The bet on Christian's experiment

[Joy Christian]

Did you check E(90, 135)?

In case anyone else is wondering, the deviation in E(90, 135) is 0.02 --- ten times smaller than what Richard would need to win the bet.

[jreed]

Joy, if I'm not mistaken your simulation is using the detector loophole to obtain the desired curve. The R code statement: good <- ca & cb selects products of the correlations where both the correlations are non-zero. I've seen this several times before in the Gisin & Gisin, Minkwe and Roth simulations when I coded them up in Mathematica.

[Joy Christian]

Joy, if I'm not mistaken your simulation is using the detector loophole to obtain the desired curve. The R code statement: good <- ca & cb selects products of the correlations where both the correlations are non-zero. I've seen this several times before in the Gisin & Gisin, Minkwe and Roth simulations when I coded them up in Mathematica.

Hi John,

The detection loophole is irrelevant both in my model and in the proposed experiment. In the model the complete or initial state of the system is a pair (e, theta), not just the vectors e by themselves. This is explained in more detail in this one page document. Therefore there is 1-to-1 correspondence between the initial state of the system and what is observed. Every state (e, theta) is detected and every detection corresponds to a state. Ultimately what appears as a detection loophole is a manifestation of the topology of the 3-sphere---i.e., the topology of the physical space we live in.

Now in my proposed experiment, the way it is set up, there cannot be any loss of detection, because we are not actually observing the "particles" along the directions a and b. What is being observed are the actual spin directions, without any reference to measurement directions a and b. So no question of detection loophole arises.

I have used the argument good <- ca & cb simply because I don't know much programming. I am certain that one can calculate correlation in the latest simulation without using the "good" line. I do not believe it will change anything. In fact it might improve the accuracy of the simulation. I am working on such improvements.

[gill1109]

The detection loophole is irrelevant both in my model and in the proposed experiment.
...

I have used the argument good <- ca & cb simply because I don't know much programming. I am certain that one can calculate correlation in the latest simulation without using the "good" line. I do not believe it will change anything. In fact it might improve the accuracy of the simulation. I am working on such improvements.

Unfortunately the detection loophole (or if you prefer to interpret it differently, the conspiracy loophole) is an important component in the Pearle model and all known variants.

Try leaving out the line "good <- ca & cb" and the subsequent selection according to the subset defined by good. It is a different subset of all the runs which have been performed for each different value of a and b, so you don't want it here, because you want to end up with just one set of runs.

You could alternatively just work with the settings which we are interested, and do something like good <- "ca1 & ca2 & cb1 & cb2" to select the same subset for all four correlations (a1,b1) etc.

[gill1109]

Zen asked me to post this code here:

Code: Select all

N <- 10^6
e <- matrix(nrow = 2, ncol = N, byrow = FALSE, data = scan("/tmp/JoyVector.txt", nlines = N))
alpha <- 90 * pi / 180
beta <- 135 * pi / 180
a <- c(cos(alpha), sin(alpha))
b <- c(cos(beta), sin(beta))
mean(sign(colSums(e * a)) * -sign(colSums(e * b)))
-cos(alpha - beta)


[Joy Christian]

Zen asked me to post this code here:

Code: Select all

N <- 10^6
e <- matrix(nrow = 2, ncol = N, byrow = FALSE, data = scan("/tmp/JoyVector.txt", nlines = N))
alpha <- 90 * pi / 180
beta <- 135 * pi / 180
a <- c(cos(alpha), sin(alpha))
b <- c(cos(beta), sin(beta))
mean(sign(colSums(e * a)) * -sign(colSums(e * b)))
-cos(alpha - beta)


Richard,

I have no idea what this code is for.

By the way, you may have noticed that the list of vector directions from my latest simulation has been made public since yesterday: angular momentum diections.

[gill1109]

Zen asked me to post this code here:

Code: Select all

N <- 10^6
e <- matrix(nrow = 2, ncol = N, byrow = FALSE, data = scan("/tmp/JoyVector.txt", nlines = N))
alpha <- 90 * pi / 180
beta <- 135 * pi / 180
a <- c(cos(alpha), sin(alpha))
b <- c(cos(beta), sin(beta))
mean(sign(colSums(e * a)) * -sign(colSums(e * b)))
-cos(alpha - beta)


Richard,
I have no idea what this code is for.
By the way, you may have noticed that the list of vector directions from my latest simulation has been made public since yesterday: angular momentum diections.

Hadn't noticed, sorry, till this morning. And now the file is taking a bloody long time to download.

I suspect that Zen is helping out with code for calculating the correlations.

So there is just one file, not two? The directions for Bob are exactly the negatives of those for Alice? The file contains x and y coordinates of vectors in the equatorial plane? Do the vectors have length 1 or are they unnormalised?

I can't do anything with your data set if you don't tell me how you mean it to be interpreted.

[Joy Christian]

Unfortunately the detection loophole (or if you prefer to interpret it differently, the conspiracy loophole) ....

As I explained to John, detection loophole or conspiracy loophole, or any loophole for that matter, is of no relevance to my model, or to my proposed experiment.

Try leaving out the line "good <- ca & cb" and the subsequent selection according to the subset defined by good. It is a different subset of all the runs which have been performed for each different value of a and b, so you don't want it here, because you want to end up with just one set of runs.

You could alternatively just work with the settings which we are interested, and do something like good <- "ca1 & ca2 & cb1 & cb2" to select the same subset for all four correlations (a1,b1) etc.

Thanks. I plan to improve my latest simulation in this and several other ways. I just wanted to submitted the crude version for 10,000 Euros from you, because it is good enough for that purpose. Without the "good" line there will be no more talk of "detection loophole." No "good line", no "detection loophole."

[gill1109]

I just wanted to submitted the crude version for 10,000 Euros from you, because it is enough for that purpose. Without the "good" line there will be no more talk of "detection loophole." No "good line", no "detection loophole."

How you create two files of the same length representing directions of vectors in R^3 is not of any concern in deciding whether or not you have won our bet. You can create two big files first and later trim then down, if that helps.

[Joy Christian]

So there is just one file, not two? The directions for Bob are exactly the negatives of those for Alice? The file contains x and y coordinates of vectors in the equatorial plane? Do the vectors have length 1 or are they unnormalised?

I can't do anything with your data set if you don't tell me how you mean it to be interpreted.

Fair enough. Ask away, and I will explain:

"So there is just one file, not two?"

As far as I understand it, yes.

"The directions for Bob are exactly the negatives of those for Alice?"

Yes, theoretically, but not necessarily in the same run (I will check the file to see if it can be split into two).

"The file contains x and y coordinates of vectors in the equatorial plane?

Yes.

"Do the vectors have length 1 or are they unnormalised?"

The vectors a and b are normalized, but the vectors e_k and -e_k are unnormalized (although they are extremely close to having length 1).

[gill1109]

I got the following four correlations from the first 10^7 directions in the file. How many are there altogether?

Code: Select all

N <- 10^7
e <- matrix(nrow = 2, ncol = N, byrow = FALSE, data = scan("JoyVector.txt", nlines = N))

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

[1] -0.7501332

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

[1] 0.2500942

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

[1] -0.7498742

alpha <- 90 * pi / 180
beta <- 135 * pi / 180
a <- c(cos(alpha), sin(alpha))
b <- c(cos(beta), sin(beta))
mean(sign(colSums(e * a)) * -sign(colSums(e * b)))
[1] -0.2498984

[gill1109]

So there is just one file, not two? The directions for Bob are exactly the negatives of those for Alice? The file contains x and y coordinates of vectors in the equatorial plane? Do the vectors have length 1 or are they unnormalised?

I can't do anything with your data set if you don't tell me how you mean it to be interpreted.

Fair enough. Ask away, and I will explain:

"So there is just one file, not two?"

As far as I understand it, yes.

"The directions for Bob are exactly the negatives of those for Alice?"

Yes, theoretically, but not necessarily in the same run (I will check the file to see if it can be split into two).

"The file contains x and y coordinates of vectors in the equatorial plane?

Yes.

"Do the vectors have length 1 or are they unnormalised?"

The vectors a and b are normalized, but the vectors e_k and -e_k are unnormalized (although they are extremely close to having length 1).

Great. Last question: how many runs?

[gill1109]

OK so this file passes the test for two pairs of directions and fails it for the other two. No 10 000 Euro today, I'm afraid.

Your correlations (in the usual order 0-45, 0-135, 90-45, 90-135) were:

-0.7501332

0.2500942

-0.7498742

-0.2498984

You might want advice whether or not I calulated the correlations correctly (ie according to your instructions as written down in our agreement). I used Zen's code with N = 10^7.

[Joy Christian]

Last question: how many runs?

10 million, or 10^7.

[gill1109]

Last question: how many runs?

10 million, or 10^7.

Yep, got it right.

The data-set fails the test.

[Joy Christian]

OK so this file passes the test for two pairs of directions and fails it for the other two. No 10 000 Euro today, I'm afraid.

Your correlations (in the usual order 0-45, 0-135, 90-45, 90-135) were:

-0.7501332

0.2500942

-0.7498742

-0.2498984

You might want advice whether or not I calulated the correlations correctly (ie according to your instructions as written down in our agreement). I used Zen's code with N = 10^7.

This is plain wrong, as one can see at once from the simulation: http://rpubs.com/jjc/16415.

Did you calculate the correlations separately without replacements as we agreed?

Ah...I know what you did wrong. Your first and third calculations are correct, but the second and fourth are incorrect. The correlations are not calculated separately.

[Heinera]

Richard's numbers are correct. Joy gets an approximate cos curve because his correlations are all computed with b fixed at (1, 0).

[Joy Christian]

Richard's numbers are correct. Joy gets an approximate cos curve because his correlations are all computed with b fixed at (1, 0).

Richard's numbers are not correct. They have nothing to do with how the experiment is supposed to have been conducted. He is not calculating the correlations separately. When calculated separately, none of the four correlations differ from the theoretical prediction by more than 0.02.

[gill1109]

I calculated each of the four correlations, separately, exactly according to how the experiment was agreed to be conducted.

Now how is this for a coincidence:

0.7501332 + 0.2500942 + 0.7498742 + 0.2498984 = 2

But Joy, if you believe you are home and dry, then you put your 10 000 Euro on the table too, and we take the data file to the adjudicators?

If 10 000 is too much risk for you, we can revert to the original 5 000

[gill1109]

Four separate calculations, according to the agreed formulas, implemented in R by Zen:

Code: Select all

N <- 10^7
e <- matrix(nrow = 2, ncol = N, byrow = FALSE, data = scan("JoyVector.txt", nlines = N))
alpha <- 0 * pi / 180
beta <- 45 * pi / 180
a <- c(cos(alpha), sin(alpha))
b <- c(cos(beta), sin(beta))
mean(sign(colSums(e * a)) * -sign(colSums(e * b)))

Code: Select all

N <- 10^7
e <- matrix(nrow = 2, ncol = N, byrow = FALSE, data = scan("JoyVector.txt", nlines = N))
alpha <- 0 * pi / 180
beta <- 135 * pi / 180
a <- c(cos(alpha), sin(alpha))
b <- c(cos(beta), sin(beta))
mean(sign(colSums(e * a)) * -sign(colSums(e * b)))

Code: Select all

N <- 10^7
e <- matrix(nrow = 2, ncol = N, byrow = FALSE, data = scan("JoyVector.txt", nlines = N))
alpha <- 90 * pi / 180
beta <- 45 * pi / 180
a <- c(cos(alpha), sin(alpha))
b <- c(cos(beta), sin(beta))
mean(sign(colSums(e * a)) * -sign(colSums(e * b)))

Code: Select all

N <- 10^7
e <- matrix(nrow = 2, ncol = N, byrow = FALSE, data = scan("JoyVector.txt", nlines = N))
alpha <- 90 * pi / 180
beta <- 135 * pi / 180
a <- c(cos(alpha), sin(alpha))
b <- c(cos(beta), sin(beta))
mean(sign(colSums(e * a)) * -sign(colSums(e * b)))

Results:

Code: Select all

[1] -0.7501332

[1] 0.2500942

[1] -0.7498742

[1] -0.2498984