How does Bell make this step in his proof?

[zonde]

QM does not posit predetermination of all outcomes. Just that a few will be perfectly correlated.

Yes, of course. To get predetermination we need perfect correlation for matching angles and we have to assume independence of measurements at any angle.

 

[rubi]

Of course we all know that Bell inequalities are violated. As far as I know there is no loophole free experiment performed for GHZ but it's not very important.

As we have seen with the Bell tests, loopholes are likely going to be closed in the future. We currently have no reason to doubt the correctness of the quantum mechanical predictions.

Important question is from what assumptions one can construct Bell inequality. So that we know what assumptions have to be questioned when we learn about experimental violation of Bell inequalities. And this question can be discussed without much reference to experiments.

This question can be discussed without reference to experiments. We would spend 20 pages discussing what's wrong with the EPR argument. But we don't need to, since GHZ already proves that the EPR reasoning must be wrong, unless one believes in a loophole and the failure of quantum mechanical predictions.

 

[DrChinese]

Yes, of course. To get predetermination we need perfect correlation for matching angles and we have to assume independence of measurements at any angle.

To deduce predetermination, we need perfect correlation for matching angles, whether measured or not.

 

[N88]

Please: What is X?

We start here:

(1) $-[A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)]$; under an integral.

IF X [a mathematical expression, explained in words; preferably something suggested by Einstein or EPR, for (in his Introduction), that's where Bell says he starts from]

THEN (by Bell's equality):

(1) $=$ (2) $=A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)-1]$; under an integral. QED.

Thank you.

 

[atyy]

Forget the EPR argument. It might be intriguing, but it is falsified experimentally by GHZ. There are no elements of reality for unobserved spin directions, not even if you don't introduce $\lambda$. However, Bell clearly uses them in his proof.

Right. That is not the classical view (EPR) that Bell was writing about. QM does not posit predetermination of all outcomes. Just that a few will be perfectly correlated. The classical view goes much further. Specifically, that outcomes are predetermined at all possible measurement settings.

I'm not fond of the EPR elements of reality argument. However, Bell does show in one of his papers that the inequality remains the same even if the outcome is not predeternined at all possible measurement settings - where possible means "possible to the experimentalist".

 

[stevendaryl]

Please: What is X?

We start here:

(1) $-[A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)]$; under an integral.

IF X (a mathematical expression, explained in words; preferably something suggested by Einstein or EPR)

THEN (by Bell's equality):

(1) $=$ (2) $=A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)-1]$; under an integral. QED.

Thank you.

X is just algebra.

Once again:

Since A(\vec{b}, \lambda) = \pm 1, then A(\vec{b}, \lambda) A(\vec{b}, \lambda) = +1. So we can write:

- (A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a},\lambda) A(\vec{c},\lambda)) = - (A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a},\lambda) A(\vec{b}, \lambda) A(\vec{b}, \lambda) A(\vec{c},\lambda))

Now, we can factor out A(\vec{a},\lambda) A(\vec{b}, \lambda):

- (A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a},\lambda) A(\vec{c},\lambda)) = - A(\vec{a}, \lambda) A(\vec{b}, \lambda) (1 - A(\vec{b}, \lambda) A(\vec{c},\lambda))

 

[dlgoff]

Just that a few will be perfectly correlated.

I'm sure this is obvious for QM experts, but could you maybe say a little about the "few"?

 

[N88]

X is just algebra.

Once again: Since

(D) A(\vec{b}, \lambda) = \pm 1, then A(\vec{b}, \lambda) A(\vec{b}, \lambda) = +1. So we can write:

(E) - (A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a},\lambda) A(\vec{c},\lambda)) = - (A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a},\lambda) A(\vec{b}, \lambda) A(\vec{b}, \lambda) A(\vec{c},\lambda))

Now, we can factor out A(\vec{a},\lambda) A(\vec{b}, \lambda):

(F) - (A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a},\lambda) A(\vec{c},\lambda)) = - A(\vec{a}, \lambda) A(\vec{b}, \lambda) (1 - A(\vec{b}, \lambda) A(\vec{c},\lambda))

So A(\vec{a}, \lambda) and B(\vec{b}, \lambda) are assumed to be ordinary functions of two parameters. His proof then just amounts to showing that there are no such functions reproducing the predictions of quantum mechanics.

In reasoning about A and B, you can forget that they correspond to measurements, and just think of them as mathematical functions.

Here's the long answer: I see no connection with EPR. Though Bell claimed his formulation was connected to EPR, their idea was NOT this simplistic.

Here's the short answer: There are no such functions of two parameters in math or physics. But let's see if we can get around that dilemma.

I start with Thank You. It's my belief that you have reproduced Bell's argument (having been-there done-that myself); and it will be interesting to see if we can improve on it.

Accepting a solution by means of functions, then there must be three parameters, else they're not functions.

Say we assume A(\vec{a}, \lambda) and B(\vec{b}, \lambda) to be ordinary functions full-stop. Since a function cannot map the same inputs to two different values, you need to accept something like this: A(\vec{b}, \lambda+)= +1; A(\vec{b}, \lambda-)= -1.

So how do you now manage the physics of the problem? How do you ensure that, when you bring random A(\vec{b}, \lambda+)= +1 and random A(\vec{b}, \lambda-)= -1 together, their product is always one?

This earlier reply has a clue to the physical problem:

But in the second equation, the first $A(\vec{b},\lambda)$ was tested by Bob when Alice's detector was set to $\vec{a}$. The second $A(\vec{b},\lambda)$ in the second equation can be tested by Bob when Alice's detector is at setting $\vec{c}$. How does Bob come to have the same particles to test to get $A(\vec{b},\lambda)^2=1$? Because doesn't that relation $A(\vec{b},\lambda)^2=1$ only hold for tests on the same λ?

Looks like there's another assumption somewhere?

So here's the new question, under the function-based analysis with A(\vec{b}, \lambda+)= +1; A(\vec{b}, \lambda-)= -1:

How many \lambdas in (E) need be identical, to allow the reduction to (F)?

 

[stevendaryl]

So how do you now manage the physics of the problem? How do you ensure that, when you bring random A(\vec{b}, \lambda+)= +1 and random A(\vec{b}, \lambda-)= -1 together, their product is always one?

The claim made is that for a FIXED value of \lambda, it must be the case that

-(A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a}, \lambda) A(\vec{c}, \lambda)) = A(\vec{a}, \lambda) A(\vec{b}, \lambda) (1 - A(\vec{a}, \lambda) A(\vec{c}, \lambda))

That has been proved. You can ask: Why was Bell interested in proving this purely mathematical fact? Well, it was a step in proving the impossibility of a certain kind of hidden-variables theory that reproduced the predictions of quantum mechanics. But that particular step is pure mathematics.

 

[zonde]

Accepting a solution by means of functions, then there must be three parameters, else they're not functions.

Say we assume A(\vec{a}, \lambda) and B(\vec{b}, \lambda) to be ordinary functions full-stop. Since a function cannot map the same inputs to two different values, you need to accept something like this: A(\vec{b}, \lambda+)= +1; A(\vec{b}, \lambda-)= -1.

So how do you now manage the physics of the problem? How do you ensure that, when you bring random A(\vec{b}, \lambda+)= +1 and random A(\vec{b}, \lambda-)= -1 together, their product is always one?

You don't multiply $A(\vec{b}, \lambda+)$ with $A(\vec{b}, \lambda-)$ that's the point. You multiply $A(\vec{b}, \lambda_1)$ with $A(\vec{b}, \lambda_1)$
and $A(\vec{b}, \lambda_2)$ with $A(\vec{b}, \lambda_2)$ and so on. You never multiply $A(\vec{b}, \lambda_1)$ with $A(\vec{b}, \lambda_3)$.

So if you object that this is possible you have to backtrack Bell's reasoning to the point where he introduces expression $A(\vec{a}, \lambda)=\pm1$

 

[zonde]

Here's the short answer: There are no such functions of two parameters in math or physics.

Yes, that's true. But we couldn't be sure about that without Bell and experiments.

 

[DrChinese]

I'm sure this is obvious for QM experts, but could you maybe say a little about the "few"?

Where theta = 0 degrees, 90 degrees, 180 degrees, 270 degrees.

 

[atyy]

It's my belief that you have reproduced Bell's argument (having been-there done-that myself); and it will be interesting to see if we can improve on it.

Many people do think Bell was not quite clear initially. For modern presentations, I like:

https://arxiv.org/abs/1503.06413
Causarum Investigatio and the Two Bell's Theorems of John Bell
Howard M. Wiseman, Eric G. Cavalcanti

https://arxiv.org/abs/1208.4119
The lesson of causal discovery algorithms for quantum correlations: Causal explanations of Bell-inequality violations require fine-tuning
Christopher J. Wood, Robert W. Spekkens

Bell was more concerned about classical reality, but a very important corollary of his theorem is that if there is no faster than light signalling, then a violation of Bell inequalities implies operational randomness. This development could, in principle, be used to certify that random numbers are truly random for any practical purpose:

https://arxiv.org/abs/0911.3427
Random Numbers Certified by Bell's Theorem
S. Pironio, A. Acin, S. Massar, A. Boyer de la Giroday, D. N. Matsukevich, P. Maunz, S. Olmschenk, D. Hayes, L. Luo, T. A. Manning, C. Monroe

 

[zonde]

To deduce predetermination, we need perfect correlation for matching angles, whether measured or not.

We speak about model of reality not reality, right?
Now what it takes to claim that in particular model measurements of A and B are independent?
My answer is that it should be possible to split from the main model two independent submodels that take as an input something that represents two separate but entangled particles and separate measurement angle for each submodel.

 

[Nicky665]

These guys, Hess/DeRaedt/Michelsen, say the major assumption in Bell's inequalities is counterfactual definitess

https://arxiv.org/pdf/1605.04889.pdf

"The major premise for the derivation of Bell’s inequality is
counterfactual definiteness, which in connection with Bell’s
use of setting variables restricts the domain of the variables
in the argument of Bell’s functions A to a subset NB of general
physical independent variables. NB does not include the
variables necessary to describe a general dynamics describing
many body interactions in the measurement equipment.
Using only the independent variables defined by NB, it is impossible
to find a violation of Bell’s inequality, which therefore
represents a demarcation between possible and impossible
experience [8], not between classical and quantum physics."

 

[Heinera]

These guys, Hess/DeRaedt/Michelsen, say the major assumption in Bell's inequalities is counterfactual definitess

Counterfactual definiteness (also known as "realism") is one of the two major assumptions in Bell's theorem, yes (the other being locality). Nothing particularly novel in that paper.