Evidence that QM does not violate Bell's inequalities

[Heinera]

It has already been shown to you in this thread that you don't violate Bell-CHSH since your expectation terms are independent. Your A in <AB> is not necessarily the same as the A in <AB'>, etc. You now need to show that in your simulation that A in <AB> (E11) is the same as the A in <AB'> (E12). But you can't.

Exactly. In my model the A in <AB> (E11) is not the same as the A in <AB'> (E12), because the model is non-local. You are now very close to understanding Bell's theroem.

Ah, you just admitted that you don't violate CHSH since if the terms are independent then the bound is 4 not 2.

And you are now close to understanding that the terms in my model are independent because the model is non-local.

[minkwe]

My model does indeed produce . Read the code again.

It absolutely does not, you have no clue do you? The subscript i represents a set of outcome pairs which you used to calculate the Expectation term. You have 4 statistically independent sets, therefore the upper bound is 4. Try a different magic trick.

If you don't believe that those are statistically independent sets save the first column of outcomes from your E11 experiment and the first column of data from your E12 experiment, then calculate the cross correlation between those two columns of data. You will find that you get close to zero. You can repeat this for all the similarly numbered pairs of columns.

In short, you have still not understood the argument from post #1, or all the other arguments we've been giving you for over a year about this.

Statistically independent sets of outcomes can not at the same time be statistically dependent.

[FrediFizzx]

Exactly. In my model the A in <AB> (E11) is not the same as the A in <AB'> (E12), because the model is non-local. You are now very close to understanding Bell's theroem.

Ah, you just admitted that you don't violate CHSH since if the terms are independent then the bound is 4 not 2.

And you are now close to understanding that the terms in my model are independent because the model is non-local.

You already admitted that the ca in E11 is not the same as the ca in E12. Therefore they are not dependent. Bound is 4 not 2. You are using the wrong inequality.

[Heinera]

My model does indeed produce . Read the code again.

It absolutely does not, you have no clue do you? The subscript i represents a set of outcome pairs which you used to calculate the Expectation term. You have 4 statistically independent sets, therefore the upper bound is 4. Try a different magic trick.

If you don't believe that those are statistically independent sets save the first column of outcomes from your E11 experiment and the first column of data from your E12 experiment, then calculate the cross correlation between those two columns of data. You will find that you get close to zero. You can repeat this for all the similarly numbered pairs of columns.

In short, you have still not understood the argument from post #1, or all the other arguments we've been giving you for over a year about this.

Statistically independent sets of outcomes can not at the same time be statistically dependent.

Read the code again. Apart from detector settings, nothing changes in the code between .

[minkwe]

My model does indeed produce . Read the code again.

It absolutely does not, you have no clue do you? The subscript i represents a set of outcome pairs which you used to calculate the Expectation term. You have 4 statistically independent sets, therefore the upper bound is 4. Try a different magic trick.

If you don't believe that those are statistically independent sets save the first column of outcomes from your E11 experiment and the first column of data from your E12 experiment, then calculate the cross correlation between those two columns of data. You will find that you get close to zero. You can repeat this for all the similarly numbered pairs of columns.

In short, you have still not understood the argument from post #1, or all the other arguments we've been giving you for over a year about this.

Statistically independent sets of outcomes can not at the same time be statistically dependent.

Read the code again. Apart from detector settings, nothing changes in the code between .

You should read my response again. You still do not get it. Here are the steps:

1. You perform 4 statistically independent experiments/calculation ,
2. Each experiment produces a 2xN spreadsheet of numbers.
3. You calculate the cross correlation between the numbers in each 2xN spreadsheet, to get the terms
4. then you combine the 4 correlations at the end.

This is what you are doing in the code. The 4 correlations are statistically independent. You can verify it using the procedure I gave you above. Calculate the cross correlation between the first columns of E11 and E12 and confirm that it is almost zero. And repeat that for all the other similarly numbered columns in the data generated by your simulation.

It is quite obvious to anyone who is not blind that your terms are statistically independent, and therefore the upper bound for the combination you have at the end is 4. Too bad you don't see it. Do the calculation of the prescribed cross correlations if you continue to deny that the terms are statistically independent.

[Heinera]

So, you guys, I will be on an expedition for the next couple of days where I probably will not have much of an internet access. In the mean time, I suggest you look at my simple model http://rpubs.com/heinera/16727. This model reproduces QM exactly, and no loopholes are exploited. Unfortunately it is non-local. But, given its simplicity, I trust it will be an easy task to convert it into a local model. See you on Wednesday, and I expect that by then one of you have made a local version available. Sayonara!

[minkwe]

So, you guys, I will be on an expedition for the next couple of days where I probably will not have much of an nternet access. In the mean time, I suggest you look at my simple model http://rpubs.com/heinera/16727. This model reproduces QM exactly, and no loopholes are exploited. Unfortunately it is non-local. But, given its simplicity, I trust it will be an easy task to convert it into a local model. See you on Wednesday, and I expect that by then one of you have made a local version available. Sayonara!

Your code is not a simulation of the terms in the CHSH as has been explained to you already. Besides multiple local realistic models of the QM expectation values have already been provided. All Bell followers are now anxiously waiting to be rescued by forever running Delft experiment. But they will be disappointed. I wonder how many Bell-test experiments have gone unpublished because the results showed the opposite of what the experimenters had hoped?

I suggest you download the following article and read it while you are away.
http://arxiv.org/pdf/0907.0767v2.pdf

Hopefully you will have an epiphany while on your expedition. At least you acheived something this week so no need to be sad. You finally admitted that Bell's inequality is meaningless in QM and therefore QM can not provide the terms in it

Bell's inequality is the following:
,

SInce this expression makes no sense in QM, and can't be computed

[FrediFizzx]

So, you guys, I will be on an expedition for the next couple of days where I probably will not have much of an internet access. In the mean time, I suggest you look at my simple model http://rpubs.com/heinera/16727. This model reproduces QM exactly, and no loopholes are exploited. Unfortunately it is non-local. But, given its simplicity, I trust it will be an easy task to convert it into a local model. See you on Wednesday, and I expect that by then one of you have made a local version available. Sayonara!

Yes your simulation does reproduce the predictions of QM but it doesn't violate Bell-CHSH with a bound of 2. It is a good example of why a lot of people believe in the non-local nonsense since that is the easiest way to reproduce the QM predictions(but there are other local ways). Since nothing can truly violate the inequalities and since it is mathematically impossible, what use are they? None that I see.

Have a good trip.

[Heinera]

So, you guys, I will be on an expedition for the next couple of days where I probably will not have much of an internet access. In the mean time, I suggest you look at my simple model http://rpubs.com/heinera/16727. This model reproduces QM exactly, and no loopholes are exploited. Unfortunately it is non-local. But, given its simplicity, I trust it will be an easy task to convert it into a local model. See you on Wednesday, and I expect that by then one of you have made a local version available. Sayonara!

Yes your simulation does reproduce the predictions of QM but it doesn't violate Bell-CHSH with a bound of 2. It is a good example of why a lot of people believe in the non-local nonsense since that is the easiest way to reproduce the QM predictions(but there are other local ways). Since nothing can truly violate the inequalities and since it is mathematically impossible, what use are they? None that I see.

Have a good trip.

Thanks! And let me return equally good wishes for you turning my model into a local one!

[Guest]

You guys keep saying in this thread that some expectations are "statistically independent". That's mathematical non-sense. It's well known that two events A and B are statistically independent if and only if P(A and B) = P(A) P(B). Also, two random variables X and Y are statistically independent if and only if P(X in A, Y in B) = P(X in A) P(Y in B) for all A and B. But the expectation E[Z] of some random variable Z is a real number. It does not make sense to say that two expectations E[Z] and E[W] are statistically independent. If you don't believe what I'm saying, please post a link to a single probability book which gives a definition of "statistically independent expectations". C'mon, guys.

[minkwe]

Somebody badly needs to recover their lost memory viewtopic.php?f=6&t=53

[FrediFizzx]

You guys keep saying in this thread that some expectations are "statistically independent". That's mathematical non-sense. It's well known that two events A and B are statistically independent if and only if P(A and B) = P(A) P(B). Also, two random variables X and Y are statistically independent if and only if P(X in A, Y in B) = P(X in A) P(Y in B) for all A and B. But the expectation E[Z] of some random variable Z is a real number. It does not make sense to say that two expectations E[Z] and E[W] are statistically independent. If you don't believe what I'm saying, please post a link to a single probability book which gives a definition of "statistically independent expectations". C'mon, guys.

Why would you need a definition for something that is clearly self evident? I suspect you are talking about something other than Heine's simulation. If you are talking about Heine's simulation, then please reference the terms he uses. IOW, simply show us mathematically how his expectations E11 and E12 are dependent on each other.

[Joy Christian]

You guys keep saying in this thread that some expectations are "statistically independent". That's mathematical non-sense. It's well known that two events A and B are statistically independent if and only if P(A and B) = P(A) P(B). Also, two random variables X and Y are statistically independent if and only if P(X in A, Y in B) = P(X in A) P(Y in B) for all A and B. But the expectation E[Z] of some random variable Z is a real number. It does not make sense to say that two expectations E[Z] and E[W] are statistically independent. If you don't believe what I'm saying, please post a link to a single probability book which gives a definition of "statistically independent expectations". C'mon, guys.

The relationship between joint probabilities and expectation values E(a, b) is elementary, straightforward, and well known. See, for example, Eq. (42) of this paper.

[FrediFizzx]

So, you guys, I will be on an expedition for the next couple of days where I probably will not have much of an internet access. In the mean time, I suggest you look at my simple model http://rpubs.com/heinera/16727. This model reproduces QM exactly, and no loopholes are exploited. Unfortunately it is non-local. But, given its simplicity, I trust it will be an easy task to convert it into a local model. See you on Wednesday, and I expect that by then one of you have made a local version available. Sayonara!

I don't see any advantage in trying to make this simulation into a local model. But I did modify it a bit in R Studio to better show how the dependency of the inequalities work.

Code: Select all

set.seed(9875)

## For reproducibility.

M <- 10^5  ## Sample size.

## Use the same, single sample of 'M' realizations of hidden variables for
## all four correlations

## The hidden variable is just a random number t between -1 and 1.  We send t
## to Alice and -t to Bob.

t <- runif(M, -1, 1)

## At both Alice's and Bob's stations the same model is used to determine
## outcomes.  We put this into a function obs(hv), that takes the hidden
## variable as the argument and returns -1 or 1.

obs <- function(hv) {
  s <- sign(hv)
  hv <- abs(hv)
  ## Since the following variable depend on both settings a and b, the model is
  ## blatantly non-local:
  L <- (1 + sum(a * b))/4
 
  if (hv < L) {
    o <- s
  } else if (hv < 2 * L) {
    o <- -s
  } else if (hv < L + 0.5) {
    o <- -1
  } else {
    o <- 1
  }
  return(o)
}

## Now we compute the four correlations in the CHSH inequality:

alpha <- 0
beta <- 45
a <- c(cos(alpha * pi/180), sin(alpha * pi/180))
b <- c(cos(beta * pi/180), sin(beta * pi/180))

## We generate the list of observations by applying the obs function to each
## element in the list of hidden variables.  First for Alice:
ca1 <- sapply(t, obs)
## Then for Bob:
cb1 <- sapply(-t, obs)

E11 <- mean(ca1 * cb1)
E11
[1] -0.70528

alpha <- 0
beta <- 135
a <- c(cos(alpha * pi/180), sin(alpha * pi/180))
b <- c(cos(beta * pi/180), sin(beta * pi/180))

ca2 <- ca1  ## this is the same as A in <AB> being the same A in <AB'>
cb2 <- sapply(-t, obs)

E12 <- mean(ca2 * cb2)
E12
[1] -0.00296

alpha <- 90
beta <- 45
a <- c(cos(alpha * pi/180), sin(alpha * pi/180))
b <- c(cos(beta * pi/180), sin(beta * pi/180))

ca3 <- sapply(t, obs)
cb3 <- cb1  ## This is the same as the B in <AB> being the same as the B in <A'B>

E21 <- mean(ca3 * cb3)
E21
[1] -0.70528

alpha <- 90
beta <- 135
a <- c(cos(alpha * pi/180), sin(alpha * pi/180))
b <- c(cos(beta * pi/180), sin(beta * pi/180))

ca4 <- ca3  ## The A' from <A'B>
cb4 <- cb2  ## The B' from <AB'>

E22 <- mean(ca4 * cb4)
E22
[1] -0.00296

## CHSH expression
-E11 + E12 - E21 - E22
[1] 1.41056


From this, it is obvious that dependent expectation terms are not going to work to violate CHSH.

[minkwe]

I don't see any advantage in trying to make this simulation into a local model. But I did modify it a bit in R Studio to better show how the dependency of the inequalities work.

Code: Select all

set.seed(9875)

## For reproducibility.

M <- 10^5  ## Sample size.

## Use the same, single sample of 'M' realizations of hidden variables for
## all four correlations

## The hidden variable is just a random number t between -1 and 1.  We send t
## to Alice and -t to Bob.

t <- runif(M, -1, 1)

## At both Alice's and Bob's stations the same model is used to determine
## outcomes.  We put this into a function obs(hv), that takes the hidden
## variable as the argument and returns -1 or 1.

obs <- function(hv) {
  s <- sign(hv)
  hv <- abs(hv)
  ## Since the following variable depend on both settings a and b, the model is
  ## blatantly non-local:
  L <- (1 + sum(a * b))/4
 
  if (hv < L) {
    o <- s
  } else if (hv < 2 * L) {
    o <- -s
  } else if (hv < L + 0.5) {
    o <- -1
  } else {
    o <- 1
  }
  return(o)
}

## Now we compute the four correlations in the CHSH inequality:

alpha <- 0
beta <- 45
a <- c(cos(alpha * pi/180), sin(alpha * pi/180))
b <- c(cos(beta * pi/180), sin(beta * pi/180))

## We generate the list of observations by applying the obs function to each
## element in the list of hidden variables.  First for Alice:
ca1 <- sapply(t, obs)
## Then for Bob:
cb1 <- sapply(-t, obs)

E11 <- mean(ca1 * cb1)
E11
[1] -0.70528

alpha <- 0
beta <- 135
a <- c(cos(alpha * pi/180), sin(alpha * pi/180))
b <- c(cos(beta * pi/180), sin(beta * pi/180))

ca2 <- ca1  ## this is the same as A in <AB> being the same A in <AB'>
cb2 <- sapply(-t, obs)

E12 <- mean(ca2 * cb2)
E12
[1] -0.00296

alpha <- 90
beta <- 45
a <- c(cos(alpha * pi/180), sin(alpha * pi/180))
b <- c(cos(beta * pi/180), sin(beta * pi/180))

ca3 <- sapply(t, obs)
cb3 <- cb1  ## This is the same as the B in <AB> being the same as the B in <A'B>

E21 <- mean(ca3 * cb3)
E21
[1] -0.70528

alpha <- 90
beta <- 135
a <- c(cos(alpha * pi/180), sin(alpha * pi/180))
b <- c(cos(beta * pi/180), sin(beta * pi/180))

ca4 <- ca3  ## The A' from <A'B>
cb4 <- cb2  ## The B' from <AB'>

E22 <- mean(ca4 * cb4)
E22
[1] -0.00296

## CHSH expression
-E11 + E12 - E21 - E22
[1] 1.41056


From this, it is obvious that dependent expectation terms are not going to work to violate CHSH.

Thanks Fred, for doing Heine's homework , the calculation confirms that the sets of outcome pairs are statistically independent. Despite Guest's confusion, the "statistically independent" description is apt.

[FrediFizzx]

Thanks Fred, for doing Heine's homework , the calculation confirms that the sets of outcome pairs are statistically independent. Despite Guest's confusion, the "statistically independent" description is apt.

You're welcome. It is pretty mind boggling that some people can't see that the Bell inequalities can't be violated by anything. What is up with that? It is so freakin' simple.

[Heinera]

So, you guys, I will be on an expedition for the next couple of days where I probably will not have much of an internet access. In the mean time, I suggest you look at my simple model http://rpubs.com/heinera/16727. This model reproduces QM exactly, and no loopholes are exploited. Unfortunately it is non-local. But, given its simplicity, I trust it will be an easy task to convert it into a local model. See you on Wednesday, and I expect that by then one of you have made a local version available. Sayonara!

I don't see any advantage in trying to make this simulation into a local model.

Of course it would be an advantage. You would then have produced a local loophole-free simulation that exactly reproduces QM. It would get you the Nobel Prize.

By the way, your modifications did turn it into a local model; unfortuantely, it also changed the correlations so they no longer agree with QM.

[FrediFizzx]

By the way, your modifications did turn it into a local model; unfortuantely, it also changed the correlations so they no longer agree with QM.

It still is using your non-local hidden variable and it is just a "proof" by negation that your simulation doesn't violate CHSH when the expectation terms are dependent. IOW, your expectation terms are independent from each other so you have shifted to an inequality with a bound of 4 instead of 2. CHSH has a bound of 2. As I said... it is pretty mind boggling.

[Heinera]

By the way, your modifications did turn it into a local model; unfortuantely, it also changed the correlations so they no longer agree with QM.

It still is using your non-local hidden variable and it is just a "proof" by negation that your simulation doesn't violate CHSH when the expectation terms are dependent. IOW, your expectation terms are independent from each other so you have shifted to an inequality with a bound of 4 instead of 2. CHSH has a bound of 2. As I said... it is pretty mind boggling.

The hidden variable is local. The nonlocality enters because outcomes in either wing depend directly on settings from both wings of the simulations. Your modifications broke that dependence, so the model is now local.

[FrediFizzx]

The hidden variable is local. The nonlocality enters because outcomes in either wing depend directly on settings from both wings of the simulations. Your modifications broke that dependence, so the model is now local.

Take another better look. The a in E11 is the same as the a in E12, etc.