by minkwe » Wed May 07, 2014 9:56 am
gill1109 wrote:delta is defined in formula (7), which refers to the population (ie to the model, not to the outcomes of a simulation experiment), and it refers to events defined in terms of counterfactual measurement of spin in both directions on both particles at once.
Richard, who are you trying to fool here? According to your paper:
The problem here is that the ensemble on which the correlations are evaluated changes
with the settings, while the original Bell inequality requires that they stay the same. In effect,
the Bell inequality only holds on the common part of the four different ensembles ΛAC′ , ΛAD′ ,
ΛBC′ , and ΛBD′ , i.e., for correlations of the form
E(AC′|ΛAC′ ∩ ΛAD′ ∩ ΛBC′ ∩ ΛBD′ ). (8)
Unfortunately our experimental data comes in the form
E(AC′|ΛAC′),
The population is Λ, the ensembles from the experiment are ΛAC′ , ΛAD′ ,ΛBC′ , and ΛBD′. Delta is calculated as:
so we need an estimate of the relation of the common part to its constituents:
δ = inf_settings P(ΛAC′ ∩ ΛAD′ ∩ ΛBC′ ∩ ΛBD′ )/P(ΛAC′ )
= inf_settings P(ΛAD′ ∩ ΛBC′ ∩ ΛBD′ |ΛAC′).
...
ΛI = ΛAC′ ∩ ΛAD′ ∩ ΛBC′ ∩ ΛBD′
This ensemble may be empty, but only when δ = 0
ΛI is the common part of the four ensembles. If your specious new explanation is correct, then the population is null when delta is zero. Which is hogwash. You can't disown your own paper !?
It is not difficult however to add the complete set of counterfactual measurement outcomes in the simulation code, and to compute the "empirical" analogue of (7).
I have extended the script. There is indeed also a violation of (7) in this particular experiment.
You must really believe I'm stupid, otherwise you will not post your new script.
You calculate rho11, rho12, rho21, rho22 from one dataset with different sample11, sample12, sample21, and sample22, then you generate 4 different datasets from which you calculate sample11, sample12, sample21, and sample22 but deliberately do not calculate rho11, rho12, rho21, rho22 for these, then you simply reuse the rho11, rho12, rho21, rho22 you calculated in the first case to calculate S, which you now use to compare with the bound calculated from the 4 different cases. Pure nonsense.
Delta must be calculated from the same dataset which you use to calculate S. Using the same sample11, sample12, sample21, and sample22. The symbols in equation (10) and equation (11) in your paper have the exact same meaning. So enough with the silly games. Here is how to fix the mess you posted:
1. Now I already showed you that your delta of zero from the first case gave you an upper bound of 4 and your S was 2.826 (not even close).
2. You now provided 4 separate calculations for each setting pair. That is fine, but we must calculate a new S for this case. I'll post the code for the first two, and anyone can complete the rest.
After the lines where the dataSet is assigned, the following three lines:
- Code: Select all
dataSet <- data.frame(settingAlice = settingAlice, spinAlice = spinAlice, timeAlice = timeAlice,
settingBob = settingBob, spinBob = spinBob, timeBob = timeBob)
delta <- abs(timeAlice - timeBob)
sample11 <- (delta < 1.5) & (settingAlice == 1) & (settingBob == 1)
rho11 <- mean((spinAlice*spinBob)[sample11])
- Code: Select all
dataSet <- data.frame(settingAlice = settingAlice, spinAlice = spinAlice, timeAlice = timeAlice,
settingBob = settingBob, spinBob = spinBob, timeBob = timeBob)
delta <- abs(timeAlice - timeBob)
sample12 <- (delta < 1.5) & (settingAlice == 1) & (settingBob == 2)
rho12 <- mean((spinAlice*spinBob)[sample12])
After doing, this for all 4 cases, you can now calculate a new S. USING THE NEW RHOs!!!!!
- Code: Select all
(S <- rho11 + rho12 + rho21 - rho22)
all <- sum(sample11 & sample12 & sample21 & sample22)
delta11 <- all / sum(sample11)
delta12 <- all / sum(sample12)
delta21 <- all / sum(sample21)
delta22 <- all / sum(sample22)
delta <- min(c(delta11, delta12, delta21, delta22))
bound <- 4 - 2 * delta
S-bound
- Code: Select all
[1] -0.005963789
Repeat this as many times as you like, with the random number seed off, and find me just one case in which S-bound is greater than 0.00001. I tried it and after 1000 iterations the biggest value for S-bound I got was -0.0041
So sorry Richard, you have failed yet again. Back to the drawing board.
Note: one cannot violate a theorem. But sample analogues can violate population inequalities, of course. To look at (7) we need to extend the simulation so as to simulate the complete set of counterfactual outcomes of all possible measurements. The model allows us to do so, easily, by definition (it is a LHV model).
Richard, we are interested in the theorem in your paper. Equation (11) and the variable delta used to calculate it, from equation (10). Stop the diversionary tactics with equation (7). We are not interested in it. Secondly, you seem to have missed the important part of the challenge:
minkwe wrote:You claim that violation of such a bound is a privileged property of QM/non-local theories/non-real theories but impossible for LHV models.
You should have been trying to find a non-local/non-real theory that violates your LG theorem, if you believed it was easier to do.
[quote="gill1109"]delta is defined in formula (7), which refers to the population (ie to the model, not to the outcomes of a simulation experiment), and it refers to events defined in terms of counterfactual measurement of spin in both directions on both particles at once.[/quote]
Richard, who are you trying to fool here? According to your paper:
[quote]The problem here is that the ensemble on which the correlations are evaluated changes
with the settings, while the original Bell inequality requires that they stay the same. In effect,
the Bell inequality only holds on the common part of the four different ensembles ΛAC′ , ΛAD′ ,
ΛBC′ , and ΛBD′ , i.e., for correlations of the form
E(AC′|ΛAC′ ∩ ΛAD′ ∩ ΛBC′ ∩ ΛBD′ ). (8)
Unfortunately our experimental data comes in the form
E(AC′|ΛAC′),
[/quote]
The population is Λ, the ensembles from the experiment are ΛAC′ , ΛAD′ ,ΛBC′ , and ΛBD′. Delta is calculated as:
[quote]so we need an estimate of the relation of the common part to its constituents:
δ = inf_settings P(ΛAC′ ∩ ΛAD′ ∩ ΛBC′ ∩ ΛBD′ )/P(ΛAC′ )
= inf_settings P(ΛAD′ ∩ ΛBC′ ∩ ΛBD′ |ΛAC′).
...
ΛI = ΛAC′ ∩ ΛAD′ ∩ ΛBC′ ∩ ΛBD′
This ensemble may be empty, but only when δ = 0
[/quote]
ΛI is the common part of the four ensembles. If your specious new explanation is correct, then the population is null when delta is zero. Which is hogwash. You can't disown your own paper !?
[quote]It is not difficult however to add the complete set of counterfactual measurement outcomes in the simulation code, and to compute the "empirical" analogue of (7).
I have extended the script. There is indeed also a violation of (7) in this particular experiment.[/quote]
You must really believe I'm stupid, otherwise you will not post your new script.
You calculate rho11, rho12, rho21, rho22 from one dataset with different sample11, sample12, sample21, and sample22, then you generate 4 different datasets from which you calculate sample11, sample12, sample21, and sample22 but deliberately do not calculate rho11, rho12, rho21, rho22 for these, then you simply reuse the rho11, rho12, rho21, rho22 you calculated in the first case to calculate S, which you now use to compare with the bound calculated from the 4 different cases. Pure nonsense.
Delta must be calculated from the same dataset which you use to calculate S. Using the same sample11, sample12, sample21, and sample22. The symbols in equation (10) and equation (11) in your paper have the exact same meaning. So enough with the silly games. Here is how to fix the mess you posted:
1. Now I already showed you that your delta of zero from the first case gave you an upper bound of 4 and your S was 2.826 (not even close).
2. You now provided 4 separate calculations for each setting pair. That is fine, but we must calculate a new S for this case. I'll post the code for the first two, and anyone can complete the rest.
After the lines where the dataSet is assigned, the following three lines:
[code]dataSet <- data.frame(settingAlice = settingAlice, spinAlice = spinAlice, timeAlice = timeAlice,
settingBob = settingBob, spinBob = spinBob, timeBob = timeBob)
delta <- abs(timeAlice - timeBob)
sample11 <- (delta < 1.5) & (settingAlice == 1) & (settingBob == 1)
rho11 <- mean((spinAlice*spinBob)[sample11])[/code]
[code]dataSet <- data.frame(settingAlice = settingAlice, spinAlice = spinAlice, timeAlice = timeAlice,
settingBob = settingBob, spinBob = spinBob, timeBob = timeBob)
delta <- abs(timeAlice - timeBob)
sample12 <- (delta < 1.5) & (settingAlice == 1) & (settingBob == 2)
rho12 <- mean((spinAlice*spinBob)[sample12])[/code]
After doing, this for all 4 cases, you can now calculate a new S. USING THE NEW RHOs!!!!!
[code](S <- rho11 + rho12 + rho21 - rho22)
all <- sum(sample11 & sample12 & sample21 & sample22)
delta11 <- all / sum(sample11)
delta12 <- all / sum(sample12)
delta21 <- all / sum(sample21)
delta22 <- all / sum(sample22)
delta <- min(c(delta11, delta12, delta21, delta22))
bound <- 4 - 2 * delta
S-bound
[/code]
[code][1] -0.005963789[/code]
Repeat this as many times as you like, with the random number seed off, and find me just one case in which S-bound is greater than 0.00001. I tried it and after 1000 iterations the biggest value for S-bound I got was -0.0041
So sorry Richard, you have failed yet again. Back to the drawing board.
[quote]Note: one cannot violate a theorem. But sample analogues can violate population inequalities, of course. To look at (7) we need to extend the simulation so as to simulate the complete set of counterfactual outcomes of all possible measurements. The model allows us to do so, easily, by definition (it is a LHV model).[/quote]
Richard, we are interested in the theorem in your paper. Equation (11) and the variable delta used to calculate it, from equation (10). Stop the diversionary tactics with equation (7). We are not interested in it. Secondly, you seem to have missed the important part of the challenge:
[quote="minkwe"]You claim that violation of such a bound is a privileged property of QM/non-local theories/non-real theories but impossible for LHV models.[/quote]
You should have been trying to find a non-local/non-real theory that violates your LG theorem, if you believed it was easier to do.