A classical challenge to Bell's Theorem?

[Gordon Watson]

Afaik, wrt optical Bell tests, λ, the hidden variable denotes an underlying polarization that's varying randomly from pair to pair.

I guess I just don't understand your treatment here. As far as I can tell it's not going to get you to a better understanding of why BIs are violated formally and experimentally, and it doesn't disprove Bell's treatment which is based on the encoding of a locality condition which, it seems, isn't, in effect, a locality condition.

And now, since I am a bit confused by your presentation, I think I will just fade back into the peanut gallery. Maybe I'll learn something.

I've responded at https://www.physicsforums.com/showthread.php?t=591572 to avoid confusing our discussion on entanglement with the classical example in the OP.

BUT note: If you carried out your "AFAIK" calculation above (with your "λ and underlying polarisation"), you would reproduce the unentangled classical example given in the OP; which is neither Bell nor Aspect, etc. But, imho, this would help with your understanding of Bell (imho). And maybe bring you back into to this thread?

 

[Gordon Watson]

It's your model, and I supplied the answers to your questions. Bell applies, and the resulting prediction is within the local realistic boundary as we would expect. Is your question how did I arrive at .375?

..
DrC, with respect: I have little idea what you are referring to! So, seeking to understand, I write in the spirit of trying to be helpful! I therefore proceed on the basis that you are sincere and not trying to muddy the waters unnecessarily.

(Please also note a very off-putting habit of yours: You continue to confuse A and B with a and b. These repeated errors are akin to your old one of identifying the particles as Alice and Bob -- which is no problem if we know what your up to: but can be quite confusing when commonly accepted conventions are abused without explanation.)

Now:

1. What is my model? Do you mean the experiment identified in the OP? I take the view that Bell tried to "model" local realism (see below).

2. To be clear about one of my questions: Can you show me, please, the A and B functions that you used in your derivation? [EDIT: DrC, Response here not necessary if the answer to the next question is No!]

3. My next question is: Do they comply with Bell's (1964), A and B = ±1?

4. If not, why not? And so what do you mean by your "Bell applies"? For, further, you could then not have used Bell's integral for E(AB)?

5. One of my questions related to the maximum value of the traditional (your term) CHSHI that you derived from the experiment in the OP. What number did you get?

6. Surely it was not 0.375?

7. I know that (in reference to your analysis) you referred to text-books having all the relevant answers in them. [EDIT: I take it that such answers are relevant to the OP?] I do not know of any such; so, alas, I must always work from first-principles. So, can you list a few such books for me, please?

8. You also mentioned (re#6 above): "The Bell lower limit is .333." What is this, please? And how derived? Is it relevant to any issue here?

9. In the interests of seeking to be very clear and precise, one aspect of this thread, in case you missed it, was addressed to you as follows:

[To DrC, responding to a very confusing post by him.]

?

Bell's (1964) analytical (and presumed local-realistic) protocol has (equivalently):

A(a, x) = ±1;

B(b, x) = ±1;

E(AB) = ∫AB ρ(x) dx.

I don't see you using any of this anywhere in your analysis of this (clearly) local-realistic experiment?

Why?

In effect: Is there some reason to follow Bell only when it suits you?

..
To rephrase my serious interest:

1: Could you explain why (in your view) Bell's protocol is not relevant to what is clearly a local-realistic experiment?

2: Would such a view indicate that Bell may not be as relevant for local-realism as you commonly suppose?

With thanks, as always, this time in anticipation,

Gordon
..

 

[DrChinese]

Gordon,

It is your model, you suggested we consider it per Bell. It's a common one, so ok, so I have. The CHSH upper limit is 2, and I don't know the exact expectation value for CHSH for this model but it is definitely less than 2. Of course the experimental value is above 2.

You asked what P(AB) would be, I indicated .375. If you want to know E(AB), just use lugita15's formula and you get E(AB)=-.25.

Not sure how I am abusing A and B, I didn't even mention them in my last post. Yes, there is a difference between the particle detected by Alice, the angle setting Alice uses, and Alice as the observer. However, I can't help you much on the point because I interchange Alice and Bob, and A and B, etc. as I think it is easier for the analogies I make. I would hope that my choice of capitalization wouldn't confuse too much, I usually try to follow what the other person is doing. In this case, I have been following you preference in this thread to use E(AB) and P(AB) which is what I am writing as well. I do not usually repeat every step of a derivation when you can see that for yourself in the source literature. If you have a specific question, I will usually answer that. Is your question how I arrived at .375? I can show that if it helps. I used the Product State statistics, which apply to the OP example, and that formula has:

P(A,B)=.25+.5(cos^2(A-B))
[Where A and B are the angle settings and there is the unspoken assumption that the example does not have some particular bias that is not mentioned, such that the source is reasonably randomized.]

I mentioned that I commonly use 0 and 120 degrees for my settings. So substituting in the above, I get .25 + .5(.25) = .375 which is the rate of matches (which is what I usually report). Keep in mind that the "proper" correlation rate subtracts mismatches so that is why you could also report .375 - .625 which is the -.25 number mentioned above. I think you are using that to be E(A,B). As long as we know what basis we are reporting on, they are really the same thing.

A bit confused about your comments about Bell not being relevant as I think. I never said the Bell protocol does not work here, and I am not sure why you keep pushing me in that direction. Further, you must keep in mind that a discussion on PhysicsForums is not going to alter what the general physics community thinks of Bell. Bell is highly regarded, in fact an interesting piece of history. By 1970, 5 years after Bell appeared in a now defunct publication, new interpretations were needing to address Bell because its logic was so powerful. See for example something cited by billschnieder:

The Statistical Interpretation of Quantum Mechanics

This devotes considerable attention to EPR, Bell and hidden variables.

 

[bohm2]

There was an article published yesterday on a related topic that some might find interesting and/or add to the confusion:

In this paper I present a stronger Bell argument which even forbids certain non-local theories. The remaining non-local theories, which can violate Bell inequalities, are characterized by the fact that at least one of the outcomes in some sense probabilistically depends both on its distant as well as on its local parameter.

A stronger Bell argument for quantum non-locality
http://philsci-archive.pitt.edu/906...er_Bell_argument_for_quantum_non-locality.pdf

 

[Gordon Watson]

There was an article published yesterday on a related topic that some might find interesting and/or add to the confusion:

A stronger Bell argument for quantum non-locality
http://philsci-archive.pitt.edu/906...er_Bell_argument_for_quantum_non-locality.pdf

..
Dear bohm2,

Welcome to the thread and many thanks for the very interesting article. I've not analysed it in detail (yet) but its sure stored for future reading. It is much appreciated!

To encourage your further involvement here: Note that the classical experiment defined in the OP is designed to minimise many complications with BT (so it is not like that article).

Rather: The OP's classical experiment raises, for me, the question: Given that Bell's protocol/system is used for analysing local-realism in the context of EPR-Bohm (from that other brilliant Bohm): Can Bell's system be used to analyse the (clearly) local-realistic experiment in the OP?

That is: We have outputs A and B that may be equated to ±1. And we have some, here presumably uniform, ρ(x): since x is randomised. These are just those ingredients that Bell (e.g., Bell 1964) focussed upon. Plus his integral: E(AB) = ∫AB ρ(x) dx !

With thanks again,

Gordon

EDIT: So (to be clear), in relation to the (clearly) realistic and Einstein-local experiment in the OP, we seek to follow Bell's protocol (e.g., from Bell 1964) and provide:

(1) A(a, x) = ±1.

(2) B(b, x) = ±1.

(3) 0 ≤ ρ(x); ∫ρ(x) dx = 1.

(4) E(AB) = ∫AB ρ(x) dx = ?

 

[Gordon Watson]

Gordon,

It is your model, you suggested we consider it per Bell. It's a common one, so ok, so I have. The CHSH upper limit is 2, and I don't know the exact expectation value for CHSH for this model but it is definitely less than 2. Of course the experimental value is above 2.

You asked what P(AB) would be, I indicated .375. If you want to know E(AB), just use lugita15's formula and you get E(AB)=-.25.

Not sure how I am abusing A and B, I didn't even mention them in my last post. Yes, there is a difference between the particle detected by Alice, the angle setting Alice uses, and Alice as the observer. However, I can't help you much on the point because I interchange Alice and Bob, and A and B, etc. as I think it is easier for the analogies I make. I would hope that my choice of capitalization wouldn't confuse too much, I usually try to follow what the other person is doing. In this case, I have been following you preference in this thread to use E(AB) and P(AB) which is what I am writing as well. I do not usually repeat every step of a derivation when you can see that for yourself in the source literature. If you have a specific question, I will usually answer that. Is your question how I arrived at .375? I can show that if it helps. I used the Product State statistics, which apply to the OP example, and that formula has:

P(A,B)=.25+.5(cos^2(A-B))
[Where A and B are the angle settings and there is the unspoken assumption that the example does not have some particular bias that is not mentioned, such that the source is reasonably randomized.]

I mentioned that I commonly use 0 and 120 degrees for my settings. So substituting in the above, I get .25 + .5(.25) = .375 which is the rate of matches (which is what I usually report). Keep in mind that the "proper" correlation rate subtracts mismatches so that is why you could also report .375 - .625 which is the -.25 number mentioned above. I think you are using that to be E(A,B). As long as we know what basis we are reporting on, they are really the same thing.

A bit confused about your comments about Bell not being relevant as I think. I never said the Bell protocol does not work here, and I am not sure why you keep pushing me in that direction. Further, you must keep in mind that a discussion on PhysicsForums is not going to alter what the general physics community thinks of Bell. Bell is highly regarded, in fact an interesting piece of history. By 1970, 5 years after Bell appeared in a now defunct publication, new interpretations were needing to address Bell because its logic was so powerful. See for example something cited by billschnieder:

The Statistical Interpretation of Quantum Mechanics

This devotes considerable attention to EPR, Bell and hidden variables.

..
Dear DrC, thanks (as always) for the new information in your response (above); it (as well as your time) is appreciated.

So, in reply: Yes please. Please show me your workings for any of the numbers or results that you have so far produced. That will surely be a step toward sorting out our differences.

But please note: Your confessed use of A & B to denote Alice & Bob, particles 1 & 2, device settings a & b is not only confusing: IT IS wrong when you consider that A & B generally convey specific understandings within the Bell literature, beginning with Bell (1964).

NB: Under any of your uses, or all of them together, this equation of yours makes NO SENSE at all: P(A,B)=.25+.5(cos^2(A-B)).

Please think about it. (Also: P generally defines a probability.)

Also: I'm not clear re lugita15's formula.

PS: I will be away from the Net for several days so I wonder: Would you mind taking your time and answering the specific questions in https://www.physicsforums.com/showpost.php?p=3840882&postcount=37 ?

The advantages that such direct answers offer for me (and maybe some others), are these:

1. All questions may be answered without any reference to QM whatsoever.

2. The experiment under discussion is easily done and is clearly Einstein-local!

3. As far as I can see, this thread, in meaningfully discussing the system that Bell uses to derive his conclusion, requires no knowledge of QM at all! (A conclusion, I'm told, that some have drawn for the whole Bell scene: Bell's theorem is NOT a property of quantum theory. So it's a property of ... ?)

Must run.

GW

 

[Gordon Watson]

...
You asked what P(AB) would be, I indicated .375. If you want to know E(AB), just use lugita15's formula and you get E(AB)=-.25.

..
DrC, I'm hoping that you'll answer my questions directly (to this point, at least; see particularly the recent posts above)? If you will not be answering, please let me know.

Also, if you are withdrawing your offer to expand on your derivations, please also let me know.

Now, re the above quote: I have no clue re "lugita15's formula" -- nor what E(AB) =-.25 relates to?

Why not give your derivation of E(AB) directly (with the explanatory notes that you offered)?

Do you not understand that E(AB) will be some function of a and b (the detector settings)?

Not sure how I am abusing A and B, I didn't even mention them in my last post. Yes, there is a difference between the particle detected by Alice, the angle setting Alice uses, and Alice as the observer. However, I can't help you much on the point because I interchange Alice and Bob, and A and B, etc. as I think it is easier for the analogies I make.

BUT your abuse of A and B continues almost immediately! See *** below!

I would hope that my choice of capitalization wouldn't confuse too much, I usually try to follow what the other person is doing. In this case, I have been following you preference in this thread to use E(AB) and P(AB) which is what I am writing as well.

Please show me where I have used P(AB) in this thread?

And what could it possibly mean?? What does it mean to you, since you clearly differentiate it from E(AB)?

I do not usually repeat every step of a derivation when you can see that for yourself in the source literature. If you have a specific question, I will usually answer that. Is your question how I arrived at .375? I can show that if it helps. I used the Product State statistics, which apply to the OP example, and that formula has:

P(A,B)=.25+.5(cos^2(A-B))
[Where A and B are the angle settings and there is the unspoken assumption that the example does not have some particular bias that is not mentioned, such that the source is reasonably randomized.]

*** So: If A and B are the angle settings, what then is P(AB), please?

I mentioned that I commonly use 0 and 120 degrees for my settings. So substituting in the above, I get .25 + .5(.25) = .375 which is the rate of matches (which is what I usually report). Keep in mind that the "proper" correlation rate subtracts mismatches so that is why you could also report .375 - .625 which is the -.25 number mentioned above. I think you are using that to be E(A,B). As long as we know what basis we are reporting on, they are really the same thing.

...

You commonly use 0 and 120? But how does that help you (in this thread) derive the maximum value attainable for the CHSH inequality for the experiment defined in the OP?

PS: I trust you will persist with your line of reasoning here: So that we might come to agree re the simple facts involved; no opinions being required (at any stage) given the simplicity of the experiment and the related maths.

..

 

[DrChinese]

Please show me where I have used P(AB) in this thread? And what could it possibly mean?? What does it mean to you, since you clearly differentiate it from E(AB)?

*** So: If A and B are the angle settings, what then is P(AB), please?

Arrgh.

Usually, P(AB)=P(A,B)=P(A-B)=P(ab)=P(a,b)=P(a-b) is the probability of matches at angles A/a and B/b. It doesn't matter what you call it, as long as you know what you are referring to. You could also call that an expectation value. Then P(AB)=E(AB), which you used in post #1. This has a range from 0 to 1.

But sometimes, either P(*) or E(*) is referring to the correlation rate, which has a range from -1 to 1. Correlations are matches less mismatches as a percentage. lugita15 discusses this point, I thought clearly, and was trying to follow his lead to make it easier. It doesn't matter to me really, but I prefer to speak in terms of probability. So a correlation of -.25 is the essentially the same thing as a match probability of 37.5%.

Your conclusions are always the same though regardless of basis. The important point is that when a-b=A-B=120 degrees, the expected match percentage is 37.5%. This is above the applicable Bell threshold, which is 33.3% (1/3). So the model respects Bell, and being explicitly classical, you would expect that.

I will be supplying the derivation as I get time, but for now just keep in mind that the formula is .25+(cos^2(a-b)) where a and b are the measurement angle settings and it is averaged over a suitable (random) population of source polarization angles (which I take as being x in your model).

It's your model in post#1, so I assume you know where you are going with this. I don't really want to spend time covering elements of this that you already know and don't have any questions on. So it would be helpful to know what that is.

 

[billschnieder]

Your conclusions are always the same though regardless of basis. The important point is that when a-b=A-B=120 degrees, the expected match percentage is 37.5%. This is above the applicable Bell threshold, which is 33.3% (1/3). So the model respects Bell, and being explicitly classical, you would expect that.

I will be supplying the derivation as I get time, but for now just keep in mind that the formula is .25+(cos^2(a-b))

I suppose then that since it is classical and obeys Bell, you will be able to factorize the formula into 2 functions which give values ±1 (according to Bell). Or better, you should be able to show how starting with such functions of only a, or b with values ±1, you arive at the result .25+(cos^2(a-b)). This is the challenge as far as I understand it.

 

[DrChinese]

I suppose then that since it is classical and obeys Bell, you will be able to factorize the formula into 2 functions which give values ±1 (according to Bell). Or better, you should be able to show how starting with such functions of only a, or b with values ±1, you arive at the result .25+(cos^2(a-b)). This is the challenge as far as I understand it.

Sure, that's it exactly. Of course, it converges on that value in actual trials. And I integrate across a range of x. If you pick only a few specific x values, your results will vary.

You don't have any question about that, do you? I am sure you have done this exercise plenty of times.

PS Somehow the formula got mangled along the way. It should be:

P(ab)=.25+.5(cos^2(a-b)) which has a range from .25 to .75

 

[billschnieder]

Sure, that's it exactly. Of course, it converges on that value in actual trials. And I integrate across a range of x. If you pick only a few specific x values, your results will vary.

You don't have any question about that, do you? I am sure you have done this exercise plenty of times.

PS Somehow the formula got mangled along the way. It should be:

P(ab)=.25+.5(cos^2(a-b)) which has a range from .25 to .75

You are misunderstanding. The challenge is for you to start out with two separate functions each depending only on either a, or b for this classical case which have values ±1, and show that you can arrive at the result P(ab)=.25+.5(cos^2(a-b)) from those functions.

In other words, show that P(ab)=.25+.5(cos^2(a-b)) is factorable into 2 separate functions A(a)=±1, B(b)=±1. If you do not want to do the math yourself, please point to an article or textbook which shows this. Note the cited source MUST be deriving the classical result P(ab)=.25+.5(cos^2(a-b)) from two separable functions of the form A(a)=±1, B(b)=±1.

This is the challenge as I understand it, nothing about trials, nothing about convergence.

 

[DrChinese]

You are misunderstanding. The challenge is for you to start out with two separate functions each depending only on either a, or b for this classical case which have values ±1, and show that you can arrive at the result P(ab)=.25+.5(cos^2(a-b)) from those functions.

In other words, show that P(ab)=.25+.5(cos^2(a-b)) is factorable into 2 separate functions A(a)=±1, B(b)=±1. If you do not want to do the math yourself, please point to an article or textbook which shows this. Note the cited source MUST be deriving the classical result P(ab)=.25+.5(cos^2(a-b)) from two separable functions of the form A(a)=±1, B(b)=±1.

This is the challenge as I understand it, nothing about trials, nothing about convergence.

Umm, OK. Are you telling me YOU don't know this already? That's what I'm asking. (I know Gordon doesn't.)

 

[billschnieder]

Umm, OK. Are you telling me YOU don't know this already? That's what I'm asking. (I know Gordon doesn't.)

Whether I know it or not is not the challenge. The challenge is for you to provide it. So stop focusing on me and provide it already, to get this thread going rather than express frustration at every turn without actually providing what is being asked.

You earlier asked for clarification of what was being asked:

It's your model in post#1, so I assume you know where you are going with this. I don't really want to spend time covering elements of this that you already know and don't have any questions on. So it would be helpful to know what that is.

I have clarified it for you so please calculate away.

 

[DrChinese]

Whether I know it or not is not the challenge. The challenge is for you to provide it. So stop focusing on me and provide it already, ... I have clarified it for you so please calculate away.

You are not the boss of me.

 

[Gordon Watson]

Umm, OK. Are you telling me YOU don't know this already? That's what I'm asking. (I know Gordon doesn't.)

Ignored. Waiting for formal response.

 

[billschnieder]

You are not the boss of me.

I wish I were, I would take out the grey matter and bleach it as it appears that's the only thing that will help. As they say, you can take a horse to the stream but if the horse is convinced that fresh-air rather than water is the cure for it's dehydration, it will die right there without drinking it. Just because the horse says "you are not the boss of me", will not change it's fate.

 

[billschnieder]

It's been 2 weeks since this challenge was posted. Still not solution from Bell supporters. The silence is deafening.

Let me help this thread along by providing a hint to the Bell proponents as they continue their search to find separable functions which give the classical result, P(ab)=.25+.5(cos^2(a-b))

cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

All you have to do is factorize the expression into a product of two functions which take only a or b but not both as variables and map them to values ±1.

 

[Gordon Watson]

It's been 2 weeks since this challenge was posted. Still not solution from Bell supporters. The silence is deafening.

Let me help this thread along by providing a hint to the Bell proponents as they continue their search to find separable functions which give the classical result, P(ab)=.25+.5(cos^2(a-b))

cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

All you have to do is factorize the expression into a product of two functions which take only a or b but not both as variables and map them to values ±1.

..
Hi Bill, and many thanks for supporting the cause. It is appreciated; as is your understanding.

But let me hurriedly add: The denouement of the challenge may be as much a challenge to some of your beliefs as it is to some others around here.

Then again: It might come to nothing:

Maybe my approach is wrong? Maybe the approach is correct but the maths is wrong? Maybe I've erred totally? That's why I was interested to see what other approaches would emerge from serious physicists (as well as those who are not physicists but Bell devotees). It can be done on the back of an envelope; and surely I'm not getting my classical physics wrong?

For those that have not followed proceedings: The challenge was initially posed in the context of "Herbert's proof" -- and was moved at the request of the OP there. When I moved it I added an EDIT with the move. The EDIT reads: I'd like to understand how physicists and mathematicians deal with the above wholly classical setting using the protocol set by Bell (1964) when arriving at his theorem.

I did expect some to think it could not be done; in those cases I was interested to see why they thought that way. After all, it is a wholly classical setting. (For them the challenge was to just do what I do: Ask questions.)

Or to reflect on this, seriously and personally: "To what extent do I understand Bell if I cannot apply his local-realistic protocol to what is clearly and definitely a classical, realistic, and Einstein-local setting!?"

With my thanks again,

PS: In the interests of clarity: The challenge requires no new physics. The consequences, however, go against accepted physics and therefore need to be discussed elsewhere. I would welcome advice as to where that might be, on the Net.

Since this thread was initiated by me, it will not be a hi-jack if readers would add suggestions as to the best place to have any errors exposed, and the consequences discussed, on the Net. Thank you.

NB: I will have spasmodic access to the Net over the next few weeks. But I'll be back 100% soon, DV!

EDIT: To be very clear: I am confident that my analysis relating to "the classical challenge" is correct. Beyond that we need not go in this thread.

Gordon
..

 

[zonde]

Let x denote any variable of your choosing. Then (as in a standard Bell-analysis) Alice's results are represented by (1) A(a, x) = ±1 where a is any analyzer orientation of her choosing; Bob's by (2) B(b, x) = ±1 where b is any analyzer orientation of his choosing; (3) 0 $\leq$ ρ(x); (4) ∫ρ(x) dx = 1.

Let x denote common polarization of both light pulses. Now intensity of light passing trough the analyzer is given by Malus law $I=I_0cos^2(x-a)$ where a is angle of analyzer.
As you are not saying how we get discretized value ±1 from intensity of light "I" I will make assumption that +1 or -1 we get probabilistically and probability of getting +1 is $P(+1,x)=I/I_0=cos^2(x-a)$
Then probability of getting coincidently +1 by both Alice and Bob is given by $P(A,B,x)=cos^2(x-a)cos^2(x-b)$
To find out what will be probability of coincidence over many runs with different x we have to find:
$$\int_{0}^{2\pi}cos^2(x-a)cos^2(x-b)dx=\frac{\pi}{2}cos^2(b-a)+\frac{\pi}{4}$$
and after normalizing this expression so that average is 0.5 we get:
$$P(A,B)=\frac{1}{2}cos^2(b-a)+\frac{1}{4}$$

 

[Gordon Watson]

Let x denote common polarization of both light pulses. Now intensity of light passing trough the analyzer is given by Malus law $I=I_0cos^2(x-a)$ where a is angle of analyzer.
As you are not saying how we get discretized value ±1 from intensity of light "I" I will make assumption that +1 or -1 we get probabilistically and probability of getting +1 is $P(+1,x)=I/I_0=cos^2(x-a)$
Then probability of getting coincidently +1 by both Alice and Bob is given by $P(A,B,x)=cos^2(x-a)cos^2(x-b)$
To find out what will be probability of coincidence over many runs with different x we have to find:
$$\int_{0}^{2\pi}cos^2(x-a)cos^2(x-b)dx=\frac{\pi}{2}cos^2(b-a)+\frac{\pi}{4}$$
and after normalizing this expression so that average is 0.5 we get:
$$P(A,B)=\frac{1}{2}cos^2(b-a)+\frac{1}{4}$$

..

Dear zonde, many thanks for having a go!

This is a bit rushed, but I will be away from the Net for about a week so I trust this helps.

(I had hoped others would have a go too, offering different solutions, and am surprised so few have. Some seem to fear a trap, but there is no trap in this classical challenge, as far as I can see. Some, like DrC, seem to think it cannot be done; or that I cannot do it.)

The key, imho, is to start with the core of Bell's protocol: A(a, λ) = ±1. B(b, x) = ±1. For these represent, imho, the healthy application of Einstein-locality; every result (±1) must derive from local variables.

I use delta-functions to yield the ±1 results; via A and B with relevant trig functions.

ρ(x) will be uniform (imho) because x is a random variable.

So E(AB) follows by carefully tracking the probabilites of the various LOCAL outcomes.

Note that you can derive E(AB) in the manner that you suggest, but you have not given that. So the main point to work on is that your method is not yet in accord with Bell's protocol.

I believe this to be important for participation in any later discussions elsewhere.

EDIT: I should have added that I use the delta-functions to represent each quantum jump (from each particle-device interaction). For I do not endorse EPR elements of physical reality: but I ensure that every relevant element of the physical reality appears in my equations.

With thanks again, and hoping these clues help,

Gordon
..

 

[morrobay]

Let x denote common polarization of both light pulses. Now intensity of light passing trough the analyzer is given by Malus law $I=I_0cos^2(x-a)$ where a is angle of analyzer.
As you are not saying how we get discretized value ±1 from intensity of light "I" I will make assumption that +1 or -1 we get probabilistically and probability of getting +1 is $P(+1,x)=I/I_0=cos^2(x-a)$
Then probability of getting coincidently +1 by both Alice and Bob is given by $P(A,B,x)=cos^2(x-a)cos^2(x-b)$
To find out what will be probability of coincidence over many runs with different x we have to find:
$$\int_{0}^{2\pi}cos^2(x-a)cos^2(x-b)dx=\frac{\pi}{2}cos^2(b-a)+\frac{\pi}{4}$$
and after normalizing this expression so that average is 0.5 we get:
$$P(A,B)=\frac{1}{2}cos^2(b-a)+\frac{1}{4}$$

Nice work zonde ! Now with this can you formulate a classical inequality from scenario in
original post . That can be violated classically with locality and realism alone ?

 

[DrChinese]

It's been 2 weeks since this challenge was posted. Still not solution from Bell supporters. The silence is deafening.

There is no challenge yet. It's just a normal classical setup. I am still waiting for the actual challenge. You may not believe this, but I don't consider it my job to do someone else's work for them on their own setup. I already provided the answer, thought I was being a nice guy to help things along. I will post the derivation of it when I have time, but if someone else wants to do this first, fine.

Gordon, if you already know this stuff, please don't waste my time playing games. There are folks here that have legitimate questions and I am assuming yours is one. If you have a point to make, make it.

Bill, you embarass yourself with your remark that the silence is deafening. Do you seriously think there is some group of us cowering in the dark?

-DrC

 

[DrChinese]

Some, like DrC, seem to think it cannot be done; or that I cannot do it.)

Do what? There is nothing here. I already gave you the answer, .375, using the formula zonde derived (which IS the standard, known, classical result). .375 is higher and well within the Bell (classical) boundary of 1/3.

So applying Bell to a classical model tells you little of interest as this is exactly what you would expect.

So again I ask, what is the question or point you are driving at?

 

[DrChinese]

Let x denote common polarization of both light pulses. Now intensity of light passing trough the analyzer is given by Malus law $I=I_0cos^2(x-a)$ where a is angle of analyzer.
As you are not saying how we get discretized value ±1 from intensity of light "I" I will make assumption that +1 or -1 we get probabilistically and probability of getting +1 is $P(+1,x)=I/I_0=cos^2(x-a)$
Then probability of getting coincidently +1 by both Alice and Bob is given by $P(A,B,x)=cos^2(x-a)cos^2(x-b)$
To find out what will be probability of coincidence over many runs with different x we have to find:
$$\int_{0}^{2\pi}cos^2(x-a)cos^2(x-b)dx=\frac{\pi}{2}cos^2(b-a)+\frac{\pi}{4}$$
and after normalizing this expression so that average is 0.5 we get:
$$P(A,B)=\frac{1}{2}cos^2(b-a)+\frac{1}{4}$$

Thanks zonde! Very helpful.

Is there a sin*sin component necessary too in addition to the cos*cos one above? That's what I was starting with...

 

[billschnieder]

As you are not saying how we get discretized value ±1 from intensity of light "I" I will make assumption that +1 or -1 we get probabilistically and probability of getting +1 is $P(+1,x)=I/I_0=cos^2(x-a)$
Then probability of getting coincidently +1 by both Alice and Bob is given by $P(A,B,x)=cos^2(x-a)cos^2(x-b)$

So what are the FUNCTIONS A(a,x) and B(a,x), please clearly state those functions. Remember they can only have values ±1. $cos^2(x-a)$ can not be a valid function according to Bell. A and B must be step functions according to Bell so you have to be integrating the product of two step functions rather than the harmonic ones you have.

and after normalizing this expression so that average is 0.5 we get:

Another gem. But something tells me, "don't go there". This is Gordon's thread so

 

[DrChinese]

So what are the FUNCTIONS A(a,x) and B(a,x), please clearly state those functions. Remember they can only have values ±1. $cos^2(x-a)$ can not be a valid function according to Bell. A and B must be step functions according to Bell so you have to be integrating the product of two step functions rather than the harmonic ones you have.

There is nothing wrong with a function like A(a,x)=cos^2(x-a) when it produces a +1 outcome or a -1 outcome. zonde assumes everyone knows this, we are looking for +1/+1 and -1/-1 permutations.

Again, this seems to be a pointless exercise. I have yet to see the slightest indication that this is anything more than a wild goose chase. Gordon, is there a point in here? If you and bill want to just a batch of baseless claims (which is all that has happened to date), we may as well stop here.

I will say again: this example follows Bell and comes to a completely expected conclusion. Here is Gordon's classical algorithm which respects Bell.

 

[Gordon Watson]

There is nothing wrong with a function like A(a,x)=cos^2(x-a) when it produces a +1 outcome or a -1 outcome. zonde assumes everyone knows this, we are looking for +1/+1 and -1/-1 permutations.

Again, this seems to be a pointless exercise. I have yet to see the slightest indication that this is anything more than a wild goose chase. Gordon, is there a point in here? If you and bill want to just a batch of baseless claims (which is all that has happened to date), we may as well stop here.

I will say again: this example follows Bell and comes to a completely expected conclusion. Here is Gordon's classical algorithm which respects Bell.

..

Uh? HERE is what? I don't see it? Please repost. Or have you done it non-locally?


MOST PEOPLE who RESPECT BELL (or at least understand his maths -- which I understand is not a strength of yours; SEE BELOW) would by now have gotten the point of the challenge!

Bill Schnieder and zonde seem to get it; or be having a go.

Note that I am (via the OP) interested in how Bell's supporters tackle the challenge.

I am not so interested in them jumping on the bandwagon of my approach.

Please read my posts again. And ask clearer questions. I'm not into gaming with Bell.

ONE POINT (for you) IS to start where BELL STARTS!

PS: And, since you did not think of doing it my way, make sure you come with your own way. BUT: For sure, your A(a,x)=cos^2(x-a) is just plain silly!

..

 

[Gordon Watson]

So what are the FUNCTIONS A(a,x) and B(a,x), please clearly state those functions. Remember they can only have values ±1. $cos^2(x-a)$ can not be a valid function according to Bell. A and B must be step functions according to Bell so you have to be integrating the product of two step functions rather than the harmonic ones you have.


Another gem. But something tells me, "don't go there". This is Gordon's thread so

Bill, you seem to understand the challenge very clearly. So PLEASE go for it here, especially while I'm away.

So: A BIG Thank-You from me.

PS: This --- "integrating the product of two step functions" -- is the hard way to go!

So look for a simpler solution that most everyone here might understand.

But you're on the ball, IMHOWMSBW!

Thanks again,

Gordon
..

..

 

[Gordon Watson]

Gordon: the point is that whatever the functions A, B and whatever the angles and whatever the probability distribution of the hidden variables, CHSH will be satisfied.

..
gill1109,

With respect: That is no point of the OP, as far I see it?

Here's an interesting real point: I had hoped (for sure), that you would have understood the challenge. For I was certainly sure that you would have seen how to proceed.

From the clues given, do you still not see it, via my way?

BUT the point was to see it via your way. Once I post my way here, I've lost the chance to see how Bell's supporters INDEPENDENTLY answered the "challenge" .. for which there are already too many clues here.

So, for me now (and maybe you could help): Where might I best discuss the consequences of this challenge on the Net?

Thanks, as always,

Gordon
..

 

[Gordon Watson]

Thanks zonde! Very helpful.

Is there a sin*sin component necessary too in addition to the cos*cos one above? That's what I was starting with...

DrC, trying to be helpful: START ANEW!

Start with BELL!

Gordon

 

[DrChinese]

MOST PEOPLE who RESPECT BELL (or at least understand his maths -- which I understand is not a strength of yours; SEE BELOW) would by now have gotten the point of the challenge!...

I certainly don't need to defend my understanding of Bell or the math behind it. You are free to think as you like.

I cannot help you further beyond telling you that a classical model will obey Bell inequalities, as I have already shown you numerous times in this thread. You have not really provided a challenge, because you do not assert that your model will provide results consistent with the predictions of quantum mechanics. That would be a necessary part of any challenge.

This is the end of my involvement in this thread as a poster. Please do not make any subsequent statements which are inconsistent with generally accepted physics, or I will report you as I will continue to monitor the thread and your posts.

Gordon, I have tried to be patient. But honestly, your recent comments have gotten unusually rude for you. (Probably due to billschnieder's involvement, as rude is his norm - and that is a kind assessment.) It is always interesting to see folks justifying their rudeness by the supposed "correctness" of their viewpoint. Well, that is sadly par for the course in all types of discourse; but I would hope that intelligent people would somehow be clever enough to make their points without being so snippy.

 

[zonde]

MOST PEOPLE who RESPECT BELL (or at least understand his maths -- which I understand is not a strength of yours; SEE BELOW) would by now have gotten the point of the challenge!

Bill Schnieder and zonde seem to get it; or be having a go.

Sorry Gordon but I don't get the point of your challenge.
I just saw DrChinese posting this formula P(ab)=.25+.5(cos^2(a-b)) and Bill asking where did he get this formula.
As I knew what chain of reasoning leads to this formula I posted it.

And I am not sure I understand (can check correctness of) math of Bell theorem so I find Nick Herbert's type of proof much more convincing.

 

[Mark M]

Gordon, I don't see what you're trying to prove here. I'm assuming you're asking for a derivation of the CHSH inequality with your given conditions. If so, I'll give it a go:

Starting with $$A(a,x)=\pm 1$$ $$B(b,x)=\pm 1$$ and$$E(A,B)= \int AB p(x) dx$$ We can write $$E(A,B)= \int A(a,x)B(b,x)p(x)dx$$ Since $a, a', b, b'$ are settings for the detector we show that $$E(a,b)-E(a,{b}')=\int [A(a,x)B(b,x)-A(a,x)B({b}'x)]p(x)dx$$ $$=\int A(a,x)B(b,x)[1 \pm A({a}'x)B({b}'x)] p(x)dx - \int A(a,x)B({b}',x)[1 \pm A({a}',x)B({b}',x)]p(x)dx$$ Using the first inequality, we know that the quantities $[1 \pm A({a}',x)B({b}',x)]p(x)$ and $[1 \pm A({a}',x)B({b},x)]p(x)$ are non-negative. Also, we will use the triangle inequality on each side $$\left | E(a,b)-E(a,{b}') \right |\leq \int [1 \pm A({a}',x)B({b}',x)]p(x)dx]+\int [1 \pm A(a,x)B(b,x)]p(x)dx$$
Since $\int p(x)=1$ we can simplify to $$\left \lfloor E(a,b)-E(a,{b}') \right \rfloor\leq 2\pm[E({a}',{b}')+E({a}',b)]$$ Which includes the CHSH inequality, with a maximum value of 2. As per usual.

QED

 

[billschnieder]

Gordon, I don't see what you're trying to prove here.

Gordon is challenging us to derive the classical result P(ab)=¼+½(cos²(a-b)) for the experiment he proposed, by starting where Bell started. With two separable functions A(a,x) and B(b,x) defined with a codomain ±1.

Zonde has provided a derivation of the above classical result by starting from the two functions:

A(a,x) = cos²(x−a)
B(b,x) = cos²(x−a)

However, this deviates from Bell because Bell insisted that A(a,x) and B(b,x) can only have values ±1, so the two functions must obey that if they are to follow Bell. In Bell's original paper, he suggested A(a,x) = sign(a · x), and B(a,x) = - sign(b · x) where a,b,x are vectors. Those functions do satisfy the A(a,x) = ±1. So the challenge is to use functions of that type or any other type which has ONLY values ±1 and derive the well known classical result for the experiment described in the OP.

 

[Gordon Watson]

I certainly don't need to defend my understanding of Bell or the math behind it. You are free to think as you like.

I cannot help you further beyond telling you that a classical model will obey Bell inequalities, as I have already shown you numerous times in this thread. You have not really provided a challenge, because you do not assert that your model will provide results consistent with the predictions of quantum mechanics. That would be a necessary part of any challenge.

This is the end of my involvement in this thread as a poster. Please do not make any subsequent statements which are inconsistent with generally accepted physics, or I will report you as I will continue to monitor the thread and your posts.

Gordon, I have tried to be patient. But honestly, your recent comments have gotten unusually rude for you. (Probably due to billschnieder's involvement, as rude is his norm - and that is a kind assessment.) It is always interesting to see folks justifying their rudeness by the supposed "correctness" of their viewpoint. Well, that is sadly par for the course in all types of discourse; but I would hope that intelligent people would somehow be clever enough to make their points without being so snippy.

..
Dear DrChinese,

So that I might apologise, I'd welcome your pointing to any rudeness on my part.

If your response has something to do with me associating you with the word silly, then I defend my position as follows:

Silly is a good Bellian word, as you no-doubt know.

I asked for a good Bellian (1964) function A(a, x) = ±1 in a specific context: the context of a simple and specific CLASSICAL experiment.

You proffered the following:

(DrC-1) A(a, x) = ±1 = cos^2 (a, x)!​

Please Note:

(a): RHS (DrC-1) can NEVER equal MINUS ONE (-1)!

(b): RHS (DrC-1) can equal +1 ONLY when a = x!

(c): HOWEVER, in the specified context, the probability that a = x (prior to the measurement interaction) is ZERO (0): P(a = x| in specifed context) = 0.


So, dear DrC: Since your proffered function cannot equal -1, and has P= 0 of equalling +1, WHEN does it EVER equal ±1?


Thus, as to 'silliness', I thought your defended mistake was worse then vN's; that's all.

Gordon

..
EDIT: My reply here was made before I saw Bill's response immediately above. To be clear, I asked for E(AB). Given the difficulties that appear to be associated with its derivation, the answer is (if my maths is correct):

(GW-1) E(AB) = (1/2) cos 2(a, b).​

..