Boole vs. Bell - the latest paper of De Raedt et al

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In summary: COMPLETE description of the paper's content would be completely wrong and would not be useful to summarise it. In summary, the De Raedt paper discusses the apparent contradictions of quantum theory and probability frameworks, and argues that these contradictions arise from incomplete considerations of the premises of the derivation of the inequalities. They present extended Boole-Bell inequalities which are binding for both classical and quantum models, and show that apparent violations of these inequalities can be explained in an Einstein local way.
  • #106
In my post #29 I wrote:

"PS, there is another intriguing remark, not sure if it is on-topic:
in contrast to the EPR-Bohm state, one can really (as EPR claimed) associate with the original EPR state a single probability measure describing incompatible quantum observables (position and momentum).
Can someone here explain what Khrennikov meant?"

Now, I think that Dr.C's remarks in post #89 are helpful:
"use just 1 datapoint [..] Obviously, if it works for the single case"

For, although that remark is wrong for such observables as polarisation, it is correct for such observables as position and momentum.

Thus Khrennikov may have meant that such a single probability distribution is valid for the data set [position, momentum] of a single entangled electron pair. This also relates to the fact that the measurements of position and momentum are mutually exclusive.

Obviously the averaging issue will not appear for a Boole-Bell like inequality for position and momentum.
 
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  • #107
harrylin said:
Now this is an important point; and Bill, I must admit that I did not expect such an outcome when you started this simple example![..]
if a high amount of random input data indeed breaks Bell's inequality for this type of test, then you managed to provide just the kind of illustration that De Raedt's paper lacks and for which I made a wishful request in post #46.

When I find the time I'll also try your example for more random inputs. :smile:

Well after all, it took only half an hour to try this on a spreadsheet. [CORRECTION: THIS WAS WRONG. See next!]
My result: with a randomly generated data set [+1 -1 +1] etc. I obtained (10 times out of 10 for 30 data points) that the Bell theorem holds for random input, as expected.

But after writing this it strikes me that I did did not exactly reproduce Bill's example. What I tested was perhaps closer to Bell's example which I already verified in the past. :rolleyes:

Bill, I thought that you gave a nearly random data set, but obviously your dataset is very non-random. I replaced random sampling by random input, but that's not the same thing and completely wrong if the input is not random... Interesting!

Is it easy to randomly sample a fixed data set in Excel or OpenOffice? I mean not by hand, but with an operator?

Thanks,
Harald
 
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  • #108
harrylin said:
Well after all, it took only half an hour to try this on a spreadsheet. [CORRECTION: THIS WAS WRONG. See next!]
My result: with a randomly generated data set [+1 -1 +1] etc. I obtained (10 times out of 10 for 30 data points) that the Bell theorem holds for random input, as expected.

But after writing this it strikes me that I did did not exactly reproduce Bill's example. What I tested was perhaps closer to Bell's example which I already verified in the past. :rolleyes:

Bill, I thought that you gave a nearly random data set, but obviously your dataset is very non-random. I replaced random sampling by random input, but that's not the same thing and completely wrong if the input is not random... Interesting!

Is it easy to randomly sample a fixed data set in Excel or OpenOffice? I mean not by hand, but with an operator?

Thanks,
Harald

OK I fixed that and took Bill's set of 30 datapoints. And added random sampling.
I calculated with both the original equation of Bell and with the equation of Dr.C. :-p

The result was the same as before: sorry Bill, I got 10 times no violation of the Bell Inequality.
Bill, if you like I can send you the spreadsheet.

Thus I am still interested if anyone can come up with an example like the one of De Raedt with doctors and patients, but that not looks like a conspiracy. :rolleyes:

Harald
 
  • #109
harrylin said:
OK I fixed that and took Bill's set of 30 datapoints. And added random sampling.
I calculated with both the original equation of Bell and with the equation of Dr.C. :-p

The result was the same as before: sorry Bill, I got 10 times no violation of the Bell Inequality.
Bill, if you like I can send you the spreadsheet.

Thus I am still interested if anyone can come up with an example like the one of De Raedt with doctors and patients, but that not looks like a conspiracy. :rolleyes:

Harald

Good work! Yes, I have run simulations in the past and it is easy once you see the method in detail to see why you will not get a violation with any reasonable samples. Vice versa, any test of any physical law could be "tricked" up to give a different result than expected. There is nothing special about Bell in that regard.

And the thing to remember about the de Raedt program: A simulation that satisfies Bell (no easy feat, I assure you) will NOT give expectation values for other tests consistent with QM. For example, it will not match Malus. Now, it is important to make a distinction between the Malus cos^2 and the Bell test cos^2. They look to be the same thing but they are not really. The Bell cos^2 coincidentally applies to 2 particle states. But, for example, it does not apply to entangled states of more than 2 particles. The underlying principle is in fact the Malus rule for a single stream. And any simulation that satisfies Bell will not be able to have the Malus rule in place. Obviously, that is a big problem for such a model because Malus was discovered over 200 years ago and is bedrock.
 
  • #110
harrylin said:
Yes indeed - for any onlookers: Dr.C misquoted me, making it appear nearly the contrary of what I wrote. :biggrin:
OK then I'll do the same here:

It's really humorous that you also believe in flying dogs. :wink:

Sorry for any messing up of what you intended to say. I DO believe in flying dogs, however. because if I open my front door, I will see mine flying out similar to the picture.

:smile:
 
  • #111
Well, I looked at the paper that started the thread. I did not finish it (way too many words) but I can see a number of serious problems with it.

For a start, section III D "Relation to Bell's work" has no relation to Bell's work whatsoever :confused: None of the original Bell's assumptions are reflected, in particular, the crucial assumption of independence of results A from settings B and vice versa is nowhere to be found. Neither is the perfect anti-correllation for the same settings of A and B (which is used quite a lot in Bell's derivation). And then the authors confuse individual outcomes of measurement with expectation values and arrive at completely wrong conclusion about triplets of data sharing the same lambda, while there are no triplets of data at all in Bell's original work, only probabilities and expectation values. And it goes downhill from there.

DK
 
  • #112
harrylin said:
In my post #29 I wrote:

"PS, there is another intriguing remark, not sure if it is on-topic:

Can someone here explain what Khrennikov meant?"

Now, I think that Dr.C's remarks in post #89 are helpful:
"use just 1 datapoint [..] Obviously, if it works for the single case"

For, although that remark is wrong for such observables as polarisation, it is correct for such observables as position and momentum.

Thus Khrennikov may have meant that such a single probability distribution is valid for the data set [position, momentum] of a single entangled electron pair. This also relates to the fact that the measurements of position and momentum are mutually exclusive.

Obviously the averaging issue will not appear for a Boole-Bell like inequality for position and momentum.

Khrennikov's remark means simply that you can not have a single probability distribution which includes mutually exclusive parts.

if A and B are mutually exclusive, it means P(AB) = 0, everywhere, so there is no probability distribution. When applied to Bell measuring two particles at three angles, P(a,b,c) can not exist since one of those angles will be mutually exclusive to the other two (since only two angles can be measured for 2 particles), Therefore the probability distribution P(a,b,c) can not exist.
 
  • #113
harrylin said:
Well after all, it took only half an hour to try this on a spreadsheet. [CORRECTION: THIS WAS WRONG. See next!]
My result: with a randomly generated data set [+1 -1 +1] etc. I obtained (10 times out of 10 for 30 data points) that the Bell theorem holds for random input, as expected.

But after writing this it strikes me that I did did not exactly reproduce Bill's example. What I tested was perhaps closer to Bell's example which I already verified in the past. :rolleyes:

Bill, I thought that you gave a nearly random data set, but obviously your dataset is very non-random. I replaced random sampling by random input, but that's not the same thing and completely wrong if the input is not random... Interesting!
None of my datasets is random although they may appear to be.

Is it easy to randomly sample a fixed data set in Excel or OpenOffice? I mean not by hand, but with an operator?

Thanks,
Harald
I usually just write code to do that.
 
  • #114
harrylin said:
OK I fixed that and took Bill's set of 30 datapoints. And added random sampling.
I calculated with both the original equation of Bell and with the equation of Dr.C. :-p

The result was the same as before: sorry Bill, I got 10 times no violation of the Bell Inequality.
Bill, if you like I can send you the spreadsheet.
You will have to describe what you mean by random sampling, because you may have misunderstood something. Which of the treatments in my post #92 were you unable to reproduce?

Please send the spreadsheet, or better, attach it.

Thus I am still interested if anyone can come up with an example like the one of De Raedt with doctors and patients, but that not looks like a conspiracy. :rolleyes:
Harald
I'm surprised you think that my example looks like a conspiracy. Remember that DrC's challenge was supposed to show that even with conspiracy, it is impossible to violate Bell's inequality. I don't think DrC was asking for a random dataset -- it kinda defeats the purpose; why would he ask for one if anybody could generate one randomly? And if it is assumed that a physical process is producing the data, why can't the physical process have behavioural patterns that are non-random?
 
  • #115
The result was the same as before: sorry Bill, I got 10 times no violation of the Bell Inequality.
Bill, if you like I can send you the spreadsheet.

I did the calculations for the random sampling case I discussed as "Treatment (3)" in post #92.

For the following dataset,

a, b, c
-----------
+1, -1, +1
-1, +1, -1
-1, +1, -1
-1, +1, -1
+1, -1, -1
-1, +1, +1
-1, +1, -1
+1, -1, +1
-1, -1, -1
-1, -1, -1
-1, -1, +1
-1, +1, +1
+1, -1, -1
+1, -1, -1
-1, -1, +1
-1, -1, -1
+1, -1, +1
-1, +1, -1
-1, +1, -1
+1, +1, +1
+1, -1, -1
-1, -1, +1
+1, -1, +1
+1, -1, +1
+1, +1, +1
-1, -1, -1
+1, +1, +1
-1, -1, -1
-1, +1, -1
+1, -1, +1

Here is the method I use. REMEMBER - you must sample without replacement. This is equivalent to randomly shuffling the dataset and then picking the first 10 of the resulting dataset for calculating ab, the next 10 for bc and the last 10 for ac, in a manner similar to treatment (2). Again to be clear, the procedure is as follows

- Randomize the sequence by shuffling it
- select the first 10 of the resulting randomized sequence, and use for the ab term, the next 10 for the bc term and the last 10 for the ac term.

This way, we are sure that every row is used NOT MORE THAN ONCE.

I did the above 10 times, randomizing everytime from the previous random sequence and got the following results:
ab=-0.4000, bc=0.0000, ac=0.4000, Violated=False
ab=-0.4000, bc=0.0000, ac=0.2000, Violated=False
ab=0.0000, bc=-0.2000, ac=0.2000, Violated=False
ab=-0.6000, bc=-0.4000, ac=0.4000, Violated=True
ab=-0.2000, bc=-0.2000, ac=0.6000, Violated=False
ab=-0.2000, bc=-0.8000, ac=0.2000, Violated=True
ab=0.0000, bc=-0.2000, ac=0.4000, Violated=False
ab=0.0000, bc=-0.4000, ac=0.2000, Violated=False
ab=-0.6000, bc=-0.4000, ac=0.0000, Violated=False
ab=-0.4000, bc=0.2000, ac=0.8000, Violated=True

Repeating this 1000, consistently gives me a violation in about 20% of the randomly sampled (without replacement) pairs.

NOTE, 1 violation is enough.
 
  • #116
billschnieder said:
Repeating this 1000, consistently gives me a violation in about 20% of the randomly sampled (without replacement) pairs.

NOTE, 1 violation is enough.
That is incorrect. You are confusing expectation values with averages. Bell's inequality is stated in terms of expectation values and you are getting averages. Expectation value of 100 fair coin tosses is 50 but the average of a particular run can be anywhere between 0 and 100. It does not mean anything.

If you want to do the job properly, then please calculate not only the mean values but also standard deviation. You will then find that Bell's inequality actually holds quite well and the occasional deviations are well within the error bars.

If you can show how to beat Bell's inequalities consistently by at least a few standard deviations (which is the case with QM), then you have the case, otherwise you would have to try harder.
 
  • #117
Delta Kilo said:
That is incorrect. You are confusing expectation values with averages. Bell's inequality is stated in terms of expectation values and you are getting averages. Expectation value of 100 fair coin tosses is 50 but the average of a particular run can be anywhere between 0 and 100. It does not mean anything.
You have no clue what you are talking about. I doubt you have made an effort to understand what this thread is about. (see http://arxiv.org/pdf/quant-ph/0211031
Foundations of Physics Letters
Volume 15, Number 5, 473-486, DOI: 10.1023/A:1023920230595
)

Bell's inequality is equivalent to saying "The sum of any 3 sides of a die will never exceed 15".
But if you measure one side from three different dice, you will get violations some of the time, even though NO single die violates the rule. You therefore can not conclude from such an experiment (3 different dice), that a single die does not have well defined values for the sides. Violation by a single case, shows that there is something wrong between correspondence of the experiment and the rule. This is what De Raedt showed.

This is the crux of the issue, which you haven't understood.

If you want to do the job properly, then please calculate not only the mean values but also standard deviation. You will then find that Bell's inequality actually holds quite well and the occasional deviations are well within the error bars.

If you can show how to beat Bell's inequalities consistently by at least a few standard deviations (which is the case with QM), then you have the case, otherwise you would have to try harder.
Bell's inequality can never be violated even for a single data point, if the mathematical operation is valid. Violation of the inequality by a single data point tells you a mathematical error has been made.
 
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  • #118
billschnieder said:

Oh, this paper is much better written than De Raedt and the cheating step is much more subtle. But it is similar.

billschnieder said:
Bell's inequality is equivalent to saying "The sum of any 3 sides of a die will never exceed 15".
No, this is not what Bell says at all. Your example is pure arithmetic and Bell's theorem is statistical. There is a big difference.

Bell basically has a triangle inequality for the expectation values of lengths. He specifically introduces [itex]P(a,b)[/itex] as expectation value in (2)*

This is different from ordinary tirangle inequality for a single triangle. Ordinary triangle inequality is valid only when all three sides belong to the same triangle (obviously). By summing over a sequence of triangles one gets similar inequality for average triangle. But there is a constraint: the same set of triangles must used when calculating averages for each side.

The constraint is removed when one transitiones from sums and averages which vary from one experiment to the next, to expectation values which are a property of the stochastic process as a whole. Doing this requires certain assumptions about the process (eg. stationary). In Bell's paper these assumptions are encoded in the probability density [itex]\rho(\lambda)[/itex] being function of [itex]\lambda[/itex] and nothing else and [itex]A(a,\lambda)[/itex] and [itex]B(b,\lambda)[/itex] being fully deterministic. These extra assumptions about the stochastic process behind the data is what allows one to estimate expectations of each side of the triangle independently.

billschnieder said:
But if you measure one side from three different dice, you will get violations some of the time, even though NO single die violates the rule. You therefore can not conclude from such an experiment (3 different dice), that a single die does not have well defined values for the sides. Violation by a single case, shows that there is something wrong between correspondence of the experiment and the rule. This is what De Raedt showed.

Both De Raedt and Sica follow similar lines of logic:
First they note that Bell's inequality is probabilistic and they also notice similar inequality which is always true for single data point.
So they have this great idea how improve on Bell and to get rid of all uncertainities once and for all. So they start by extending formula for single data point to a sequence. While doing that they discover that they no longer need [itex]\rho(\lambda)[/itex] and they happily get rid of it, thereby throwing baby out with the bathwater.

But then when it comes to the experiment, the absence of [itex]\rho(\lambda)[/itex] assumption comes back and bites them in the nose: in Bell's paper factorization between (14) and (15) comes naturally from the math thanks to the shape of the formula (14).

De Raedt derives his "EBBI" (which he claims are equivalent to but better then Bell's) then discovers they cannot be easily factrorized because of non-commuting measurements, creates a big fuss out of it and claims it all to be Bell's fault.

Sica also claims to derive Bell's inequality in section 1.2. He says:
If the numerical correlation estimates in (11) approach ensemble average limits, as [itex]N\rightarrow\infty[/itex], then replacing the estimates in (11) with these limits results in the usual form of Bell's inequality.
However he does not actually do this step and so he does not spell the assumptions that have to be made. Later also gets in trouble with factorization and non-commuting masurements. Since he cannot get 3 datapoints from one pair of measurements, he resorts to a a bit of cheating: he re-arranges the data so that measirements of [itex]<A_{i}B_{i}>[/itex] and [itex]<A'_{i}B_{i}>[/itex] with the same [itex]i[/itex] have the same value for [itex]B_{i}[/itex] and then calculates [itex]<A_{i}A'_{i}>[/itex] by going through [itex]<A_{i}B_{i}>[/itex] and [itex]<A'_{i}B_{i}>[/itex]. Somehow he is not worried at all that (12) and (16) have different shape even though they must to be the same thing from symmetry point of view. Then of course he gets wrong result for (21), again different from (22) and (23) from which he claims that QM satisfies Bell's inequality.

SUMMARY: How to disprove Bell's inequality:
  1. Take Bell's inequality
  2. "Improve it" by throwing vital bits out
  3. Run into trouble
  4. Blame Bell
  5. ...?
  6. Publish!
billschnieder said:
Bell's inequality can never be violated even for a single data point, if the mathematical operation is valid. Violation of the inequality by a single data point tells you a mathematical error has been made.

If you refer to formulas with sums rather than expectations in them (those that are supposed to be true in arithmetic rather than statistical sense), please do not call them "Bell's inequalities". Call them "De Raedt" or "Sica" or "Bill's inequalities" if you wish.

Again, I stress, Bell's inequality is for expectations and expectations are not averages. One cannot simply plug experimental averages into the formula for expectations and expect it to work 100%. At the very least one has to compute std deviation and define error bars.

*) J.S.Bell. "On Enstein Podolsky Rosen paradox" 1964

DK
 
  • #119
Delta Kilo said:
Oh, this paper is much better written than De Raedt and the cheating step is much more subtle. But it is similar.


No, this is not what Bell says at all. Your example is pure arithmetic and Bell's theorem is statistical. There is a big difference.
I disagree, Bell's inequality is a pure arithmetic identity applied to statistics. You can not derive the inequality if you start from statistics. If you like, we can go through the derivation step by step, to convince you that the derivation is pure arithmetic.


This is different from ordinary tirangle inequality for a single triangle. Ordinary triangle inequality is valid only when all three sides belong to the same triangle (obviously). By summing over a sequence of triangles one gets similar inequality for average triangle. But there is a constraint: the same set of triangles must used when calculating averages for each side.

Interesting that you mention this. See post #152 in the related thread (https://www.physicsforums.com/showpost.php?p=3308861&postcount=152) where I discussed this. as follows

I suppose you know about the triangle inequality which says for any triangle with sides labeled x, y, z where x, y, z represents the lengths of the sides

z <= x + y

Note that this inequality applies to a single triangle. What if you could only measure one side at a time. Assume that for each measurement you set the label of the side your instrument should measure and it measured the length destroying the triangle in the process. So you performed a large number of measurements on different triangles. Measuring <z> for the first run, <x> for the next and <y> for the next.

Do you believe the inequality
<z> <= <x> + <y>

Is valid? In other words, you believe it is legitimate to use those averages in your inequality to verify its validity?


Funny thing, your statement in bold says, exactly the same set of triangles must be used to calculate the terms. Isn't it disingenuous then for you to suggest something different in Bell's case?

In other words -- Do you agree that for Bell's inequality to be guaranteed to be obeyed, the same set of photons must be used to calculate all three expectation values used in the inequalities ? Please I need a Yes/No answer here.

If you agree, then you have conceded Sica's point, and De Raedt's point and my point.
If you disagree, see the next point.

The constraint is removed when one transitiones from sums and averages which vary from one experiment to the next, to expectation values which are a property of the stochastic process as a whole. Doing this requires certain assumptions about the process (eg. stationary).

It is your claim therefore that you do not need to use the same set of particles, because the process generating the particles is stationary?
I need a Yes/No answer here.

If you agree, then I suppose you have proof that that it is stationary. If I were to provide evidence that process producing the photons is not stationary, will you concede therefore that expectation values from such a process is not compatible with Bell's inequality?? I need a Yes/No answer here.

In Bell's paper these assumptions are encoded in the probability density [itex]\rho(\lambda)[/itex] being function of [itex]\lambda[/itex] and nothing else and [itex]A(a,\lambda)[/itex] and [itex]B(b,\lambda)[/itex] being fully deterministic. These extra assumptions about the stochastic process behind the data is what allows one to estimate expectations of each side of the triangle independently.

I disagree, it is the factorization mentioned in equation (5) of Sica's paper above is the crucial step which introduces the assumption of stationary. That step is equivalent to going from the universally valid arithmetic inequality:

<z> <= <x + y>

To the statistical inequality

<x> <= <x> + <y>

Which is only obeyed when the process generating the triangles is stationary

Both De Raedt and Sica follow similar lines of logic:
First they note that Bell's inequality is probabilistic and they also notice similar inequality which is always true for single data point.
So they have this great idea how improve on Bell and to get rid of all uncertainities once and for all. So they start by extending formula for single data point to a sequence. While doing that they discover that they no longer need [itex]\rho(\lambda)[/itex] and they happily get rid of it, thereby throwing baby out with the bathwater.
This is a mischaracterization of their work, which I don't think you have understood yet. They are not trying to improve Bell. They are explaining why datasets from experiments/QM are not compatible with Bell's inequality.

But then when it comes to the experiment, the absence of [itex]\rho(\lambda)[/itex] assumption comes back and bites them in the nose: in Bell's paper factorization between (14) and (15) comes naturally from the math thanks to the shape of the formula (14).
I think you are just handwaving here. Experimenters do not know or care about lambda.

De Raedt derives his "EBBI" (which he claims are equivalent to but better then Bell's) then discovers they cannot be easily factrorized because of non-commuting measurements, creates a big fuss out of it and claims it all to be Bell's fault.
Non-commuting measurements are not compatible with stationarity as Sica explains. Therefore you can not use expectation values from QM/Experiments as valid terms for Bell's inequality. That is the point, you have admitted by stating that stationarity is a pre-requisite. If Bell introduces stationarity as a requirement without cause, that is is problem. If Bell fails to realize that the stationarity requirement is incompatible with QM to start with, then that is his fault.

Sica also claims to derive Bell's inequality in section 1.2. He says: However he does not actually do this step and so he does not spell the assumptions that have to be made. Later also gets in trouble with factorization and non-commuting masurements. Since he cannot get 3 datapoints from one pair of measurements, he resorts to a a bit of cheating: he re-arranges the data so that measirements of [itex]<A_{i}B_{i}>[/itex] and [itex]<A'_{i}B_{i}>[/itex] with the same [itex]i[/itex] have the same value for [itex]B_{i}[/itex] and then calculates [itex]<A_{i}A'_{i}>[/itex] by going through [itex]<A_{i}B_{i}>[/itex] and [itex]<A'_{i}B_{i}>[/itex]. Somehow he is not worried at all that (12) and (16) have different shape even though they must to be the same thing from symmetry point of view. Then of course he gets wrong result for (21), again different from (22) and (23) from which he claims that QM satisfies Bell's inequality.

You do not understand it at all. In Bell test experiments, 3 sub "experiments" are done in which the following measurements are made:

"a1 b1"
"b2 c2"
"a3 c3"

Sica essentially says if the process producing the data is stationary, it should be possible to sort the datasets such that the number and pattern of switching between +1 and -1 in b1 and b2 are identical in the first two runs, and after doing that, you could *factor* out the a1 list from run 1 and the c2 list from run 2, recombine them and create a counterfactual "a1 c2" run. Therefore you do not need to measure run 3 at all. You will have:

"a1 b1"
"b2 c2"
"a1 c2"

Which will never violate the inequality. This is exactly the type of factorization which Bell did in equations (15) to (16).

If however, it is not possible to sort the data from experiment runs 1 and 2 as outlined above, your stationarity assumption fails and the inequality is not applicable to the data. Do you agree?


SUMMARY: How to disprove Bell's inequality:
  1. Take Bell's inequality
  2. "Improve it" by throwing vital bits out
  3. Run into trouble
  4. Blame Bell
  5. ...?
  6. Publish!
You are not being serious.
 
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  • #120
I also draw your attention to the follow-up paper by L. Sica which addresses the stationarity issue further.


Bell's inequality violation due to misidentification of spatially non stationary random processes
Journal of Modern Optics, 2003, Vol. 50, No. 15-17, 2465-2474
http://arxiv.org/abs/quant-ph/0305071v1

Correlations for the Bell gedankenexperiment are constructed using probabilities given by quantum mechanics, and nonlocal information. They satisfy Bell's inequality and exhibit spatial non stationarity in angle. Correlations for three successive local spin measurements on one particle are computed as well. These correlations also exhibit non stationarity, and satisfy the Bell inequality. In both cases, the mistaken assumption that the underlying process is wide-sense-stationary in angle results in violation of Bell's inequality. These results directly challenge the wide-spread belief that violation of Bell's inequality is a decisive test for nonlocality.
 
  • #121
billschnieder said:
z <= x + y

Note that this inequality applies to a single triangle. What if you could only measure one side at a time. Assume that for each measurement you set the label of the side your instrument should measure and it measured the length destroying the triangle in the process. So you performed a large number of measurements on different triangles. Measuring <z> for the first run, <x> for the next and <y> for the next.

Do you believe the inequality
<z> <= <x> + <y>

Is valid? In other words, you believe it is legitimate to use those averages in your inequality to verify its validity?

It depends. Do the values <x>, <y> and <z> represent experimental averages or expectation values? It it is the former than the answer is NO. If it is the latter then the answer is "it depends on the properties of stochastic process generating the triangles", with a further twist that one cannot not obtain exact expectation value from a finite run, no matter how long. One can only get estimation of true expectation value which will contain some errors that would have to be accounted for.

billschnieder said:
Funny thing, your statement in bold says, exactly the same set of triangles must be used to calculate the terms. Isn't it disingenuous then for you to suggest something different in Bell's case?
It is a different case from Bell's. Bell works with expectations rather than sums and has extra assumptions about the process.

billschnieder said:
In other words -- Do you agree that for Bell's inequality to be guaranteed to be obeyed, the same set of photons must be used to calculate all three expectation values used in the inequalities ? Please I need a Yes/No answer here.
The answer to that is NO.

billschnieder said:
It is your claim therefore that you do not need to use the same set of particles, because the process generating the particles is stationary?
I need a Yes/No answer here.
The answer is YES. Being stationary basically tells you there ia s well defined expectation value for P(a,b) which does not change with time and so one can estimate it using different sets of data ans still get the same result.

billschnieder said:
If you agree, then I suppose you have proof that that it is stationary.
I quote myself:
In Bell's paper these assumptions are encoded in the probability density ρ(λ) being function of λ and nothing else and A(a,λ) and B(b,λ) being fully deterministic. These extra assumptions about the stochastic process behind the data is what allows one to estimate expectations of each side of the triangle independently.

If I were to provide evidence that process producing the photons is not stationary, will you concede therefore that expectation values from such a process is not compatible with Bell's inequality?? I need a Yes/No answer here.
The answer is YES. Such a process would violate the assumptions of Bell's theorem therefore the result would not be applicable. However I do not believe you can provide such evidence.

I disagree, it is the factorization mentioned in equation (5) of Sica's paper above is the crucial step which introduces the assumption of stationary. That step is equivalent to going from the universally valid arithmetic inequality:

<z> <= <x + y>

To the statistical inequality

<x> <= <x> + <y>

Which is only obeyed when the process generating the triangles is stationary
Look up the definition of stationary process and see for yourself its connection to ρ(λ). In case of Bell, this property is explicitly encoded in ρ(λ), and that is what makes factorization possible. In your case the assumption of stationary process is not encoded anywhere in the math. When the authors reach factorization step they discover that something is amiss here and go into handwawing mode with disastorous results.

They are not trying to improve Bell. They are explaining why datasets from experiments/QM are not compatible with Bell's inequality.
How come they use their own derivations instead of Bell's for that. Why don't they identify an error in the original derivation for a change?

Experimenters do not know or care about lambda.
Not the λ itself, no, but that's what not I said. To compute estinations for expectation values they compute means and std. deviations and that just doesn't quite work if probability densiity is not stationary. And that's what ρ(λ) assumption is about.

Non-commuting measurements are not compatible with stationarity as Sica explains. Therefore you can not use expectation values from QM/Experiments as valid terms for Bell's inequality.
Where does he says that? These are two different things. The correlation P(a,b) is stationary but the individual measurements [itex]A_{i}, B_{i}[/itex] are non-commuting.

If Bell introduces stationarity as a requirement without cause, that is is problem. If Bell fails to realize that the stationarity requirement is incompatible with QM to start with, then that is his fault.
Curiosier and curiosier. Can you quote the exact words (or better yet, formulas) to that effect?
Sica essentially says if the process producing the data is stationary, it should be possible to sort the datasets such that the number and pattern of switching between +1 and -1 in b1 and b2 are identical in the first two runs, and after doing that, you could *factor* out the a1 list from run 1 and the c2 list from run 2, recombine them and create a counterfactual "a1 c2" run. Therefore you do not need to measure run 3 at all.
This is just data massaging gone wrong. I see you didn't comment on my statement that [STRIKE]there is another elephant in this room[/STRIKE] equation (16) and therefore (21) are so obviously wrong, they violate both theory and practice and even basic symmetry.
Bell says here is the relationship between thee correlation numbers. So if you want to check it experimentally you go and measure said three numbers, what's so difficult about that? Why would you calculate the third number using the wrong formula instead?

If however, it is not possible to sort the data from experiment runs 1 and 2 as outlined above, your stationarity assumption fails and the inequality is not applicable to the data. Do you agree?
NO. Stationary assumption does not fail. Instead the author fails to incorporate it properly into his math.

DK
 
  • #122
Delta Kilo said:
It is a different case from Bell's. Bell works with expectations rather than sums and has extra assumptions about the process.
You make an artificial distinction between Bell's integral, and a sum which does not exist. An expectation value is a weighted sum of all possible values, it will be an integral if the values are continuous. In an experiment in which all values are representatively realized, the expectation value IS the average. In such a case you do NOT need an infinite run. In any case, this is a red-herring to the issue being discussed in this thread, which has to do with the compatibility of Bell's inequality to Bell-test experiments and QM.

The bottom line is, you have admitted that:

stationarity of the system is a requirement for Bell's inequality to be applicable to the system

We will come back to this. I asked you to provide proof that the system being tested in actual Bell-test experiments is stationary, and you said:

I quote myself:

In Bell's paper these assumptions are encoded in the probability density ρ(λ) being function of λ and nothing else and A(a,λ) and B(b,λ) being fully deterministic. These extra assumptions about the stochastic process behind the data is what allows one to estimate expectations of each side of the triangle independently.

Are you serious, the fact that bell assumed stationarity is proof that the EPR system is stationary? Maybe you misunderstood the question.


Such a process would violate the assumptions of Bell's theorem therefore the result would not be applicable.
So then you admit that violation of Bell's inequality by a system *could* mean simply that they system is not stationary, and therefore Bell's inequality does not apply. YES/NO.

In case of Bell, this property is explicitly encoded in ρ(λ), and that is what makes factorization possible.
Certainly the stationarity assumption was implied but I object that it was *explicitly* introduced.

On a related but very important note relevant to this thread, I assume you will also admit that, Bell's derivation also introduces the assumption that there also exists an expectation value P(a,b,c)?? In other words, Bell implicitly assumes ('explicitly' if you prefer) that there exists a probability distribution ρ(a,b,c)?? Please I need an answer to this question.

In your case the assumption of stationary process is not encoded anywhere in the math. When the authors reach factorization step they discover that something is amiss here and go into handwawing mode with disastorous results.
This is a mischaracterization of their work. Please be honest about representing their work. When I asked you if <z> <= <x> + <y> was valid, and you answered that only if the system producing the triangles is stationary, in what way is stationarity encoded in the math? Don't be disingenuous. Bell's inequality is not valid unless stationarity is assumed and the authors show that clearly.

How come they use their own derivations instead of Bell's for that. Why don't they identify an error in the original derivation for a change?
There is no error in the original derivation. None of the authors claim that. They use their own derivations in order to isolate all the assumptions implicit but not explicitly mention by Bell, such as "stationarity". They use their own derivations to show that the assumptions which most people focus on such as "locality" and "reality" are peripheral to the derivation. In other words, you do not need those assumptions to obtain the inequality which the authors admit to be valid inequalities.

Not the λ itself, no, but that's what not I said. To compute estinations for expectation values they compute means and std. deviations and that just doesn't quite work if probability densiity is not stationary. And that's what ρ(λ) assumption is about.
Again, this thread and the articles being discussed is concerned with the compatibility of the actual Bell-test experiments, and QM with Bell's inequalities. You have admitted that stationarity is a requirement. Which means if Bell's inequallity is violated by an experiment, it *could* be because the system of the experiment is not stationary.

L. Sica said:
Bell's inequality necessarily constrains the single function characterizing the correlation of periodic, spatially stationary stochastic processes, and that such processes cannot produce the cosine correlation of the singlet state.
Processes that are both more general than spatially stationary, and interesting from the perspective of Bell correlation characterization might initially be thought to constitute an empty set. However, the purpose of the present paper is to show that when the data of the customary real experiments are ordered so as to be consistent with the derivation of Bell's inequality, the set of correlation functions that result conforms to the definition of a spatially non-stationary process.

Where does he says that? These are two different things. The correlation P(a,b) is stationary but the individual measurements [itex]A_{i}, B_{i}[/itex] are non-commuting.
QM does not predict individual events.
L. Sica said:
It must be observed that the assumption that A(c,λ ) may be read at any number of different
values of angle c for a given λ represents a dramatic violation of quantum mechanical principles.
This is due to the fact that the operators representing A(a) and A(a' ) for the same particle do not
commute. Non-commutation in quantum mechanics is interpreted to mean that a sequence of
operations to measure A(a), A(a'), A(a) does not necessarily return the same value for A(a) in both
instances which is unlike the case of the stochastic process defined above. However, measurement
A(a) commutes with B(b) since these represent operations on two different particles.

This is just data massaging gone wrong.
What are you talking about? You have thrown negative words about their articles but you have nothing substantial against their argument. These are peer-reviewed articles in respected journals. If you think you have found a flaw in the argument, you will have to be mature enough to actually present a coherent argument or better, write a rebuttal and submit for publication. All the ad-hominem is getting old and stale.

I see you didn't comment on my statement that [STRIKE]there is another elephant in this room[/STRIKE] equation (16) and therefore (21) are so obviously wrong, they violate both theory and practice and even basic symmetry.
Just more unsubstantiated claims. Are you ever going to state clearly what you claim is wrong with those equations?

Bell says here is the relationship between thee correlation numbers. So if you want to check it experimentally you go and measure said three numbers, what's so difficult about that?

After you answer the question I asked you above "you will also admit that, Bell's derivation also introduces the assumption that there also exists an expectation value P(a,b,c)?? In other words, Bell implicitly assumes ('explicitly' if you prefer) that there exists a probability distribution ρ(a,b,c)??" you might begin to understand.

Why would you calculate the third number using the wrong formula instead?
You have not been paying attention.
- Experimenters perform three runs and obtain three sequences of outcome pairs (a1, b1), (b2, c2), (a3, c3)? Yes or No?
- Experimenters calculate expectation values (or their estimates if you like), P(a1, b1), P(b2, c2), P(a3, c3)? Yes or No?
- Experimenters then plug those expectation values into Bell's inequality and obtain a violation? Yes or No?

Now can you please explain to me why it will be wrong for experimenters to also calculate P(a1,c2), and use that in the inequality instead of the third run P(a3, c3). Don't you agree that for a stationary system, P(a1,c2) and P(a3, c3) should be the same? Yes or No? How come then that when you do that, the inequality is not violated? This is the essence of what Sica is showing in the paper.

NO. Stationary assumption does not fail. Instead the author fails to incorporate it properly into his math.
Further unsubstantiated claims. Can you show me a published bell-test experiment in which the experimenters made sure the system generating the particles was stationary? In other words, do you have any evidence (note "evidence" means something different from "assumption"), that the systems producing the particles in actual Bell-test experiments is stationary?

Let me rephrase the question: Do you have any evidence that ρ(λ) is spatially and temporally uniform in actual Bell-test experiments?
 
  • #123
billschnieder said:
I did the calculations for the random sampling case I discussed as "Treatment (3)" in post #92.

For the following dataset,

a, b, c
-----------
+1, -1, +1
-1, +1, -1
[...]
-1, -1, -1
-1, +1, -1
+1, -1, +1

Here is the method I use. REMEMBER - you must sample without replacement. This is equivalent to randomly shuffling the dataset and then picking the first 10 of the resulting dataset for calculating ab, the next 10 for bc and the last 10 for ac, in a manner similar to treatment (2). Again to be clear, the procedure is as follows

- Randomize the sequence by shuffling it
- select the first 10 of the resulting randomized sequence, and use for the ab term, the next 10 for the bc term and the last 10 for the ac term.

This way, we are sure that every row is used NOT MORE THAN ONCE.

I did the above 10 times, randomizing everytime from the previous random sequence and got the following results:
ab=-0.4000, bc=0.0000, ac=0.4000, Violated=False
ab=-0.4000, bc=0.0000, ac=0.2000, Violated=False
ab=0.0000, bc=-0.2000, ac=0.2000, Violated=False
ab=-0.6000, bc=-0.4000, ac=0.4000, Violated=True
ab=-0.2000, bc=-0.2000, ac=0.6000, Violated=False
ab=-0.2000, bc=-0.8000, ac=0.2000, Violated=True
ab=0.0000, bc=-0.2000, ac=0.4000, Violated=False
ab=0.0000, bc=-0.4000, ac=0.2000, Violated=False
ab=-0.6000, bc=-0.4000, ac=0.0000, Violated=False
ab=-0.4000, bc=0.2000, ac=0.8000, Violated=True

Repeating this 1000, consistently gives me a violation in about 20% of the randomly sampled (without replacement) pairs.

NOTE, 1 violation is enough.

OK I did roughly the same only I calculated ab, bc, and ac for the first three data and then again for the next three, etc. I shuffled the rows by using the data sort function on a column with random numbers, see the attached file (I added what I think are the inequalities of Bertlmann's socks and DrC).

By chance I got no violation the first 10 times and I have the impression that I get a violation less than 10% of the time (Both for Bell-1 as for DrC).

Anyway, even 20% of the time is not good enough: for it means that usually you do not get a violation. It's pretty sure that the more you average, the less often you will get a violation.

Harald
 

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  • #124
Currently I cannot reply to everything. Just a few comments:
billschnieder said:
[...]
Are you serious, the fact that bell assumed stationarity is proof that the EPR system is stationary? Maybe you misunderstood the question.
This appears to be a key point: Bell's Theorem seems to be based on a circular argument. That impression can be avoided if someone can refer to a published paper that proofs that Bell's averaging method must be valid for any possible "local" theory of QM.
[...] There is no error in the original derivation. None of the authors claim that.[..]

In fact they do, although implicitly - as I motivated in one or two earlier posts in this thread. They state in effect that Bell's derivation is not valid for its purpose. An invalid use of math is an error.
 
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  • #125
Delta Kilo said:
SUMMARY: How to disprove Bell's inequality:
  1. Take Bell's inequality
  2. "Improve it" by throwing vital bits out
  3. Run into trouble
  4. Blame Bell
  5. ...?
  6. Publish!

So true! :biggrin:

A note about Bill: Everyone else is an idiot, and he has brilliantly discovered things no one else has ever considered. Oh, and the entire scientific community is involved in a Bell conspiracy. Bill, does that about sum it up?
 
  • #126
harrylin said:
This appears to be a key point: Bell's Theorem seems to be based on a circular argument.

And what would that be, do tell?

Circular: If you assume A, then A is proved.
 
  • #127
billschnieder said:
These are peer-reviewed articles in respected journals.

That's a stretch. These articles are NOT seriously considered as disproofs of Bell, and you are being disingenuous to imply otherwise.

But of course, you stretch as well as you bob and weave. :-p
 
  • #128
DrChinese said:
So true! :biggrin:

A note about Bill: Everyone else is an idiot, and he has brilliantly discovered things no one else has ever considered. Oh, and the entire scientific community is involved in a Bell conspiracy. Bill, does that about sum it up?

For someone trying to understand Bell’s theorem, I find it essential to read papers on both sides of the argument. Of course, the problem here is that probability theory is not easily understood by the average reader and the experts seem to be able to rationalize away the other person’s arguments all too easy and not reach a consensus.

By the way, hasn't Bill supplied sufficient references to each and every major argument presented by him? I don't understand, where has Bill claimed these are his discoveries? Also, try publishing a paper in one of the journals referenced in this thread. Good Luck!

I do like your quote by Einstein: "Why 100? If I were wrong, one would have been enough." Rather than taking countless unprofessional jab's at someone for their assertions wouldn't one good rebuttal be enough?
 
  • #129
rlduncan said:
For someone trying to understand Bell’s theorem, I find it essential to read papers on both sides of the argument. Of course, the problem here is that probability theory is not easily understood by the average reader and the experts seem to be able to rationalize away the other person’s arguments all too easy and not reach a consensus.

By the way, hasn't Bill supplied sufficient references to each and every major argument presented by him? I don't understand, where has Bill claimed these are his discoveries? Also, try publishing a paper in one of the journals referenced in this thread. Good Luck!

I do like your quote by Einstein: "Why 100? If I were wrong, one would have been enough." Rather than taking countless unprofessional jab's at someone for their assertions wouldn't one good rebuttal be enough?

One? How about 100 rebuttals in the various threads. Besides, my jabs are not so much unprofessional as being in fun. Bill gets a bit serious, and it would be easier on me if he would take his pills on time. Or join me for a glass of wine. :smile:

The issue here is that this is NOT the place to debate standard, generally accepted science. It is the place to learn it. I see (and sometimes read) about 5 "disproofs" Bell a month, so one more or less is not going to make much difference to me. I probably have read more on the anti-Bell side than most. But there are a lot of readers here who will not be aware, after reading Bill, that there are essentially NO serious questions about Bell at this point. Anyone familiar with the body of literature on this subject will easily understand why - just reade the other 995 papers published each month advancing entanglement theory and experiment. So I usually try to keep the record straight for those readers.

P.S. All local realists, like Bill, essentially DENY the existence of entanglement as a physical state (Einstein's "spooky action at a distance"). That should give pause to anyone flirting with such ideas. They have now found entanglement in hundreds of different contexts. So I really don't see what the question is here other than a philosophical exercise. It is a moot point.
 
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  • #130
DrChinese said:
One? How about 100 rebuttals in the various threads. Besides, my jabs are not so much unprofessional as being in fun. Bill gets a bit serious, and it would be easier on me if he would take his pills on time. Or join me for a glass of wine. :smile:

The issue here is that this is NOT the place to debate standard, generally accepted science. It is the place to learn it. I see (and sometimes read) about 5 "disproofs" Bell a month, so one more or less is not going to make much difference to me. I probably have read more on the anti-Bell side than most. But there are a lot of readers here who will not be aware, after reading Bill, that there are essentially NO serious questions about Bell at this point. Anyone familiar with the body of literature on this subject will easily understand why - just reade the other 995 papers published each month advancing entanglement theory and experiment. So I usually try to keep the record straight for those readers.

P.S. All local realists, like Bill, essentially DENY the existence of entanglement as a physical state (Einstein's "spooky action at a distance"). That should give pause to anyone flirting with such ideas. They have now found entanglement in hundreds of different contexts. So I really don't see what the question is here other than a philosophical exercise. It is a moot point.

When all the relevant facts cannot be know, wouldn’t you agree that the main point is that in order to prove Bell’s inequality is applicable to the EPRB experiments, that it should be consistent with the known facts in a large number of random trials, without exception?
 
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  • #131
billschnieder said:
You make an artificial distinction between Bell's integral, and a sum which does not exist. An expectation value is a weighted sum of all possible values, it will be an integral if the values are continuous. In an experiment in which all values are representatively realized, the expectation value IS the average. In such a case you do NOT need an infinite run.
Please get a clue first. Say, you have two independed random variables X and Y uniformly distributed on (-1,1). Consider the function A = { 1 if x2+y2<1, else 0 }. The expectation of A is [itex]\pi/4[/itex]. Just how many experiments do you need to get your "representative realization"?

Another example: you have random variable X uniformly distributed on (-1,1). You also have Y = X + 0.001. Consider if the following statements are universally true:
[itex]X_{i}<Y_{i}[/itex] - Yes
[itex]X_{i}<Y_{j},i\neq j[/itex] - No
[itex]\frac{1}{N}\sum_{i=1}^{N}X_{i} < \frac{1}{N}\sum_{i=1}^{N}Y_{i}[/itex] - Yes
[itex]\frac{1}{N}\sum_{i=1}^{N}X_{i} < \frac{1}{N}\sum_{i=M}^{N+M}Y_{i}[/itex] - No
[itex]E[X] < E[Y][/itex] - Yes
Now do you see the difference?

billschnieder said:
The bottom line is, you have admitted that:

stationarity of the system is a requirement for Bell's inequality to be applicable to the system
Now before we all get worked up, let's check the definition, shall we? http://books.google.com/books?id=mwW-iODttSQC&q=stationary#v=snippet&q=stationary&f=false"So basically it's just an invariance with regads to time translation. Nothing special about it.

billschnieder said:
Are you serious, the fact that bell assumed stationarity is proof that the EPR system is stationary? Maybe you misunderstood the question.
Well, sorry for that. I was of course referring to the Bell's model of EPR experiment, which as we all know has been experimentally violated.
billschnieder said:
So then you admit that violation of Bell's inequality by a system *could* mean simply that they system is not stationary, and therefore Bell's inequality does not apply. YES/NO.
Yes, it could. Bell's violation shows that one or more underlying assumptions are wrong, but it does not tell us which one. Unortodox QR tells us that ti is the non-locality assumption (independence of A(a,l) from b and vice versa) that does not hold. There is no reason to suspect that it is the stationary assumption that is violated.

billschnieder said:
On a related but very important note relevant to this thread, I assume you will also admit that, Bell's derivation also introduces the assumption that there also exists an expectation value P(a,b,c)?? In other words, Bell implicitly assumes ('explicitly' if you prefer) that there exists a probability distribution ρ(a,b,c)?? Please I need an answer to this question.
You have to be more specific. What does P(a,b,c) mean? In Bell's paper P(a,b) is clearly defined to be an expectation value of the product of the results produced by detectors A and B on opposite sides of the apparatus in the same experiment, with detector A set to angle a and detector B set to angle b respectively. Since the apparatus has only two sides P(a,b,c) does no make any sense in this context.

In Bell's paper P(a,b) is introduced as
[itex]P(a,b) = \int A(a,\lambda)B(b,\lambda)\rho(\lambda)d\lambda[/itex] (2)
Bell obviously considered it so obvious that he did not need to spell out the details. Let's go through this again. The outcome [itex]A_{i}, B_{i}[/itex] of the i-th experiment of the run with the angles a and b is assumed to depend only on the value of hidden parameter [itex]\lambda_{i}[/itex]: [itex]A_{i} = A(a,\lambda_{i}), B_{i} = B(b,\lambda_{i})[/itex]. Product [itex]C_{i} = A_{i}B_{i} = A(a, \lambda_{i})B(b,\lambda_{i}) = C(a,b,\lambda_{i}).[/itex] The expectation [itex]P(a,b) = \int C(a,b,\lambda)\rho(\lambda)d\lambda = \int A(a,\lambda)B(b,\lambda)\rho(\lambda)d\lambda[/itex] where [itex]\rho(\lambda)[/itex] is the probability density independent of a and b (or anything else for that matter).

When I asked you if <z> <= <x> + <y> was valid, and you answered that only if the system producing the triangles is stationary, in what way is stationarity encoded in the math?
It is not encoded. That's why I gave you conditional answer. If it was encoded I might me able to say "yeah, it's valid" with no strings attached.
But on the other hand there is a difference between one's inability to demonstrate stationary condition and the system being truly non-stationary.

There is no error in the original derivation. None of the authors claim that.
Thanks goodness for that.
They use their own derivations in order to isolate all the assumptions implicit but not explicitly mention by Bell, such as "stationarity". They use their own derivations to show that the assumptions which most people focus on such as "locality" and "reality" are peripheral to the derivation. In other words, you do not need those assumptions to obtain the inequality which the authors admit to be valid inequalities.
Right, but how does that disprove Bell or make it not applicable? What exactly are these assumptions that Bell makes and the other people don't? That "stationary" thing, again? Are you saying it is not stationary? Care to demonstrate it perharps?

Again, this thread and the articles being discussed is concerned with the compatibility of the actual Bell-test experiments, and QM with Bell's inequalities. You have admitted that stationarity is a requirement. Which means if Bell's inequallity is violated by an experiment, it *could* be because the system of the experiment is not stationary.
Yes it could, but so it could for another reason (see above).

I see you didn't comment on my statement that [STRIKE]there is another elephant in this room[/STRIKE] equation (16) and therefore (21) are so obviously wrong, they violate both theory and practice and even basic symmetry.
Just more unsubstantiated claims. Are you ever going to state clearly what you claim is wrong with those equations?
Sure. Just compare Eq (12)
[itex]<AB>=\sum_{A,B}A(\Theta_{A})B(\Theta_{B})p( A( \Theta _{A}), B(\Theta_{B}) )[/itex]
and Eq (16)
[itex]<AA'>=\sum_{A,A',B}A(\Theta_{A})A'(\Theta_{A'})p( A( \Theta _{A}|B(\Theta_{B}))p(A'( \Theta _{A'}|B(\Theta_{B}) )p(B(\Theta_{B}) )[/itex]
and tell me if they look the same to you (with appropiate variable substitution).
Let A = person's height, A' = weight and B = birthday
So <AB> according to eq (12) means correlation between height and birthday and one would expect <AA'> to mean correlation between weight and height. But the author discovers that he cannot measure weight and height for the same person. So what he does instead, he says, oh bugger, let's make a correlation between one persons height and the weight of another person with the same birthday, and use it in place of the correlation between weight and height. Because that's exactly what eq (16) means.

With regards to eq (21) he's got:
[itex]<AA'>=cos(\Theta_{A}-\Theta_{B})cos(\Theta_{A'}-\Theta_{B})[/itex]
What the heck is [itex]\Theta_{B}[/itex] doing in the expression for [itex]<AA'>[/itex]? Where does it come from? And, this is presented as be the prediction of QM.
Now you please tell me, are the equations (16) and (21) correct estimations for the correlation <AA'>? YES or NO please.

After you answer the question I asked you above "you will also admit that, Bell's derivation also introduces the assumption that there also exists an expectation value P(a,b,c)?? In other words, Bell implicitly assumes ('explicitly' if you prefer) that there exists a probability distribution ρ(a,b,c)??" you might begin to understand.
There is no [STRIKE]spoon[/STRIKE] P(a,b,c). It does not make sense. See above.


- Experimenters perform three runs and obtain three sequences of outcome pairs (a1, b1), (b2, c2), (a3, c3)? Yes or No?
- Experimenters calculate expectation values (or their estimates if you like), P(a1, b1), P(b2, c2), P(a3, c3)? Yes or No?
- Experimenters then plug those expectation values into Bell's inequality and obtain a violation? Yes or No?
Yes to all 3

Now can you please explain to me why it will be wrong for experimenters to also calculate P(a1,c2), and use that in the inequality instead of the third run P(a3, c3).
Because you calculate it wrong. Or to be exact you calculate the wrong thing.

Don't you agree that for a stationary system, P(a1,c2) and P(a3, c3) should be the same? Yes or No?
No. The formula for P(a1,C2) is plain wrong. See above.

How come then that when you do that, the inequality is not violated?
Because he is using the wrong formula?

Can you show me a published bell-test experiment in which the experimenters made sure the system generating the particles was stationary? In other words, do you have any evidence (note "evidence" means something different from "assumption"), that the systems producing the particles in actual Bell-test experiments is stationary?
Well, last I heard they showed the violation up to 30 sigmas (sorry don't have a link handy). Don't you think they would not have noticed?

Let me rephrase the question: Do you have any evidence that ρ(λ) is spatially and temporally uniform in actual Bell-test experiments?
Are you suggesting that it is not? One would think it would be noticed by now. That would basically mean violation of rotational or time-translational symmetry. Are you prepared to go that far to defend this heresy?

DK
 
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  • #132
Delta Kilo said:
Please get a clue first. Say, you have two independed random variables X and Y uniformly distributed on (-1,1). Consider the function A = { 1 if x2+y2<1, else 0 }. The expectation of A is [itex]\pi/4[/itex]. Just how many experiments do you need to get your "representative realization"?
No you get a clue. Read up on the meaning of "expectation value", and save yourself the embarrassment that this is becoming.

http://mathworld.wolfram.com/ExpectationValue.html
http://en.wikipedia.org/wiki/Expected_value
http://planetmath.org/encyclopedia/Expectation.html

The expectation value "For discrete random variables this is equivalent to the probability-weighted sum of the possible values. For continuous random variables with a density function it is the probability density-weighted integral of the possible values."

There is nothing about number of trials, in the definition of expectation value, Period! Your efforts to defend the ridiculous claim that "one cannot not obtain exact expectation value from a finite run, no matter how long." is clearly wrong and misinformed.

Now before we all get worked up, let's check the definition, shall we? http://books.google.com/books?id=mwW-iODttSQC&q=stationary#v=snippet&q=stationary&f=false"So basically it's just an invariance with regads to time translation. Nothing special about it.
Apparently you are uninformed that there is such a thing as spatial stationarity. Stationarity is not limitted to the time domain. But I'm not surprised that you had no clue about this.

Bell's violation shows that one or more underlying assumptions are wrong, but it does not tell us which one. Unortodox QR tells us that ti is the non-locality assumption (independence of A(a,l) from b and vice versa) that does not hold. There is no reason to suspect that it is the stationary assumption that is violated.
I suppose you are also unaware (or ignored) the other article I have quoted to you, which demonstrates that precisely this point. You can't make it disappear by sticking your head in the sand. Here they are:

http://arxiv.org/abs/quant-ph/0305071v1
Journal of Modern Optics, 2003, Vol. 50, No. 15-17, 2465-2474
Bell's inequality violation due to misidentification of spatially non stationary random processes
Correlations for the Bell gedankenexperiment are constructed using probabilities given by quantum mechanics, and nonlocal information. They satisfy Bell's inequality and exhibit spatial non stationarity in angle. Correlations for three successive local spin measurements on one particle are computed as well. These correlations also exhibit non stationarity, and satisfy the Bell inequality. In both cases, the mistaken assumption that the underlying process is wide-sense-stationary in angle results in violation of Bell's inequality. These results directly challenge the wide-spread belief that violation of Bell's inequality is a decisive test for nonlocality.


You have to be more specific. What does P(a,b,c) mean? In Bell's paper P(a,b) is clearly defined to be an expectation value of the product of the results produced by detectors A and B on opposite sides of the apparatus in the same experiment, with detector A set to angle a and detector B set to angle b respectively. Since the apparatus has only two sides P(a,b,c) does no make any sense in this context.
There you have it. P(a,b,c) does not make sense, but
|P(a,b) - P(a,c)| <= 1 + P(b,c)

Makes sense? You just conceded Sica, and DeRaedt's argument. You can not have one make sense but not the other. Need I remind you that Bell's P(b,c) term was never introduced in the derivation but rather, factored out from both P(a,b) and P(a,c). Your recent epiphany that P(a,b,c) does not make any sense will also imply that you can not produce a P(b,c) term from P(a,b) and P(a,c). Anyone in doubt should verify how the P(b,c) term is obtained in Bell's original paper, equations 14-15.

For example, consider the following derivation which is equivalent to the arithmetic operations caried out by Bell in equations 14-15:

ab − ac = a (b − c ) : "Factor out a"
a (b − c ) = ab (1− bc) : "factor out b noting that 1/b = b, since b = +/- 1

NOTE:
1) how the bc term has appeared by arithmetic manipulation of the ab and ac terms
2) Note that arithmetic manipulations rely on being able to divide/multiple "a" AND "c" with "b"
3) If it doesn't make sense to multiply "a"*"b"*"c", then the mathematical operations required to obtain the bc term from ab and ac, are invalid.

THEREFORE, there must be, according to Bell, a P(a,b,c) expectation value which Doctor Kilo now says does not make sense. In summary, you have affirmed the point being made by Sica, and the DeRaedt paper, under the guise of disagreement. Go figure.

But on the other hand there is a difference between one's inability to demonstrate stationary condition and the system being truly non-stationary.
To conclude that the reason for violation is due to failure of "locality" or "realism", while knowing fully well that you have not excluded "non-stationarity" is intellectual dishonesty. So those making such claims MUST demonstrate that they have proof of the system being truly stationary.

Are you saying it is not stationary? Care to demonstrate it perhaps?
See the article I mentioned a while back which you ignored. If you have even a single published article which attempts to justify that the systems are stationary, present it. Note that an assumption that it is, is not enough.

Sure. Just compare Eq (12)
[itex]<AB>=\sum_{A,B}A(\Theta_{A})B(\Theta_{B})p( A( \Theta _{A}), B(\Theta_{B}) )[/itex]
and Eq (16)
[itex]<AA'>=\sum_{A,A',B}A(\Theta_{A})A'(\Theta_{A'})p( A( \Theta _{A}|B(\Theta_{B}))p(A'( \Theta _{A'}|B(\Theta_{B}) )p(B(\Theta_{B}) )[/itex]
and tell me if they look the same to you (with appropiate variable substitution).
As expected, you did not bother to read what they actually said, and simply drew a false conclusion after glancing at the page for a second.

Let me simplify it to you, even though I did that already. Equation 16 is obtained not by variable substitution, but by factoring out as described in section 2.1 of the paper.

Essentially, instead of measuring a1b1, a2c2, b3c3 in 3 runs, you measure a1b1, a2c2 just like Bell started his derivation. And then you do the same operations which Bell use to obtain the bc term as follows by factoring out b1 from a1b1 and c2 from a2c2 and recombining them to obtain b1c2 which you then use as the third expectation value. That is how equation 12 is obtained, not by "variable substition". Please read the description in section 2.1 before you mumble any further about this. When you do it this way, which is consistent with the way Bell derived the original inequalities, you get no violation.

Let A = person's height, A' = weight and B = birthday
So <AB> according to eq (12) means correlation between height and birthday and one would expect <AA'> to mean correlation between weight and height. But the author discovers that he cannot measure weight and height for the same person. So what he does instead, he says, oh bugger, let's make a correlation between one persons height and the weight of another person with the same birthday, and use it in place of the correlation between weight and height. Because that's exactly what eq (16) means.
Funny, you are arguing against yourself. What you describe is what is done in Bell test experiments and
the opposite of what Sica is doing here. Sica says if you measure height, weight for one group of people, and then weight and birthday for the second group, and you sort the results such that weight's match exactly, then you will be able to obtain a counterfactual dataset of "height" and "birthday" by factoring and recombining the columns as Bell did in deriving his inequalities. And if you do that, you get no violation. Note, you can not argue against this because the stationarity assumption implies that the statistics of the first group and the second group are stationary. So by arguing that this operation is wrong, you are implying that the stationarity assumption is also wrong, which is essentially Sica's point!


Because you calculate it wrong. Or to be exact you calculate the wrong thing.

No. The formula for P(a1,C2) is plain wrong. See above.
If it is wrong, then either Bell's derivation is also wrong, or the stationarity assumption fails. You can not eat your cake and have it.

Are you suggesting that it is not?
See the quoted article above and be prepared to present one (just one) article demonstrating that it is. Again, an assumption is not enough.
 
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  • #133
billschnieder said:
...lots of comments...
I'm not going to reply to each comment separately, otherwise people reading this would have difficulty following it. Instead I'll just go through one of the papers in details.

First, I'm not going to say much about DeRaedt's paper. I've said everything I wanted in this post. If you want to argue about it, please address those points first. I wouldn't recommend it though, it's a mess :smile:

Let's now turn to Sica's papers. Start with this one. Up to eq. (3) everything is soft of all right. The author says
Intrinsic to the process so far specified is the fact that measurement values exist at multiple angular settings of A and B for each [itex]\lambda[/itex] i.e. for each realization of [itex]A(a,\lambda)[/itex]
Well, that's what hidden variables assumption is about. We assume that the internal state of the pair is represented by [itex]\lambda[/itex] and the interaction with detector A is represented by deterministic function [itex]A(a,\lambda)[/itex], which is defined for all possible states [itex]A(a,\lambda)[/itex].

It must be observed that the assumption that [itex]A(a,\lambda)[/itex] may be read at any number of different values of angle c for a given [itex]\lambda[/itex] represents a dramatic violation of quantum mechanical principles.
This is where it starts to go wrong. There is no such implied assumption. Yes, there is value for every possible [itex]A(a,\lambda)[/itex] but no, it cannot be read for a given [itex]\lambda[/itex] precisely because [itex]\lambda[/itex] cannot be given. That's exactly why it is called "hidden" variable.

Thus, for the stochastic process Bell defined, he could compute a difference of correlations involving real and counterfactual variables
This is not so. Equation (4) does not contain any "counterfactual variables". The confusion comes mostly from the fact that the same symbol [itex]\lambda[/itex] is used for two different purposes: when we talk about results of individual experiments, [itex]\lambda[/itex] denotes random value [itex]\lambda \in \{ \lambda_{i} \}[/itex] representing a hidden state associated with this particular experiment.

In contrast, [itex]\lambda[/itex] in eqs (4)-(6) is merely a variable being intergated over, a loop variable which has a perfectly defined value at all times. In fact, it would be perfectly valid (and a lot less confusing) to rename [itex]\lambda[/itex] to [itex]u[/itex] or [itex]v[/itex] or something else neutral in eqs (4)-(6). Eq (4)-(6) do not refer to any particular experiment or series of experiments. They simply follow directly from the definition of P(a,b) given by eq (3) and the fact that [itex]B(a,\lambda)=-A(a,\lambda)[/itex].

In any case, eq (7) directly follows from definition (3), assumption (2), [itex]B(a,\lambda)=-A(a,\lambda)[/itex] and [itex]\int \rho(\lambda)d\lambda=1[/itex] by application of simple math.

Note that the correlations of real data pairs, and the correlations of counterfactual and real data pairs, are assumed to be given by the same (quantum mechanical) correlation function depending only on the angular differences of detector settings.
This is incorrect. All correlations in eq. (8) are real. QM tells you nothing at all about counterfactual measurements.

Several assumptions have now been identified in the construction of the Bell theorem. A question may be raised as to their necessity. It may be answered by re-deriving the inequality with fewer assumptions.

1.2 New Derivation of Bell's Inequality

Assume the existence of three lists, a, b, and b' of length N composed of elements [itex]a_{i}[/itex], [itex]a'_{i}[/itex], [itex]b_{i}[/itex], each equal to ±1.
Well, talk about fewer assumptions! There are no lists of triples in the original Bell's work.

The following derivation is obvious, except the assumptions under which the estimates converge are not spelled out. And it should be clearly understood that, while the shape of the resultant inequality (25) is the same, the assumptions on which it is based are different.

In section 1.3 the author claims:
Independent trials at each pair of detector settings implies that the crucial factoring step used in (4) and (10) no longer holds.
This is not correct. The factoring step follows directly from the definition with very simple math. It does not require or depend on the results of any experiments. In fact, the reason why factorization works here (and does not work quite so well for "improved derivation") is because all the randomness has already been factored out into [itex]\rho(\lambda)[/itex] in the definition (3), and what's left are simple deterministic functions A() and B().

Now let's jump straight to formulas in section 2.3. It all goes well upto and including eq (15). But then we have:

Since A and A' are measured with B in separate independent experiments, they are statistically independent except for their conditional dependence on B.
So here we have eq (16) and the right side assumes that A and A' are from differet experiments. But but but the notation <AA'> used on the left side implies the correlation of results of the same experiment! This is how it is used in eq (25), eqs (12)-(14), and also this is what you have as P(b,c) in the original Bell's notation.

So (15) is plain wrong. It would be correct if you change the notation on the left side from <AA'> (which is reserved for results of the same experiment) to something else. But then you won't be able to plug it into eq. (25).

As a result, eq (21) is also wrong. The rest is "rubbish in - rubbish out".


The other paper of the same author is built on top of this one, so there is no need to discuss it until we get this one out of the way.

DK
 
  • #134
billschnieder said:
Delta Kilo said:
Let A = person's height, A' = weight and B = birthday
So <AB> according to eq (12) means correlation between height and birthday and one would expect <AA'> to mean correlation between weight and height. But the author discovers that he cannot measure weight and height for the same person. So what he does instead, he says, oh bugger, let's make a correlation between one persons height and the weight of another person with the same birthday, and use it in place of the correlation between weight and height. Because that's exactly what eq (16) means.
Funny, you are arguing against yourself. What you describe is what is done in Bell test experiments and the opposite of what Sica is doing here. Sica says if you measure height, weight for one group of people, and then weight and birthday for the second group, and you sort the results such that weight's match exactly, then you will be able to obtain a counterfactual dataset of "height" and "birthday" by factoring and recombining the columns as Bell did in deriving his inequalities.

If you re-read what I wrote and what you wrote very carefully you will hopefully realize that you are saying exactly the same thing as I do except the labels "height" and "birthday" are swapped (which is of no consequence since they are just arbitrary labels for some angles a, b and c). Well, let me help you:
Sica says if you measure height, [STRIKE]weight[/STRIKE] birthday for one group of people, and then [STRIKE]weight[/STRIKE] birthday and [STRIKE]birthday[/STRIKE] weight for the second group, and you sort the results such that [STRIKE]weight[/STRIKE] birthday's match exactly, then you will be able to obtain a counterfactual dataset of "height" and "[STRIKE]birthday[/STRIKE] weight" by factoring and recombining the columns as Bell did in deriving his inequalities.
Lets say this again: measure height of one guy, find a guy from another group with the same birthday and measure his weight. Then plot a chart height vs. weight. Does it make any sense to you at all?

And no, Bell certainly did not do this in his derivation. And Bell tests measure each of the three (or more) correllations directly by accumulating simultaneously produced pairs.

DK
 
  • #135
After I wrote,
"This appears to be a key point: Bell's Theorem seems to be based on a circular argument.":
DrChinese said:
And what would that be, do tell?
Circular: If you assume A, then A is proved.

Exactly :biggrin:
Is it really necessary to elaborate? OK then - and I'll expand on it.

It looks as if, based on certain assumptions, Bell tried in vain to come up with a local theory; and when he couldn't, he turned his failed attempts into the theorem that it can't be done. It's Bell who claimed to have proved that it can't be done; in principle it's not up to his audience to "fix" that claim or to prove the contrary.

Thus in their latest paper, De Raedt et al pointed out that in 1964 Bell based his derivation on the assumption that all measurement results due to any possible local theory can be described with a single probability distribution. De Raedt's paper demonstrates that this assumption is not generally true.

However, in his 1980 talk on Bertlmann's socks, Bell added the random sampling hypothesis which he had not mentioned in his original paper; and at first sight that could indeed "fix" that issue (at least De Raedt's counter example with patients seems to be inapplicable with that added assumption). Still, it looks like an ad hoc fix to me; and Bell did not formally prove it.

Thus I asked if someone can refer to a published paper that proves that Bell's averaging method is valid for any possible "local" theory of QM.

Indeed, it's quite useless to simply assume to be true that what you claim to prove.

Harald
 
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  • #136
Delta Kilo said:
Well, I looked at the paper that started the thread. I did not finish it (way too many words) but I can see a number of serious problems with it.

For a start, section III D "Relation to Bell's work" has no relation to Bell's work whatsoever :confused: None of the original Bell's assumptions are reflected, in particular, the crucial assumption of independence of results A from settings B and vice versa is nowhere to be found. Neither is the perfect anti-correllation for the same settings of A and B (which is used quite a lot in Bell's derivation). And then the authors confuse individual outcomes of measurement with expectation values and arrive at completely wrong conclusion about triplets of data sharing the same lambda, while there are no triplets of data at all in Bell's original work, only probabilities and expectation values. And it goes downhill from there.

DK

Hi DK, the relation of Boole's inequalities with Bell's work is that they used similar inequalities. The "class of probabilistic models that form the core of Bell’s work" follows directly from Boole. Isn't that clear from the text? As I mentioned before, even Bell knew this as he indicated by his mention of Lille and Lyon in his talk on Bertlmann's socks.

What you seem to have missed is that De Raedt et al take a fresh look at those inequalities, and so they do not follow Bell. The independence of results at A from conditions at B does not really matter for Boole's inequalities ("the inequalities of Section II [are] independent of the details of the physical or arithmetic processes that produce the data"). Anyway, this is the case in Boole's example of patients in Lille, Lyon and Paris (VII A). Perhaps you did not read far enough. :wink:

You also wonder why the anti-correlation at certain settings is not discussed. That is besides the point: the issue is Bell's assumption of a single probability distribution. Thus they write: "The models that we consider in this subsection do not pretend to account for the correlations of two spin-1/2 particles in the singlet state". However, they do discuss "dichotomic variables", isn't that more or less the same as "anti-correlation"? Please enlighten me, I'm not familiar with such jargon!

Your two remaining issues are intriguing. Please clarify:

- where does the paper confuse individual outcomes of measurement with expectation values?
- where does the paper claim that there are triplets of data described in Bell-1964?

Note that the paper stresses: "there are no Boole inequalities Eq. (13) for the corresponding pair correlations unless we make the hypotheses that there is an underlying process of triples that gives rise to the data."

Thanks,
Harald
 
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  • #137
harrylin said:
Hi DK, <...many valid points...>
First I must concede that I did not read the paper carefully enough. I assumed that there is a connection between [itex]\Upsilon^{(3)}[/itex] of section II and [itex]f^{(3)}[/itex] of section III etc. Apparently there is no such connection. There is a lot of talk about the significance of the results of section II but the but the actual eq (13) is not used anywhere in the subsequent derivations.

Now I'm going to pull out my argument. It's a long and heavy iron bar with "1" permanently etched on one end and "-1" on another. When measured it always produces the same pair of values (1,-1) for any values of parameters. This sophisticated lab equipment fulfills all assumptions of the original Bell's theorem. All probabilities and expectation values exist at all times, P(a,b) = P(a,c) = P(b,c) = -1 and Bell's inequality is satisfied.

Let's apply the argument to section II of the paper in question. To do this we need to find some triples {a,b,c} so that the pairs (a,b), (a,c) and (b,c) simultaneously match experimental data (1,-1). Clearly no such triple can be constructed. What does it mean? It means that the assumption of section II (the existence of [itex]\Upsilon^{(3)}[/itex]) is not satisfied. Which means the assumption of [itex]\Upsilon^{(3)}[/itex] is in fact more restrictive than the original Bell's assumptions, no matter what the paper makes it sound like.

Now, armed with the same heavy argument, we approach section III. So we want to see if the "EBBI" eq (30) are applicable. To do this, we try to construct a joint probability distribution [itex]f^{(3)}(s1,s2,s3)[/itex] according to theorem II (or IV, whichever). Well, what a surprise, the distribution [itex]f^{(3)}(s1,s2,s3)=0 \forall {s1,s2,s3}[/itex] since none of the triples s1,s2,s3 match experimental data. So again EBBI are not applicable to a perfectly good classical apparatus while the original Bell's theorem works just fine.

But but but, according to section III D, Bell's inequality is just a special case of the mighty EBBI, how come? Here comes a bit of cheating: equations (49) are subtly different from Bells original derivation. Namely, there's a hidden symmetry assumption. The three equations (49) represent the probabilities of getting outcome (s,s') in three measurements: (_,^), (_,~) and (^,~) respectively. Note that measurement (^) is done on the right side of stick in the first equation and on the left side of stick in the third. Of course one can re-shuffle the order but there is no consistent way to assign "left" and "right" ends to measurements (_,^,~), at least one of them has to be done on both ends of the stick. Note that the same function [itex]\widehat{f}^{(1)}[/itex] is used for probabilities on both ends of the stick. That implies the symmetry with respect to measurement (^). On other words, if we make the same measurement on both sides, the correlation will always be non-negative.

In Bell's case when the same measurement is done on both ends, the outcomes are perfectly anti-correlated, that is P(a,a) = -1. This was made explicit at the very beginning: [itex]A(a,\lambda)=-B(a,\lambda)[/itex]. If we try this, eq (50) will have '+' when used on the left side of the stick and '-' when used on the right side, and instead of (49) we will get something like [itex]f^{(2)}(S,S')=\frac{1-SE_{1}^{(2)}+S'E_{2}^{(2)}-SS'E^{(2)}}{4}[/itex] (note the minuses) and the construction of non-negative [itex]f^{(3)}(S,S',S'')[/itex] no longer works.

Regarding section VII A (doctors/patients): the example is meant to be a model of Bell's experiment and to show the apparent violation of Bell's inequation (assuming that patients a,b and c correspond to angle settings, locations 1 and 2 correspond to the two ends of the apparatus and day of week is [itex]\lambda[/itex]. However the setup violates the anti-correlation assumption [itex]A(a,\lambda)=-B(a,\lambda)[/itex] as can be seen from the truth table, therefore Bell's inequality is not applicable.

The same applies to section section VII B. (Factorizable model): the expectation [itex]E^{(2)}(a,a)[/itex] is neither 1 nor -1, which means measurements with the same settings on both ends are neither perfectly correlated nor perfectly anti-correlated. This violates the assumption of Bell's derivation so Bell's inequation is not applicable.

So the net result is that EBBI use different set of assumptions from Bell's. There are cases when Bell is applicable and EBBI is not and vice versa. In particular, in case of EPR experiment Bell's inequality applies while EBBI does not. Same thing happens if the whole EPR setup is replaced with iron bar argument.

Overall, while the math is a bit flakey but sort of OK, the conclusions drawn from it are just outright wrong.

Regards
DK
 
  • #138
Delta Kilo said:
[..] Now I'm going to pull out my argument. It's a long and heavy iron bar with "1" permanently etched on one end and "-1" on another. When measured it always produces the same pair of values (1,-1) for any values of parameters. This sophisticated lab equipment fulfills all assumptions of the original Bell's theorem. All probabilities and expectation values exist at all times, P(a,b) = P(a,c) = P(b,c) = -1 and Bell's inequality is satisfied.
As you already indicate, your iron bar model is certain to fail to reproduce the predictions of QM. Perhaps it's like the first model that Bell attempted (it certainly is like the toy model that I tried a long time ago). What do you want to achieve with an example that, as we all know, doesn't work? Or do you suggest that Bell only considered that type of failing model?
Let's apply the argument to section II of the paper in question. To do this we need to find some triples {a,b,c} so that the pairs (a,b), (a,c) and (b,c) simultaneously match experimental data (1,-1). Clearly no such triple can be constructed. What does it mean? It means that the assumption of section II (the existence of [itex]\Upsilon^{(3)}[/itex]) is not satisfied. Which means the assumption of [itex]\Upsilon^{(3)}[/itex] is in fact more restrictive than the original Bell's assumptions, no matter what the paper makes it sound like.
I'm not sure about your experimental data but obviously your iron bar model doesn't produce it. And there's something more interesting to remark here. Section 2 is supposed to be Boollean logic, similar to 1+1=2. If you say that according to experimental data 1+1=/=2 (or more precisely, eq.16 is not obeyed) then logically there is a something wrong with your data. The only alternative is that De Raedt's derivation is erroneous so that his eq.16 is not similar to 1+1=2 (then where is the error?).
Now, armed with the same heavy argument, we approach section III. So we want to see if the "EBBI" eq (30) are applicable. To do this, we try to construct a joint probability distribution [itex]f^{(3)}(s1,s2,s3)[/itex] according to theorem II (or IV, whichever). Well, what a surprise, the distribution [itex]f^{(3)}(s1,s2,s3)=0 \forall {s1,s2,s3}[/itex] since none of the triples s1,s2,s3 match experimental data. So again EBBI are not applicable to a perfectly good classical apparatus while the original Bell's theorem works just fine.
Same as above. :-p
And if Bell's theorem worked just fine for an iron bar model, then there would be no puzzle to solve. :rolleyes:
But but but, according to section III D, Bell's inequality is just a special case of the mighty EBBI, how come? Here comes a bit of cheating: equations (49) are subtly different from Bells original derivation. Namely, there's a hidden symmetry assumption. The three equations (49) represent the probabilities of getting outcome (s,s') in three measurements: (_,^), (_,~) and (^,~) respectively. Note that measurement (^) is done on the right side of stick in the first equation and on the left side of stick in the third. Of course one can re-shuffle the order but there is no consistent way to assign "left" and "right" ends to measurements (_,^,~), at least one of them has to be done on both ends of the stick. Note that the same function [itex]\widehat{f}^{(1)}[/itex] is used for probabilities on both ends of the stick. That implies the symmetry with respect to measurement (^). On other words, if we make the same measurement on both sides, the correlation will always be non-negative.

In Bell's case when the same measurement is done on both ends, the outcomes are perfectly anti-correlated, that is P(a,a) = -1. This was made explicit at the very beginning: [itex]A(a,\lambda)=-B(a,\lambda)[/itex]. If we try this, eq (50) will have '+' when used on the left side of the stick and '-' when used on the right side, and instead of (49) we will get something like [itex]f^{(2)}(S,S')=\frac{1-SE_{1}^{(2)}+S'E_{2}^{(2)}-SS'E^{(2)}}{4}[/itex] (note the minuses) and the construction of non-negative [itex]f^{(3)}(S,S',S'')[/itex] no longer works.
That's a good one, I think that you are right about that!
However, in most discussions of Bell's Theorem this is not regarded as essential for the inequality, because an appropriate rotation of reference angle on one side turns a perfect anti-correlation into a perfect correlation. Also in the little debate in this thread between Billschneider and DrC they agreed on an example with a perfect correlation. You could emulate that for your iron bar example with a +/- inverter on one side.
Therefore I doubt that it's pertinent, perhaps someone else can comment?
[rearranging:] Regarding section VII A (doctors/patients): the example is meant to be a model of Bell's experiment and to show the apparent violation of Bell's inequation (assuming that patients a,b and c correspond to angle settings, locations 1 and 2 correspond to the two ends of the apparatus and day of week is [itex]\lambda[/itex]. However the setup violates the anti-correlation assumption [itex]A(a,\lambda)=-B(a,\lambda)[/itex] as can be seen from the truth table, therefore Bell's inequality is not applicable.
From the truth table of this example and following their second variation, I see a perfect anti-correlation in Lille and Lyon for each patient. That is in agreement with the text:

"After lengthy discussions they conclude that there must be some influence
at a distance going on and the outcomes depend on the exams in both Lille and
Lyon such that a single outcome manifests itself randomly in one city and that
the outcome in the other city is then always of opposite sign.
"
The same applies to section section VII B. (Factorizable model): the expectation [itex]E^{(2)}(a,a)[/itex] is neither 1 nor -1, which means measurements with the same settings on both ends are neither perfectly correlated nor perfectly anti-correlated. This violates the assumption of Bell's derivation so Bell's inequation is not applicable.
It sounds to me that you give here a precision to their introductory disclaimer that "The models that we consider in this subsection do not pretend to account for the correlations of two spin-1/2 particles in the singlet state but provide further illustrations of the ideas presented above."
So the net result is that EBBI use different set of assumptions from Bell's. There are cases when Bell is applicable and EBBI is not and vice versa. In particular, in case of EPR experiment Bell's inequality applies while EBBI does not. Same thing happens if the whole EPR setup is replaced with iron bar argument. Overall, while the math is a bit flakey but sort of OK, the conclusions drawn from it are just outright wrong. [..]
What De Raedt et al argue, I think, is that Bell's inequalities are a restriction of EBBI (eq.16) and that EBBI can never be broken. You could convince me that that is wrong by giving real detailed data that break EBBI and not Bell's inequality.

Regards,
Harald
 
  • #139
Delta Kilo said:
Now I'm going to pull out my argument. It's a long and heavy iron bar with "1" permanently etched on one end and "-1" on another. When measured it always produces the same pair of values (1,-1) for any values of parameters. This sophisticated lab equipment fulfills all assumptions of the original Bell's theorem. All probabilities and expectation values exist at all times, P(a,b) = P(a,c) = P(b,c) = -1 and Bell's inequality is satisfied.

Let's apply the argument to section II of the paper in question. To do this we need to find some triples {a,b,c} so that the pairs (a,b), (a,c) and (b,c) simultaneously match experimental data (1,-1). Clearly no such triple can be constructed. ...

Not sure I follow this line of thinking. The Alice triple for your iron bar is {1, 1, 1} and the Bob triple is {-1. -1, -1}. That is what I would call a realistic data set. I don't think this changes any of your conclusions though.
 
  • #140
DrChinese said:
Not sure I follow this line of thinking. The Alice triple for your iron bar is {1, 1, 1} and the Bob triple is {-1. -1, -1}. That is what I would call a realistic data set. I don't think this changes any of your conclusions though.

Well, no. There is no Alice and no Bob. [itex]\{ S_{1,\alpha}, S_{2,\alpha}, S_{3,\alpha} \}[/itex] is supposed to be a triple of "counterfactual" outcomes form a single experiment. The reasoning goes like that: we measure [itex]1[/itex] and [itex]2[/itex] getting [itex]S_{1,\alpha}[/itex] and [itex]S_{2,\alpha}[/itex], but if we had measured [itex]1[/itex] and [itex]3[/itex] instead we would have got [itex]S_{1,\alpha}[/itex] and [itex]S_{3,\alpha}[/itex] with the same [itex]\alpha[/itex].

Eg. let's look at the paper, section II eq(1), (11) and (13):

[itex]\Gamma^{(n)}=\{(S_{1,\alpha},...,S_{n,\alpha})| \alpha=1,...,M\}[/itex]

[itex]F^{(3)}_{ij} = \frac{1}{M} \sum^{M}_{\alpha=1} S_{i,\alpha}S_{j,\alpha}[/itex]

[itex]\left| F^{(3)}_{ij} \pm F^{(3)}_{ik} \right| \leq 1 \pm F^{(3)}_{jk} [/itex]

and compare it to Bell's:

[itex]P(\vec{a},\vec{b}) = \int d \lambda \rho(\lambda) A(\vec{a},\lambda) B(\vec{b},\lambda) = -\int d \lambda \rho(\lambda) A(\vec{a},\lambda) A(\vec{b},\lambda)[/itex]

[itex]1 + P(\vec{b},\vec{c}) \geq \left|P(\vec{a},\vec{b}) - P(\vec{a},\vec{c}) \right| [/itex]

Well, ignoring the sign difference for the moment (due to Bell's case being anti-symmetrical), we can see that [itex]F^{(3)}_{ij}[/itex] corresponds to [itex]P(\vec{a},\vec{b})[/itex]. [itex]F^{(3)}_{ij}[/itex] is an average of products [itex]S_{i,\alpha}S_{j,\alpha}[/itex] for all [itex]\alpha[/itex]. Similarly [itex]P(\vec{a},\vec{b})[/itex] is an average of [itex]A(\vec{a},\lambda) A(\vec{b},\lambda)[/itex] over all [itex]\rho(\lambda)d\lambda[/itex]. If we push it further, we'll find that [itex]\{ S_{1,\alpha}, S_{2,\alpha}, S_{3,\alpha} \}[/itex] for a given [itex]\alpha[/itex] should be identified with something like [itex]\{ A(\vec{a},\lambda), A(\vec{b},\lambda), A(\vec{c},\lambda) \}[/itex] for a given [itex]\lambda[/itex]. Well, the whole thing seem to revolve around the question: "does such a triple of counterfactual values, which can never be obtained simultaneously, make sense"?

Here the math stops and vague verbose arguments start. The authors have this to say:
The relation of Bell’s work to Theorems II and IV shows the mathematical solidity and strength of Bell’s work. It also shows, however, the Achilles heel of Bell’s interpretations:
Because [itex]\lambda[/itex] has a physical interpretation representing an element of reality, Eq. (49) implies that in the actual experiments identical [itex]\lambda[/itex]’s are available for each of the data pairs (1,2), (1,3), (2,3). This means that all of Bell’s derivations assume from the start that ordering the data into triples as well as into pairs must be appropriate and commensurate with the physics.

This “hidden” assumption was never discussed by Bell and his followers5 and has “invaded” the mathematics in an innocuous way. Once it is made, however, the inequalities Eq. (30) apply and even influences at a distance cannot change this.
Well, these statements are simply not true, especially the last one. The authors have missed an elephant in the room here. The existence (in theory) of [itex]\{ A(\vec{a},\lambda), A(\vec{b},\lambda), A(\vec{c},\lambda) \}[/itex] follows directly from Bell's assumption of local realism, that is from the definition of [itex]A(\vec{a},\lambda)[/itex]. Basically independence of [itex]A(a,\lambda)[/itex] from settings on the other end means we can measure [itex]b[/itex] but could have measured [itex]c[/itex] and the result for [itex]a[/itex] would have been the same. As a result we can write joint probability distribution [itex]\rho(s_{1},s_{2},s_{3}|abc)=\int \rho(\lambda) \frac{1+s_{1} A(a,\lambda)}{2}\frac{1+s_{2} A(b,\lambda)}{2}\frac{1+s_{3} A(c,\lambda)}{2} d\lambda[/itex].
Again, this is exactly what local realism is all about. It is not a hidden assumption and it is not true in general if local realism condition is violated. Testing this assumption is exactly what Bell tests are about.

At the same time it is essential that [itex]\{ S_{1,\alpha}, S_{2,\alpha}, S_{3,\alpha} \}[/itex] and [itex]\rho(s_{1},s_{2},s_{3}|abc)[/itex] are purely mathematical artefacts with no physical meaning attached to them. The triples, which might exist in theory, can never be measured (nor do they need to be). The result of Bell's derivation contains only expectations of pairwise correlations P(a,b), P(a,c) and P(b,c), and these can be easily measured in separate experiments.

What this all means is in general EBBI are not appicable to all experiments with 2 outcomes exactly because the triples are not always guaranteed to exist together. It is Bell's local realism assumption (plus hidden symmetry assumption) which provides sufficient condition for such triples to exist.

However the authors failed to realize this connection. They show numerous examples where either the symmetry or the local realism assumptions are violated and where EBBI are violated as a result.

Since the authors [STRIKE]blindly refuse to see[/STRIKE] do not see [STRIKE]the elephant in the room[/STRIKE] the connection between local realism assumption and the separability of outcomes/existence of triples/joint probability, the austors wrongly conclude that Bell is just a consequence of EBBI and since EBBI are sometimes violated, then Bell's inequality violation is also not such a big deal.

Here is another quote:
Last but not least we note that John Bell4 based his famous theorem on two assumptions: (a) Bell assumed in his original paper by the algebraic operations of his Eqs. (14) – (22) and the additional assumption that his [itex]\lambda[/itex] represents elements of reality a clear grouping into triples because he implies the existence of identical elements of reality for each of the three pairs. (b) By the same operations Bell assumed that he deals with dichotomic variables that follow the algebra of integers. From our work above it is then an immediate corollary that Bell’s inequalities cannot be violated; not even by influences at a distance.
Again the authors just didn't get it. It is not the assumption of [itex]\lambda[/itex] that makes a difference, it is local realism, that is the shape of [itex]A(a,\lambda)[/itex]. And the potential grouping into triplets is not an additional hidden assumption of Bell's but direct algebraic consequence of local realism. And Bell's inequalities are most certainly violated by influences at a distance. QM provides such a model.


, Doc, you forced me to go through this paper again. It's a mess. It hurts my brain.

More later
DK
 
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