itex] and [itex] P_B(\beta, \lambda)[/itex] are functions generating +1 or -1, not probabilities. You should check Bell's original Paper.

The Expectation value for the paired product at two stations is necessarily factorable whether or not the processes generating the outcomes are local or non-local.
 

stevendaryl

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And which of them will you be substituting into Bell's inequalities to demonstrate violation?
Bell's inequalities are not about probabilities, they are about correlations. The correlation [itex]C(\alpha, \beta)[/itex] is equal to:

[itex]P_{both}(\alpha, \beta) + P_{neither}(\alpha, \beta) - P_{Alice}(\alpha, \beta) - P_{Bob}(\alpha, \beta)[/itex]

where [itex]P_{both}(\alpha, \beta)[/itex] is the probability that both Alice and Bob measure spin-up, [itex]P_{neither}(\alpha, \beta)[/itex] is the probability than neither measure spin-up,
[itex]P_{Alice}(\alpha, \beta)[/itex] is the probability that just Alice measures spin-up, and [itex]P_{Bob}(\alpha, \beta)[/itex] is the probability that just Bob measures spin-up. Assuming perfect detection, the predictions of QM are:

[itex]P_{both}(\alpha, \beta) = \frac{1}{2} sin^2(\frac{1}{2}(\beta - \alpha))[/itex]

[itex]P_{neither}(\alpha, \beta) = \frac{1}{2} sin^2(\frac{1}{2}(\beta - \alpha))[/itex]

[itex]P_{Alice}(\alpha, \beta) = \frac{1}{2} cos^2(\frac{1}{2}(\beta - \alpha))[/itex]

[itex]P_{Bob}(\alpha, \beta) = \frac{1}{2} cos^2(\frac{1}{2}(\beta - \alpha))[/itex]

So the prediction of QM is:
[itex]C(\alpha, \beta) = sin^2(\frac{1}{2}(\beta - \alpha)) - cos^2(\frac{1}{2}(\beta - \alpha)) = 1 - 2 cos^2(\frac{1}{2}(\beta - \alpha))[/itex]

What is measured in tests of Bell's inequality is [itex]C(\alpha, \beta)[/itex].
 
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The dataset you are asking for is NOT what is measured in experiments. We already know ahead of time that there is no such dataset, so there's no point in testing that. What is measured in experimental is the correlations between Alice's and Bob's measurements.
Huh? You do not know what you are talking about. Correlations are CALCULATED from the measured dataset not directly measured. What is measured are clicks at given detectors for individual photons.
 

stevendaryl

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I didn't think so. See posts #424 and #492 in this thread.

I didn't want to get into this but you've made the error repeatedly. You do realize that the expression
[itex]\int d\lambda P_A(\alpha, \lambda) P_B(\beta, \lambda) P_L(\lambda)[/itex]

is not a conditional probability statement but a statement for the expectation value of the paired product of outcomes at A and B, don't you. Where [itex]P_A(\alpha, \lambda)[/itex] and [itex] P_B(\beta, \lambda)[/itex] are functions generating +1 or -1, not probabilities. You should check Bell's original Paper.
I told you that I was NOT following Bell, but instead making a different, but related, probability claim. The perfect correlations for anti-aligned detectors shows that the probabilities [itex]P_A(\alpha, \lambda)[/itex] and [itex]P_B(\beta, \lambda)[/itex] must be 0 or 1. That means that the spin-up versus spin-down is a deterministic function of lambda, which is what Bell assumed. But you don't have to assume it, it's forced by the perfect anti-correlations.
 
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stevendaryl

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Correlations are CALCULATED from the measured dataset not directly measured. What is measured are clicks at given detectors for individual photons.
Whatever. The distinction between measuring and calculating from measurements is not the critical point. The correlation function is "measured" by computing:

[itex]\frac{1}{N} \sum S_{A,n} S_{B,n} [/itex]

where [itex]S_{A,n} = +1[/itex] if Alice measures spin-up on run [itex]n[/itex] and [itex]S_A = -1[/itex] if Alice measures spin-down on run [itex]n[/itex], and similarly for Bob, and [itex]N[/itex] is the total number of runs.

But this data set is not the data set that one can easily prove does not exist (the factored probabilities).
 
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What is measured in tests of Bell's inequality is [itex]C(\alpha, \beta)[/itex].
Please you really should read an experimental description for an EPR type experiment. Using DA's diagram, what is measured is a series of time-tagged pluses and minuses depending on which detector fires D+ or D- at each station. So for a given angle setting "a", Alice has a long list of +1s and -1s which appear random each with a time tag indicating when the detector fired, and Bob has a similar list for setting "b". Then at the end of the experiment, the time tags are compared to figure out which ones were "coincident", then the results of each pair that belong to a "coincident" pair are multiplied to each other to obtain the product and then the average is calculated to obtain the expectation value of the paired product at the two stations, also known as the correlation.

So in bell test experiments, what you call [itex]C(\alpha, \beta)[/itex] is actuall <ab> where a represents the outcomes at angle α and b represents the outcomes at angle β.
 
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Whatever. The distinction between measuring and calculating from measurements is not the critical point. The correlation function is "measured" by computing:

[itex]\frac{1}{N} \sum S_{A,n} S_{B,n} [/itex]

where [itex]S_{A,n} = +1[/itex] if Alice measures spin-up on run [itex]n[/itex] and [itex]S_A = -1[/itex] if Alice measures spin-down on run [itex]n[/itex], and similarly for Bob, and [itex]N[/itex] is the total number of runs.

But this data set is not the data set that one can easily prove does not exist (the factored probabilities).
You are still arguing that experiments produce something which is impossible. You can not argue that something is impossible and then also claim that non-local experiments produce it, whatever it is. Besides, the expectation values calculated from the experiment is clearly factorable yet the experiments violate the inequality. Go figure.
 

stevendaryl

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Please you really should read an experimental description for an EPR type experiment. Using DA's diagram, what is measured is a series of time-tagged pluses and minuses depending on which detector fires D+ or D- at each station. So for a given angle setting "a", Alice has a long list of +1s and -1s which appear random each with a time tag indicating when the detector fired, and Bob has a similar list for setting "b". Then at the end of the experiment, the time tags are compared to figure out which ones were "coincident", then the results of each pair that belong to a "coincident" pair are multiplied to each other to obtain the product and then the average is calculated to obtain the expectation value of the paired product at the two stations, also known as the correlation.

So in bell test experiments, what you call [itex]C(\alpha, \beta)[/itex] is actuall <ab> where a represents the outcomes at angle α and b represents the outcomes at angle β.
How is that different from what I said?
 

stevendaryl

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You are still arguing that experiments produce something which is impossible. You can not argue that something is impossible and then also claim that non-local experiments produce it, whatever it is.
I'm not arguing that the correlations predicted by quantum mechanics are impossible, I'm arguing that it is impossible to achieve those correlations using a local hidden variables theory. You seem deeply confused about this point. As I said, the quantum joint probabilities
[itex]P(\alpha, \beta)[/itex] are certainly possible, and the correlations are calculated from joint probabilities. But you cannot express the joint probability as a factored probability, which is what you would expect from a local hidden variables theory. Bell's inequalities are not impossible to violate, they are impossible to violate using factored probabilities.
 

stevendaryl

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Let me try to clarify what, exactly, is impossible in a local hidden variables theory.

First, what is the correlation function, which I'm calling [itex]C(\alpha, \beta)[/itex]?

If you have a list, or run, of 4-tuples of numbers: [itex]\alpha_n, A_n, \beta_n, B_n[/itex] where for each [itex]n[/itex], [itex]\alpha_n[/itex] and [itex]\beta_n[/itex] are angles, and [itex]A_n[/itex] and [itex]B_n[/itex] are each either [itex]+1[/itex] or [itex]-1[/itex], then you can compute a correlation [itex]C(\alpha, \beta)[/itex] as follows:

[itex]C(\alpha, \beta) = \frac{1}{N_{\alpha \beta}} \sum A_n B_n[/itex]
where the sum is over those runs [itex]n[/itex] such that [itex]\alpha_n = \alpha[/itex] and [itex]\beta_n = \beta[/itex], and where [itex]N_{\alpha \beta}[/itex] is the total number of such runs.

Suppose that we generate such a list as follows: We create a sequence of spin-1/2 twin pairs. On run [itex]n[/itex], one particle is detected by Alice using a spin-measurement device aligned in the x-y plane at angle [itex]\alpha_n[/itex] away from the x-axis, and the other is detected by Bob at an angle [itex]\beta_n[/itex]. If Alice measures spin-up, then [itex]A_n = +1[/itex]. If Alice measures spin-down, then [itex]A_n = -1[/itex]. If Bob measures spin-up, then [itex]B_n = +1[/itex]. If Bob measures spin-down, then [itex]B_n = -1[/itex].

The prediction of quantum mechanics is that in the limit as the number of trials at each angle goes to infinity, is that

[itex]C(\alpha, \beta) = - cos(\beta - \alpha)[/itex]

At this point, let's specialize to specific values for [itex]\alpha[/itex] and [itex]\beta[/itex]. Assume that [itex]\alpha[/itex] and [itex]\beta[/itex] are always chosen to be from the set
{ 0°, 120°, 240°}. A way to explain the correlations using deterministic local hidden variables is to assume that corresponding to run number [itex]n[/itex] there is a hidden variable [itex]\lambda_n[/itex], which is (or determines) a triple of values [itex]\lambda_n = \langle A_{0\ n}, A_{120\ n}, A_{240\ n}\rangle[/itex]. Then [itex]A_n[/itex] and [itex]B_n[/itex] are deterministic functions of the angles [itex]\alpha[/itex] and [itex]\beta[/itex] and the "hidden variable" [itex]\lambda_n[/itex]:

[itex]A_n = A_{\alpha_n\ n}[/itex]
[itex]B_n = B_{\beta_n\ n} \equiv - A_{\beta_n\ n}[/itex]

Now, here's where the impossibility claim arises: If we have a sequence of triples [itex]\langle A_{0\ n}, A_{120\ n}, A_{240\ n}\rangle[/itex], then we can compute correlation functions as follows:

[itex]C'(\alpha \beta) = \frac{1}{N} \sum A_{\alpha\ n} B_{\beta\ n}[/itex]

where [itex]N[/itex] is the total number of runs.

I'm using a prime to distinguish this correlation function from the previous. The difference between the two is that [itex]C(\alpha \beta)[/itex] is computed using those runs in which Alice happens to choose detector angle [itex]\alpha[/itex], and Bob happens to choose detector angle [itex]\beta[/itex]. In contrast, [itex]C'(\alpha \beta)[/itex] is computed using all runs, since by assumption, [itex]A_{\alpha\ n}, B_{\beta\ n}[/itex] determines what Alice and Bob would have gotten on run n had they chosen settings [itex]\alpha[/itex] and [itex]\beta[/itex].

The impossibility claim is that there is no sequence of triples [itex]\langle A_{0\ n}, A_{120\ n}, A_{240\ n}\rangle[/itex] such that the corresponding [itex]C'(\alpha, \beta)[/itex] agrees with the quantum mechanical prediction for the correlation.

This impossibility claim is NOT contradicted by actual experiments, because an actual experiment cannot measure the triple [itex]\langle A_{0\ n}, A_{120\ n}, A_{240\ n}\rangle[/itex]; it can only measure two of the three values.
 
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Let me try to clarify what, exactly, is impossible in a local hidden variables theory.
...
You argument supports my point (which apparently you still have not considered carefully) not yours. The reason being -- why would any sane person expect the EPR experiment to provide a list of triples rather than 4 lists of doubles?

Forget about probabilities and separability. We start from the inequalities already derived, OK! We have terms 4 correlation terms in the CHSH inequality.

(1) we calculate the terms from QM and obtain a violation - this is the origin of Bell's theorem
(2) we measure the terms from an experiment and obtain a violation. We also realize that the results match QM.

Do you agree with this? If you do then we can dig deeper into the origin of the violation.
 
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stevendaryl

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You argument supports my point (which apparently you still have not considered carefully) not yours.
Well, I don't understand what your point is. I don't think you've been very clear.

The reason being -- why would any sane person expect the EPR experiment to provide a list of triples rather than 4 lists of doubles?
Because of perfect anti-correlations between the two detectors when they are set at the same detector angle, we can assume, under a local hidden-variables theory, that the outcome is a deterministic function of the hidden variable [itex]\lambda[/itex]. That means that there is a function [itex]F(\alpha, \lambda)[/itex] returning +1 or -1 for whether Alice will measure spin-up or spin-down at angle [itex]\alpha[/itex]. So if Alice has three possible angles to choose from, then there are three relevant quantities:

[itex]F(a,\lambda), F(b,\lambda), F(c,\lambda)[/itex]

The first gives the result if Alice happened to choose angle [itex]a[/itex], the second if Alice happened to choose angle [itex]b[/itex], etc.

Bob's results are anti-correlated with Alice's, so we can get Bob's results by using
[itex]F'(\alpha,\lambda) = 1 - F(\alpha,\lambda)[/itex]


Forget about probabilities and separability. We start from the inequalities already derived, OK!
No. I don't want to start there.
 
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Well, I don't understand what your point is. I don't think you've been very clear.

No. I don't want to start there.
Then you are deliberately refusing to see my point that's why you keep getting confused about what I'm saying. I'm not arguing with you about how the inequalities are derived so you are wasting your time trying to demonstrate separability etc.

You agreed earlier the correlations calculated from experiments is
[itex]\frac{1}{N} \sum S_{A,n} S_{B,n} [/itex]
You admit that the above correlation matches the QM prediction for the experiment. Although you admit that the above correlation is separable as can be seen from the equation, you turn around and contradict yourself by saying it is impossible to obtain the QM correlation in separable form. Don't you see your error?! If it is impossible to obtain the QM correlation in separable form, then it is impossible for the experiment to match QM as you claim!

...the outcome is a deterministic function of the hidden variable [itex]\lambda[/itex]. That means that there is a function [itex]F(\alpha, \lambda)[/itex] returning +1 or -1 for whether Alice will measure spin-up or spin-down at angle [itex]\alpha[/itex]. So if Alice has three possible angles to choose from, then there are three relevant quantities:

[itex]F(a,\lambda), F(b,\lambda), F(c,\lambda)[/itex]
Yes, Yes, Yes! For Alice, the outcomes are:

- F(a,λ) if she measures at angle (a)
- F(b,λ) if she had measured at angle (b) instead of at the (a) at which she actually measured
- F(c,λ) if she had measured at angle (c) instead of at the (a) at which she actually measured

Don't you see that if Alice in fact measured at (a), then the last two F(b,λ), F(c,λ) MUST be counterfactual! We are not talking about measuring any random photon, we are talking about what she would have obtained were it possible for her to rewind time and measure the same photon at a different angle! Don't you see that those are not the outcomes measured in any real experiment.
 
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stevendaryl

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Then you are deliberately refusing to see my point that's why you keep getting confused about what I'm saying.
No, I just don't think you've been very clear.

You agreed earlier the correlations calculated from experiments is
[itex]\frac{1}{N} \sum S_{A,n} S_{B,n} [/itex]
You admit that the above correlation matches the QM prediction for the experiment.
It's not that I admit it, I'm pointing it out.

Although you admit that the above correlation is separable as can be seen from the equation.
No, I don't agree with that. The kind of separability that I'm talking about is separability of the joint probabilities, not correlations. Bell's assumption is that if all interactions are local, and

[itex]P(\alpha \wedge \beta) \neq P(\alpha) P(\beta)[/itex]

then there must be some "hidden variable" [itex]\lambda[/itex] such that

[itex]P(\alpha \wedge \beta) = \int d\lambda P(\lambda) P(\alpha | \lambda) P(\beta | \lambda)[/itex]

The correlations predicted by quantum mechanics cannot be generated by factorable probabilities of this form.

Don't you see that if Alice in fact measured at (a), then the last two F(b,λ), F(c,λ) MUST be counterfactual!
Of course. A hidden variables theory implies that counterfactuals have definite values.
 
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It's not that I admit it, I'm pointing it out.
Huh? You are pointing out but without agreeing with the way the correlations are calculated in experiments? Do you or do you not agree that those correlations as calculated in the experiments are separable? The equation you provided yourself shows that they are!!!

The kind of separability that I'm talking about is separability of the joint probabilities, not correlations.
...
[itex]P(\alpha \wedge \beta) = \int d\lambda P(\lambda) P(\alpha | \lambda) P(\beta | \lambda)[/itex]
The correlations predicted by quantum mechanics cannot be generated by factorable probabilities of this form.
You must be very confused then. Is [itex]P(\alpha \wedge \beta)[/itex] a joint probability or a correlation? Are you saying the correlations are separable but the joint probabilities are not !? When I demonstrate to you that the correlations from the experiment are separable, you argue that you were talking about probabilities not correlations and then turn around and use the two terms synonymously.
 

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Don't you see that if Alice in fact measured at (a), then the last two F(b,λ), F(c,λ) MUST be counterfactual
Yes.
And if we proceed under the assumption that these counterfactuals have real numerical values, we end up drawing conclusions that are not supported by experimental results.

We then have two possibilities:
1) Because of some flaw in their design or execution, the experiments also do not falsify our conclusion. Appeals to the fair-sampling loophole would fall in this category.
2) The experiments are not fatally flawed so they do falsify our conclusion; therefore the assumption leading to the conclusion must be false.

Are you going somewhere else with this line of argument?
 
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Yes.
And if we proceed under the assumption that these counterfactuals have real numerical values, we end up drawing conclusions that are not supported by experimental results.
You are jumping the gun. If you proceed under the assumption of counterfactual terms in the inequality, then the terms in the inequality must be interdependent and therefore can never be measured in any experiment. So you don't even get to any experiment because a genuine experiment to test the inequality is impossible to perform.

We then have two possibilities:
1) Because of some flaw in their design or execution, the experiments also do not falsify our conclusion. Appeals to the fair-sampling loophole would fall in this category.
2) The experiments are not fatally flawed so they do falsify our conclusion; therefore the assumption leading to the conclusion must be false.
There no mention of loopholes in my argument. My argument does not rely or use any loopholes. Once you realize that the experiment is impossible, it becomes nonsensical to even suggest that the experimental result disagrees with the inequality because the experiment you are talking about would be testing something else not Bell's inequality.
 
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[itex]P(\alpha \wedge \beta) = \int d\lambda P(\lambda) P(\alpha | \lambda) P(\beta | \lambda)[/itex]
BTW I hope you realize that for an EPR experiment in which we have coincidence counting the correct probability expression should be

[itex]P(\alpha \wedge \beta) = \int d\lambda P(\lambda) P(\alpha | \lambda) P(\beta | \alpha, \lambda)[/itex]
 

Nugatory

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Once you realize that the experiment is impossible, it becomes nonsensical to even suggest that the experimental result disagrees with the inequality because the experiment you are talking about would be testing something else not Bell's inequality.
I think that falls under #1 - the experiment that has been done is so flawed in its execution or interpretation as to neither confirm or falsify the conclusion.

If that's what you mean, at least it's a coherent statement whose merits can be discussed.
(It's also a statement that I disagree with, but let's try to identify what we're arguing about before we have the argument :smile:)
 
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I think that falls under #1 - the experiment that has been done is so flawed in its execution or interpretation as to neither confirm or falsify the conclusion.

If that's what you mean, at least it's a coherent statement whose merits can be discussed.
(It's also a statement that I disagree with, but let's try to identify what we're arguing about before we have the argument :smile:)
Fair enough. Why do you disagree with it? Here is the argument again, please explain what part of it you disagree with:

QM violates the inequalities because the terms that people calculate from QM and substitute into the inequalities in order to obtain violation, are not the correct terms.

They've calculated and used the following three terms (scenario X):
C(a,b) = correlation for what we would get if we measure (a,b)
C(b,c) = correlation for what we would get if we measure (b,c)
C(a,c) = correlation for what we would get if we measure (a,c)

When Bell's inequalities DEMAND that the correlations should be (scenario Y):
C(a,b) = correlation for what we would get if we measure (a,b)
C(a,c) = correlation for what we would have gotten had we measured (a,c) instead of (a,b)
C(b,c) = correlation for what we would have gotten had we measured (b,c) instead of (a,b)

Experiments violate the inequalities because the correlations measured correspond to scenario X not the correct scenario Y. QM and Experiments agree with each other because they both refer to scenario X not Bell's scenario Y. In other words, if you insist on using the terms from QM and experiment to compare with the inequality, then you are making an extra assumption that the correlations in scenario X and Y are equivalent. Now once you obtain a violation, it is this assumption that should be thrown out. As I have demonstrated already, those two scenarios are different without any non-locality or conspiracy, and such an assumption should never even be introduced if reasoning correctly.
 
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Now let me explain again why Scenario X is different from Scenario Y, photon by photon.

As stevendaryl explained the correlation is calculated as:
[itex]C(\alpha, \beta) = \frac{1}{N} \sum F_{\alpha n} F'_{\beta n}[/itex]

Let us start with the first photon pair arriving at Alice and Bob respectively, and assume that the outcome was +1 for Alice and -1 for Bob for the angle pair (a,b). In other words, the first outcome which goes into the C(a,b) calculation is F(a,λ1)=+1 for Alice, and F'(b, λ1)=-1.

For Scenario Y, then, the first outcome which goes into calculating the remaining two terms is immediately restricted by that result to F(a,λ1)=+1, and F'(c, λ1) = ? for calculating C(a,c) and F(c,λ1)= ?, and F'(b, λ1) = -1 for calculating C(c, b). This is obviously the case because, if Alice got F(a,λ1)=+1, she would not have gotten anything other than what she got had she measured at the same angle which she did in fact measure at and if Bob obtained F'(b, λ1) = -1 he wouldn't have gotten anything other than what he got, had he measured at the same angle at which he measured at. Therefore the outcomes used to calculate C(a,c), and C(c,b) are not independent of those used to calculate C(a,b). We can then go to the next photon pair, and the next and the same conditions apply for the whole set used to calculate the correlation.

However for Scenario X, we are not dealing with counter-factuals and there is no restriction to specific photons. We can measure any random photon for any of the correlations. We are allowed to obtain F(a,λ1)=+1 for Alice for the first outcome used to calculate the C(a,b) correlation and F(a,λ1) = -1 for Alice for the first outcome used to calculate the C(a,c) correlation etc. As you can see, Scenario X has many more degrees of freedom than Scenario Y. The terms in scenario X are independent of each other contrary to scenario Y.
 
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Now let us simulate all the possibilities for each scenario and demonstrate that while scenario Y would never result in a violation, violations can be expected for scenario X.

Note I'm using the shorthand a,b,c to represent Fa, Fb, Fc which are outcomes not angles. I'm using the 3-term Bell inequality |ab + ac| - bc <= 1 in which each term shares outcomes with the other two terms.
Scenario Y:
Code:
a,b,c = (+1,+1,+1): |(+1) + (+1)| - (+1) <= 1, obeyed=True
a,b,c = (+1,+1,-1): |(+1) + (-1)| - (-1) <= 1, obeyed=True
a,b,c = (+1,-1,+1): |(-1) + (+1)| - (-1) <= 1, obeyed=True
a,b,c = (+1,-1,-1): |(-1) + (-1)| - (+1) <= 1, obeyed=True
a,b,c = (-1,+1,+1): |(-1) + (-1)| - (+1) <= 1, obeyed=True
a,b,c = (-1,+1,-1): |(-1) + (+1)| - (-1) <= 1, obeyed=True
a,b,c = (-1,-1,+1): |(+1) + (-1)| - (-1) <= 1, obeyed=True
a,b,c = (-1,-1,-1): |(+1) + (+1)| - (+1) <= 1, obeyed=True
* Note that no violation is ever obtained for any individual pair of outcomes, and consequently no violation is possible for the correlations which are essentially averages of paired products |<ab> + <ac>| - <bc> <= 1
* Note that there are only 8 distinct possible outcome combinations for this scenario each of which always satisfies the inequality

Now for Scenario X, we measure C(a,b) from one set of photons, C(c,b) from a different set of photons and and C(a,c) from yet another set of photons. Really what we are measuring is C(a1,b1), C(a2,c2) and C(c3,b3) and if we substitute in the inequality, we actually have
|a1b1 + a2c2| - b3c3 <= 1 in which there is no dependency between any of the terms. No two terms share the same outcome contrary to scenario Y. Simulating this, we get

(see next post)
 
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808
10
Code:
a1,a2,b1,b3,c2,c3 = (+1,-1,+1,+1,-1,-1): |(+1) + (+1)| - (-1) <= 1, obeyed=False
a1,a2,b1,b3,c2,c3 = (+1,-1,+1,-1,-1,+1): |(+1) + (+1)| - (-1) <= 1, obeyed=False
a1,a2,b1,b3,c2,c3 = (+1,-1,+1,-1,-1,-1): |(+1) + (+1)| - (+1) <= 1, obeyed=True
a1,a2,b1,b3,c2,c3 = (+1,+1,-1,+1,+1,+1): |(-1) + (+1)| - (+1) <= 1, obeyed=True
a1,a2,b1,b3,c2,c3 = (+1,+1,-1,+1,+1,-1): |(-1) + (+1)| - (-1) <= 1, obeyed=True
...
a1,a2,b1,b3,c2,c3 = (-1,+1,-1,-1,-1,+1): |(+1) + (-1)| - (-1) <= 1, obeyed=True
a1,a2,b1,b3,c2,c3 = (-1,+1,-1,-1,-1,-1): |(+1) + (-1)| - (+1) <= 1, obeyed=True
a1,a2,b1,b3,c2,c3 = (-1,-1,-1,+1,-1,+1): |(+1) + (+1)| - (+1) <= 1, obeyed=True
a1,a2,b1,b3,c2,c3 = (-1,-1,-1,+1,-1,-1): |(+1) + (+1)| - (-1) <= 1, obeyed=False
a1,a2,b1,b3,c2,c3 = (-1,-1,-1,-1,-1,+1): |(+1) + (+1)| - (-1) <= 1, obeyed=False
a1,a2,b1,b3,c2,c3 = (-1,-1,-1,-1,-1,-1): |(+1) + (+1)| - (+1) <= 1, obeyed=True
(see post 125 for the full simulation results)

* Note that there are 64 distinct possible outcome combinations in Scenario X as opposed to just 8 in Scenario Y.
* Note also that the inequality is violated many times (in 1/4 of the cases).
* Note that when a1=a2 and b1=b3 and c2=c3 as required by scenario X, the inequality is NEVER violated.
 
Last edited:

DevilsAvocado

Gold Member
751
90
I don't like dragging free will into discussions about science, because I just don't think it has any relevance. "Free" choices that humans make could very well be determined by conditions at a microscopic level, and that wouldn't make much difference, in practice. What I thought was the difference between determinism and superdeterminism is this:

  • A theory is deterministic if a past state uniquely singles out one possible future state.
  • A theory is superdeterministic if there is only one possible past, as well.
Agreed 100% :approve:

The reason for bringing in free will in this, is Bell’s statement on superdeterminism, but to me it’s the same dish as all the other “not so very bright” loopholes.
 

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