Third Loophole Against Entanglement Eliminated

[billschnieder]

Well, for the spin-1/2 twin pair EPR experiment, the prediction of QM is

C(a,b) = -cos(b-a)

So that inequality becomes:

|-cos(b-a) + cos(c-a)| ≤ 1 - cos(b-c)

But for the twin pair, according to QM, C(a,b) does not commute with C(c,a) nor does it commute with C(b,c). In other words, the three correlations are incompatible. Therefore, although each of the correlations standing alone is a valid QM calculation, combining the three into a single expression gives an invalid QM expression. Unless you introduce the added assumption that commuting observables can be substituted into an inequality based on non-commuting terms without regard for the non-commutatitivity. Another way of see the error is that, you are using three correlations calculated on three different wavefunctions to draw conclusions about three correlations that would have been obtained from a single wavefunction, were it possible to measure them.

This subtle error is the source of the violation, not non-locality or any other spooky business.

That is why I asked the following questions:
1) Do you agree that there are two scenarios involved in this discussion:

Scenario X, involving the three correlations:

C(a,b) = correlation for what we would get if we measure (a,b)
C(b,c) = correlation for what we would get if we measure (b,c)
C(a,c) = correlation for what we would get if we measure (a,c)​

Scenario Y, involving the three correlations:

C(a,b) = correlation for what we would get if we measure (a,b)
C(a,c) = correlation for what we would have gotten had we measured (a,c) instead of (a,b)
C(b,c) = correlation for what we would have gotten had we measured (b,c) instead of (a,b)​

2) Do you agree that Scenario X is different from Scenario Y?
3) Do you agree that the correlations in Bell's inequalities correspond to Scenario Y NOT Scenario X?
4) Do you agree that correlations calculated from QM correspond to Scenario X not Scenario Y?
5) Do you agree that correlations measured in experiments correspond to Scenario X not Scenario Y?

While the correlations in Scenario X all commute, those in Scenario Y do not all commute.

 

[billschnieder]

You claimed that my error was hidden by the use of inequalities, so I switched to equations. And then your criticism seemed to be that I was being selective in my application of no-conspiracy to only certain probabilities. But that's no mystery: the no-conspiracy condition states that if a statement is meaningful in both "scenarios" (to use your terminology), i.e. if it is meaningful whether you restrict yourself to factual statements or whether you also allow counterfactuals, then the probability of the statement being true is the same in both scenarios. If you'd like, I can explain again the justification for this no-conspiracy condition.

Your no-conspiracy condition is essentially that Scenario X and Scenario Y (from above) are exactly the same, in other words, your no-conspiracy condition is equivalent to saying, the QM result from a single wavefunction must be the same as the QM result from three different wavefunctions.

your argument was:

OK, let me be more explicit in my logic and not use inequalities at all.
1. P(C|w)=P(A|w)+P(B|w)-2P(A & B|w)
2. P(A|x)=.25, P(B|y)=.25, P(C|z)=.75
3. P(A|x)=P(A|w), P(B|y)=P(B|w), P(C|z)=P(C|w) (From no-conspiracy.)
4. P(A|w)=.25, P(B|w)=.25, P(C|w)=.75 (From 2 and 3)
5. .75 = .25 + .25 -2P(A & B|w) (From 1 and 4)
6. P(A & B|w) = -.125 (From 5)

.

And I showed you that step (3) was incomplete, Step (3) What does no-conspiracy say about P(AB|w). According to your logic, no-conspiracy also implies that P(AB|w)=P(AB|x,y). But x and y are two different sets of photons, which means P(AB|x,y) is undefined/meaningless. All you have proven is the triviality that the joint probablity distribution P(ABC|x,y,z) for outcomes from three different sets of photons (x,y,z) is undefined, although the joint probability distribution P(ABC|w) from the single set of photons (w) is well defined. So not unlike what I explained in my previous post, the error is to assume that non-commuting observables can be mixed as if there were commuting observables in the inequality. It is a mathematical error.

 

[DrChinese]

...

That is why I asked the following questions:
1) Do you agree that there are two scenarios involved in this discussion:

Scenario X, involving the three correlations:

C(a,b) = correlation for what we would get if we measure (a,b)
C(b,c) = correlation for what we would get if we measure (b,c)
C(a,c) = correlation for what we would get if we measure (a,c)​

Scenario Y, involving the three correlations:

C(a,b) = correlation for what we would get if we measure (a,b)
C(a,c) = correlation for what we would have gotten had we measured (a,c) instead of (a,b)
C(b,c) = correlation for what we would have gotten had we measured (b,c) instead of (a,b)​

2) Do you agree that Scenario X is different from Scenario Y?
3) Do you agree that the correlations in Bell's inequalities correspond to Scenario Y NOT Scenario X?
4) Do you agree that correlations calculated from QM correspond to Scenario X not Scenario Y?
5) Do you agree that correlations measured in experiments correspond to Scenario X not Scenario Y?

While the correlations in Scenario X all commute, those in Scenario Y do not all commute.

Bill, you already posted this last line of reasoning in another thread - in fact this is verbatim. That thread is closed for moderation. Please do not continue to post your personal pet arguments that you cannot support with suitable* references. Else you will again be reported. PhysicsForums' Quantum Physics is NOT the place to make arguments that go counter to generally accepted physics. Start your own blog or get one of your papers published. This is an educational** forum first and foremost.

*By PF standards.
** I realize you believe you are educating people by sharing your version of the truth, but that is not how it is defined here.

 

[stevendaryl]

But for the twin pair, according to QM, C(a,b) does not commute with C(c,a) nor does it commute with C(b,c).

C(a,b) is an expectation value, a real number, not an operator. Real numbers always commute.

 

[billschnieder]

C(a,b) is an expectation value, a real number, not an operator. Real numbers always commute.

Huh? it is the result of an operation, the operations do not commute. If the time to put on socks is a real number Ta, and the time it takes to put on shoes is another real number Tb. Won't it be silly to suggest that becase Ta, and Tb are real numbers, then you should say the time to put on Shoes and then put on socks must be Tb + Ta? Don't you realize that because the two operations do not commute you can not simply add the times like that?

 

[stevendaryl]

Huh? it is the result of an operation, the operations do not commute.

Expectation values always commute.

C(a,b) = -cos(b-a)

That is not an operator, it is a real number, for every choice of a and b.

 

[stevendaryl]

Huh? it is the result of an operation, the operations do not commute. If the time to put on socks is a real number Ta, and the time it takes to put on shoes is another real number Tb. Won't it be silly to suggest that becase Ta, and Tb are real numbers, then you should say the time to put on Shoes and then put on socks must be Tb + Ta? Don't you realize that because the two operations do not commute you can not simply add the times like that?

I think you're drifting off into territory where I have nothing to say. You asked me how I would calculate the QM prediction for correlations, and I gave you an answer. I don't think there is anything more for me to say. Correlations are real numbers, and they can be added in any order, and you get the same answers. It's a separate question of what the meaning of an arithmetical expression.

 

[billschnieder]

Correlations are real numbers, and they can be added in any order, and you get the same answers. It's a separate question of what the meaning of an arithmetical expression.

That is where you are mistaken. Do you have a reference for that claim? I just gave you and example with shoes and socks which clearly demonstrates that you are wrong.

On the Problem of Hidden Variables in Quantum Mechanics
John S. Bell. Reviews of Modern Physics, Vol 38, Number 3, (1966)

Page 448:
"A quantum mechanical "system" is supposed to have "observables" represented by Hermitian operators in a complex linear vector space. Every "measurement" of an observable yields one of the eigenvalues of the corresponding operator. Observables with commuting operators can be measured simultaneously."
...
"Any real linear combination of any two Hermitian operators represents an observable, and the same linear combination of expectation values is the expectation value of the combination".

Page 449
"It was not the objective measurable predictions of quantum mechanics which ruled out hidden variables. It was the arbitrary assumption of a particular (and impossible) relation between the results of incompatible measurements either of which might be made on a given occasion but only one of which can in fact be made."

 

[stevendaryl]

That is where you are mistaken. Do you have a reference for that claim?

The claim that real numbers commute? You want a reference for that?

I just gave you and example with shoes and socks which clearly demonstrates that you are wrong.

No, it doesn't. If T1 = 5 and T2 = 3, then T1 + T2 = 8. There is no issue of whether operators commute.

You've gone off the deep end, here.

 

[DrChinese]

... I just gave you and example with shoes and socks which clearly demonstrates that you are wrong.

Great, Bill. Assuming we have settled some point, none of this has anything to do with the topic of this thread - which is closing another Bell test "loophole".

You don't even seem to object to Bell tests on the grounds of "loopholes" anyway, so I am not sure what you really have to contribute here. Please do not turn this into a discussion of the soundness of Bell's Theorem. If you feel the need to continue, perhaps you could start a thread on the point you have a question about.

 

[billschnieder]

The claim that real numbers commute?

No, the claim that you can add results from non-commuting operations as if they were commuting operations and expect to have an expression that is meaningful in any way. BTW you had exactly nothing to say about Bell's argument which I quoted to you.

 

[stevendaryl]

No, the claim that you can add results from non-commuting operations as if they were commuting operations and expect to have an expression that is meaningful in any way.

If the results are real numbers, then you can add them.

BTW you had exactly nothing to say about Bell's argument which I quoted to you.

That's right, I had exactly nothing to say about it.

 

[DevilsAvocado]

1) Do you agree that there are two scenarios involved in this discussion:

Scenario X, involving the three correlations:

C(a,b) = correlation for what we would get if we measure (a,b)
C(b,c) = correlation for what we would get if we measure (b,c)
C(a,c) = correlation for what we would get if we measure (a,c)​

Scenario Y, involving the three correlations:

C(a,b) = correlation for what we would get if we measure (a,b)
C(a,c) = correlation for what we would have gotten had we measured (a,c) instead of (a,b)
C(b,c) = correlation for what we would have gotten had we measured (b,c) instead of (a,b)​

For the first time it looks like there’s something coherent...

Scenario X => QM theory => EPR-Bell experiments => what actually works
Scenario Y => assumptions of Local Realism => definite values all the time => CFD => not working

For clarity we exchange Scenario X = QM theory and Scenario Y = Local Realism, and then it’s easy to see where it goes wrong.

2) Do you agree that Scenario X is different from Scenario Y?

Yes of course, it’s big difference between QM theory and Local Realism.

3) Do you agree that the correlations in Bell's inequalities correspond to Scenario Y NOT Scenario X?

Yes of course, Bell's inequalities sets the upper limit for what any model of Local Realism can achieve.

4) Do you agree that correlations calculated from QM correspond to Scenario X not Scenario Y?

Yes of course, calculated correlations of QM theory naturally will correspond to QM theory.

5) Do you agree that correlations measured in experiments correspond to Scenario X not Scenario Y?

Yes of course, when we perform QM experiments in form of EPR-Bell test experiments, it will naturally correspond to QM theory.

Do you now see the issue?

No I don’t see the issue, but I can guess what windmills you are tilting.

Let me try to “read your mind”:

To make Bell's inequalities a valid test of Local Realism, we need to be able to perform experiments that delivers *real data* for the inequalities of Local Realism, i.e. we do not necessarily need to violate Bell's inequalities, but just get the data in our hands. If we can’t do this, then it’s impossible to say that we have violated Bell's inequalities for Local Realism, because there is absolutely nothing there to violate, just a vacuum of buffoonery and mysticism!​

A lot of people have spent a lot of time trying to show you exactly how erroneous this “counterfactual logic” of yours really is. This is what you fail to see:

1. All modern science builds on the framework; Idea -> Mathematical model -> Empirical test -> Yes/No. Do you get this??
   
   
2. We can build a mathematical model of the properties that Local Realism must possesses to be a model of local realism, and one of these properties is that every value (LHV) in the model naturally has to be definite from the beginning. Do you get this??
   
   
3. This definite property of LHV unconditionally leads to Counterfactual definiteness (CFD). Do you get this??
   
   
4. In QM, non-commuting operators cannot simultaneously be known precisely, which mean that definiteness of non-commuting values [before measurement] is impossible. Do you get this??
   
   
5. According to the proponents of local realism the problem with non-commuting operators and the [current] inability to simultaneously/precisely measure counterfactual values could be due to the incompleteness of QM and/or [current] shortcomings in measurement techniques. Do you get this??
   
   
6. Bell's inequalities is a mathematical model for setting the upper boundaries what Local Realism and LHV/CFD could ever produce when it comes to particle correlations between space-like separated measurements. Do you get this??
   
   
7. There is nothing in Bell’s theorem, or any other place, that states that we must be able to simultaneously measure all the values that counterfactual definiteness represents in Local Realism. In fact, this would require the potential to measure non-commuting operators precisely, which would mean that the measurement would refute QM, which would mean that if someone could actually perform this measurement he/she would get an instant Nobel Prize in Physics. Do you get this??

Sum up:

• You are tilting windmills while painting yourself into a corner.
• If you rebut counterfactual definiteness, you also rebut local realism.
• If you admit counterfactual definiteness, your “counterfactual logic” leads to the bizarre situation where local realism could never be scientifically tested, which automatically puts local realism in the metaphysical/philosophical/religious domain.
• Your corner is getting narrower and narrower and narrower...

DO YOU GET THIS??

 

[billschnieder]

DA, This discussion has to continue in a different thread if you are interested.

 

[lugita15]

Your no-conspiracy condition is essentially that Scenario X and Scenario Y (from above) are exactly the same,

Yes, the no-conspiracy condition says that if a statement is meaningful in both scenarios, then the probability is equal for both scenarios.

in other words, your no-conspiracy condition is equivalent to saying, the QM result from a single wavefunction must be the same as the QM result from three different wavefunctions.

Where in the world did you get that from? We're talking about different possible measurements we could perform on a system with the same wavefunction. We're not talking about different wavefunctions.

And I showed you that step (3) was incomplete, Step (3) What does no-conspiracy say about P(AB|w). According to your logic, no-conspiracy also implies that P(AB|w)=P(AB|x,y).

No-conspiracy states that if a statement S is meaningful in both x and w, then P(S|w)=P(S|x) (and similarly for y and z). But A & B is not meaningful in x, so no-conspiracy doesn't tell you anything in this case.

Also, what do you mean by P(A & B|x,y)? Do you mean a combined space which is the union of x and y? Well, my reasoning doesn't talk about combined spaces like that. It only discusses x, y, z, and w.

But x and y are two different sets of photons, which means P(AB|x,y) is undefined/meaningless

Again, I didn't say anything about P(A & B|x,y).

. All you have proven is the triviality that the joint probablity distribution P(ABC|x,y,z) for outcomes from three different sets of photons (x,y,z) is undefined, although the joint probability distribution P(ABC|w) from the single set of photons (w) is well defined.

I didn't say anything about P(A & B & C|x,y,z). And again, since A & B & C is meaningless in x, y, or z, the no-conspiracy condition says nothing in this case either.

 

[harrylin]

Upon my comments concerning Bell's sweeping claims:

What is the more general conception of locality you have in mind? Regardless, it's possible to give a proof of Bell's theorem with this meager definition of locality: events can only influence events within their future light cone.

Yes, I just replied to DevilAvocado. To sum up, in principle the term "counterfactual definiteness" COULD refer to something more general, but for the purposes of Bell's theorem all we need is the meaningfullness of asking what a measurement that you didn't make would yield if you had made it. [..]

I had in mind to reply in a more general thread, but I see that today Dadface has started a new topic on just this same issue:
https://www.physicsforums.com/showthread.php?t=697939

I hope to comment there soon, after contemplating on a few more papers.