Scholarpedia article on Bell's Theorem

[ThomasT]

Hi TT, What is it about my argument you do not understand?

Hi Bill, ok, first, what is your argument? Are you saying that Bell's math is wrong? Or are you saying that concluding that nature is nonlocal from Bell's math (and experimental violations of BIs) is unwarranted? If the latter, then we're sort of on the same page. But maybe not really, because it seems that you're approaching the consideration in a different way than I am.

Which is not to say that there aren't multiple legitimate ways of approaching the question. It's just that I don't think I fully understand your approach.

My current hypothesis is that Bell's locality assumption/condition isn't, strictly speaking, as it's encoded in LR models of entanglement, exclusively and uniquely a locality assumption. Which, if true, entails that BI violations don't necessarily inform wrt the underlying reality.

 

[billschnieder]

Hi Bill, ok, first, what is your argument? Are you saying that Bell's math is wrong?

This is what I said back in post #104:

And just to be clear, I do not believe there is an error in the proof. The are two errors:

1 - Thinking that the terms from QM could be meaningfully plugged into the LHS of the CHSH.
2 - Thinking that the terms from Experiments could be meaningfully plugged into the LHS of the CHSH.

Now if Bell proponents insist that Bell's math is attempting to model exactly the way the experiments are performed, then one could argue that the math will be the wrong model of the experiment unless addition assumptions are made. ie that a1=a2, b1=b3, c2=c3, and also that ρ(λ) = ρ(a,b,λ). In that case, I would argue that these assumptions are unreasonable.

So the argument goes like this:

1) Bell's math as derived is essentially correct mathematically, but does not correspond to the QM or experimental situation
2) Even if we grant (for argument sake) that the math corresponds to QM and the experimental situation, you will need to make the unreasonable assumptions that a1=a2, b1=b3, c2=c3, and ρ(λ) = ρ(a,b,λ)

Therefore no matter how you look at it, you can not draw any conclusion from violation of Bell's math by QM and Experiments about the real physical situation happening in the experiments.

Or are you saying that concluding that nature is nonlocal from Bell's math (and experimental violations of BIs) is unwarranted? If the latter, then we're sort of on the same page.

I think we are on the same page.

 

[ThomasT]

This is what I said back in post #104:

And just to be clear, I do not believe there is an error in the proof. The are two errors:

1 - Thinking that the terms from QM could be meaningfully plugged into the LHS of the CHSH.
2 - Thinking that the terms from Experiments could be meaningfully plugged into the LHS of the CHSH.

Now if Bell proponents insist that Bell's math is attempting to model exactly the way the experiments are performed, then one could argue that the math will be the wrong model of the experiment unless addition assumptions are made. ie that a1=a2, b1=b3, c2=c3, and also that ρ(λ) = ρ(a,b,λ). In that case, I would argue that these assumptions are unreasonable.

So the argument goes like this:

1) Bell's math as derived is essentially correct mathematically, but does not correspond to the QM or experimental situation
2) Even if we grant (for argument sake) that the math corresponds to QM and the experimental situation, you will need to make the unreasonable assumptions that a1=a2, b1=b3, c2=c3, and ρ(λ) = ρ(a,b,λ)

Therefore no matter how you look at it, you can not draw any conclusion from violation of Bell's math by QM and Experiments about the real physical situation happening in the experiments.

I think we are on the same page.

Maybe, maybe not. Why is the assumption that ρ(λ) = ρ(a,b,λ) unreasonable? It seems reasonable to me. Also, a, b and c are just polarizer settings, aren't they? So, why is it unreasonable to say that a1 (the polarizer setting in one run) is the same as a2 (the same polarizer setting in another run) ... and the same for b and c?

My focus is currently on the locality condition. Is it, necessarily, a locality condition? Jarrett apparently doesn't think so. And until Dr. Norsen tells me why I shouldn't agree with Jarrett's parsing and conclusion, then that's how I'm thinking about this. In other words, there is an effective cause wrt BI violations, and it has to do with the formal explication of the locality (independence) assumption, wrt which BIs are effectively violated due to a disparagement between the LR formalism and the design and execution of optical Bell tests vis the encoding of the locality assumption/condition, and therefore it can't be concluded from experimental BI violations that nature is nonlocal. In other words, wrt the question of whether or not nature is nonlocal, Bell test results are ambiguous because of the ambiguity of the encoded locality assumption/condition.

 

[lugita15]

If you haven't become too tired of this, then might you simplify and synopsize (preferably in ordinary language) how/why the formalized Bell locality condition/assumption can only be violated due to the fact that nature is nonlocal, and not due to some other, more mundane, reason (such as a more or less trivial incompatibility between the formalization of the locality assumption, and the design and execution of Bell tests)?

ThomasT, I've been trying to do this for you for a while now, with my restatement of Herbert's proof. What step do you disagree with now? I have said that the only remotely nontrivial step in the argument is the step from 1 to 2. You have disagreed with step 3, but as I explained it follows directly from step 2 and the transitive property of equality. So we're back to step 2, and the only criticism you've leveled against it is that its wording is too anthropomorphic, but I responded that you can easily replace "the particles agree in advance what angles to go through and not go through" with "it is determined in advance what angles the two particles will go through or not go through". With that rewording, do you still disagree in the logic from step 1 to step 2? (You may want to reply back in the other thread, to keep this thread uncluttered).

 

[Delta Kilo]

Now if Bell proponents insist that Bell's math is attempting to model exactly the way the experiments are performed, then one could argue that the math will be the wrong model of the experiment unless addition assumptions are made.

But the proof is not supposed to "model exactly the way the experiments are performed". It is a black box. If you agree with assumptions, you have to accept the conclusions. You cannot argue against it because some intermediate step in derivation does not have a clear physical meaning. Yes, there is a step in Bell's proof where there is a product of A(a,λ)A(b,λ)A(c,λ) under the integral, so what? It is just an intermediate step, it does not have to have physical meaning. Specifically, it does not introduce any new physical assumptions. The only assumptions are those used to formulate initial steps of the derivation.

Second, some people apparently confuse the requirement of 'having the same ρ(λ)' with 'having the same λ' when carrying out measurements C(a,b), C(a,c), C(b,c). Bell requires the former, but not the latter.

Third, some people confuse expectation value $E[X] = \sum \rho_i X_i$ and arithmetic mean $\overline{X}=\frac{1}{N}\sum X_i$. Bell's inequality applies only to the former and not the latter. Which means you do NOT apply Bell's inequality to individual measurement outcomes. Instead, you use experimental data to estimate expectation values, and you plug those expectation values (along with their estimated standard deviations) into Bell's inequality.

PS: We've been through that several times before.

 

[ThomasT]

ThomasT, I've been trying to do this for you for a while now, with my restatement of Herbert's proof. What step do you disagree with now? I have said that the only remotely nontrivial step in the argument is the step from 1 to 2. You have disagreed with step 3, but as I explained it follows directly from step 2 and the transitive property of equality. So we're back to step 2, and the only criticism you've leveled against it is that its wording is too anthropomorphic, but I responded that you can easily replace "the particles agree in advance what angles to go through and not go through" with "it is determined in advance what angles the two particles will go through or not go through". With that rewording, do you still disagree in the logic from step 1 to step 2? (You may want to reply back in the other thread, to keep this thread uncluttered).

Hi lugita. How do you go from step 2 to step 3? Honestly, I don't understand how Herbert concludes that nature is nonlocal. In one sentence he's saying that it's assumed that events at A and B are independent. And then he proceeds to calculate the expected results in a way that I can't connect to his assumption of locality (independence). Same for your restatement of Herbert's proof.

Not insignificantly, Herbert makes the statement that no local reality can explain the facts of optical Bell tests, and therefore reality must be nonlocal.

But that statement is misleading. It should be phrased that no Bell-LR model of quantum entanglement (not "no local reality") can correctly predict the results of optical Bell tests.

 

[ThomasT]

If you agree with assumptions, you have to accept the conclusions.

This is the crux of the matter, imo. The assumption of locality is, as Bell and Dr. Norsen have indicated, the crucial assumption. This assumption is encoded/formalized in a certain way. It's been shown by Jarrett and others that this formalization is ambiguous wrt the assumption of locality, that is, this form also contradicts the design and execution of optical Bell tests in a way that might have nothing to do with whether nature is local or nonlocal.

 

[lugita15]

Hi lugita. How do you go from step 2 to step 3? Honestly, I don't understand how Herbert concludes that nature is nonlocal. In one sentence he's saying that it's assumed that events at A and B are independent. And then he proceeds to calculate the expected results in a way that I can't connect to his assumption of locality (independence). Same for your restatement of Herbert's proof.

Not insignificantly, Herbert makes the statement that no local reality can explain the facts of optical Bell tests, and therefore reality must be nonlocal.

But that statement is misleading. It should be phrased that no Bell-LR model of quantum entanglement (not "no local reality") can correctly predict the results of optical Bell tests.

I've replied back in the other thread here, so that we don't clutter this thread that's supposed to be about ttn's article.

 

[harrylin]

Well, now it's my turn to quibble about terminology. If "dogma" means "anything different from what the herd endlessly repeats", then yes, our article is "dogmatic". [..]

Exactly - and I expect better from a good encyclopedia than proclaiming the POV of the herd, instead I require a fair and neutral presentation of POV's. But perhaps you hold that we already have Wikipedia for that purpose.

[..] you can still appreciate that it's a good thing that there now exists a thorough and careful treatment of Bell's theorem from this particular POV.

Certainly! Each encyclopedia (even each article) has its strong and weak points, and it suffices to be aware of them - for example Wikipedia is generally a mess, but happily it exists.

As to von Neumann, I think it's fair to say that if anybody had scrutinized his theorem as carefully as we scrutinize, in our article, the reasoning involved in Bell's theorem, the theorem would have been "disproved" much earlier. (Of course, really it's not the theorem that was disproved -- just its significance vis a vis "hidden variables".) [..].

I interpret that as a clear overestimation of your group's abilities; which fits with the impression that I already got from the introduction.
Anyway, it looks like a nice overview from that point of view, and I'll read on!

 

[ttn]

My focus is currently on the locality condition. Is it, necessarily, a locality condition? Jarrett apparently doesn't think so. And until Dr. Norsen tells me why I shouldn't agree with Jarrett's parsing and conclusion, then that's how I'm thinking about this.

http://arxiv.org/abs/0808.2178

 

[mattt]

First of all, the inequality is dealing the total numbers of mismatches not averages

This must be a joke, right?

 

[mattt]

In the CHSH-Bell Inequality Theorem:

In a experiment, at both ends each team may choose to measure a,b,c or d to each particle that is coming. The measured result of each measure is +1 or -1.

At the end, they may have recorded, let's say, 1000000 of measured pairs (1,-1), (1,1), (1,1), (-1,-1), (-1,1), (1,-1), (1,-1), (-1,1),...

It may be the case that 105000 of the 1000000 measured pairs correspond to the setting (a,b), other 99000 to the setting (a,c), other 85000 to the setting (d,b), other 450000 to the setting (d,c), and the 261000 remaining measured pairs correspond to the other possible settings ( (a,a), (b,a), (b,b), (c,a), (c,c)...).

If the 105000 measured pairs that correspond to the setting (a,b) are (1,-1), (-1,1), (1,-1), (1,1), (-1,-1),... then C(a,b)=(1(-1)+(-1)1+1(-1)+1*1+(-1)(-1)+...)/105000

And the same with the other terms C(a,c), C(d,b), C(d,c).

 

[DrChinese]

Where I actually listed all the posibilities for the 6 lists obtained in Bell test experiments and showed that violations of Bell's inequality were obtained 25% of the time. It doesn't matter if you sample 1million or 1 billion times, you will still see a violation 25% of the time. Maybe you think 75% is overwhelmingly large enough for you to declare that the inequality is obeyed but for anyone with any training in basic math, ONE counter example is enough to reject a mathematical theorem -- ONE.

That is unbelievable, even coming from you. You think there is something special about arbitrarily calling a trial one where n=10? Guess what, if you make n=1, then every trial that is a match is counter-evidence by your bizarre standards.

Similarly, you would probably reject the idea that the chance of heads after a coin flip is 50%. Any result would disprove it!

 

[rlduncan]

You and several others seem to miss some key points here. PhysicsForums is not intended as a spot to debate your original or nonstandard ideas, which clearly describes your assertion. The key idea here is to share useful knowledge, answer normal questions, etc.

I understand your concern and I have tried not to incorporate my own ideas or theories in the postings.

The examples of experiments and ideas presented in this thread are not original to me. I must give credit to others, particularly the inventor of Boolean logic, George Boole. And of course the published works of others pertaining to his ideas in which references have provided in this thread.

I would not consider Boole’s ideas to be nonstandard but the standard! This discussion via the computer is possible because of his concepts.

Best regards

 

[DrChinese]

I understand your concern and I have tried not to incorporate my own ideas or theories in the postings.

The examples of experiments and ideas presented in this thread are not original to me. I must give credit to others, particularly the inventor of Boolean logic, George Boole. And of course the published works of others pertaining to his ideas in which references have provided in this thread.

I would not consider Boole’s ideas to be nonstandard but the standard! This discussion via the computer is possible because of his concepts.

Best regards

I like Aristotle, Newton and Bohr too. But referencing those whose shoulders we stand on is not an acceptable substitute for today's generally accepted science. You can't really talk about the Bohr model when something better has come along. And If Boole had known of Bell, I am certain he would tip his hat.

Similarly for published references. A lot of things have been published, that does not make it a suitable reference here.

So please, stay with the program here. Identify as such points you are making which are not standard so other readers will understand the difference between your non-standard opinion and that which is generally accepted. I believe you clearly know the difference. Even billschnieder does, he often points out that he is right and everyone else is wrong.

 

[billschnieder]

Why is the assumption that ρ(λ) = ρ(a,b,λ) unreasonable? It seems reasonable to me. Also, a, b and c are just polarizer settings, aren't they?

ρ(λ) = ρ(a,b,λ) is unreasonable because it implies that the distribution of λ which corresponds to the measured outcomes does not depend on the angular settings on either arm of the setup (Alice or Bob). But this can not be true where coincidence counting is done.

To be clear, imagine that every pair of outcomes (Ai, Bi) obtained, corresponds to a specific λi value, ie A(ai,λi) = Ai = ±1, B(bi,λi) = Bi = ±1.

To say that ρ(λ) = ρ(a,b,λ), implies that if you would take all the λs corresponding to all the measured outcomes, their distribution will be exactly the same irrespective of whether we measured at angles (a,b) or (b,c) or (a,c). With coincidence counting and what we know classically about Malus law, this is unreasonable.

To be clear even further, it is equivalent to saying that for the setup of a single stream of particles and 2 polarizers A,B in sequence set at angles a, b, the distribution of hidden polarization parameter λ of the photons that pass through both polarizers is independent of the angles at which both are set, ie ρ(λ) = ρ(a,b,λ)

You do not need conspiracy to realize that the assumption is unreasonable in classical case of two polarizers in sequence. Coincidence counting for two separate arms does effectively the same thing because it only allows photons to be considered on one side if they passed through the other side.

 

[billschnieder]

Also, a, b and c are just polarizer settings, aren't they? So, why is it unreasonable to say that a1 (the polarizer setting in one run) is the same as a2 (the same polarizer setting in another run) ... and the same for b and c?

a1 is complete context for run1 when the angle "a" was set. a2 is the complete context for run2 when angle "a" was set. "a1" is (angle setting "a" + everything else that makes run 1 different from run 2, including time, the complete microscopic state of the device etc.). When Bell's inequality is written as

|<ab> - <ac>| <= 1 + <bc>

"a" in that expression does not represent the angle, it is a label for the list of outcomes. Similarly, "a1" in this case represents the list of outcomes when the angle was set to "a" under context 1 (run 1), and "a2" is the list of outcomes when the angle was set to "a2" for run 2. The above inequality is not valid unless the list of outcomes labeled "a1" is identical to the list of outcomes labeled "a1". In other words, the six lists of outcomes a1,a2,b1,b3,c2,c3 MUST be sortable such that we end up with just three lists a,b,c (ie, a1=a2, b1=b3,c2=c3). Note we are talking about lists of outcomes here.

My focus is currently on the locality condition.

I think it is peripheral. How else will you explain Boole's derivation, he made no locality assumption. There has been a lot of misunderstanding and confusion caused by mixing up functions with probabilities. Some people think that Bell was trying to calculate a joint probability in his equation (2) of his original paper, but he was not. He wrote:

P(a,b) = ∫A(a,λ)B(b,λ)ρ(λ)

Some have misunderstood this to be a joint probability equation which it is not. First of all A(a,λ), B(b,λ) can take up negative values contrary to probabilities. What Bell was calculating was an expectation value for the paired product of the outcome at Alice and Bob. So there is no locality here. The separability of the expectation value is simply due to the fact that a paired product is necessarily separable.

As I showed in post #123, you can derive the same inequality if you start with 3 dichotomous variables and calculate their paired products, or 3 list containing values ±1, without any additional assumption. So any "Blah blah bla" that gives you paired products of 3 dichotomous variables can be used to fool you into thinking the "Blah blah bla" is important for the inequality. It is not.

Earlier in the thread (post #101), I got ttn to admit that λ could be non-local hidden variables and you will still obtain the inequalities.

 

[billschnieder]

In the CHSH-Bell Inequality Theorem:

In a experiment, at both ends each team may choose to measure a,b,c or d to each particle that is coming. The measured result of each measure is +1 or -1.

At the end, they may have recorded, let's say, 1000000 of measured pairs (1,-1), (1,1), (1,1), (-1,-1), (-1,1), (1,-1), (1,-1), (-1,1),...

It may be the case that 105000 of the 1000000 measured pairs correspond to the setting (a,b), other 99000 to the setting (a,c), other 85000 to the setting (d,b), other 450000 to the setting (d,c), and the 261000 remaining measured pairs correspond to the other possible settings ( (a,a), (b,a), (b,b), (c,a), (c,c)...).

If the 105000 measured pairs that correspond to the setting (a,b) are (1,-1), (-1,1), (1,-1), (1,1), (-1,-1),... then C(a,b)=(1(-1)+(-1)1+1(-1)+1*1+(-1)(-1)+...)/105000

And the same with the other terms C(a,c), C(d,b), C(d,c).

But this is not how C(a,b), C(a,c), C(d,b) and C(d,c) are defined within the inequality. They are defined such that the ABSOLUTELY CRUCIAL integration variable λ, is identical for all the terms. In other words, if you take all the individual lambda values from all cases in which the setting pair was (a,b) and all the individual lambda values from all the cases in which the setting pair was (b,c) etc, they will be identical from setting pair to setting pair. ONLY under such conditions can the inequality be derived and ONLY under this scenario are the terms you measured equivalent to the terms in CHSH inequality.

Now I ask you, is it a reasonable assumption to say that the distribution of lambda values for MEASURED pairs is IDENTICAL from setting pair to setting pair.

 

[ThomasT]

ρ(λ) = ρ(a,b,λ) is unreasonable because it implies that the distribution of λ which corresponds to the measured outcomes does not depend on the angular settings on either arm of the setup (Alice or Bob). But this can not be true where coincidence counting is done.

To be clear, imagine that every pair of outcomes (Ai, Bi) obtained, corresponds to a specific λi value, ie A(ai,λi) = Ai = ±1, B(bi,λi) = Bi = ±1.

To say that ρ(λ) = ρ(a,b,λ), implies that if you would take all the λs corresponding to all the measured outcomes, their distribution will be exactly the same irrespective of whether we measured at angles (a,b) or (b,c) or (a,c). With coincidence counting and what we know classically about Malus law, this is unreasonable.

To be clear even further, it is equivalent to saying that for the setup of a single stream of particles and 2 polarizers A,B in sequence set at angles a, b, the distribution of hidden polarization parameter λ of the photons that pass through both polarizers is independent of the angles at which both are set, ie ρ(λ) = ρ(a,b,λ)

You do not need conspiracy to realize that the assumption is unreasonable in classical case of two polarizers in sequence. Coincidence counting for two separate arms does effectively the same thing because it only allows photons to be considered on one side if they passed through the other side.

Ok. Thanks for explaining Bill. I understand and agree with the above.

I'll be reading ttn's paper, http://arxiv.org/abs/0808.2178 , to see if I understand and agree with how he deals with Jarrett's analysis.

But first, I'll deal with your post #192.

 

[billschnieder]

But the proof is not supposed to "model exactly the way the experiments are performed". It is a black box. If you agree with assumptions, you have to accept the conclusions. You cannot argue against it because some intermediate step in derivation does not have a clear physical meaning. Yes, there is a step in Bell's proof where there is a product of A(a,λ)A(b,λ)A(c,λ) under the integral, so what? It is just an intermediate step, it does not have to have physical meaning. Specifically, it does not introduce any new physical assumptions. The only assumptions are those used to formulate initial steps of the derivation.

Second, some people apparently confuse the requirement of 'having the same ρ(λ)' with 'having the same λ' when carrying out measurements C(a,b), C(a,c), C(b,c). Bell requires the former, but not the latter.

Third, some people confuse expectation value $E[X] = \sum \rho_i X_i$ and arithmetic mean $\overline{X}=\frac{1}{N}\sum X_i$. Bell's inequality applies only to the former and not the latter. Which means you do NOT apply Bell's inequality to individual measurement outcomes. Instead, you use experimental data to estimate expectation values, and you plug those expectation values (along with their estimated standard deviations) into Bell's inequality.

PS: We've been through that several times before.

It seems like you are responding to me but there is nothing in your post that is actually a response to anything I've actually said.

 

[billschnieder]

First of all, the inequality is dealing the total numbers of mismatches not averages

This must be a joke, right?

Eh NO! Did you read the context of the statement? Or have you gone non-contextual on me

In case you were not paying attention, in was in the context of the coin toss example:

3 coins (a,b,c), where nAB represents number of MISMATCHES between the outcomes of a and b.

Inequality: nAB + nAC >= nBC
a= THHHTHTH
b= HHHTTTHH
c= TTTHHHHT
4 + 5 >= 7 : Obeyed (ONLY 3 lists of outcomes)

a1= HTHTTHHT
b1= HHTTTTHT

a2= TTHHTTHT
c2= THHHTTTT

b3= HTHHTTHH
c3= THTTTHTT
3 + 2 >= 7 Disobeyed (6 lists of outcomes)

ttn's response was to change the inequality to a different one between averages. Which was a non-response because he was effectively saying: "I have no explanation why the inequality you provided is violated, but look here I have a different one that is always obeyed".

 

[ThomasT]

a1 is complete context for run1 when the angle "a" was set. a2 is the complete context for run2 when angle "a" was set. "a1" is (angle setting "a" + everything else that makes run 1 different from run 2, including time, the complete microscopic state of the device etc.). When Bell's inequality is written as

|<ab> - <ac>| <= 1 + <bc>

"a" in that expression does not represent the angle, it is a label for the list of outcomes. Similarly, "a1" in this case represents the list of outcomes when the angle was set to "a" under context 1 (run 1), and "a2" is the list of outcomes when the angle was set to "a2" for run 2. The above inequality is not valid unless the list of outcomes labeled "a1" is identical to the list of outcomes labeled "a1". In other words, the six lists of outcomes a1,a2,b1,b3,c2,c3 MUST be sortable such that we end up with just three lists a,b,c (ie, a1=a2, b1=b3,c2=c3). Note we are talking about lists of outcomes here.


I think it is peripheral. How else will you explain Boole's derivation, he made no locality assumption. There has been a lot of misunderstanding and confusion caused by mixing up functions with probabilities. Some people think that Bell was trying to calculate a joint probability in his equation (2) of his original paper, but he was not. He wrote:

P(a,b) = ∫A(a,λ)B(b,λ)ρ(λ)

Some have misunderstood this to be a joint probability equation which it is not. First of all A(a,λ), B(b,λ) can take up negative values contrary to probabilities. What Bell was calculating was an expectation value for the paired product of the outcome at Alice and Bob. So there is no locality here. The separability of the expectation value is simply due to the fact that a paired product is necessarily separable.

As I showed in post #123, you can derive the same inequality if you start with 3 dichotomous variables and calculate their paired products, or 3 list containing values ±1, without any additional assumption. So any "Blah blah bla" that gives you paired products of 3 dichotomous variables can be used to fool you into thinking the "Blah blah bla" is important for the inequality. It is not.

Earlier in the thread (post #101), I got ttn to admit that λ could be non-local hidden variables and you will still obtain the inequalities.

Ok, thanks Bill. I actually think I understand your argument now. It seems to make sense. I'm going to have to reread this and other threads to understand exactly why some people are objecting to it.

 

[billschnieder]

Ok, thanks Bill. I actually think I understand your argument now. It seems to make sense. I'm going to have to reread this and other threads to understand exactly why some people are objecting to it.

Thank you for your patience. Glad to help anytime.

 

[ttn]

Congrats to Thomas for figuring out exactly how to probe Bill to get him to confess openly what I've been saying all along he thinks:

ρ(λ) = ρ(a,b,λ) is unreasonable because it implies that the distribution of λ which corresponds to the measured outcomes does not depend on the angular settings on either arm of the setup (Alice or Bob).

In short, he thinks the "no conspiracy" assumption is unreasonable.

Note also the highly illuminating reasons given:

With coincidence counting and what we know classically about Malus law, this is unreasonable.

To be clear even further, it is equivalent to saying that for the setup of a single stream of particles and 2 polarizers A,B in sequence set at angles a, b, the distribution of hidden polarization parameter λ of the photons that pass through both polarizers is independent of the angles at which both are set, ie ρ(λ) = ρ(a,b,λ)

It is equivalent to that, if (in both cases) the λ refers to the state of the particles *before any measurements are made on them*. That, of course, is precisely what λ means in all the derivations. But why in the world should ρ(λ) -- the distribution of states of an ensemble of particles that have just been shot toward some polarizers -- depend on the orientation of the polarizers? The answer is: there's no reason it should depend on that, not in the Bell type case and not in the Malus type case Bill has in mind here.

My guess is that Bill is wrongly thinking of λ as referring to the state of the particles after some or all of the polarization measurements have been made. If that's what you think λ means, then -- no doubt -- ρ(λ) should probably depend on the polarizer orientations! But ... that's simply not what λ refers to.

You do not need conspiracy to realize that the assumption is unreasonable in classical case of two polarizers in sequence.

No, the assumption is completely reasonable in this "classical case". It only seems unreasonable if you misunderstand what is being claimed -- in particular, if you misunderstand ρ(λ) to mean the distribution of particle states *after the measurements*. But if you correctly understand ρ(λ) to mean the distribution of particle states *prior to any measurements* -- i.e., the distribution of particle states *emitted by the particle source* -- it starts to seem pretty darn reasonable that this should be the same, no matter what somebody is going to decide (later and independently) to measure.

Coincidence counting for two separate arms does effectively the same thing because it only allows photons to be considered on one side if they passed through the other side.

Note another misunderstanding here. The modern experiments use 2-channel polarizers. It isn't true that a pair only gets counted if both particles "pass" their polarizers. Yes, there are detection efficiency issues in the experiments, but basically each photon is subjected to a measurement in which one detector clicks if the photon is "horizontal" (relative to the axis a) and a different detector clicks if the photon is instead "vertical" (relative to a). That is, each photon is detected either way. The alternative to "passing through the polarizer" is not getting absorbed and hence never seen and hence never counted, but is rather getting counted instead by the other detector.

Not that this particular issue is all that relevant to the main discussion here, except insofar as it shows another way in which Bill doesn't know what he's talking about.

 

[mattt]

Travis, I think now I have a problem with your "CHSH-Bell Inequality Theorem".

The mathematical theorem is correct, i.e. the mathematical proof is correct (in my opinion at least).

The problem I think I have now is with respect to the mathematical setup (the \lambda distribution meaning) and the factorizability condition (4) (concretely about how it is supposed to codify a necessary condition for locality).

Imagine I have a theory that predict the outcomes of the experiment the following mathematical way (kind of a hidden variable stochastic theory) :

My theory says that the process that produces the pairs of particles doesn't produce exactly the same "kind of pair" everytime, but there are "different groups" of pairs (following a probability distribution), and my theory says that pairs from the same group will have the exact same (stochastic) outcomes prediction for the experiment (but pairs from different "groups" will have different (stochastic) outcomes prediction for the experiment).

Mathematically:

The Probability Space $$(\Lambda, P)$$ will account for the different "groups" of pairs.

Then, for each $$\lambda=\lambda_0$$, $$\alpha_1=a$$, $$\alpha_2=b$$, my theory says that the outcomes will follow a distribution we will call $$P_{a,b}(A_1,A_2|\lambda_0)$$

Imagine one of these distributions prediction outcomes is the following one:

$$P_{a,b}(A_1=1,A_2=1|\lambda_0)=0.9$$

$$P_{a,b}(A_1=1,A_2=-1|\lambda_0)=0.04$$

$$P_{a,b}(A_1=-1,A_2=1|\lambda_0)=0.03$$

$$P_{a,b}(A_1=-1,A_2=-1|\lambda_0)=0.03$$


This "theory" does NOT satisfy your factorizability condition (4), but now I don't see why someone who believes in "locality" would call this "theory" non-local.

ETA: if someone does not understand, I will put it into words:

This "theory" says that the source produce "different types of pairs" and the theory says that if we only pay attention to pairs of one type (called \lambda_0 ) then, if the setting is a,b then 90% of the outcomes (of pairs from exactly this group) will be (1,1), 4% will be (1,-1), 3% will be (-1,1) and 3% will be (-1,-1).

Of course this "theory" has to give another stochastic/probabilistic prediction for any other setting and any other "type/group" of pairs.

The thing is that this "theory" does not satisfy Travis's "necessary condition for locality" (4) and what I ask is if everybody would call this "theory" non-local (taking into account the meaning of "locality" anyone of you may have).

 

[mattt]

But this is not how C(a,b), C(a,c), C(d,b) and C(d,c) are defined within the inequality. They are defined such that the ABSOLUTELY CRUCIAL integration variable λ, is identical for all the terms. In other words, if you take all the individual lambda values from all cases in which the setting pair was (a,b) and all the individual lambda values from all the cases in which the setting pair was (b,c) etc, they will be identical from setting pair to setting pair. ONLY under such conditions can the inequality be derived and ONLY under this scenario are the terms you measured equivalent to the terms in CHSH inequality.

Now I ask you, is it a reasonable assumption to say that the distribution of lambda values for MEASURED pairs is IDENTICAL from setting pair to setting pair.

No, I don't understand his mathematical expressions to mean that.

What I understand from his mathematical expressions is that, if you run the experiment for long enough time, recording millions of measured pairs in total, it doesn't even matter that there may be different number of pairs corresponding to the a,b setting than corresponding to any other setting a,c or a,d or a,a or b,a ...(like my own previous example tried to show).

As long as there are enough pairs corresponding to each setting (millions of pairs for example, for every setting chosen), we reasonably can think that the \lambda variable has appeared such high number of times (one for each pair) for each setting chosen, that it follows its own theoretical distribution frecuencies with enough accuracy. Just that.

 

[ThomasT]

What Bell was calculating was an expectation value for the paired product of the outcome at Alice and Bob. So there is no locality here. The separability of the expectation value is simply due to the fact that a paired product is necessarily separable.

Bell's math as derived is essentially correct mathematically, but does not correspond to the QM or experimental situation.

So, the reason it doesn't correspond to QM or the experimental situation is due to the separability of the paired product form. Right?

But it's the separability of the paired product form that, supposedly, represents locality vis independence. Isn't it?

As I showed in post #123, you can derive the same inequality if you start with 3 dichotomous variables and calculate their paired products, or 3 list containing values ±1, without any additional assumption.

I understand that, but does that, by itself, imply that the separability of the paired product form doesn't represent locality? Or, does the relationship between the separability of the paired product form and the experimental situation need to be analysed further?

 

[billschnieder]

No, I don't understand his mathematical expressions to mean that.

What I understand from his mathematical expressions is that, if you run the experiment for long enough time, recording millions of measured pairs in total, it doesn't even matter that there may be different number of pairs corresponding to the a,b setting than corresponding to any other setting a,c or a,d or a,a or b,a ...(like my own previous example tried to show).

As long as there are enough pairs corresponding to each setting (millions of pairs for example, for every setting chosen), we reasonably can think that the \lambda variable has appeared such high number of times (one for each pair) for each setting chosen, that it follows its own theoretical distribution frecuencies with enough accuracy. Just that.

I think you misunderstood. Performing a million or even a billion pairs for each setting pair does not automatically cause the lambda distribution of actually measured pairs to be the same for the (a,b) pairs or as for the (b,c) pairs. Remember that actual experiments use coincidence counting. The only distribution of lambda that matters is the one corresponding to the measured outcomes (ie, filtered through coincidence circuitary).

 

[ThomasT]

It is equivalent to that, if (in both cases) the λ refers to the state of the particles *before any measurements are made on them*. That, of course, is precisely what λ means in all the derivations. But why in the world should ρ(λ) -- the distribution of states of an ensemble of particles that have just been shot toward some polarizers -- depend on the orientation of the polarizers? The answer is: there's no reason it should depend on that, not in the Bell type case and not in the Malus type case Bill has in mind here.

Thanks for clarifying. The way Bill was describing it, I was thinking of ρ(λ) as referring to the distribution prior to detection, and ρ(a,b,λ) as referring to the distribution after a detection. But if ρ(a,b,λ) refers to prior detection, then ρ(λ)=ρ(a,b,λ) makes sense.

Is it true that ρ(λ) for a given pair is altered by the registration of a detection at either end? That is, a detection at A alters the sample space at B, and vice versa. What I'm getting at is the outcome indpendence part of the locality condition.

 

[harrylin]

OK I came at the first for me interesting point in your article.

The introduction mentions correctly, but with lack of clarification, that EPR formulated their argument in terms of position and momentum and Bohm reformulated it in terms of spin. The clarification that I think would be good is if the qualitative difference is important or not, and why not: position vs. momentum is directly and obviously related to the uncertainty principle, but up spin vs down spin not, or at least not in the same way. Maybe it's somewhere further in the article, but then still a mention of it there would be useful.

Thanks,
Harald

 

[mattt]

I think you misunderstood. Performing a million or even a billion pairs for each setting pair does not automatically cause the lambda distribution of actually measured pairs to be the same for the (a,b) pairs or as for the (b,c) pairs. Remember that actual experiments use coincidence counting. The only distribution of lambda that matters is the one corresponding to the measured outcomes (ie, filtered through coincidence circuitary).

Read my previous message about the "hypothetical theory". In that case, each \lambda would label each "type/group" of pairs (inside each group all the pairs will produce the exact same stochastic prediction, that is how this "theory" works).

Imagine there are 10 "groups" and the source produces (in average) 15% pairs from the first group, 10% pairs from the second type/group,...

If we have measured 2 million pairs with the a,b setting, in average 15% of this 2 million will be "of the first type/group", 10% of this 2 million will be "of the second type/group", ...

If we have measured 3 million pairs with the a,c setting, in average 15% of this 3 million will be "of the first type/group", 10% of this 3 million will be "of the second type/group", ...

That is why if the numbers are big enough, the lambda experimental distribution (even if they are not actually known) of relative frecuencies will be the same for any setting, because it will resemble more and more the theoretical distribution (15% from the first type, 10% from the second type...) that the Theory establishes.

For each setting a,b there will be millions of pairs from all the different "types" (lambdas), that is why later on you have to integrate wrt the lambdas (to take into account their own theoretical distribution that your "theory" establishes) to be able to get the predicted value of C(a,b).

 

[ttn]

More quick debunking of Bill:

Some people think that Bell was trying to calculate a joint probability in his equation (2) of his original paper, but he was not. He wrote:

P(a,b) = ∫A(a,λ)B(b,λ)ρ(λ)

Some have misunderstood this to be a joint probability equation which it is not. First of all A(a,λ), B(b,λ) can take up negative values contrary to probabilities. What Bell was calculating was an expectation value for the paired product of the outcome at Alice and Bob.

All correct so far. But then...

So there is no locality here.

Locality is there, in the assumption that the outcome on the one side can only depend on "stuff" that is locally accessible there: A can depend on the local polarizer setting (a) and the state of the particles (λ) ... but it cannot depend on the distant polarizer setting (b). That is, A = A(a,λ). Likewise, B = B(b,λ). (B cannot depend on the setting that is distant to it.) That is locality.

As I showed in post #123, you can derive the same inequality if you start with 3 dichotomous variables and calculate their paired products, or 3 list containing values ±1, without any additional assumption. So any "Blah blah bla" that gives you paired products of 3 dichotomous variables can be used to fool you into thinking the "Blah blah bla" is important for the inequality. It is not.

This confusion is addressed directly in section 10.6 of the article.

http://www.scholarpedia.org/article/Bell%27s_theorem#Locality_versus_non-contextual_hidden_variables

Earlier in the thread (post #101), I got ttn to admit that λ could be non-local hidden variables and you will still obtain the inequalities.

I guess you didn't understand something I was taking for granted. One needs to distinguish two distinctions: local vs. non-local *beables* (i.e., physically real stuff that can be unproblematically assigned a location in regular physical space or spacetime, vs. stuff -- like QM wave functions if they are "stuff" -- for which this is not the case) and then local vs. non-local *causality* (i.e., causal influences going exclusively slower vs. sometimes faster than light). The point is that the λ in Bell's derivation (which remember is a complete description of some of the physically real stuff, some of the beables, involved) needn't be just local beables. It can be, or include, nonlocal beables as well -- and still the derivation goes through. What's excluded in the derivation, though, is non-local *causation*. That's what is meant when we say that "locality" is assumed as a premise -- and what we mean when we say that the violation of the inequality in actual experiments proves that there exists non-locality.

In short, I was not at all "confessing" what you seem to think you got me to confess -- that somehow you can derive the inequality still even if you *don't* assume locality. Although actually, you can. There are lots of different sets of assumptions that imply the inequality. But the point is that most of those other assumptions aren't all that interesting. For example, we already know (from various no-hidden-variable theorems) that non-contextual hidden variable theories can't be right. So the fact that you can derive a Bell inequality from *that* assumption is old news. yawn. But the fact that you can derive a Bell inequality from *locality* (basically alone) is huge news. It means there exist in nature causal influences that violate what everybody took to be relativity's speed limit! Wow!

 

[billschnieder]

So, the reason it doesn't correspond to QM or the experimental situation is due to the separability of the paired product form. Right?

No. It is simply because the terms from QM and experiments are not equivalent to the terms in Bell's inequality. There is nothing preventing you from calculating expectation values of paired-products from experiments. The problem is that you have 6 lists of outcomes from experiments, so the expectation values of paired products you calculate are not comparable to the ones you would have calculated from only 3 lists. They are apples and oranges. See post #125.

So it is a matter of factorizability rather than separability, ie:

You can do the factorization |ab-ac| = |a(b-c)| , but you can not do the factorization |a1b1 - a2c2| = |a1(b1-c2)| UNLESS a1 = a2. This is why the inequality is violated, even though the paired products involved are still separable, simply because factorization operations such as these are relied on to derive the inequalities from paired products.

For QM, the three expectation values calculated are independent terms not cyclically linked the way the inequality implied. QM is predicting exactly what the experiments are measuring -- ie independent terms, not the dependent terms present in the inequality. Just so that the word "independent" is not misconstrued, this is what I mean:

in the inequality |<ab> - <ac> | <= 1+ <bc>, once <ab> and <ac> are determined, <bc> automatically follows and when <ac> and <bc> are determined, <ab> automatically follows etc. So we don't have 3 separate independent terms here, we have three cyclically dependent terms.

But it's the separability of the paired product form that, supposedly, represents locality vis independence. Isn't it?

No. As I explained, you can still calculate any expectation value of a paired product from any results you like. Even if you had spooky-action at a distance, the paired product of the outcomes at the two stations will still be separable. Independence only comes in because the way the inequality is derived by factorization, you can do the factorization

|E(a|λ)E(b|λ) - E(a|λ)E(c|λ)| = |E(a|λ)[E(b|λ) - E(c|λ)]|

but you cannot do the factorization

|E(a|λ1)E(b|λ1) - E(a|λ2)E(c|λ2)| = |E(a|λ1)[E(b|λ1) - E(c|λ2)]| UNLESS ρ(a, b, λ1) = ρ(a, c, λ2) = ρ(λ2) = ρ(λ), which is an unreasonable assumption for the Bell-test experiments given what we know from classical physics.

I understand that, but does that, by itself, imply that the separability of the paired product form doesn't represent locality?

It can represent locality and it can represent a million other things. Describe any experiment, local or non-local and the expectation value of the paired product will be separable. There is nothing uniquely local about the expectation of the paired product, it is by definition separable.

Or, does the relationship between the separability of the paired product form and the experimental situation need to be analyzed further?

I believe it is a red-herring as far as Bell's theorem is concerned but that doesn't mean it is not worthy of study.

 

[lugita15]

It means there exist in nature causal influences that violate what everybody took to be relativity's speed limit! Wow!

Is such a conclusion really inevitable? What if you had an acausal theory?

 

[billschnieder]

Thanks for clarifying. The way Bill was describing it, I was thinking of ρ(λ) as referring to the distribution prior to detection, and ρ(a,b,λ) as referring to the distribution after a detection. But if ρ(a,b,λ) refers to prior detection, then ρ(λ)=ρ(a,b,λ) makes sense

Is it true that ρ(λ) for a given pair is altered by the registration of a detection at either end? That is, a detection at A alters the sample space at B, and vice versa. What I'm getting at is the outcome indpendence part of the locality condition.

Look again at Bell's equation and note the following.
P(a,b) = ∫A(a,λ)B(b,λ)ρ(λ)

1) We are calculating an expectation value for a paired product of "OUTCOMES" not released photons.
2) Not all photons are ever detected so if the expression is for ALL photons Bell's initial functions A(a,λ)=±1 would be wrong since it does not account for non detection but since he was talking about outcomes, his equation is correct.
3) To calculate the expectation, you take each OUTCOME pair, multiply them together, and multiply them with the probability of the λ which resulted in it, and then integrate over all the λs for which you have outcomes. The ONLY relevant λ under discussion is the λ for measured outcomes.

So ttn is simply wrong here.