SciPhysicsForums.com

by **Heinera** » Thu Jun 11, 2015 10:29 am

AnotherGuest wrote:
minkwe wrote:There is absolutely nothing wrong with
$P(a,b)=\int d\lambda_{ab}\rho(\lambda_{ab})A_a(\lambda_{ab})B_b(\lambda_{ab})$
and even if you still see any problem in the above integral, there should be absolutely no complain about
$P(a,b)=\int d\gamma\rho(\gamma)A_a(\gamma)B_b(\gamma)$ , where $\gamma$ is a non-local hidden variable.

What is important is that there is a fixed set over which these integrals are taken.

Minkwe is right that it doesn't matter (for Bell's subsequent analysis) what *name* you give to the variable of integration, as long as it varies over a fixed set (fixed means: not depending on a or b).
.

But since the intregation variable $d\gamma$ now depends on a and b, the integration is no longer over a fixed set, but over a subset that depends on a and b. This is equivalent to saying that the probability measure $\rho$ depends on a and b.

If the hidden variable depends on a and b, you can just as well say that the source (which produces the hidden variable) knows about a and b, and adjusts its distribution $\ro$ of generated lambdas accordingly. Bell's proof does not work in such a case.

by **minkwe** » Thu Jun 11, 2015 11:41 am

Heinera wrote:But since the intregation variable $d\gamma$ now depends on a and b, the integration is no longer over a fixed set, but over a subset that depends on a and b.

Again, $d\gamma$ does not depend on "a" and "b", if you've been following what exacly $d\gamma$ represents.
Secondly, "a" and "b" are not variables on the RHS of equation (2). They are fixed, they do not vary. The $A(a, \lambda)$ notation was a poor choice on Bell's part. It should have been something like $A_a(\lambda)$
Thirdly, I've given you other valid probability measures which do not involve any hidden variables, local or non-local, with expressions to calculate the same expectation value. The point being that, there is nothing in equation (2) that should be unique for local hidden variables. It is simply a weighted average of Alice's outcome multiplied by the corresponding Bob's outcome for all the unique possibilities in any measure. In experiments, which some nowadays claim as evidence of non-locality, such averages are routinely calculated without fanfare. I've not see any complains that their calculations are wrong.

There is no physical content in equation (2), it is a mathematical expression for the expectation value of the product of two outcomes irrespective of whether those outcomes are obtained locally or not.

Are you suggesting that Bell's derivation will fail to proceed if we replaced equation (2) with:

$P(a,b) = \frac{1}{N} \sum_{i}^{N} A_a^i B_b^i, \: N \to \infty$ , or $\langle A_aB_b \rangle$

This expression is theory-agnostic.

by **Heinera** » Thu Jun 11, 2015 11:52 am

minkwe wrote:Are you suggesting that Bell's derivation will fail to proceed if we replaced equation (2) with:

$P(a,b) = \frac{1}{N} \sum_{i}^{N} A_a^i B_b^i, \: N \to \infty$ , or $\langle A_aB_b \rangle$

Since there is no $\lambda$ nor any $\rho$ in that expression, Bell's derivation would obviously fail to proceed.

by **AnotherGuest** » Thu Jun 11, 2015 11:58 am

Heinera wrote:
AnotherGuest wrote:
minkwe wrote:There is absolutely nothing wrong with
$P(a,b)=\int d\lambda_{ab}\rho(\lambda_{ab})A_a(\lambda_{ab})B_b(\lambda_{ab})$
and even if you still see any problem in the above integral, there should be absolutely no complain about
$P(a,b)=\int d\gamma\rho(\gamma)A_a(\gamma)B_b(\gamma)$ , where $\gamma$ is a non-local hidden variable.

What is important is that there is a fixed set over which these integrals are taken.

Minkwe is right that it doesn't matter (for Bell's subsequent analysis) what *name* you give to the variable of integration, as long as it varies over a fixed set (fixed means: not depending on a or b).
.

But since the intregation variable $d\gamma$ now depends on a and b, the integration is no longer over a fixed set, but over a subset that depends on a and b. This is equivalent to saying that the probability measure $\rho$ depends on a and b.

If the hidden variable depends on a and b, you can just as well say that the source (which produces the hidden variable) knows about a and b, and adjusts its distribution $\rho$ of generated lambdas accordingly. Bell's proof does not work in such a case.

As far as I know mathematics, if $f$ is some function defined on a set $X$ , then in the expression $\int_{x\in X}f(x) d x$ you can give the variable $x$ any other name you like, as long as it is not simultaneously in use for something else. You can give it the name $\lambda_{ab}$ if you like. So you could also write $\int_{\lambda_{ab}\in X}f(\lambda_{ab}) d \lambda_{ab}$ if you want to. Nothing has changed. What's in a name? The important thing is that the function $f$ is the same, and the set over which we integrate, $X$ , is the same. One could also just write $\int_{X}f$ . It stands for the same thing.

by **minkwe** » Thu Jun 11, 2015 12:05 pm

Heinera wrote:
minkwe wrote:Are you suggesting that Bell's derivation will fail to proceed if we replaced equation (2) with:

$P(a,b) = \frac{1}{N} \sum_{i}^{N} A_a^i B_b^i, \: N \to \infty$ , or $\langle A_aB_b \rangle$

Since there is no $\lambda$ nor any $\rho$ in that expression, Bell's derivation would obviously fail to proceed.

Needless to say, your statement is obviously and demonstrably false. Do you want me to derive Bell's inequalities from that expression or are you ready to withdraw your statement.

by **Heinera** » Thu Jun 11, 2015 12:24 pm

minkwe wrote:
Heinera wrote:
minkwe wrote:Are you suggesting that Bell's derivation will fail to proceed if we replaced equation (2) with:

$P(a,b) = \frac{1}{N} \sum_{i}^{N} A_a^i B_b^i, \: N \to \infty$ , or $\langle A_aB_b \rangle$

Since there is no $\lambda$ nor any $\rho$ in that expression, Bell's derivation would obviously fail to proceed.

Needless to say, your statement is obviously and demonstrably false. Do you want me to derive Bell's inequalities from that expression or are you ready to withdraw your statement.

Please go ahead.

by **Joy Christian** » Thu Jun 11, 2015 12:26 pm

minkwe wrote:Needless to say, your statement is obviously and demonstrably false. Do you want me to derive Bell's inequalities from that expression or are you ready to withdraw your statement.

No need to waste your time. The trivial derivation can be found in Eqs. (11) to (14) of this paper: http://arxiv.org/abs/1501.03393.

by **AnotherGuest** » Thu Jun 11, 2015 12:33 pm

Minkwe is right, yet again!

Take rho to be the probability measure putting equal mass 1/N on each i = 1, ..., N

So we define Lambda = {1, ..., N}

A^i_a = +/- 1 is now a function of i and a.

The "local hidden variable" lambda is now the index "i".

P_N(a, b) = 1/N sum A^i_a B^i_b is yet another instance of Bell's integral expression for P(a, b).

So you can derive Bell's inequality for it. Then let N go to infinity.

By the way, I don't see any difference at all between the notations $A_a(\lambda)$ and $A(a, \lambda)$ . I mean: you can use either, it is a matter of taste. Do you want to think of A as a function of two variables, or an indexed family of functions of one variable? You can think either way, it comes down to the same thing.

by **minkwe** » Thu Jun 11, 2015 1:49 pm

Joy Christian wrote:
minkwe wrote:Needless to say, your statement is obviously and demonstrably false. Do you want me to derive Bell's inequalities from that expression or are you ready to withdraw your statement.

No need to waste your time. The trivial derivation can be found in Eqs. (11) to (14) of this paper: http://arxiv.org/abs/1501.03393.

Thanks Joy, I was hoping Heine will flip a couple of pages to equations (13) to (15) of Bell's paper and verify that $\lambda$ plays no role in the derivation of the inequality anyone can can follow along from Bells equation (13):

$P(a,b) - P(a,c) = \frac{1}{N} \sum_{i}^{N} A_a^i B_b^i - \frac{1}{N} \sum_{i}^{N} A_a^i B_c^i$
$= \frac{1}{N} \sum_{i}^{N} [A_a^i B_b^i - A_a^i B_c^i]$
$= -\frac{1}{N} \sum_{i}^{N} [A_a^i A_b^i - A_a^i A_c^i]$ , Since $B_b^i = -A_b^i$
and after factorizing out $A_a^iB_b^i$ , remembering that $A_b^i = \frac{1}{A_b^i}$ we get
$= -\frac{1}{N} \sum_{i}^{N} A_a^iA_b^i[1 - A_b^iA_c^i]$
$|P(a,b) - P(a,c)| \leq \frac{1}{N} \sum_{i}^{N} [1 - A_b^iA_c^i]$
The second term on the right is $P(b,c)$ .
Do I need to continue? We now see that we get Bell's inequality without any physics whatsoever. All we need is theory-agnostic outcomes $A_a = \pm 1, B_b = \pm 1, A_a = -B_a$ and the expectation value of the product of the paired outcomes $E(A_aB_b)$ , or as Bell would say $P(a,b)$ . The inequality is therefore theory-agnostic and has nothing whatsoever to do with locality.

by **minkwe** » Thu Jun 11, 2015 3:14 pm

AnotherGuest wrote:Minkwe is right, yet again!

Thanks!

by **AnotherGuest** » Thu Jun 11, 2015 8:56 pm

minkwe wrote:The inequality is therefore theory-agnostic and has nothing whatsoever to do with locality.

Here I believe Minkwe is wrong.

(1) The assumption of mathematical existence of outcomes $A_a(\lambda)$ and $B_b(\lambda)$ in the two wings of the experiment for all possible settings $a$ and $b$ simultaneously, is exactly what most writers call "local realism". It is the combination of "counterfactual definiteness" and "local causality", see http://en.citizendium.org/wiki/Entanglement_(physics). It states that Bob's outcome when he uses setting "b" would have been the same, whatever setting "a" is used by Alice.

These are good old EPR characterisations of realism and locality. Einstein considered that these assumptions do have everything to do with realism and locality. (OK so maybe Einstein was wrong and Bohr was right - the Copenhagen dogma asserts that there is no hidden layer behind quantum mechanics, there are no "event-based" functions "A" and "B" at all.)

(2) Under these assumptions, Bell arrives at an inequality. The inequality would not in general need to be true - it relies on the assumptions of local realism, and of perfect anti-correlation when the settings are equal (and of course, on +/-1 valued outcomes). So the inequality is not theory-agnostic.

The assumptions of perfect anti-correlation when the settings are equal and of +/-1 valued outcomes come from the standard QM analysis of the singlet state; which of course also leads to the singlet correlations.

by **Schmelzer** » Fri Jun 12, 2015 12:16 am

minkwe wrote:Do I need to continue? We now see that we get Bell's inequality without any physics whatsoever. All we need is theory-agnostic outcomes $A_a = \pm 1, B_b = \pm 1, A_a = -B_a$ and the expectation value of the product of the paired outcomes $E(A_aB_b)$ , or as Bell would say $P(a,b)$ . The inequality is therefore theory-agnostic and has nothing whatsoever to do with locality.

Wow, this was indeed unexpected for me.

But this, of course, explains the heavy emotional reaction against my rejection to accept his attempts to present formulas (1) and (2) as if they contain no nontrivial information, while, in fact, they contain the first half (and the conceptually difficult half) of the whole theorem, namely the EPR argument.

It is, indeed, quite obvious that the equations (1), (2) are far away from being theory-agnostic descriptions of $A_a = \pm 1, B_b = \pm 1, A_a = -B_a$ , which is, as I have explained, is already an expression which requires nontrivial context (namely that $A_a$ is not a function defined for all experiments of this type, which would be simply wrong in QM, but a denotation for a fixed single experiment).

This was, initially, not more than a minor quibble from my side, I simply wanted to avoid inaccurate formulations open to misinterpretations. But it hit the point.

And, even if we ignore the point that the very existence of a function $A(a,\lambda)$ for the description of the measurement results already contains nontrivial hypotheses (the existence of whatever hidden variables, which are questioned by those who want to preserve Einstein locality by giving up realism), all one can get without locality, thus, theory-independent, is only
$E(AB|a,b) = \int A(a,b, \lambda) B(a,b,\lambda) \rho(a,b,\lambda) d\lambda$
which does not allow to prove Bell's inequality.

by **Heinera** » Fri Jun 12, 2015 3:44 am

minkwe wrote:Thanks Joy, I was hoping Heine will flip a couple of pages to equations (13) to (15) of Bell's paper and verify that $\lambda$ plays no role in the derivation of the inequality anyone can can follow along from Bells equation (13):

$P(a,b) - P(a,c) = \frac{1}{N} \sum_{i}^{N} A_a^i B_b^i - \frac{1}{N} \sum_{i}^{N} A_a^i B_c^i$
$= \frac{1}{N} \sum_{i}^{N} [A_a^i B_b^i - A_a^i B_c^i]$
$= -\frac{1}{N} \sum_{i}^{N} [A_a^i A_b^i - A_a^i A_c^i]$ , Since $B_b^i = -A_b^i$
and after factorizing out $A_a^iB_b^i$ , remembering that $A_b^i = \frac{1}{A_b^i}$ we get
$= -\frac{1}{N} \sum_{i}^{N} A_a^iA_b^i[1 - A_b^iA_c^i]$
$|P(a,b) - P(a,c)| \leq \frac{1}{N} \sum_{i}^{N} [1 - A_b^iA_c^i]$
The second term on the right is $P(b,c)$ .

Here you use the integers from 1 to N as lambda, and 1/N as rho. Fine.

Do I need to continue? We now see that we get Bell's inequality without any physics whatsoever. All we need is theory-agnostic outcomes $A_a = \pm 1, B_b = \pm 1, A_a = -B_a$ and the expectation value of the product of the paired outcomes $E(A_aB_b)$ , or as Bell would say $P(a,b)$ . The inequality is therefore theory-agnostic and has nothing whatsoever to do with locality.

No. You use exactly the same locality assumptions as Bell. The fact that you assign values to both $A_a^i$ and $A_b^i$ means that you postulate the possibility of counterfactual outcomes. The fact that you in this expression

$P(a,b) - P(a,c) = \frac{1}{N} \sum_{i}^{N} A_a^i B_b^i - \frac{1}{N} \sum_{i}^{N} A_a^i B_c^i$

use the same values $A^i_a$ in both sums means the theory assumes locality; the value of $A^i_a$ is independent of the setting in the other station.

by **Heinera** » Fri Jun 12, 2015 3:52 am

AnotherGuest wrote:As far as I know mathematics, if $f$ is some function defined on a set $X$ , then in the expression $\int_{x\in X}f(x) d x$ you can give the variable $x$ any other name you like, as long as it is not simultaneously in use for something else. You can give it the name $\lambda_{ab}$ if you like. So you could also write $\int_{\lambda_{ab}\in X}f(\lambda_{ab}) d \lambda_{ab}$ if you want to. Nothing has changed. What's in a name? The important thing is that the function $f$ is the same, and the set over which we integrate, $X$ , is the same. One could also just write $\int_{X}f$ . It stands for the same thing.

Yes, what's in a name? The point is that $\lambda_{ab}$ is no longer an independent variable, but a function of $a$ and $b$ . The space you integrate over will thus be different for different values of $a$ and $b$ . And as you correctly stated earlier, this space must be the same for all $a,b$ for Bell's theorem to hold.

by **AnotherGuest** » Fri Jun 12, 2015 4:35 am

Heinera wrote:
AnotherGuest wrote:As far as I know mathematics, if $f$ is some function defined on a set $X$ , then in the expression $\int_{x\in X}f(x) d x$ you can give the variable $x$ any other name you like, as long as it is not simultaneously in use for something else. You can give it the name $\lambda_{ab}$ if you like. So you could also write $\int_{\lambda_{ab}\in X}f(\lambda_{ab}) d \lambda_{ab}$ if you want to. Nothing has changed. What's in a name? The important thing is that the function $f$ is the same, and the set over which we integrate, $X$ , is the same. One could also just write $\int_{X}f$ . It stands for the same thing.

Yes, what's in a name? The point is that $\lambda_{ab}$ is no longer an independent variable, but a function of $a$ and $b$ . The space you integrate over will thus be different for different values of $a$ and $b$ . And as you correctly stated earlier, this space must be the same for all $a,b$ for Bell's theorem to hold.

If $\lambda_{ab}$ is a supposed to stand for a function instead of a variable, you cannot integrate with respect to it. If it is meant to stand for a variable, then its name is irrelevant. In the first case, the expression $\int_{\lambda_{ab}\in X}f(\lambda_{ab}) d \lambda_{ab}$ is meaningless. In the second case, it means exactly the same as $\int_{\lambda\in X}f(\lambda) d \lambda$

by **minkwe** » Fri Jun 12, 2015 4:55 am

Heinera wrote:Here you use the integers from 1 to N as lambda, and 1/N as rho. Fine.

I absolutely do not. $i \in [1..N]$ is a summation index representing particles, not lambda.

No. You use exactly the same locality assumptions as Bell. The fact that you assign values to both $A_a^i$ and $A_b^i$ means that you postulate the possibility of counterfactual outcomes. The fact that you in this expression

Just because you proclaim it doesn't make it true. There is not locality assumption in it. Please review my previous posts which I have explicity defined $\gamma$ as non-local, and yet the derivation will proceed exactly the same, even using Bell's notation, simply replace $\lambda$ with $\gamma$

use the same values $A^i_a$ in both sums means the theory assumes locality; the value of $A^i_a$ is independent of the setting in the other station.

False. There is no locality in that. There may be counterfactual definiteness, but absolutely no locality assumption.

by **minkwe** » Fri Jun 12, 2015 5:01 am

AnotherGuest wrote:(1) The assumption of mathematical existence of outcomes $A_a(\lambda)$ and $B_b(\lambda)$ in the two wings of the experiment for all possible settings $a$ and $b$ simultaneously, is exactly what most writers call "local realism".

Most writers are wrong here. There is no locality in that assumption. It is counterfactual definiteness which is a different thing completely from locality. As anyone can see by considering $A_a(\gamma)$ and $B_b(\gamma)$ . Until now I haven't said anything about counterfactual definiteness. But I believe I've shown convincingly that there is absolutely no locality assumption required to obtain the inequalities.

(2) Under these assumptions, Bell arrives at an inequality. The inequality would not in general need to be true - it relies on the assumptions of local realism, and of perfect anti-correlation when the settings are equal (and of course, on +/-1 valued outcomes). So the inequality is not theory-agnostic.

You guys keep saying "local realism" which I understand is common verbiage, but the derivation is devoid of locality. Perhaps you think counterfactual definiteness applies only to certain theories. We shall find out next, but there is absolutely no locality in my derivation so far.

by **Heinera** » Fri Jun 12, 2015 5:07 am

minkwe wrote:
Heinera wrote:
use the same values $A^i_a$ in both sums means the theory assumes locality; the value of $A^i_a$ is independent of the setting in the other station.

False. There is no locality in that. There may be counterfactual definiteness, but absolutely no locality assumption.

What I wrote is the very definition of locality.

by **AnotherGuest** » Fri Jun 12, 2015 7:01 am

minkwe wrote:
AnotherGuest wrote:(1) The assumption of mathematical existence of outcomes $A_a(\lambda)$ and $B_b(\lambda)$ in the two wings of the experiment for all possible settings $a$ and $b$ simultaneously, is exactly what most writers call "local realism".

Most writers are wrong here. There is no locality in that assumption. It is counterfactual definiteness which is a different thing completely from locality. As anyone can see by considering $A_a(\gamma)$ and $B_b(\gamma)$ . Until now I haven't said anything about counterfactual definiteness. But I believe I've shown convincingly that there is absolutely no locality assumption required to obtain the inequalities.

(2) Under these assumptions, Bell arrives at an inequality. The inequality would not in general need to be true - it relies on the assumptions of local realism, and of perfect anti-correlation when the settings are equal (and of course, on +/-1 valued outcomes). So the inequality is not theory-agnostic.

You guys keep saying "local realism" which I understand is common verbiage, but the derivation is devoid of locality. Perhaps you think counterfactual definiteness applies only to certain theories. We shall find out next, but there is absolutely no locality in my derivation so far.

Counterfactual definiteness would give us the existence of $A_{ab}(\lambda)$ and $B_{ab}(\lambda)$ ; relativistic causality adds to this: $A_{ab}(\lambda)$ does not depend on $b$ , $B_{ab}(\lambda)$ does not depend on $a$ . Thus the two together give us existence of $A_a(\lambda)$ and $B_b(\lambda)$ . Which many, many writers call "local realism". But everyone is free to give it another name. It doesn't matter for the ensuing derivation "where" $\lambda$ is thought to be located. If you want to think of $\lambda$ as being "all around us, everywhere" then you are also including some models that you would call "non-local" as well. So be it.

You are not allowing arbitrary non-local models: you get Bell's inequality, which puts restrictions on the correlation function.

EPR's motivation for (essentially) coming up with $A_{a}(\lambda)$ and $B_{b}(\lambda)$ , where the settings are "measure position" and "measure momentum" and in their minds, $\lambda$ is coming from the source of the two particles, were the combination of relativistic causality, counterfactual definiteness, and "perfect anti-correlation when the two settings are the same". Exactly what Bell has assumed so far, too (not surprisingly).

Of course we are also assuming that experimenters can choose settings freely ("no conspiracy").

by **minkwe** » Fri Jun 12, 2015 7:09 am

Heinera wrote:What I wrote is the very definition of locality.

It absolutely is not, despite your claims. The definition of locality is:
- Locality means the outcome depends only on information inside the light-cone of the station.
- Non-locality means the outcome depends on information outside the light-cone of the station.

I have shown that if with a $\gamma$ represents information outside of the light-cone of the station, then the same equations apply as if you use $\lambda$ , which is restricted to information only inside the light-cone of the station. In fact, I have shown that you can derive the inequality without invoking any hidden variables at all.

I could have started by saying:

Assume that at given moment in time, 3 outcomes $A_a, A_b, A_c \in {\pm1}$ spontaneously appear at at three distant stations where settings "a", "b", and "c" are in effect. Let successive measurements of these outcomes be made $i \in [1 \to N]$ Note that no hidden variables are involved, yet, I can derive the inequalities in exactly the same way, without any reference to locality, or even any hidden variables.

$P(a,b) - P(a,c) = \frac{1}{N} \sum_{i}^{N} A_a^i A_b^i - \frac{1}{N} \sum_{i}^{N} A_a^i A_c^i$

The suggestion that locality or even hidden variables (local or non-local) is required is simply indefensible. The only assumption required is the assumption that we are simultaneously considering three numbers $A_a, A_b, A_c \in [-1, +1]$ from any source whatsoever, Some people call this "counterfactual definiteness", and others call it "realism". But there is absolutely no locality assumption here, whatsoever.

Once we've put this so-called "locality assumption" in its grave, we can proceed to examine the "counterfactual definiteness/realism" assumption.

SciPhysicsForums.com

Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Re: Thoughts about Bell, Bohm, Christian, et al.

Who is online