Did Bell's Theory Contain a Mathematical Flaw?

[billschnieder]

Is there a TeX WRAP command?

I think it is double back-slash, but it does not work in the forum. You may have to use two separate tex blocks.
(14)
P''(G, G'|H , a, b') = \sum_{k=1}^N \frac{1}{4}[(\lambda_{k} \hookrightarrow \lambda_{a}, \lambda'_{k} \hookrightarrow \lambda'_{a}) + (\lambda_{k} \hookrightarrow \lambda_{ai}, \lambda'_{k} \hookrightarrow \lambda'_{ai}) + (\lambda'_{k} \hookrightarrow \lambda'_{b}, \lambda_{k} \hookrightarrow \lambda_{b})
<br /> + (\lambda&#039;_{k} \hookrightarrow \lambda&#039;_{bi}, \lambda_{k} \hookrightarrow \lambda_{bi})] * (cos^2(a, \lambda_{k}) | H , a, \lambda_{k}) * (cos^2(b&#039;, \lambda&#039;_{k}) | H , b&#039;, \lambda&#039;_{k}) = \frac{1}{2} cos^2 (a, b&#039;).

 

[Gordon Watson]

I think it is double back-slash, but it does not work in the forum. You may have to use two separate tex blocks.
(14)
P&#039;&#039;(G, G&#039;|H , a, b&#039;) = \sum_{k=1}^N \frac{1}{4}[(\lambda_{k} \hookrightarrow \lambda_{a}, \lambda&#039;_{k} \hookrightarrow \lambda&#039;_{a}) + (\lambda_{k} \hookrightarrow \lambda_{ai}, \lambda&#039;_{k} \hookrightarrow \lambda&#039;_{ai}) + (\lambda&#039;_{k} \hookrightarrow \lambda&#039;_{b}, \lambda_{k} \hookrightarrow \lambda_{b})
<br /> + (\lambda&#039;_{k} \hookrightarrow \lambda&#039;_{bi}, \lambda_{k} \hookrightarrow \lambda_{bi})] * (cos^2(a, \lambda_{k}) | H , a, \lambda_{k}) * (cos^2(b&#039;, \lambda&#039;_{k}) | H , b&#039;, \lambda&#039;_{k}) = \frac{1}{2} cos^2 (a, b&#039;).

Dear billschnieder,

Great, many thanks. I love it when great minds think alike. I HAD that in the early submissions but deleted it and more when the vbar problem kept occurring. &&^%$. But I am a beginner - - in TeX.

Question: Are you happy with me using P in all the equations? Rather than E in some?

I am thinking that Eqn (2) should stay with a P but change its intro to Joint probability defined?

Note: I hope I avoid the need for E anywhere here, actually; whereas Bell should have used E in his 1964 (2). But his P arguments were over A and B, not G and G', like mine?

E of course comes later when we combine the 4 P-s with their signs.

Also: Once we have one correct P, consistent with QM -- then we know the E will be correct too?

EDIT: I am not yet brave enough to re-edit Post #49 again, so will tidy it up in a new Post after some more comments are received.

Thank you again,

JenniT

 

[billschnieder]

Dear billschnieder,
Question: Are you happy with me using P in all the equations? Rather than E in some?
I am thinking that Eqn (2) should stay with a P but change its intro to Joint probability defined?

You are correct, (2) is a marginalization with respect to (λ,λ'), not an expectation. Not sure about what name is appropriate since (1) is also a joint probability.

Note: I hope I avoid the need for E anywhere here, actually; whereas Bell should have used E in his 1964 (2). But his P arguments were over A and B, not G and G', like mine?

Bells (2) is an Expectation is because he is multiplying the functions A(.), B(.) with the probability P(λ) and integrating/Summing over all λ. A(.), and B(.) are not probability distributions but two-valued functions (+1, -1). In your case, G, G' are also two valued functions (0,1). Is this correct?

E of course comes later when we combine the 4 P-s with their signs.

Are you referring to equation (9). I do not understand what you are doing in equation (9) within the square brackets. I can see it lists all the possibilities for how a specific pair of (λ, λ') can be transformed during the interaction but is not clear to me what you are adding up within the square brackets, number of case instances?

 

[Gordon Watson]

You are correct, (2) is a marginalization with respect to (λ,λ'), not an expectation. Not sure about what name is appropriate since (1) is also a joint probability.

Looks like we're out of my comfort zone and right into yours -- which is good.

Please be insistent here.

I'd like to be accurate and reflect the standards accepted by the relevant disciplines.

What about: Name (1) = Local realism defined via stochastic independence?

What about: Name (2) = Local realism defined via marginalization over local beables?

Bells (2) is an Expectation is because he is multiplying the functions A(.), B(.) with the probability P(λ) and integrating/Summing over all λ. A(.), and B(.) are not probability distributions but two-valued functions (+1, -1). In your case, G, G' are also two valued functions (0,1). Is this correct?

I think not. G, G' are discrete outcomes.

Are you referring to equation (9). I do not understand what you are doing in equation (9) within the square brackets. I can see it lists all the possibilities for how a specific pair of (λ, λ') can be transformed during the interaction but is not clear to me what you are adding up within the square brackets, number of case instances?

(9) uses JesseM's notation re P3. I'm OK with that. I personally use rho(λ, λ'), it being a normalized probability distribution (in my perhaps loose terms)? Is rho reserved for continuous distributions only?

[x], x = all the equiprobable [edit: hence the 1/2 out front] mutually-exclusive collectively-exhaustive beables. The beauty of the theory is that's all there are ... as far as Alice is concerned. Yes?

Are you OK with the use of Bell's term beables?

EDIT: In co-variant (15) there are 4 equiprobable [hence the 1/4 out front] mutually-exclusive collectively-exhaustive beables. OK?

Thanks,

JenniT

 

[billschnieder]

I'd like to be accurate and reflect the standards accepted by the relevant disciplines.

What about: Name (1) = Local realism defined via stochastic independence?

What about: Name (2) = Local realism defined via marginalization over local beables?

Not trying to be pendantic but what about
(1) = Local realism defined ...
(2) = Marginalized over local beables

I think not. G, G' are discrete outcomes.

Ah, I see.

(9) uses JesseM's notation re P3. I'm OK with that. I personally use rho(λ, λ'), it being a normalized probability distribution (in my perhaps loose terms)? Is rho reserved for continuous distributions only?

[x], x = all the equiprobable [edit: hence the 1/2 out front] mutually-exclusive collectively-exhaustive beables. The beauty of the theory is that's all there are ... as far as Alice is concerned. Yes?

Still having problems. Let's use a coin analogy, we have two coins (CC') with mutually exclusive equi-probable outcomes HH', HT', TH', TT'

I can understand why you may write P(H,H') = 1/4,
What I don't understand is why you would write it as 1/4[(H,H')+(H,T') +(T,H') + (T,T')]
If the terms within [.] represent probabilities, then each term in the square bracket is 1/4 and they necessarily sum up to 1 no? Are you trying to list all the equi-probable outcomes? If so it could be made clearer if you put that in a sentence and simply state that P(H,H') = 1/4 and be done with it.

Are you OK with the use of Bell's term beables?

I welcome the term.

 

[Gordon Watson]

Not trying to be pendantic but what about
(1) = Local realism defined ...
(2) = Marginalized over local beables

1. You will never hear me object to pedantry.

2. I usually go with Bell's Cuisine idea that these equations are consequences of local causality. Any objection or pedantic improvement? Please let me know.

Ah, I see.

Good.

Still having problems. Let's use a coin analogy, we have two coins (CC') with mutually exclusive equi-probable outcomes HH', HT', TH', TT'

OK.

I can understand why you may write P(H,H') = 1/4,
What I don't understand is why you would write it as 1/4[(H,H')+(H,T') +(T,H') + (T,T')]

It is a density; a normalised probability distribution?

If the terms within [.] represent probabilities, then each term in the square bracket is 1/4 and they necessarily sum up to 1 no?

Stop. What is in [.] are not probabilities. They are the 4 possible outcomes, each equiprobable, so put the 1/4 in front of [.]. Then you have your normalized distribution.

Perhaps I am using terms incorrectly?[/QUOTE]

Are you trying to list all the equi-probable outcomes? If so it could be made clearer if you put that in a sentence and simply state that P(H,H') = 1/4 and be done with it.

Yes, that is what I am trying to do.

NB: BUT if I do not list them all, no one seems to see the consequent picture?

Incidentally. Ignore that Alice and Bob stuff here.

I need a co-variant distribution over the whole sample space, and I need readers to see that I have such. So it is the God view that is important. Alice and Bob are but angels in this business.

I welcome the term.

Great to hear that.

Thanks,

JenniT

 

[JesseM]

Hi Jenni, sorry about the delay, I ended up being away from my apt. most of this week...

RE-SUBMISSION OF POST# 44 to minimize server/TeX clash re \vbars. \uparrow substituted.

Some small clarifications added
Dear Jesse, Thank you for this very thoughtful note. Welcome back:

--------------------------------------------------A study based on Mermin, Spooky actions at a distance: Mysteries of the quantum theory. Encyclopedia Britannica, Chicago - The Great Ideas Today 1988.

NB: The singlet state (used in most EPRB-Bell tests) is invariant under rotations, not just about the line-of-flight axis.

--------------------------------------------------(1) Local realism defined:

P&#039;&#039;(G, G&#039; \uparrow H, a, b&#039;, \lambda_{k}, \lambda&#039;_{k}) = P(G \uparrow H, a, \lambda_{k})*P&#039;(G&#039; \uparrow H, b&#039;, \lambda&#039;_{k}).

(2) Expectation defined:

P&#039;&#039;(G, G&#039; \uparrow H , a, b&#039;) = \sum_{k=1}^N P_3(\lambda_{k}, \lambda&#039;_{k})*P(G \uparrow H , a, \lambda_{k})*P&#039;(G&#039; \uparrow H , b&#039;, \lambda&#039;_{k}).

--------------------------------------------------

Scene 1: Alice and Bob -- far far apart -- happily await the arrival of N twins: pairwise correlated photons -- one twin at a time to each, respectively bearing the properties \lambda_{k} and \lambda&#039;_{k} with k = 1 - N. Note: k is a number; other subscripts -- such as a, ai, b, bi -- define spin-orientations in 3-space.

Alice has a dichotomic polarizer-analyzer [a/ai] which respectively signals G or R [often termed V or H] as tested photons enter the related output channel. Bob likewise has [b'/bi'] signaling respectively G' or R' [perhaps termed V' or H']. In this experiment, a is an orientation orthogonal to the line-of-flight axis, ai is orthogonal to both a and the line-of-flight axis; primes (') denote elements of reality in Bob's locale; etc.

Alice, with her elegant simplicity, supposes fundamental particles to be magic gyroscopes with perfect rotational symmetries and related binding spin trajectories. She supposes their polarizers to be devices which burn off extrinsic spin and re-orient the intrinsic spin.

Alice supposes that \lambda_{k} and \lambda&#039;_{k} are spin vectors, unconstrained as to length and orientation -- but pairwise correlated by the relevant conservation laws attending their birth. That is, for this experiment, they are pairwise equal and not otherwise.

Recognizing the rotational symmetry of the cosine function, Alice supposes the lambdas to be such that, if only she knew more about them: Without in any way disturbing a photon, she could predict with certainty the positive outcome of any spin-test upon it. Alice writes:

(3)
P(G \uparrow H , a, \lambda_{k}) = (cos^2 (a, \lambda_{k}) \uparrow H, a, \lambda_{k}) = 0\oplus1 = 0 xor 1.

I'm not sure I understand, are you saying the angle between a and the spin vector is always going to be either 90 or 0, so that cos^2 of the angle between them is always 0 or 1? This would seem to require that the source "know in advance" what setting Alice was going to choose so it always created the particle with a spin vector parallel or orthogonal to Alice's setting, a violation of the "no-conspiracy condition" which appears in rigorous versions of Bell's proof (this condition says that P(λ) = P(λ|a), i.e. there is no correlation between hidden variables and the choice of detector setting). I suppose your condition that a and ai are orthogonal implies that the source doesn't have to know which specific choice Alice makes on each specific trial (since a lambda parallel/orthogonal to a will automatically be orthogonal/parallel to ai), but it still has to "know" the two possible angles Alice is choosing from and make sure it only emits lambdas parallel/orthogonal to them. And of course Bell's own proof does not restrict the choice of detector angles to all be parallel/orthogonal to one another, you could have an experiment with three settings a1=0, a2=60, and a3=120 for example.

Alice, supposing that elements of physical reality mediate all her assured (certain) positive outcomes, writes:

(4)
P(G \uparrow H , a, \lambda_{k}\hookrightarrow \lambda_{a}) = (cos^2 (a, \lambda_{k}) \uparrow H, a, \lambda_{k} \hookrightarrow \lambda_{a}) = cos^2 (a, \lambda_{a}) = 1.

(5)
P(R \uparrow H , ai, \lambda_{k}\hookrightarrow \lambda_{ai}) = (cos^2 (ai, \lambda_{k}) \uparrow H, ai, \lambda_{k} \hookrightarrow \lambda_{ai}) = cos^2 (ai, \lambda_{ai}) = 1.

OK, so λ_a just means a spin vector parallel to Alice's detector setting a, right? Your notation is odd--I think P(G \uparrow H , a, \lambda_{k}\hookrightarrow \lambda_{a}) means "the probability of result G given experiment H, detector setting a, and a value of k such that \lambda_k = \lambda_a"...if that's right, I think it would be more in line with standard notation conventions to write it as P(G | H, a, \lambda_k = \lambda_a )

In set notation, Alice defines elements of physical reality as CFCs (counterfactual conditionals) -- binding spin trajectories associated with the perfect rotational symmetries of pristine fundamental particles:

(6)
\{\lambda_{k} \hookrightarrow \lambda_{a} \uparrow \,If \,\lambda_{k} \rightarrow [a/ai] \,then \,\lambda_{k} \rightarrow \lambda_{a}\};

(7)
\{\lambda_{k} \hookrightarrow \lambda_{ai} \uparrow \,If \,\lambda_{k} \rightarrow [a/ai] \,then \,\lambda_{k} \rightarrow \lambda_{ai}\}.

I don't follow this. What "set notation" do you mean? What does the uparrow represent? Earlier you seemed to just replace the | denoting conditional probability with an uparrow, but these equations don't seem to be expressing conditional probabilities. Can you explain in words what these equations are saying?

So, in Alice's frame of reference:

(8)
P&#039;&#039;(G, G&#039; \uparrow H , a, b&#039;) = \sum_{k=1}^N P_3(\lambda_{k}, \lambda&#039;_{k})*P(G \uparrow H , a, \lambda_{k})*P&#039;(G&#039; \uparrow H , b&#039;, \lambda&#039;_{k}).

(9)
P_3(\lambda_{k}, \lambda&#039;_{k}) = (1/2)[(\lambda_{k} \hookrightarrow \lambda_{a}, \lambda&#039;_{k} \hookrightarrow \lambda&#039;_{a}) + (\lambda_{k} \hookrightarrow \lambda_{ai}, \lambda&#039;_{k} \hookrightarrow \lambda&#039;_{ai})].

Is the right side of equation 9 supposed to involve probabilities? If so I think it would make more sense to write it as P_3(\lambda_{k}, \lambda&#039;_{k}) = (1/2)[P(\lambda_{k} = \lambda_{a}, \lambda&#039;_{k} = \lambda&#039;_{a}) + P(\lambda_{k} = \lambda_{ai}, \lambda&#039;_{k} = \lambda&#039;_{ai})]...is that OK? Even so the equation is a little unclear, does the 1/2 just mean you assume P(\lambda_{k} = \lambda_{a}, \lambda&#039;_{k} = \lambda&#039;_{a}) = 1/2 and P(\lambda_{k} = \lambda_{ai}, \lambda&#039;_{k} = \lambda&#039;_{ai}) = 1/2, i.e. these are the only two possible ways the source can emit spin vectors and both ways are equally probable? Then the sum of the two would be 1, and you'd have to multiply that by 1/2 to get P_3(\lambda_{k}, \lambda&#039;_{k}) for any specific value of k (k=a or k=ai). If that's the idea, I think it would be a lot less convoluted to just say that k can only take one of two values, k=a or k=ai, (which could perhaps be represented by the numerals 1 and 2, so the sum would just be \sum_{k=1}^2) and that either way P_3(\lambda_{k}, \lambda&#039;_{k}) = 1/2. In that case we could go from equation (8) to a simplified version of (10):

P&#039;&#039;(G,G&#039;|H,a,b&#039;) = \sum_{k=1}^2 (1/2)*(cos^2(a,\lambda_k)|H,a,\lambda_k)*(cos^2(b&#039;,\lambda_k)|H,b,\lambda_k)

Let me know what you think of my comments/questions/suggested changes to notation so far, then we can proceed.

 

[Gordon Watson]

Hi Jenni, sorry about the delay, I ended up being away from my apt. most of this week...

I'm not sure I understand, are you saying the angle between a and the spin vector is always going to be either 90 or 0, so that cos^2 of the angle between them is always 0 or 1? This would seem to require that the source "know in advance" what setting Alice was going to choose so it always created the particle with a spin vector parallel or orthogonal to Alice's setting, a violation of the "no-conspiracy condition" which appears in rigorous versions of Bell's proof (this condition says that P(λ) = P(λ|a), i.e. there is no correlation between hidden variables and the choice of detector setting). I suppose your condition that a and ai are orthogonal implies that the source doesn't have to know which specific choice Alice makes on each specific trial (since a lambda parallel/orthogonal to a will automatically be orthogonal/parallel to ai), but it still has to "know" the two possible angles Alice is choosing from and make sure it only emits lambdas parallel/orthogonal to them. And of course Bell's own proof does not restrict the choice of detector angles to all be parallel/orthogonal to one another, you could have an experiment with three settings a1=0, a2=60, and a3=120 for example.

The angle between a and the spin vector is always going to be either 90 or 0, so that cos^2 of the angle between them is always 0 or 1. That is what polarizers do: They orient the randomly arriving (and so randomly specified k) particles in just that way.

So I believe that your questions should move to the random specifications? Yes?

Especially as you fix my notation below with an impossibility. Which is not a criticism of you at all -- just an excellent way to show what I cannot say. AND MY notation sure could do with some help -- to make it more "standard" to the disciplines of maths and physics.

OK, so λ_a just means a spin vector parallel to Alice's detector setting a, right? Your notation is odd--I think P(G \uparrow H , a, \lambda_{k}\hookrightarrow \lambda_{a}) means "the probability of result G given experiment H, detector setting a, and a value of k such that \lambda_k = \lambda_a"...if that's right, I think it would be more in line with standard notation conventions to write it as P(G | H, a, \lambda_k = \lambda_a )

Oh NO, goodness NO!

Could I get away with that!?

Really, that would be too easy, surely? Yes?

As I see it: This is the standard trick in QM -- as I say, as I see it.

You are assigning to an unmeasured photon a property that has probability zero prior to any measurement. It at the same time being a property that the subject photons will certainly have after the measurement interaction?

You have here the crux of my case, I think?

I don't follow this. What "set notation" do you mean? What does the uparrow represent? Earlier you seemed to just replace the | denoting conditional probability with an uparrow, but these equations don't seem to be expressing conditional probabilities. Can you explain in words what these equations are saying?

See note given earlier, below. Is that OK? How would you word it?

Is the right side of equation 9 supposed to involve probabilities? If so I think it would make more sense to write it as P_3(\lambda_{k}, \lambda&#039;_{k}) = (1/2)[P(\lambda_{k} = \lambda_{a}, \lambda&#039;_{k} = \lambda&#039;_{a}) + P(\lambda_{k} = \lambda_{ai}, \lambda&#039;_{k} = \lambda&#039;_{ai})]...is that OK? Even so the equation is a little unclear, does the 1/2 just mean you assume P(\lambda_{k} = \lambda_{a}, \lambda&#039;_{k} = \lambda&#039;_{a}) = 1/2 and P(\lambda_{k} = \lambda_{ai}, \lambda&#039;_{k} = \lambda&#039;_{ai}) = 1/2, i.e. these are the only two possible ways the source can emit spin vectors and both ways are equally probable? Then the sum of the two would be 1, and you'd have to multiply that by 1/2 to get P_3(\lambda_{k}, \lambda&#039;_{k}) for any specific value of k (k=a or k=ai). If that's the idea, I think it would be a lot less convoluted to just say that k can only take one of two values, k=a or k=ai, (which could perhaps be represented by the numerals 1 and 2, so the sum would just be \sum_{k=1}^2) and that either way P_3(\lambda_{k}, \lambda&#039;_{k}) = 1/2. In that case we could go from equation (8) to a simplified version of (10):

P&#039;&#039;(G,G&#039;|H,a,b&#039;) = \sum_{k=1}^2 (1/2)*(cos^2(a,\lambda_k)|H,a,\lambda_k)*(cos^2(b&#039;,\lambda_k)|H,b,\lambda_k)

NOW please see here, and note please: Any specific value of k (k=a or k=ai) has probability zero before a test and equiprobable certainty after a test. This simple dichotomy being the basis of our 2 lovely equivalence classes.

How would you put this FACT in your words and notation, please?

Also: Avoiding that zero-probability clash -- how about this?

What if we said something like this?

''Whereas before k was a number -- k = 1- N -- we now let k = 1 be the proxy for all the twins in equivalence class 1, and k = 2 be the proxy all the twins in equivalence class 2.''

How might you word that to be compatible with your specialist discipline?

Let me know what you think of my comments/questions/suggested changes to notation so far, then we can proceed.

Dear Jesse, so glad to learn that you are back home, safe and sound, and in good form -- from what I see here.

This is just a short reply to say "many thanks" again. And to knock off some rough edges.

1. To the last para: OK, will do.

2. The \uparrow was used to by-pass my big-time server clashes with too many \vbars. It is to be read as \vbar throughout (despite how much I like it as it is).

3. The set notation piece is simply saying that we are defining (just two will do) disjoint sets by giving the condition that the elements of each set must satisfy.

4. I think I see again the lovely simplification that billschnieder pointed out. Thanks to you too for that.

Pardon my caution in not replying too quickly: I need to be sure that absolutely nothing is lost or presumed when we simplify stuff.

PS: Some EDITS inserted above by me might help you comment progressively? I hope they do not confuse you?

More soon,

XXOOXXOO

Jenni

 

[JesseM]

The angle between a and the spin vector is always going to be either 90 or 0, so that cos^2 of the angle between them is always 0 or 1. That is what polarizers do: They orient the randomly arriving (and so randomly specified k) particles in just that way.

So do you mean the λ_k in cos^2(a, λ_k) refers to the angle of the particle's spin vector after it's been reoriented by the polarizer (when the angle between λ_k and a is always guaranteed to be 0 or 90), not the angle of the spin vector given to the particle when it's first sent out by the source, before any such reorientation? Is cos^2 supposed to be the probability it will pass through the polarizer (so the polarizer is imagined to first reorient, then allow to pass through or not) or is it the probability of something else happening after passing through it, and if so what?

Also, do the polarizers reorient the spin vectors in a deterministic way, so that if both polarizers are at the same angle and both spin vectors start out at the same k before passing through the polarizer, they are guaranteed to both end up parallel or both end up orthogonal, with no chance of one ending up parallel to the polarizer and one ending up orthogonal? If so we can say that at the moment the two particles first begin their journey from the source, their initial spin vectors should give them a predetermined answer to what direction they would end up if they met a polarizer at any given angle, right? For example, we might have a situation where one particle's initial spin vector (the spin vector assigned to it by the source, prior to encountering the polarizers) is such that it if it encountered a polarizer at a=30 it would be predetermined to end up parallel to it, predetermined to come out orthogonal a polarizer at a=60, and predetermined to come out parallel to a polarizer at a=120. Is this how you see things, or do you not agree that the angle of the initial spin vector plus the angle of the polarizer should deterministically generate the angle of the spin vector after being reoriented by the polarizer?

OK, so λ_a just means a spin vector parallel to Alice's detector setting a, right? Your notation is odd--I think P(G \uparrow H , a, \lambda_{k}\hookrightarrow \lambda_{a}) means "the probability of result G given experiment H, detector setting a, and a value of k such that \lambda_k = \lambda_a"...if that's right, I think it would be more in line with standard notation conventions to write it as P(G | H, a, \lambda_k = \lambda_a )

Oh NO, goodness NO!

"No" to what? Is my English summary of the meaning of your equation as "the probability of result G given experiment H, detector setting a, and a value of k such that \lambda_k = \lambda_a" incorrect? And given your point about "reorientation" here which I hadn't caught before, does λ_k in this equation represent the angle of the spin vector after it's been reoriented by the spin vector (when it's guaranteed to be parallel or orthogonal to a) or to the spin vector prior to reorientation? Also, does λ_a mean a spin vector parallel to a, with λ_ai representing a spin vector orthogonal to a?

You are assigning to an unmeasured photon a property that has probability zero prior to any measurement.

What property are you talking about?

I'll wait for your answers before continuing...

 

[Gordon Watson]

So do you mean the λ_k in cos^2(a, λ_k) refers to the angle of the particle's spin vector after it's been reoriented by the polarizer (when the angle between λ_k and a is always guaranteed to be 0 or 90), not the angle of the spin vector given to the particle when it's first sent out by the source, before any such reorientation?

As I see it, we are dealing with Probability Functions. They map a subset of the Sample Space to [0, 1].

λ_k is the angle of the spin vector given to the particle when it is first sent out by the source.

If that were not the case, it would be written λ_X, or something.

Any change in the sample space relating to λ_k will be reflected in the mapping.

It is my understanding that polarizers deterministically re-orient the spin.

So the Sample Space changes ... so the mapping changes correspondingly.

Is cos^2 supposed to be the probability it will pass through the polarizer (so the polarizer is imagined to first reorient, then allow to pass through or not) or is it the probability of something else happening after passing through it, and if so what?

Doesn't it have to be the first?

Also, do the polarizers reorient the spin vectors in a deterministic way, so that if both polarizers are at the same angle and both spin vectors start out at the same k before passing through the polarizer, they are guaranteed to both end up parallel or both end up orthogonal, with no chance of one ending up parallel to the polarizer and one ending up orthogonal?

This is Mermin's baby. In my view the answer here is Yes.

If it is something else, like a different singlet state, we just change the correlation and proceed.

If so we can say that at the moment the two particles first begin their journey from the source, their initial spin vectors should give them a predetermined answer to what direction they would end up if they met a polarizer at any given angle, right?

That is correct.

Is that what is required?

For example, we might have a situation where one particle's initial spin vector (the spin vector assigned to it by the source, prior to encountering the polarizers) is such that it if it encountered a polarizer at a=30 it would be predetermined to end up parallel to it, predetermined to come out orthogonal a polarizer at a=60, and predetermined to come out parallel to a polarizer at a=120. Is this how you see things,...

Yes.

IS THAT HOW IT SHOULD BE? Sorry for caps - hit wrong key.

... or do you not agree that the angle of the initial spin vector plus the angle of the polarizer should deterministically generate the angle of the spin vector after being reoriented by the polarizer?

Does that question make sense?

The angle of the initial spin vector plus the angle of the polarizer deterministically generate the NEW angle of the spin vector WITHIN THE the polarizer. THE PHOTON THEN ENTERS THE RELEVANT ANALYZER CHANNEL.

"No" to what? Is my English summary of the meaning of your equation as "the probability of result G given experiment H, detector setting a, and a value of k such that \lambda_k = \lambda_a" incorrect?

Yes. This property ... \lambda_k = \lambda_a ... has probability zero before the test!

Do you see that?

And given your point about "reorientation" here which I hadn't caught before, does λ_k in this equation represent the angle of the spin vector after it's been reoriented by the spin vector [SIC] POLARIZER Yes?

(when it's guaranteed to be parallel or orthogonal to a) or to the spin vector prior to reorientation?

No -- see above re Sample Space. You can't change λ_k in the trig argument except via the SS.

Also, does λ_a mean a spin vector parallel to a, with λ_ai representing a spin vector orthogonal to a?

Yes.

What property are you talking about?

This property \lambda_k = \lambda_a has probability zero before the test!

I'll wait for your answers before continuing...

Regards,

JenniT

 

[Gordon Watson]

Is the right side of equation 9 supposed to involve probabilities?

No, (9) is meant to represent a discrete normalized probability distribution. So LHS says what it is and RHS gives the normalized distribution.

If this were a continuous case then RHS would be written with delta-functions.

Does this last comment help?

If so I think it would make more sense to write it as P_3(\lambda_{k}, \lambda&#039;_{k}) = (1/2)[P(\lambda_{k} = \lambda_{a}, \lambda&#039;_{k} = \lambda&#039;_{a}) + P(\lambda_{k} = \lambda_{ai}, \lambda&#039;_{k} = \lambda&#039;_{ai})]...is that OK?

As explained in earlier reply: These equalities have probability zero prior to measurement.

Do you now agree?

Even so the equation is a little unclear, does the 1/2 just mean you assume P(\lambda_{k} = \lambda_{a}, \lambda&#039;_{k} = \lambda&#039;_{a}) = 1/2 and P(\lambda_{k} = \lambda_{ai}, \lambda&#039;_{k} = \lambda&#039;_{ai}) = 1/2, i.e. these are the only two possible ways the source can emit spin vectors and both ways are equally probable?

Yes, but you have to see that we simply cannot go with those equality signs.

We have to go with the characteristic spin-related transformation symmetries that define the equiprobable etc. equivalence classes (EC).

Then the sum of the two would be 1, and you'd have to multiply that by 1/2 to get P_3(\lambda_{k}, \lambda&#039;_{k}) for any specific value of k (k=a or k=ai). If that's the idea, I think it would be a lot less convoluted to just say that k can only take one of two values, k=a or k=ai, (which could perhaps be represented by the numerals 1 and 2, so the sum would just be \sum_{k=1}^2) and that either way P_3(\lambda_{k}, \lambda&#039;_{k}) = 1/2.

Check what I said in earlier reply, please.

k_1 now denotes the defining condition for EC_1, k_2 the defining condition for EC_2.

Do you go with that?

In that case we could go from equation (8) to a simplified version of (10):

P&#039;&#039;(G,G&#039;|H,a,b&#039;) = \sum_{k=1}^2 (1/2)*(cos^2(a,\lambda_k)|H,a,\lambda_k)*(cos^2(b&#039;,\lambda_k)|H,b,\lambda_k)

Well I don't see how this gives any output??

If you change it to this, well OK:

P&#039;&#039;(G,G&#039;|H,a,b&#039;) = \sum_{k=1}^2 (1/2)*(cos^2(a,\lambda_k)|H,a,\lambda_k = \lambda_1)*(cos^2(b&#039;,\lambda_k)|H,b, \lambda_k = \lambda_1,

where lambda_1 is defined by the \uparrow business -- which is a discriminator between the ECs.

And, for completeness you include the equivalent expression for lambda_2 -- which is also is defined by the \uparrow business -- it being also a discriminator between the ECs. Though this addition = 0, on summation, it is needed for completeness.

Which is all getting messy for me, and surely for you?

Thanks, as always,

Jenni

 

[JesseM]

As I see it, we are dealing with Probability Functions. They map a subset of the Sample Space to [0, 1].

λ_k is the angle of the spin vector given to the particle when it is first sent out by the source.

OK, so you were using the horizontal arrow in P(G \uparrow H , a, \lambda_{k}\hookrightarrow \lambda_{a}) to represent the possibility that the initial spin vector λ_k was changed to the new spin vector λ_a as the particle passed through the polarizer, correct? If so my earlier interpretation of the equation was wrong so much of my subsequent discussion of your equations was based on wrong premises, we should step back and start from here.

If so we can say that at the moment the two particles first begin their journey from the source, their initial spin vectors should give them a predetermined answer to what direction they would end up if they met a polarizer at any given angle, right?

That is correct.

Is that what is required?

Yes, that's what Bell would assume must be true under local realism where the experimenters choose detector angles randomly and always find that same detector angle=same result...it must be true (under local realism and the given experimental conditions) that the particles had identical predetermined responses for each detector angle (so presumably the source always emits pairs of particles with hidden variables that give them identical predetermined responses).

For example, we might have a situation where one particle's initial spin vector (the spin vector assigned to it by the source, prior to encountering the polarizers) is such that it if it encountered a polarizer at a=30 it would be predetermined to end up parallel to it, predetermined to come out orthogonal a polarizer at a=60, and predetermined to come out parallel to a polarizer at a=120. Is this how you see things,...

Yes.

IS THAT HOW IT SHOULD BE? Sorry for caps - hit wrong key.

Yes, according to local realism. OK, so let's focus on this issue for now. Given that each λ_k gives predetermined responses to each polarizer angle, would you agree that if the experimenters are choosing between three polarizer angles 0, 60, and 120, we can define a new variable L_i with only 8 possible values, like this:

L₁ means the original λ_k was such that the particle would be predetermined to pass through setting 0, to pass through setting 60, and to pass through setting 120. We can denote this 0+, 60+, 120+

L₂ means the original λ_k was such that the particle would be predetermined to pass through setting 0, to pass through setting 60, but not to pass through setting 120. We can denote this 0+, 60+, 120-

L₃: 0+, 60-, 120+

L₄: 0+, 60-, 120-

L₅: 0-, 60+, 120+

L₆: 0-, 60+, 120-

L₇: 0-, 60-, 120+

L₈: 0-, 60-, 120-

So no matter what direction λ_k is on any given trial, it must be associated with a single value of L_i, one of the 8 possible values above. Agreed? And to explain the fact that the experimenters always get identical results when they both choose the same polarizer angle, we must assume that on each trial, each particle has an identical value of L_i...e.g. if the particle sent to Alice has an initial spin vector λ_k that can be classified as L₅, then the particle sent to Bob must also have an initial spin vector that can be classified as L₅. Would you agree with this so far?

 

[Gordon Watson]

OK, so you were using the horizontal arrow in P(G \uparrow H , a, \lambda_{k}\hookrightarrow \lambda_{a}) to represent the possibility that the initial spin vector λ_k was changed to the new spin vector λ_a as the particle passed through the polarizer, correct? If so my earlier interpretation of the equation was wrong so much of my subsequent discussion of your equations was based on wrong premises, we should step back and start from here.


Yes, that's what Bell would assume must be true under local realism where the experimenters choose detector angles randomly and always find that same detector angle=same result...it must be true (under local realism and the given experimental conditions) that the particles had identical predetermined responses for each detector angle (so presumably the source always emits pairs of particles with hidden variables that give them identical predetermined responses).


Yes, according to local realism. OK, so let's focus on this issue for now. Given that each λ_k gives predetermined responses to each polarizer angle, would you agree that if the experimenters are choosing between three polarizer angles 0, 60, and 120, we can define a new variable L_i with only 8 possible values, like this:

L₁ means the original λ_k was such that the particle would be predetermined to pass through setting 0, to pass through setting 60, and to pass through setting 120. We can denote this 0+, 60+, 120+

L₂ means the original λ_k was such that the particle would be predetermined to pass through setting 0, to pass through setting 60, but not to pass through setting 120. We can denote this 0+, 60+, 120-

L₃: 0+, 60-, 120+

L₄: 0+, 60-, 120-

L₅: 0-, 60+, 120+

L₆: 0-, 60+, 120-

L₇: 0-, 60-, 120+

L₈: 0-, 60-, 120-

So no matter what direction λ_k is on any given trial, it must be associated with a single value of L_i, one of the 8 possible values above. Agreed? And to explain the fact that the experimenters always get identical results when they both choose the same polarizer angle, we must assume that on each trial, each particle has an identical value of L_i...e.g. if the particle sent to Alice has an initial spin vector λ_k that can be classified as L₅, then the particle sent to Bob must also have an initial spin vector that can be classified as L₅. Would you agree with this so far?

OK, thank you very much for these clarifications.

I think I am saying Yes to all the above; but maybe you should explain this small piece:

if the experimenters are choosing between three polarizer angles 0, 60, and 120, we can define a new variable L_i with only 8 possible values, like this

Q1: Why only 8?

Q2: Would it be better to discuss GHZ?

PS: Like, have you checked your mail-box?

Thank you,

JenniT

 

[JesseM]

OK, thank you very much for these clarifications.

I think I am saying Yes to all the above; but maybe you should explain this small piece:

if the experimenters are choosing between three polarizer angles 0, 60, and 120, we can define a new variable L_i with only 8 possible values, like this

Q1: Why only 8?

Because the point of L_i is just to specify which of two possible binary measurement results (+ or -) would be predetermined to occur for each of the three polarizer settings. Eight is the total number of possible combinations of +'s and -'s for the three angles:

0+, 60+, 120+
0+, 60+, 120-
0+, 60-, 120+
0+, 60-, 120-
0-, 60+, 120+
0-, 60+, 120-
0-, 60-, 120+
0-, 60-, 120-

Q2: Would it be better to discuss GHZ?

The point I'm getting at is just about whether cos^2(a-b) could be possible for the types of hidden variables theory you're assuming. I want to show that, given the assumption that every λ_k falls into one of the above 8 categories of predetermined responses above, it's impossible to get cos^2(a-b) for the probability they both get the same response when they choose different polarizer angles (note that for the polarizer angles 0, 60, and 120, if Alice and Bob choose different angles, cos^2(a-b) will always be equal to 0.25)

PS: Like, have you checked your mail-box?

Yup, sorry for the delay on that!

 

[Gordon Watson]

Because the point of L_i is just to specify which of two possible binary measurement results (+ or -) would be predetermined to occur for each of the three polarizer settings. Eight is the total number of possible combinations of +'s and -'s for the three angles:

0+, 60+, 120+
0+, 60+, 120-
0+, 60-, 120+
0+, 60-, 120-
0-, 60+, 120+
0-, 60+, 120-
0-, 60-, 120+
0-, 60-, 120-

The point I'm getting at is just about whether cos^2(a-b) could be possible for the types of hidden variables theory you're assuming. I want to show that, given the assumption that every λ_k falls into one of the above 8 categories of predetermined responses above, it's impossible to get cos^2(a-b) for the probability they both get the same response when they choose different polarizer angles (note that for the polarizer angles 0, 60, and 120, if Alice and Bob choose different angles, cos^2(a-b) will always be equal to 0.25)

Yup, sorry for the delay on that!

I'm also sorry for the delay too.

But note that lambda has to deliver results at any combination of any angular settings; so OK; please go ahead and demonstrate the point. If it fails at 8, I guess you'd agree it fails at a countable infinity of angular settings also. Right?

PS: On a related point, could you briefly explain what I should understand by spin? For example, photons are spin 1; electrons spin 1/2. What does this difference mean? What are its consequences, please? If someone said that electrons were spin 1, photons spin 2, would the world crumble? Why would they be wrong? Could an EPR-Bohm test shoot them down? Or some other test?

You guessed: I'm not very much into QM; mainly maths. Hope you can help; thanks.

Jennifer

 

[Gordon Watson]

I want to show that, given the assumption that every λ_k falls into one of the above 8 categories of predetermined responses above, it's impossible to get cos^2(a-b) for the probability they both get the same response when they choose different polarizer angles (note that for the polarizer angles 0, 60, and 120, if Alice and Bob choose different angles, cos^2(a-b) will always be equal to 0.25)

OK, good. So that I understand what is going on: Please be sure to show the QM result for however you arrive at your conclusion with the above assumptions, please.

Must run, J

 

[JesseM]

I'm also sorry for the delay too.

But note that lambda has to deliver results at any combination of any angular settings; so OK; please go ahead and demonstrate the point. If it fails at 8, I guess you'd agree it fails at a countable infinity of angular settings also. Right?

Right, although any given Bell test will involve only a small number of possible angular settings for the experimenters to choose from, like the three possible settings 0, 60 and 120 that I assumed are available to Alice and Bob. Of course, it's arbitrary what set of angles we choose to make available, I just chose 0, 60 and 120 for convenience.

So, assume that on each trial Alice and Bob choose their settings randomly, and independently. Then in the limit as the number of trials goes to infinity, each possible combination of settings will occur equally frequently--for example (Alice:60, Bob:120) occurs on the same fraction of trials as (Alice:120, Bob:0).

On each trial, both particles have the same value for Li (for example on any trial where Alice's particle has a λ_k corresponding to L₅, Bob's particle also has a λ_k corresponding to L₅). This means that in the subset of trials where Alice and Bob both chose the same detector setting, they always get the same measurement result with probability 1. But what if we look at the subset of trials where they chose different settings, what is the probability they got the same measurement result in that case? Here we want the conditional probability that they get the same result given that they chose different settings, which we can write as P(same measurement result | Alice&Bob chose different measurement settings).

Now, under the type of hidden variables theory you are proposing, would you agree we can break this up into the following sum?

P(same measurement result | Alice&Bob chose different measurement settings) =
P(same measurement result | Alice&Bob chose different measurement settings AND L₁)*P(L₁) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₂)*P(L₂) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₃)*P(L₃) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₄)*P(L₄) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₅)*P(L₅) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₆)*P(L₆) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₇)*P(L₇) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₈)*P(L₈)

If you agree this sum is valid, consider an individual term like P(same measurement result | Alice&Bob chose different measurement settings AND L₅). L₅ was 0-, 60+, 120+. Would you agree that since two predetermined results are + and one is -, the only way they can both get the same measurement result is if they both chose settings with predetermined result +? And would you agree that if we know Alice chose her setting randomly, there should be a 2/3 chance she chose a setting with predetermined result + (i.e. either 60 or 120), and if we know Bob also chose randomly and we are dealing with a trial where his setting was different from Alice's, there is a 1/2 chance he chose the other setting with predetermined result +? In other words, do you agree that P(same measurement result | Alice&Bob chose different measurement settings AND L₅) = (2/3)*(1/2) = 1/3? Another way of seeing this is just by considering all possible ways they can choose different settings, and what their results will be given particles in state L₅:
(Alice:0, Bob:60): different measurement results (Alice-, Bob+)
(Alice:0, Bob:120): different measurement results (Alice-, Bob+)
(Alice:60, Bob:0): different measurement results (Alice+, Bob-)
(Alice:60, Bob:120): same measurement results (Alice+, Bob+)
(Alice:120, Bob:0) different measurement results (Alice+, Bob-)
(Alice:120, Bob:60) same measurement results (Alice+, Bob+)

So, in 2 out of the 6 possible cases where they choose different settings, they get the same result, meaning P(same measurement result | Alice&Bob chose different measurement settings AND L₅) = 2/6 = 1/3.

If you agree with this, it's not hard to see why P(same measurement result | Alice&Bob chose different measurement settings AND L₂) would also equal 1/3 (since here you also have two + and one - for the predetermined results), and likewise for L₃, L₄, L₆ and L₇, which we might call the "inhomogenous states" since they each involve two predetermined results of one type and one of the other. On the other hand, for the "homogenous" states L₁ and L₈ where all three predetermined results are the same, then even if they choose different settings they are guaranteed to get the same result, so P(same measurement result | Alice&Bob chose different measurement settings AND L₁) = P(same measurement result | Alice&Bob chose different measurement settings AND L₈) = 1.

Let me know whether you agree with all of this so far, and if so I'll continue, if not I'll try to elaborate on whatever you're unclear about in the above.

 

[Gordon Watson]

Right, although any given Bell test will involve only a small number of possible angular settings for the experimenters to choose from, like the three possible settings 0, 60 and 120 that I assumed are available to Alice and Bob. Of course, it's arbitrary what set of angles we choose to make available, I just chose 0, 60 and 120 for convenience.

So, assume that on each trial Alice and Bob choose their settings randomly, and independently. Then in the limit as the number of trials goes to infinity, each possible combination of settings will occur equally frequently--for example (Alice:60, Bob:120) occurs on the same fraction of trials as (Alice:120, Bob:0).

On each trial, both particles have the same value for Li (for example on any trial where Alice's particle has a λ_k corresponding to L₅, Bob's particle also has a λ_k corresponding to L₅). This means that in the subset of trials where Alice and Bob both chose the same detector setting, they always get the same measurement result with probability 1. But what if we look at the subset of trials where they chose different settings, what is the probability they got the same measurement result in that case? Here we want the conditional probability that they get the same result given that they chose different settings, which we can write as P(same measurement result | Alice&Bob chose different measurement settings).

Now, under the type of hidden variables theory you are proposing, would you agree we can break this up into the following sum?

P(same measurement result | Alice&Bob chose different measurement settings) =
P(same measurement result | Alice&Bob chose different measurement settings AND L₁)*P(L₁) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₂)*P(L₂) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₃)*P(L₃) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₄)*P(L₄) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₅)*P(L₅) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₆)*P(L₆) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₇)*P(L₇) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₈)*P(L₈)

If you agree this sum is valid, consider an individual term like P(same measurement result | Alice&Bob chose different measurement settings AND L₅). L₅ was 0-, 60+, 120+. Would you agree that since two predetermined results are + and one is -, the only way they can both get the same measurement result is if they both chose settings with predetermined result +? And would you agree that if we know Alice chose her setting randomly, there should be a 2/3 chance she chose a setting with predetermined result + (i.e. either 60 or 120), and if we know Bob also chose randomly and we are dealing with a trial where his setting was different from Alice's, there is a 1/2 chance he chose the other setting with predetermined result +? In other words, do you agree that P(same measurement result | Alice&Bob chose different measurement settings AND L₅) = (2/3)*(1/2) = 1/3? Another way of seeing this is just by considering all possible ways they can choose different settings, and what their results will be given particles in state L₅:
(Alice:0, Bob:60): different measurement results (Alice-, Bob+)
(Alice:0, Bob:120): different measurement results (Alice-, Bob+)
(Alice:60, Bob:0): different measurement results (Alice+, Bob-)
(Alice:60, Bob:120): same measurement results (Alice+, Bob+)
(Alice:120, Bob:0) different measurement results (Alice+, Bob-)
(Alice:120, Bob:60) same measurement results (Alice+, Bob+)

So, in 2 out of the 6 possible cases where they choose different settings, they get the same result, meaning P(same measurement result | Alice&Bob chose different measurement settings AND L₅) = 2/6 = 1/3.

If you agree with this, it's not hard to see why P(same measurement result | Alice&Bob chose different measurement settings AND L₂) would also equal 1/3 (since here you also have two + and one - for the predetermined results), and likewise for L₃, L₄, L₆ and L₇, which we might call the "inhomogenous states" since they each involve two predetermined results of one type and one of the other. On the other hand, for the "homogenous" states L₁ and L₈ where all three predetermined results are the same, then even if they choose different settings they are guaranteed to get the same result, so P(same measurement result | Alice&Bob chose different measurement settings AND L₁) = P(same measurement result | Alice&Bob chose different measurement settings AND L₈) = 1.

Let me know whether you agree with all of this so far, and if so I'll continue, if not I'll try to elaborate on whatever you're unclear about in the above.

Pray, good sir; please continue -- this is getting interesting -- must rush; J xxoo

 

[JesseM]

OK, if you agreed that P(same measurement result | Alice&Bob chose different measurement settings AND L_i) would be equal to 1/3 if L_i was L₂, L₃, L₄, L₅, L₆ or L₇, and you also agreed that it would be equal to 1 if L_i was L₁ or L₈, then this equation:

P(same measurement result | Alice&Bob chose different measurement settings) =
P(same measurement result | Alice&Bob chose different measurement settings AND L₁)*P(L₁) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₂)*P(L₂) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₃)*P(L₃) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₄)*P(L₄) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₅)*P(L₅) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₆)*P(L₆) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₇)*P(L₇) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₈)*P(L₈)

...would reduce to this:

P(same measurement result | Alice&Bob chose different measurement settings) = 1/3*[P(L₂) + P(L₃) + P(L₄) + P(L₅) + P(L₆) + P(L₇)] + 1*[P(L₁) + P(L₈)]

Now, we don't know the value of each P(L_i), they represent the frequencies that the source sends out pairs of particles with different values of L_i, which would depend on how the source works in the context of our hidden-variables theory. But we do know that the sum of all 8 probabilities must be 1 since those are the only possible values for L_i, so [P(L₂) + P(L₃) + P(L₄) + P(L₅) + P(L₆) + P(L₇)] + [P(L₁ + P(L₈)] = 1.

If [P(L₂) + P(L₃) + P(L₄) + P(L₅) + P(L₆) + P(L₇)] = 1 while [P(L₁ + P(L₈)] = 0, then that would mean P(same measurement result | Alice&Bob chose different measurement settings) = 1/3. If [P(L₂) + P(L₃) + P(L₄) + P(L₅) + P(L₆) + P(L₇)] = 0 while [P(L₁ + P(L₈)] = 1, that would mean P(same measurement result | Alice&Bob chose different measurement settings) = 1. If both [P(L₂) + P(L₃) + P(L₄) + P(L₅) + P(L₆) + P(L₇)] and [P(L₁ + P(L₈)] have values between 0 and 1 (which together add up to one), then that would mean P(same measurement result | Alice&Bob chose different measurement settings) is somewhere between 1/3 and 1. In any case, according to this type of hidden variables theory it must be true that P(same measurement result | Alice&Bob chose different measurement settings) is larger than or equal to 1/3. Do you agree?

If so, it becomes easy to see why this type of hidden variables theory can't reproduce the cos^2 relationship predicted by QM. If the three angles are 0, 60, and 120, then on any trial where Alice and Bob pick different detector angles, the probability of getting the same result according to QM will always be 0.25, since cos^2(120 - 0) = cos^2(120 - 60) = cos^2(60 - 0) = cos^2(60 - 120) = cos^2(0 - 60) = cos^2(0 - 120) = 0.25, which is smaller than 1/3.