Bell's Inequality Explanation for Intelligent Non-Scientist

[ep1987]

I have literally spent all day reading and am still very much in the dark.

First off does anyone have a link to detailed blow by blow account that doesn't assume an understanding of advanced maths and physics concepts and notations but will actually address the issue in depth?

Here are some specifics which I'm struggling with:

1. How do you 'read' a unitary matrix?

2. Isn't introducing square roots basically introducing an indeterminate?

3. Is there an explanation or theory of why experiments are returning results which favour QM at certain 'angles'?

4. Can someone explain this, specifically amplitudes "So, cancellation between positive and negative amplitudes can be seen as the source of all quantum weirdness - the one thing that makes quantum mechanics different from classical probability theory"?

 

[Nugatory]

Our own DrChinese maintains a <understatement>pretty decent</understatement> web page on Bell's Theorem: http://www.drchinese.com/Bells_Theorem.htm

Try the "Bell's Theorem with easy math" section...

 

[DrChinese]

3. Is there an explanation or theory of why experiments are returning results which favour QM at certain 'angles'?

The theory is QM itself. If it is correct, then it SHOULD return correct results. So I will go out on a big limb: QM is a great theory as is.

There were classical attempts (EPR 1935) to reproduce the same results as QM. These met with some success until Bell (1964) came along. Only then was it discovered that there was any difference between the classical attempts and QM. And those differences related to correlations of entangled particle pairs. Little was known about these, as there hadn't been much actual experimental research - no one knew how to produce and harness them. Advances in optics filled in the gaps. It wasn't until the 1970's that detail research became available. After Aspect, things really broke open.

There are hundreds of new experiments on entanglement performed annually. They all focus on the differences between classical vs quantum predictions. The differences are found in far more areas than just a few angle settings for polarization. There are now literally hundreds of Bell-type inequalities, covering anything which can be entangled (which is a lot!).

 

[Fredrik]

1-2. I think you will have to be more specific about what you mean.

3. This would have to be a theory of physics that's better than QM at making predictions about the same experiments. No such theory is known.

4. The words "positive and negative" don't seem to belong there, but I think he's talking about how in QM, instead of adding up probabilities for mutually exclusive possibilities to get the total probability, you add up complex numbers called "amplitudes" associated with the classical possibilities, to get a total amplitude. Then you compute the probability as the square of the absolute value of the amplitude. If you want a better answer, you will at least have to post a reference to where you found that quote.

 

[Avodyne]

First off does anyone have a link to detailed blow by blow account that doesn't assume an understanding of advanced maths and physics concepts and notations but will actually address the issue in depth?

David Mermin's 1985 article in Physics Today, "Is the moon there when nobody looks? Reality and the quantum theory", is still the best explanation with the least math: https://cp3.irmp.ucl.ac.be/~maltoni/PHY1222/mermin_moon.pdf

I recommend skipping the historical intro and beginning at "A gedanken demonstration" on page 4.

 

[ep1987]

Thanks for the great replies so far; i'll read those articles later tonight.

My earlier thoughts were obviously garbled (qm will do that to a novice i guess).

With regards to clarification

3. http://en.wikipedia.org/wiki/Bell's_theorem The correlation graph there; why do experiments produce the correlations that QM predicts?

 

[atyy]

3. http://en.wikipedia.org/wiki/Bell's_theorem The correlation graph there; why do experiments produce the correlations that QM predicts?

Because QM is correct.

 

[Fredrik]

why do experiments produce the correlations that QM predicts?

No one knows why.

Don't make the mistake to think that this is a problem. The question "Why does theory X make such accurate predictions about these experiments?" can only be answered when theory X is inferior to some other theory and that theory answers the question.

The only "problem" is that QM is superior to every other theory.

 

[bahamagreen]

Quantum mechanics is definitive; ("reality" is frequently incorrect).

 

[.Scott]

3. Is there an explanation or theory of why experiments are returning results which favour QM at certain 'angles'?

The explanation is exceedingly simple: QM is a description based on experimental results. In other words, someone has already done this experiment and figured out what equations fit the results. QM is simply the equations that have been discovered through such experimentation.

The "common sense" (but wrong) way of thinking about two entangled particles is that at the time they split from each other, they each got a separate copy of a common attribute, for example a "spin angle", and when a spin angle measurement is made on each particle, it is completely independent of what has happened or will happen to the other entangled particle.

What's important about the Bell Inequality is that it describes how QM (those equations based on experiments) makes predictions that cannot be explained by the "common sense" model described above. Instead, something non-local is happening - and the angle measured on one particle has an influence on the measured results from the other - even when those measurements occur meters or kilometers apart. (very non-local)

So the point is that, not only is QM right, but it is also non-local. That is, not only does it contradict classical mechanics, it also violates any set of rules based on that same notion that effects are entirely the result of local causes.

 

[DrChinese]

...The "common sense" (but wrong) way of thinking about two entangled particles is that at the time they split from each other, they each got a separate copy of a common attribute, for example a "spin angle", and when a spin angle measurement is made on each particle, it is completely independent of what has happened or will happen to the other entangled particle.

...

So the point is that, not only is QM right, but it is also non-local.

Sometimes this is called "quantum non-locality" to cover non-realistic interpretations as well as non-local interpretations. One of the Bell assumptions is that the choice of measurement by Alice does not affect the results of Bob, and vice versa. There could be other ("non-realistic or non-causal") mechanisms that violate this assumption that do in fact respect c.

 

[ep1987]

After reading the links to Dr. Chinese explanation things are a lot clearer (i've also refreshed my understanding of quantum superposition and the like).

There are two things which I'm still having trouble with:

1. Where does the math which predicts the correlations which result in CSHS inequalities come from? What do the experimental results tell us about the nature of particles at the quantum level? When testing the spin of entangled particles doe the results not indicate some violation of the conservation of angular momentum?

2. There seems to be a logical jump from classic probability not matching experimental results to local realism being discredited.

 

[Nugatory]

2. There seems to be a logical jump from classic probability not matching experimental results to local realism being discredited.

Classical probability matches the experimental results just fine - you just say that the probability of photon absorption or transmission at one detector is a function of the setting at both detectors and you can get the QM prediction that matches experimental results.

Bell's theorem shows that all theories in which the probability is not a function of the position of both detectors (which is, loosely speaking, local and realistic theories) will necessarily make predictions that obey Bell's inequality.

But either way, we're applying classic probability theory, just starting with different assumptions about what might affect the probability of detection at one of the detectors.

 

[atyy]

After reading the links to Dr. Chinese explanation things are a lot clearer (i've also refreshed my understanding of quantum superposition and the like).

There are two things which I'm still having trouble with:

1. Where does the math which predicts the correlations which result in CSHS inequalities come from? What do the experimental results tell us about the nature of particles at the quantum level? When testing the spin of entangled particles doe the results not indicate some violation of the conservation of angular momentum?

2. There seems to be a logical jump from classic probability not matching experimental results to local realism being discredited.

Alice can set a measurement setting $a$ to one of a number of possibilities, and she can get a result $A$ which can take anyone of several possibilities. Similarly, Bob can set a measurement setting $b$ to one of a number of possibilities, and she can get a result $B$ which can take anyone of several possibilities. We also assume that there are variables $\lambda$ which describe the entire past of Alice and Bob. So in general there will be some probability distribution of the joint results $A$ and $B$, ie. $P(A,B|a,b,\lambda)$.

The locality assumption is that because the variable $\lambda$ specifies the complete past of Alice and Bob, to specify the probability of Alice's outcome $A$ we only need $\lambda$ and her local measurement setting $a$. In other words, in a local theory we don't need Bob's result and measurement setting to know the probabilities of Alice's result, if we already have $\lambda$, because $\lambda$ also controls Bob's results, ie. $P(A,B|a,b,\lambda) = P(A|a,\lambda)P(B|b,\lambda)$

Actually there are some other assumptions needed to go from "factorizability" to "nonlocality" and you can read about them at
http://arxiv.org/abs/1303.3081
http://arxiv.org/abs/1303.2849
http://www.scholarpedia.org/article/Bell's_theorem

See especially Fig 2 of the scholarpedia article: "Figure 2: Spacetime diagram for EPR–Bell type experiment. Region 3 closes off the past light cones of both regions 1 and 2 and shields each of those regions off from their overlapping past light cones. Hence, according to Bell's definition of locality, a complete specification of local beables in 3 should render information about goings-on in 2 (resp., 1) irrelevant for predictions about 1 (resp., 2)."

 

[DrChinese]

2. There seems to be a logical jump from classic probability not matching experimental results to local realism being discredited.

Hopefully you have a fair understanding of EPR (1935). This laid out the definitions generally accepted for local realism:

a) Locality = no spooky action at a distance.
b) Elements of reality (ie "realism") which need not be simultaneously observable to be simultaneously real (by assumption).

Although my Bell pages may sadly and poorly represent Bell's argument, it was precisely these two things which were addressed in Bell's 1964 paper. Keep in mind that in 1935, there was no detail analysis of quantum entanglement and in fact the word had just been coined.

If there is a jump involved, I think it is more accepting that realism has specific implications not present in QM. Once you say what being realistic means, and apply Bell, you should be able to accept the contradiction with QM.

 

[billschnieder]

The locality assumption is that because the variable $\lambda$ specifies the complete past of Alice and Bob, to specify the probability of Alice's outcome $A$ we only need $\lambda$ and her local measurement setting $a$. In other words, in a local theory we don't need Bob's result and measurement setting to know the probabilities of Alice's result, if we already have $\lambda$, because $\lambda$ also controls Bob's results, ie. $P(A,B|a,b,\lambda) = P(A|a,\lambda)P(B|b,\lambda)$

According to probability theory the correct joint probability equation is $P(A,B|a,b,\lambda) = P(A|a,b,\lambda)P(B|A, a, b,\lambda)$ This equation is universally valid, no?

However in the case when $B,b$ do not add provide any other information, the expression, and only in that case will $P(B|A, a, b,\lambda) = P(B b,\lambda)$ and therefore in that case $P(A,B|a,b,\lambda) = P(A|a,\lambda)P(B|b,\lambda)$ is also a valid expression for the joint probability. But my point is this, even in this case the former full expression is still correct and should give you the exact same result if you are calculating correctly. Agreed?

In Bell test experiments, the P(A,B|a,b,λ) is calculated from coincidence data. In this case $a,\lambda$ is not enough information.

 

[johana]

According to probability theory the correct joint probability equation is $P(A,B|a,b,\lambda) = P(A|a,b,\lambda)P(B|A, a, b,\lambda)$ This equation is universally valid, no?

Isn't $\lambda$ supposed to be a function rather than just a number? So that variables are:

unknown A & B photon polarization (?a, ?b),
and polarizer A & B angles (a,b), then for QM:


$E(a,b) = \lambda(a,b) = cos^2(a-b) - sin^2(a-b)$

...and for local theory:

$E(a,b) = E(\lambda(?a, a), \lambda(?b, b)) = cos^2(a-b) - sin^2(a-b)$

 

[billschnieder]

Isn't $\lambda$ supposed to be a function rather than just a number? So that variables are:

$\lambda$ represents all hidden variables. Its exact nature is unspecied. $A, B$ represent outcomes at Alice and Bob respectively, with possible values +1/-1. $a,b$ represent Alice and Bobs settings, not polarisation. Polarisation should be accounted for by $\lambda$.

 

[billschnieder]

hen for QM:


$E(a,b) = \lambda(a,b) = cos^2(a-b) - sin^2(a-b)$

...and for local theory:

$E(a,b) = E(\lambda(?a, a), \lambda(?b, b)) = cos^2(a-b) - sin^2(a-b)$

No. $\lambda(a,b)$ is not correct. Remember that $E(a,b)$ is simply the expectation value of the paired product of Alice and Bobs result. That is the limit of the average <A×B> as N approaches infinity.

 

[johana]

$\lambda$ represents all hidden variables.

It either represents, or is a function of. Do you agree there is a difference between variable number and constant function that gives variable result depending on variable input?

Its exact nature is unspecied.

What do you mean? We have our input parameters (variables): photons A & B polarization which is unknown, and polarizer A & B polarization angles which are known. And we know the result we are supposed to get. What more could you wish for?

So I say what we are looking for now is not a variable, but CONSTANT function which will give correct result that varies according to those initial input variables, i.e. one function which works for any (a-b) angle combination.

But if you say we are not after a function and instead looking for some variable that is independent number by itself, then what is the function this new variable needs to be plugged into to get the expected result?

Polarisation should be accounted for by $\lambda$.

How can $\lambda$ account for photon polarization if it is not a function of photon polarization to begin with?

So what I'm saying is that this is may be complete expression for QM:

$P(A,B|a,b,\lambda) = P(A|a,\lambda)P(B|b,\lambda)$


... but for local theory $\lambda$ is SEPARATE and may be different for a and b, so it should be:

$P(A,B|a,b,\lambda_a,\lambda_b) = P(A|a,\lambda_a)P(B|b,\lambda_b)$

 

[billschnieder]

It either represents, or is a function of. Do you agree there is a difference between variable number and constant function that gives variable result depending on variable input?

Yes there is a difference but that difference is irrelevant for Bell's theorem since they both fit the description of $\lambda$ used by Bell. In fact he says in his paper:

It is a matter of indifference in the following whether $\lambda$ represents a single variable, or a set, or even a set of functions, and whether the variables are discrete or continuous.

What do you mean? We have our input parameters (variables): photons A & B polarization which is unknown, and polarizer A & B polarization angles which are known. And we know the result we are supposed to get. What more could you wish for?

No. Bell does not use polarisation explicitly, as separate from $\lambda$. In discussions involving polarisation, everyone understands polarisation is just an example of $\lambda$.

So I say what we are looking for now is not a variable, but CONSTANT function which will give correct result that varies according to those initial input variables, i.e. one function which works for any (a-b) angle combination.But if you say we are not after a function and instead looking for some variable that is independent number by itself, then what is the function this new variable needs to be plugged into to get the expected result?

This is all clearly explained in Bell's original paper. The functions you are talking about are $A(a,\lambda), B(b,\lambda)$ which produce the results {+1,-1}. When you see expressions using $A,B$, it is understood that those are short-hand for those functions.

So what I'm saying is that this is may be complete expression for QM:

$P(A,B|a,b,\lambda) = P(A|a,\lambda)P(B|b,\lambda)$... but for local theory $\lambda$ is SEPARATE and may be different for a and b, so it should be:

$P(A,B|a,b,\lambda_a,\lambda_b) = P(A|a,\lambda_a)P(B|b,\lambda_b)$

No, you are confusing probability notation with function notation. $P(A|a,\lambda)$ is not a function.

 

[johana]

This is all clearly explained in Bell's original paper. The functions you are talking about are $A(a,\lambda), B(b,\lambda)$ which produce the results {+1,-1}.

Do you think LOCAL hidden variable $\lambda$ must be the same for both sides?

When you see expressions using $A,B$, it is understood that those are short-hand for those functions.

Those are only symbolic containers, real function we can actually evaluate for QM looks like this:

$E(a,b) = cos^2(a-b) - sin^2(a-b)$
$E(60,30) = cos^2(30) - sin^2(30) = 0.5$

So if I told you when a=60 and b=30, local hidden variable $\lambda$ is 42, then in what equation would you plug it into to check if it yields correct result?

 

[billschnieder]

Do you think LOCAL hidden variable $\lambda$ must be the same for both sides?

They can be the same or different. It doesn't matter to Bell's derivation. The point is that it is allowed for the particle at Alice to know about the hidden parameters of the particle at Bob even if they are different because both particles obtained those parameters from the source to start with so it is irrelevant to the Bell discussion whether they are the same or different.

Those are only symbolic containers, real function we can actually evaluate for QM looks like this:

$E(a,b) = cos^2(a-b) - sin^2(a-b)$
$E(60,30) = cos^2(30) - sin^2(30) = 0.5$

So if I told you when a=60 and b=30, local hidden variable $\lambda$ is 42, then in what equation would you plug it into to check if it yields correct result?

There is a difference between an expectation value and a probability. E(a,b) is not a probability but an expectation value. E(a,b) is different from the A and B functions.

$E(a,b) = \int_{\Lambda} A(a,\lambda)B(b,\lambda) P(\lambda) d\lambda$

This is equation (2) of Bell's original paper, have you read it? Bell used the notation P(a,b) instead of E(a,b) which should not be confused with probability. E(a,b) does not apply to a specific $\lambda$ but rather is a probability-weighted average over all possible $\lambda's$ $(\Lambda)$. In experiments it is a result of averaging the product of all the paired results obtained for the given pair of angles.

Your 42 example above is just one pair of results and since A, B can only take values of +1 or -1. It is impossible to obtain a paired product different from +1 or -1 for that specific lambda. The cosine relationship only applies to averages from many different pairs with many different lambdas. Bell's claim is that it is impossible to find two functions $A(a,\lambda), B(b,\lambda)$ which when averaged (or integrated) in this manner will reproduce the QM prediction for E(a,b).

 

[johana]

They can be the same or different. It doesn't matter to Bell's derivation. The point is that it is allowed for the particle at Alice to know about the hidden parameters of the particle at Bob even if they are different because both particles obtained those parameters from the source to start with so it is irrelevant to the Bell discussion whether they are the same or different.

$A(a,\lambda); B(b,\lambda) = A(a,\lambda_A); B(b,\lambda_B)$

Perhaps it's irrelevant to Bell, but for LOCAL hidden variable do you think these two expressions are really equal or that left one can adequately represent the right one?

There is a difference between an expectation value and a probability. E(a,b) is not a probability but an expectation value. E(a,b) is different from the A and B functions.

It's a function that holds the final result we need to compare with in order to check if any local hidden variable function can match or not. Isn't it?

Your 42 example above is just one pair of results and since A, B can only take values of +1 or -1. It is impossible to obtain a paired product different from +1 or -1 for that specific lambda.

Do you think $\lambda$ can not be the same for each pair? What if $\lambda$ is a probability, the same probability for each pair and yet produces different +1/-1 results from pair to pair?

The cosine relationship only applies to averages from many different pairs with many different lambdas. Bell's claim is that it is impossible to find two functions $A(a,\lambda), B(b,\lambda)$ which when averaged (or integrated) in this manner will reproduce the QM prediction for E(a,b).

How about $A(a,\lambda) = cos^2((a-b)/2); B(b,\lambda) = cos^2((a-b)/2)$?

 

[billschnieder]

$A(a,\lambda); B(b,\lambda) = A(a,\lambda_A); B(b,\lambda_B)$

Perhaps it's irrelevant to Bell, but for LOCAL hidden variable do you think these two expressions are really equal or that left one can adequately represent the right one?

$\lambda =\{\lambda_A, \lambda_B\}$ so yes it can represent it.

It's a function that holds the final result we need to compare with in order to check if any local hidden variable function can match or not. Isn't it?

Yes.

Do you think $\lambda$ can not be the same for each pair? What if $\lambda$ is a probability, the same probability for each pair and yet produces different +1/-1 results from pair to pair?

It doesn't matter either way.

How about $A(a,\lambda) = cos^2((a-b)/2); B(b,\lambda) = cos^2((a-b)/2)$?

No. $A(a,\lambda), B(b,\lambda)$ can have only 2 possible values $\pm 1$.
$cos^2$ is a continuous function with only possitive values between 0 and 1.

 

[johana]

No. $A(a,\lambda), B(b,\lambda)$ can have only 2 possible values $\pm 1$.
$cos^2$ is a continuous function with only possitive values between 0 and 1.

I should have said it's a probability function, where 0 to 1 correspond to 0% - 100% the function will return +1, and otherwise -1. Something like this, for example:

$A(a,\lambda) = cos^2((a-b)/2)$
$B(b,\lambda) = cos^2((a-b)/2)$

$A(60, \lambda) = P(+) = cos^2(30)$ = 0.75
$B(30, \lambda) = P(+) = cos^2(30)$ = 0.75

$P(++ | --)$ = (0.75 * 0.75) + (0.25 * 0.25) = 0.625
$P(+- | -+)$ = (0.75 * 0.25) + (0.25 * 0.75) = 0.375

$E(60,30) = P_{++} + P_{--} - P_{+-} - P_{-+}$ = 0.625 - 0.375 = 0.25

 

[Nugatory]

I should have said it's a probability function, where 0 to 1 correspond to 0% - 100% the function will return +1, and otherwise -1. Something like this, for example:

$A(a,\lambda) = cos^2((a-b)/2)$
$B(b,\lambda) = cos^2((a-b)/2)$

Right... and those are both excluded by Bell's assumption as $A$ and $B$ are functions of both $a$ and $b$. There is no way of writing them in such a way that $A$ is a function of $a$ but not $b$ and $B$ is a function of $b$ but not $a$.

 

[johana]

Right... and those are both excluded by Bell's assumption as $A$ and $B$ are functions of both $a$ and $b$. There is no way of writing them in such a way that $A$ is a function of $a$ but not $b$ and $B$ is a function of $b$ but not $a$.

Yes, unless there is no rotational invariance, in which case a and b become relative to absolute rotational variance average instead of to each other.

 

[Nugatory]

Yes, unless there is no rotational invariance, in which case a and b become relative to absolute rotational variance average instead of to each other.

Even if there is no rotational invariance (that is, there is some anistropy in the environment such that the result at detector A(B) depend on the value of a(b) and some third angle c that represents the anistropy) you still cannot allow the result at A(B) to also be a function of b(a).

 

[billschnieder]

I should have said it's a probability function, where 0 to 1 correspond to 0% - 100% the function will return +1, and otherwise -1. Something like this, for example:

$A(a,\lambda) = cos^2((a-b)/2)$
$B(b,\lambda) = cos^2((a-b)/2)$

$A(60, \lambda) = P(+) = cos^2(30)$ = 0.75
$B(30, \lambda) = P(+) = cos^2(30)$ = 0.75

$P(++ | --)$ = (0.75 * 0.75) + (0.25 * 0.25) = 0.625
$P(+- | -+)$ = (0.75 * 0.25) + (0.25 * 0.75) = 0.375

$E(60,30) = P_{++} + P_{--} - P_{+-} - P_{-+}$ = 0.625 - 0.375 = 0.25

Then that has no relationship to Bell's A and B which by definition can only have values $\pm 1$. Your notation is recipe for confusion. $A(a,\lambda) = cos^2((a-b)/2)$ You have b in the a function. Please read Bell's paper.

Believe me when I tell you, you will not understand Bell if you go down this road, and this is coming from someone who believes Bell made a mistake. But you have to correctly understand Bell first.

 

[johana]

Then that has no relationship to Bell's A and B which by definition can only have values $\pm 1$.

I did say it's a probability function and that it does indeed return only $\pm 1$ values, as is evident from the equations. It's just like a coin has its probability function and it only returns heads or tails.

You are missing the point my example is illustrating, which is that hidden variables is a function, and not just any function, but probability function. Think about it. What else can it be? There is no other way to obtain meaningful result, is there?

Your notation is recipe for confusion. $A(a,\lambda) = cos^2((a-b)/2)$ You have b in the a function. Please read Bell's paper.

The important thing is that it actually calculates comparable result. Is this better:

$A(a,\lambda_A) = cos^2(a)$
$B(b,\lambda_B) = cos^2(b)$

$A(60, \lambda_A) = P(+) = cos^2(60)$ = 0.25
$B(30, \lambda_B) = P(+) = cos^2(30)$ = 0.75

$P(++ | --)$ = (0.25 * 0.75) + (0.75 * 0.25) = 0.375
$P(+- | -+)$ = (0.25 * 0.25) + (0.75 * 0.75) = 0.625

$E(60,30) = P_{++} + P_{--} - P_{+-} - P_{-+}$ = 0.375 - 0.625 = -0.25

 

[Avodyne]

No, it is not better. $A(a,\lambda_A)$ can have the values $+1$ or $-1$ only. Therefore it cannot equal $\cos^2(a)$.

 

[johana]

No, it is not better. $A(a,\lambda_A)$ can have the values $+1$ or $-1$ only. Therefore it cannot equal $\cos^2(a)$.

$A(a,\lambda_A) = cos^2(a), S$={+1,-1}

It's a probability function with sample space +1 and -1. What's the problem? Is there anything else I need to do with the notation to make that more clear?

 

[Avodyne]

The problem is that $A(a,\lambda_A)$ is most definitely not a probability function. It is the actual result that one would get if the polarizer was set to $a$ and the incoming photon carried hidden information $\lambda_A$. This actual result can have only the values $+1$ or $-1$.

The whole point of hidden-variable theory is that there is always an actual result that could be specified for every possible measurement, whether that measurement was made or not, and that we could predict those results if only we knew the values of the hidden variables.

The values of the hidden variables are assigned a probability distribution $\rho(\lambda_A,\lambda_B)$, but the results of measurements $A(a,\lambda_A)=\pm 1$ and $B(b,\lambda_B)=\pm 1$ are always definite.

 

[DrChinese]

$A(a,\lambda_A) = cos^2(a), S$={+1,-1}

It's a probability function with sample space +1 and -1. What's the problem? Is there anything else I need to do with the notation to make that more clear?

1. Since the above doesn't match up with anything we can make sense of, yes.

The function you have above doesn't return +1 or -1. It is not rotationally invariant either, since it depends only on a.2. Also regarding the formula you presented:

E(60,30)=P+++P−−−P+−−P−+ = 0.375 - 0.625 = -0.25

I can see this is supposed to be some kind of correlation related to settings of 60 and 30 degree measurement settings. But:

a) If this is Type II entangled pairs, it is not correct as the correlation of those will be sin^2(theta=30)-cos^2(theta=30)= -.5.

b) If the source stream of pairs is polarized at 0 degrees (and therefore not entangled), then the correlation formula is correct as you have it.

c) And in fact if the source stream of pairs is such that every pair is polarized identically at some random angle, the resulting expectation value would be:

(cos^2(theta=30)-sin^2(theta=30))/2= -.25.

Note that b) and c) are different from the correlated expectation value for entangled pairs a). Here is an example of a local realistic theory that fails (ie one in which entangled pairs are supposed to be randomly polarized at some unknown angle). I think this is what you were trying to say, but it is not clear (and maybe you were saying something else entirely).