Scholarpedia article on Bell's Theorem

  • #501
billschnieder said:
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: P(\alpha, \beta) = \frac{1}{2} sin^2(\frac{1}{2} (\beta - \alpha)). 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 P(\alpha, \beta) by a "factored" distribution of the form

\int d\lambda P_A(\alpha, \lambda) P_B(\beta, \lambda) P_L(\lambda)

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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  • #502
stevendaryl said:
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: P(\alpha, \beta) = \frac{1}{2} sin^2(\frac{1}{2} (\beta - \alpha)). 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 P(\alpha, \beta) by a "factored" distribution of the form

\int d\lambda P_A(\alpha, \lambda) P_B(\beta, \lambda) P_L(\lambda)

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
\int d\lambda P_A(\alpha, \lambda) P_B(\beta, \lambda) P_L(\lambda)

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 P_A(\alpha, \lambda) and P_B(\beta, \lambda) 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.
 
  • #503
billschnieder said:
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 C(\alpha, \beta) is equal to:

P_{both}(\alpha, \beta) + P_{neither}(\alpha, \beta) - P_{Alice}(\alpha, \beta) - P_{Bob}(\alpha, \beta)

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

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

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

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

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

So the prediction of QM is:
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))

What is measured in tests of Bell's inequality is C(\alpha, \beta).
 
  • #504
stevendaryl said:
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.
 
  • #505
billschnieder said:
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
\int d\lambda P_A(\alpha, \lambda) P_B(\beta, \lambda) P_L(\lambda)

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 P_A(\alpha, \lambda) and P_B(\beta, \lambda) 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 P_A(\alpha, \lambda) and P_B(\beta, \lambda) 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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  • #506
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:

\frac{1}{N} \sum S_{A,n} S_{B,n}

where S_{A,n} = +1 if Alice measures spin-up on run n and S_A = -1 if Alice measures spin-down on run n, and similarly for Bob, and N 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).
 
  • #507
stevendaryl said:
What is measured in tests of Bell's inequality is C(\alpha, \beta).
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 C(\alpha, \beta) is actuall <ab> where a represents the outcomes at angle α and b represents the outcomes at angle β.
 
  • #508
stevendaryl said:
Whatever. The distinction between measuring and calculating from measurements is not the critical point. The correlation function is "measured" by computing:

\frac{1}{N} \sum S_{A,n} S_{B,n}

where S_{A,n} = +1 if Alice measures spin-up on run n and S_A = -1 if Alice measures spin-down on run n, and similarly for Bob, and N 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.
 
  • #509
billschnieder said:
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 C(\alpha, \beta) 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?
 
  • #510
billschnieder said:
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
P(\alpha, \beta) 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.
 
  • #511
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 C(\alpha, \beta)?

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

C(\alpha, \beta) = \frac{1}{N_{\alpha \beta}} \sum A_n B_n
where the sum is over those runs n such that \alpha_n = \alpha and \beta_n = \beta, and where N_{\alpha \beta} 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 n, one particle is detected by Alice using a spin-measurement device aligned in the x-y plane at angle \alpha_n away from the x-axis, and the other is detected by Bob at an angle \beta_n. If Alice measures spin-up, then A_n = +1. If Alice measures spin-down, then A_n = -1. If Bob measures spin-up, then B_n = +1. If Bob measures spin-down, then B_n = -1.

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

C(\alpha, \beta) = - cos(\beta - \alpha)

At this point, let's specialize to specific values for \alpha and \beta. Assume that \alpha and \beta 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 n there is a hidden variable \lambda_n, which is (or determines) a triple of values \lambda_n = \langle A_{0\ n}, A_{120\ n}, A_{240\ n}\rangle. Then A_n and B_n are deterministic functions of the angles \alpha and \beta and the "hidden variable" \lambda_n:

A_n = A_{\alpha_n\ n}
B_n = B_{\beta_n\ n} \equiv - A_{\beta_n\ n}

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

C&#039;(\alpha \beta) = \frac{1}{N} \sum A_{\alpha\ n} B_{\beta\ n}

where N 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 C(\alpha \beta) is computed using those runs in which Alice happens to choose detector angle \alpha, and Bob happens to choose detector angle \beta. In contrast, C&#039;(\alpha \beta) is computed using all runs, since by assumption, A_{\alpha\ n}, B_{\beta\ n} determines what Alice and Bob would have gotten on run n had they chosen settings \alpha and \beta.

The impossibility claim is that there is no sequence of triples \langle A_{0\ n}, A_{120\ n}, A_{240\ n}\rangle such that the corresponding C&#039;(\alpha, \beta) 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 \langle A_{0\ n}, A_{120\ n}, A_{240\ n}\rangle; it can only measure two of the three values.
 
  • #512
stevendaryl said:
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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  • #513
billschnieder said:
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 \lambda. That means that there is a function F(\alpha, \lambda) returning +1 or -1 for whether Alice will measure spin-up or spin-down at angle \alpha. So if Alice has three possible angles to choose from, then there are three relevant quantities:

F(a,\lambda), F(b,\lambda), F(c,\lambda)

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

Bob's results are anti-correlated with Alice's, so we can get Bob's results by using
F&#039;(\alpha,\lambda) = 1 - F(\alpha,\lambda)


Forget about probabilities and separability. We start from the inequalities already derived, OK!

No. I don't want to start there.
 
  • #514
stevendaryl said:
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
\frac{1}{N} \sum S_{A,n} S_{B,n}
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 \lambda. That means that there is a function F(\alpha, \lambda) returning +1 or -1 for whether Alice will measure spin-up or spin-down at angle \alpha. So if Alice has three possible angles to choose from, then there are three relevant quantities:

F(a,\lambda), F(b,\lambda), F(c,\lambda)
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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  • #515
billschnieder said:
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
\frac{1}{N} \sum S_{A,n} S_{B,n}
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

P(\alpha \wedge \beta) \neq P(\alpha) P(\beta)

then there must be some "hidden variable" \lambda such that

P(\alpha \wedge \beta) = \int d\lambda P(\lambda) P(\alpha | \lambda) P(\beta | \lambda)

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.
 
  • #516
stevendaryl said:
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.
...
P(\alpha \wedge \beta) = \int d\lambda P(\lambda) P(\alpha | \lambda) P(\beta | \lambda)
The correlations predicted by quantum mechanics cannot be generated by factorable probabilities of this form.
You must be very confused then. Is P(\alpha \wedge \beta) 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.
 
  • #517
billschnieder said:
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?
 
  • #518
Nugatory said:
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.
 
  • #519
stevendaryl said:
P(\alpha \wedge \beta) = \int d\lambda P(\lambda) P(\alpha | \lambda) P(\beta | \lambda)

BTW I hope you realize that for an EPR experiment in which we have coincidence counting the correct probability expression should be

P(\alpha \wedge \beta) = \int d\lambda P(\lambda) P(\alpha | \lambda) P(\beta | \alpha, \lambda)
 
  • #520
billschnieder said:
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:)
 
  • #521
Nugatory said:
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.
 
  • #522
Now let me explain again why Scenario X is different from Scenario Y, photon by photon.

As stevendaryl explained the correlation is calculated as:
C(\alpha, \beta) = \frac{1}{N} \sum F_{\alpha n} F&#039;_{\beta n}

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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  • #523
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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  • #524
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.
 
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  • #525
stevendaryl said:
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.
 
  • #526
billschnieder said:
It is only when Alice has collected all her pluses and minuses and Bob has done the same that they start comparing time-tags to see coincidences and then they can figure out what the angular difference was at the moment of detection!

Not the angular difference, but the relative angle (a-b).

[all bolding mine]

billschnieder said:
The photon gives no rodent's behind whether Alice's angle was chosen randomly or not! All it meets is a polarizer at a given angle resulting in an OUTCOME of (+, -, or non-detection).

Misunderstanding; yes the angle is chosen randomly, but the experimental QM outcome/result is also always random 50/50 (+/-).

billschnieder said:
I'm afraid you have seriously misunderstood. Nobody is assuming static/predetermined results. The result is random for a given photon, but once Alice has measured and obtained +1 for that photon at 67.5°, it is a mathematical/logical error to say Alice would have obtained -1 had she measure that specific photon at the same specific angle 67.5°! You can't set a realism assumption and them immediately gut it and expect it to stay put.

More misunderstandings; we all know that it’s impossible to measure both +1 and -1 for one photon in one measurement, and most agree that if we repeat the measurement at the same angle – the result is 100% random.

Most also agree that if some dude comes up with a theorem that shows “1 + 1 = 9” we don’t have to prove this theorem to be correct, before proving it to be incorrect. I think it is called common sense.

billschnieder said:
Huh? I just show you that it is impossible to derive Bell's theorem without using CFD

Your derivation is quite strange since you are claiming that to be able to use CFD in any theorem we are obligated to actually measure these values in experiments, which of course most understand is impossible:

billschnieder said:
We do not need to sacrifice anything, because there is nothing there there to start with. The terms in Bell's inequality and the CHSH can never be tested experimentally, if reasoning correctly. The inequalities can never be violated if reasoning correctly. So I guess what has to be sacrificed is buffoonery.

billschnieder said:
We have three terms here C(a,b), C(a,c), C(b,c). Those terms can never be all factual as far as the EPR experiment is concerned. At least two of them MUST be counterfactual! There is no other way. Thinking otherwise is just buffoonery. The inequalities can NOT be derived UNLESS the other two terms are counterfactual. As soon as you see that, you realize immediately that NO EXPERIMENT can ever measure them all! None! You can measure one but not the other two. Is that clear enough?

No comment, speaks for itself.

billschnieder said:
So then, we are left with a lot of experimentalists who do not know what they are doing, publishing in lofty journals whose editors and reviewers do not know what they are doing, a many who love mysticism regurgitating what they've read without thinking for themselves. No news here.

I could be wrong, but to me it’s actually more mysticism in regarding all these highly educated/skilled peoples rigorously scrutinizing the mathematical and experimental result of Bell’s theorem – and not one see what you see...

Why thousands? If you were wrong, then one would have been enough!

A lot of ignorance out there, including Schrödinger et al., right? :biggrin:

billschnieder said:
Having eliminated all the experiments, we now have QM left. How come then that QM can violate 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) = QM correlation for what we would get if we measure (a,b)
C(b,c) = QM correlation for what we would get if we measure (b,c)
C(a,c) = QM 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) = QM correlation for what we would get if we measure (a,b)
C(a,c) = QM correlation for what we would have gotten had we measured (a,c) instead of (a,b)
C(b,c) = QM correlation for what we would have gotten had we measured (b,c) instead of (a,b)

More counterfactual confusion.

billschnieder said:
Now what is the probability that we would have obtained + at Alice if Alice and Bob had measured at angles (a,c) instead of (a,b). Note this is counterfactual. If you answer 1/3 you need to learn some probability theory. The correct answer is 1, we already know that measuring the photon at angle a gives +, where is the conspiracy in that?! Knowing what was obtained in the factual experiment, changes the probability we calculate for the counterfactual situation, nothing spooky involved. Now we can carry this all the way and include coincidences and you will see that using scenario X correlations in Bell's inequality is deeply flawed.

Well, to me all the above is good example of flawed conspiracy spookiness. You mix performed experimental outcomes with mathematical counterfactual speculations. I have never seen anything like it. You need to learn some QM theory.

billschnieder said:
QM gives the correct answer for the experiments performed, but neither QM nor the experiments can provide the right answers for substitution in the inequalities.

So what are the actual measured correlations? I hope you understand the experimental difference between entangled and non-entangled photons?

billschnieder said:
Are you serious? You must have misunderstood something very fundamental about the EPR experiment. For each photon that leaves the source and heads towards Alice, you have a single outcome, (+, -, or non-detection). Alice's polarizer is set to a specific angle say 67.5° for that photon. Same thing for Bob.

I can see from the discussion that there is in fact some serious confusion about the word outcome:

1. a final product or end result; consequence; issue.
2. a conclusion reached through a process of logical thinking.​

You seem to wobble between 1 & 2 without any specific notion on what you actually mean. Let me give you some examples:

billschnieder said:
It still does not change the fact that we have outcomes at 4 angles. For each angle there is an outcome.

billschnieder said:
Are you sure? If you insist, I suppose you can provide a NON-LOCAL dataset of outcomes for three angles a, b, c which violates the inequalities.

billschnieder said:
The only condition is there are 3 outcomes for 3 angles for each photon measured.

billschnieder said:
For each photon that leaves the source and heads towards Alice, you have a single outcome, (+, -, or non-detection).

billschnieder said:
The photon gives no rodent's behind whether Alice's angle was chosen randomly or not! All it meets is a polarizer at a given angle resulting in an OUTCOME of (+, -, or non-detection).

billschnieder said:
So? Who said any thing about the stream of outcomes appearing at Alice or Bob appearing other than random. That does not change the fact that there is an outcome. I have the files from Weihs' experiment and there is one outcome for each photon detected.

billschnieder said:
As soon as you see that, you realize immediately that NO EXPERIMENT can ever measure them all! None! You can measure one but not the other two. Is that clear enough?

billschnieder said:
Consider a pair of photons heading toward Alice and Bob resp, with polarizesr set to the angles (a,b). Let us say the possible outcomes are (+, -, 0 for nondetection) for each side and they are all equaly likely.

billschnieder said:
is not a conditional probability statement but a statement for the expectation value of the paired product of outcomes at A and B,

billschnieder said:
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.

billschnieder said:
is actuall <ab> where a represents the outcomes at angle α and b represents the outcomes at angle β.

billschnieder said:
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

billschnieder said:
Don't you see that those are not the outcomes measured in any real experiment.

billschnieder said:
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).

billschnieder said:
Therefore the outcomes used to calculate C(a,c), and C(c,b) are not independent of those used to calculate C(a,b).

billschnieder said:
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.

billschnieder said:
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.

billschnieder said:
* 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

billschnieder said:
No two terms share the same outcome contrary to scenario Y.

billschnieder said:
* Note that there are 64 distinct possible outcome combinations in Scenario X as opposed to just 8 in Scenario Y.

Get the point? How can we ever discuss this when you are jumping freely between:

  • outcome = experimental result
  • outcome = logical derivation
  • outcome = calculated expectation
  • outcome = probabilities
And on top of this you strangely enough couple counterfactual definiteness + CHSH to the impossibility of experimentally verify the outcome one would have obtained if one had measured a different angle:

  • outcome = counterfactual paradox
?

Care to straighten out some question marks?


Counterfactual definiteness - Wikipedia said:
Counterfactual definiteness is a basic assumption, which, together with locality, leads to Bell inequalities. In their derivation it is explicitly assumed that every possible measurement, even if not performed, can be included in statistical calculations. The calculation involves averaging over sets of outcomes which cannot all be simultaneously factual—if some are assumed to be factual outcomes of an experiment others have to be assumed counterfactual. (Which ones are designated as factual is determined by the experimenter: the outcomes of the measurements he actually performs become factual by virtue of his choice to do so, the outcomes of the measurements he doesn't perform are counterfactual.) Bell's Theorem actually proves that every type of quantum theory must necessarily violate either locality or CFD.
 
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  • #527
DevilsAvocado said:
Not the angular difference, but the relative angle (a-b).
:smile: Now I know you are not being serious at all.
Please read this article http://arxiv.org/pdf/quant-ph/9810080v1.pdf and pay attention at figure 2 of the above article, and notice the table of outcomes?

we all know that it’s impossible to measure both +1 and -1 for one photon in one measurement, and most agree that if we repeat the measurement at the same angle – the result is 100% random.
Huh? But it is impossible to repeat the same measurement! This is the whole point. You probably mean if you repeat a similar measurement for many different photons, the results appear random. Of course but Bell's inequalities are derived for the same set of photons (remember realism? remember CFD?) The whole point which you still do not get is that repeating it for different photons does not get you the same result as what you would have gotten, were it possible to repeat for the same photon. While the former may be random, the latter must not be.

Most also agree that if some dude comes up with a theorem that shows “1 + 1 = 9” we don’t have to prove this theorem to be correct, before proving it to be incorrect. I think it is called common sense.
I doubt that. If that were the case, we won't have wasted more than half a century on Bell's theorem. Since it is equivalent to "1 + 1 = 9". But I bet, it will be another 50 years before the buffoonery stops.

Your derivation is quite strange since you are claiming that to be able to use CFD in any theorem we are obligated to actually measure these values in experiments, which of course most understand is impossible
No. The use of CFD in the inequalities places that requirement on the experiment, not me. See the simulations in my previous post where the reason is clearly illustrated. Violation of this requirement is enough to obtain violation of the inequalities without any non-locality!
 
  • #528
Well, to me all the above is good example of flawed conspiracy spookiness. You mix performed experimental outcomes with mathematical counterfactual speculations. I have never seen anything like it.
You mean you have never read how Bell's theorem is derived. You have made my argument succintly. Bell's theorem is derived (at least the so-called experimental violations of the inequalities) by mixing performed experimental outcomes with mathematical counterfactual speculations. This is exactly my argument. I'm happy you now see it :smile:. Please say it one more time and let is sink in:

Bell's theorem and it's so-called experimental confirmation are obtained by mixing performed experimental outcomes with mathematical counterfactual speculations.

If you think I've used outcome inconsistently in those quotes, then it is you who has reading comprehension issues.
 
  • #529
billschnieder said:
BTW I hope you realize that for an EPR experiment in which we have coincidence counting the correct probability expression should be

P(\alpha \wedge \beta) = \int d\lambda P(\lambda) P(\alpha | \lambda) P(\beta | \alpha, \lambda)

No, absolutely not. Not according to a local realistic model. That's the whole point of Bell's argument, is that the probability of Bob getting a spin-up result cannot depend on Alice's device setting, which can be changed "in flight".
 
  • #530
DevilsAvocado said:
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.

I don't think it's a matter of "loopholes". It's a matter of what is the meaning of the nonlocal correlations in quantum mechanics. I agree that one way of looking at superdeterminism is pretty unappetizing: the quantum correlations come about through a "conspiracy". On the other hand, if superdeterminism arises in an "organic" way (I mentioned, either on this thread or another the possibility that the actual history of the world is forced on us by self-consistency), I think that would be cool...if there's actually some mathematics to play with, as opposed to philosophical speculation.
 
  • #531
billschnieder said:
Huh? You are pointing out but without agreeing with the way the correlations are calculated in experiments?

I didn't disagree with the way that correlations are calculated in experiments.

Do you or do you not agree that those correlations as calculated in the experiments are separable?

I didn't bring up the word "separable correlations". I brought up the notion of a "factorable probability distribution", and I gave a definition of that. Bell proves that the correlations predicted by quantum mechanics cannot arise from such a factorable probability distributions. Now, it is the correlations which are measured, not the probability distribution (I actually don't know why the calculations are done in terms of correlations, rather than probabilities), but the theoretical predictions for the correlations are computed from the probability distributions. Bell's inequality is derived from a certain assumed form of the probability distribution, and his proof shows that any probability distribution of that form cannot lead to a violation of that inequality. He didn't prove that the inequality cannot be violated--of course it can. He proved that it isn't violated in any theory that predicts a probability distribution of a certain form, the so-called "local realistic" theories.
 
  • #532
stevendaryl said:
No, absolutely not. Not according to a local realistic model. That's the whole point of Bell's argument, is that the probability of Bob getting a spin-up result cannot depend on Alice's device setting, which can be changed "in flight".
If you have studied any probability theory, you would not say that. You are confused between outcome functions F(a,L) and Probabilities. In the EPR experiment with coincidence counting, you only consider outcomes at Bob for which there was an outcome at Alice.

P(b|a,L) means probability of an outcome at Bob given that an outcome was measured at Alice for the given Lambda.
 
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  • #533
billschnieder said:
:smile: Now I know you are not being serious at all.
Please read this article http://arxiv.org/pdf/quant-ph/9810080v1.pdf and pay attention at figure 2 of the above article, and notice the table of outcomes?

I’m glad to see you in a good mood Bill, but maybe instead of rofl:ing and looking at fancy pictures, you should actually read the paper?? :biggrin:

[all bolding mine]

http://arxiv.org/pdf/quant-ph/9810080v1.pdf - page 3 said:
Quantum theory predicts a sinusoidal dependence for the coincidence rate C^{qm}_{++}(\alpha,\beta) \propto sin^{2}(\beta - \alpha) on the difference angle of the analyzer directions in Alice’s and Bob’s experiments. The same behavior can also be seen in the correlation function E^{qm}(\alpha,\beta) = - cos(2(\beta - \alpha)).


billschnieder said:
Huh? But it is impossible to repeat the same measurement! This is the whole point.

I did say “if we repeat the measurement at the same angle, that is of course not the same measurement. But I think these words is the key to all this confusion – you require for all counterfactual values to be realized in the real experiment [which you also know is impossible] – if not, you think you have proved something wrong, which of course is very wrong.

billschnieder said:
You probably mean if you repeat a similar measurement for many different photons, the results appear random.

Lot of very strange statements lately... I sure hope you’re not claiming that QM is now a deterministic theory??

billschnieder said:
I doubt that. If that were the case, we won't have wasted more than half a century on Bell's theorem. Since it is equivalent to "1 + 1 = 9". But I bet, it will be another 50 years before the buffoonery stops.

Really?? :eek::confused::bugeye: ...okay, a completely new definition of scientific refutability... well... let’s see, if you in this brand new epic light want to refute The Flat Earth Society... eh... you must (did I really get that right?? ) first prove that Earth is flat??



(and of course you also have to prove Bell’ theorem correct, before proving it wrong!)

:smile::smile::smile:

billschnieder said:
No. The use of CFD in the inequalities places that requirement on the experiment, not me. See the simulations in my previous post where the reason is clearly illustrated. Violation of this requirement is enough to obtain violation of the inequalities without any non-locality!

Okay, you have a working model of Local Realism that violates Bell’s inequalities?? I guess we’re going to help two guys getting to Stockholm, collecting their rightfully reward of two gold medals and $1.4 million.

Quantum Randi Challenge: Help Perimeter Physicist Joy Christian To Collect The Nobel Prize

billschnieder said:
Bell's theorem and it's so-called experimental confirmation are obtained by mixing performed experimental outcomes with mathematical counterfactual speculations.

Papers, reference, names please, to anyone but you making this hair-raising claim.

billschnieder said:
If you think I've used outcome inconsistently in those quotes, then it is you who has reading comprehension issues.

Now I know you are not being serious at all:

billschnieder said:
It still does not change the fact that we have outcomes at 4 angles. For each angle there is an outcome.

billschnieder said:
The only condition is there are 3 outcomes for 3 angles for each photon measured.

billschnieder said:
For each photon that leaves the source and heads towards Alice, you have a single outcome, (+, -, or non-detection).

billschnieder said:
Don't you see that those are not the outcomes measured in any real experiment.

billschnieder said:
As soon as you see that, you realize immediately that NO EXPERIMENT can ever measure them all! None! You can measure one but not the other two. Is that clear enough?

Totally unclear = contradictory = impossible to discuss

Sorry Bill, you need to state your claims clearly.
 
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  • #534
stevendaryl said:
I think that would be cool...if there's actually some mathematics to play with, as opposed to philosophical speculation.

Nooo Steven... you’re way too smart for superdeterminism... everything is a waste, including this discussion (well... some other stuff in this thread actually could be... :smile:):

Anton Zeilinger said:
We always implicitly assume the freedom of the experimentalist... This fundamental assumption is essential to doing science. If this were not true, then, I suggest, it would make no sense at all to ask nature questions in an experiment, since then nature could determine what our questions are, and that could guide our questions such that we arrive at a false picture of nature.
 
  • #535
billschnieder said:
If you have studied any probability theory, you would not say that. You are confused between outcome functions F(a,L) and Probabilities.

That distinction has nothing to do with anything I've said. It might have something to do with what you're talking about, but I have no idea--you're not being very clear.

Let me try to make a couple of claims that I believe are true, and you can say definitively whether you agree or disagree with those claims.

1. Bell proved that for all theories of a certain type, the correlations predicted by those theories obey a certain inequality.

2. The correlations predicted by quantum mechanics do not obey that inequality.

3. Therefore, the correlations predicted by quantum mechanics cannot be explained by such a theory.

4. Experimentally, the correlations confirm the predictions of quantum mechanics.

Do you agree with statements 1-4? If so, then what are we arguing about? If not, which one? There is one simplification that is made in the analysis, which is that the quantum mechanical prediction is most easily made in terms of unachievable perfect detections: That is, the assumption that for every pair produced, Alice detects one particle and Bob detects the other. That's an oversimplification, and your point is that this oversimplification makes Bell's result supect, then I don't have much to argue against you. I would have to spend more time thinking about what the implications of non-detection are for Bell's argument.
 
  • #536
DevilsAvocado said:
Nooo Steven... you’re way too smart for superdeterminism... everything is a waste, including this discussion (well... some other stuff in this thread actually could be... :smile:):

I view theoretical physics as ultimately entertainment. It has some practical consequences, but those don't really depend on any of the outstanding questions in theoretical physics: the meaning of quantum mechanics, quantum gravity, the origin of time asymmetry, the information paradox of black holes, etc. For all those questions, I just consider it to be a puzzle to be solved for our amusement. It's a matter of taste which solutions seem like cheats. But that isn't important; not everyone laughs at the same jokes, either.
 
  • #537
DevilsAvocado said:
We always implicitly assume the freedom of the experimentalist... This fundamental assumption is essential to doing science. If this were not true, then, I suggest, it would make no sense at all to ask nature questions in an experiment, since then nature could determine what our questions are, and that could guide our questions such that we arrive at a false picture of nature.

That's a philosophical issue that doesn't really worry me. If nature can answer the questions that I actually think of asking, then that's good enough for me. To worry about whether there are questions that I could have asked, but didn't is too meta for me.

I don't agree that superdeterminism makes science not worth doing. It's worth doing if we enjoy doing it.
 
  • #538
stevendaryl said:
...
4. Experimentally, the correlations confirm the predictions of quantum mechanics.

Do you agree with statements 1-4? If so, then what are we arguing about? If not, which one? There is one simplification that is made in the analysis, which is that the quantum mechanical prediction is most easily made in terms of unachievable perfect detections: That is, the assumption that for every pair produced, Alice detects one particle and Bob detects the other. That's an oversimplification, and your point is that this oversimplification makes Bell's result suspect, then I don't have much to argue against you.

Based on the exchange in #517, #518, #520, #521, I believe that BillSchneider is indeed rejecting #4, but for a different reason - there's another assumption built into the experiments, one that's not just a simplification but necessary for them to actually falsify the Bell inequality.

As an aside, it's not Bell's result that is suspect in any case. It's a theorem, and if the conclusion follows from the premises it's a valid theorem regardless of the truth of the premises: "If A then B" can be true even if A is false.
 
  • #539
Nugatory said:
Based on the exchange in #517, #518, #520, #521, I believe that BillSchneider is indeed rejecting #4, but for a different reason - there's another assumption built into the experiments, one that's not just a simplification but necessary for them to actually falsify the Bell inequality.

Well, don't leave me hanging---what's the assumption? I didn't really get it from Bill's posts.
 
  • #540
DevilsAvocado said:
I’m glad to see you in a good mood Bill, ...

DA, you do understand the difference between a QM prediction for an experiment and the outcomes actually measured in an experiment don't you, because you seem utterly confused by the difference. Pay attention:

1) QM makes a prediction for the expectation value for measuring a large number of photon pairs at angles α and β for Alice and Bob respectively. It is this expectation value that is -cos 2(α-β). QM says absolutely nothing, and makes absolutely no prediction about the outcome of measuring a single photon pair.

2) In experiments, each photon has an outcome \ni (+1, -1). Hundreds of thousands of photons are measured and long tables are recorded with one entry for each photon measured. Each entry is time-tagged, the time tags are compared, and only coincident entries are considered togeter to calculate <FαF'β> which is then found to match the QM correlation. The table of outcomes at Bob's end is a separate list from the table of outcomes at Alice's end, Bob does not know or case about any "relative angle" nor does Alice. Get it?

DA said:
I did say “if we repeat the measurement at the same angle, that is of course not the same measurement. But I think these words is the key to all this confusion – you require for all counterfactual values to be realized in the real experiment [which you also know is impossible]
It is not me who derived the inequalities, it is Bell. It is not me who requires them to be measured in the same experiment. It is sound logic and the use of CFD in the inequalities which require them to be measured in the same experiment if anyone claims they are trying to test the inequalities. I said from the beginning that a genuine Bell test experiment is impossible. It is up to anyone who disagrees with that to make sure they are measured in the same experiment, otherwise they not testing Bell's inequalities or the CHSH no matter what they claim. This is common sense.
 
  • #541
stevendaryl said:
That distinction has nothing to do with anything I've said.
On the contrary, it is at the root of your misunderstanding. When I suggested
P(\alpha \wedge \beta) = \int d\lambda P(\lambda) P(\alpha | \lambda) P(\beta | \alpha, \lambda)

You said
No, absolutely not. Not according to a local realistic model. That's the whole point of Bell's argument, is that the probability of Bob getting a spin-up result cannot depend on Alice's device setting
Apparently unaware that if you are right that the Probability at Bob does not depend on the setting at Alice (not that you are), then

P(\beta | \lambda) = P(\beta | \alpha, \lambda)
The only time when those two are not equal is when Bob's probability is dependent of Alice's setting. In other words, the equation I gave is ALWAYS CORRECT, but yours in ONLY CORRECT WHEN THERE IS INDEPENDENCE.

But you are thinking that the outcome at Bob does not depend on Alice's setting. In other words, the outcome at Bob is a function of β and λ only ie F'(β,λ). And then you get confused by assuming that this means the probability of Bob's result is independent of the setting at Alice's detector. As I have explained, just because Bob's outcome does not depend on Alice's setting does not mean the probability calculated for Bob's outcome does not depend on Alice's setting. In fact, it must depend on Alice setting if you rely on any kind of coincidence counting.
 
  • #542
stevendaryl said:
Well, don't leave me hanging---what's the assumption? I didn't really get it from Bill's posts.

coming... :smile:
 
  • #543
billschnieder said:
... I said from the beginning that a genuine Bell test experiment is impossible. It is up to anyone who disagrees with that to make sure they are measured in the same experiment, otherwise they not testing Bell's inequalities or the CHSH no matter what they claim. This is common sense.

Per usual, your ongoing misunderstanding of Bell goes completely against the grain of nearly everything written about the subject. Bell tests are merely experiments to show that the QM prediction is correct for selected angles (or various other observables), nothing more. Hundreds have been performed and published in peer-reviewed journals.

Bell's Theorem, on the other hand, shows that QM's predictions for (the same) selected angles are incompatible with the assumptions of local realism. Therefore Bell tests support QM, which by Bell's Theorem proves one of the local realistic assumptions wrong.
 
  • #544
billschnieder said:
On the contrary, it is at the root of your misunderstanding.

No, I think you are confused.
 
  • #545
Bill:

Please publish your pet personal ideas rather than bring them here. We've been through this time and time again. PhysicsForums is for generally accepted science. If you have a suitable published reference for your statement about Bell tests not being valid, provide it. Else I will report your post.

-DrC
 
  • #546
stevendaryl said:
Let me try to make a couple of claims that I believe are true, and you can say definitively whether you agree or disagree with those claims.

1. Bell proved that for all theories of a certain type, the correlations predicted by those theories obey a certain inequality.
I agree that Bell derived certain inequalities. But I do not necessarily agree that the key assumptions required to obtain the inequalities are the ones you think they are. However, for the purpose of the discussions here, I do not care about the derivation, the inequalities are valid and we can start from there as I've told you previously, although I'll be happy to discuss in another thread why those inequalities are more general than you think.
2. The correlations predicted by quantum mechanics do not obey that inequality.
What correlations? This is one of the issues. Please spell out how you have arrived at this conclusion. Write down the inequality and write down the correlations which violate the inequality, term by term.
3. Therefore, the correlations predicted by quantum mechanics cannot be explained by such a theory.
We do not even reach this point yet. We have to address #2.

4. Experimentally, the correlations confirm the predictions of quantum mechanics.
Yes, the correlations from the experiments match QM.

There is one simplification that is made in the analysis, which is that the quantum mechanical prediction is most easily made in terms of unachievable perfect detections
My argument does not depend or rely on any loopholes.
 
  • #547
billschnieder said:
It depend on the setting at Alice (not that you are), then

P(\beta | \lambda) = P(\beta | \alpha, \lambda)
The only time when those two are not equal is when Bob's probability is dependent of Alice's setting. In other words, the equation I gave is ALWAYS CORRECT, but yours in ONLY CORRECT WHEN THERE IS INDEPENDENCE.

I KNOW that. That's what you get from assuming local realism. Bell's theorem is about locally realistic theories. The assumption that Bell was making is that the correlation between Alice's result and Bob's result is due entirely to the shared hidden variable \lambda. That is, the correlation goes away completely once you fix \lambda.

So what Bell proved was that

IF the joint probability P(\alpha \wedge \beta) for Alice detecting spin-up at angle \alpha and Bob detecting spin-up at angle \beta has the form

P(\alpha \wedge \beta) = \int d\lambda P_L(\lambda) P_A(\alpha | \lambda) P_B(\beta | \lambda)

THEN the correlation between Alice's result and Bob's result will obey a certain inequality. It's certainly possible that the joint probability distribution doesn't have that form. As you point out, the general case is:

P(\alpha \wedge \beta) = \int d\lambda P_L(\lambda) P_A(\alpha | \lambda) P_B(\beta | \lambda \wedge \alpha)

Bell isn't talking about the general case. He's talking about the case in which the correlation between Alice's result and Bob's result is completely due to the presence of the hidden variable \lambda
 
  • #548
Nugatory said:
Based on the exchange in #517, #518, #520, #521, I believe that BillSchneider is indeed rejecting #4
This is wrong. The experiments match QM. I do not reject #4.
 
  • #549
stevendaryl said:
No, I think you are confused.
I think you are confused.
stevendaryl said:
Bell isn't talking about the general case. He's talking about the case in which the correlation between Alice's result and Bob's result is completely due to the presence of the hidden variable \lambda

Bell isn't talking about joint probability distributions. But using Expectation values for the paired product of outcomes at two stations, where each outcome is a deterministic function of the setting at the station and the lambda in play with possible values +1/-1. I told you to forget about probabilities but you decided to keep getting confused by them.
 
  • #550
billschnieder said:
nugatory said:
Based on the exchange in #517, #518, #520, #521, I believe that BillSchneider is indeed rejecting #4
This is wrong. The experiments match QM. I do not reject #4.

Fair enough... I am interpreting #4 as "the experiments match QM and therefore falsify Bell's inequality" because I'm pretty sure that's what stevendaryl meant, but yes, that's a slightly different statement than his #4.

If I could ask you how you interpret the outcome of those four posts (#517, #518, #520, #521), suggest a concise statement that does capture the area of disagreement to your satisfaction?
 

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