Heinera wrote:So if the original correlations (all computed on the whole set) didn't violate the CHSH inequality (CHSH<2), and the correlations computed on four disjoint random subset would not change much, we can now conclude that the four latter correlations would still not significantly violate the CHSH inequality, since term by term, they are approximately equal to the original correlations?
gill1109 wrote:Yes Heinera, you are home. They might violate it a little, but in all probability they won't violate it by much
If we measure <A> on a fair sample, of the population, we should get almost the same result as the population.
If we measure <B> on a fair sample, of the population, we should get almost the same result as the population.
Just because we measured <A> on one fair sample, and <B> on a different fair sample, does not change the fact that E(A) + E(B) = 0 for the population. You might violate it a little due to experimental error, but it shouldn't matter whether we use a single population or two disjoint fair samples of the population. Then the expectation values from the two coins should still not significantly violate the expression we obtained from the single coin
Minkwe wrote:Do you guys get it now?
gill1109 wrote:Minkwe wrote:Do you guys get it now?
No, because I nowhere make any erroneous substitution. So everything you say is pretty irrelevant. Anyway, I am interested in an experiment which has nothing to do with CHSH but which will either give Joy and the experimenter the Nobel prize, or all three of us the igNobel prize. And I'm looking for people who will help that experiment to happen.
minkwe wrote:Richard,
What part of it do you not understand. Point it out and I will explain.
minkwe wrote:The argument I have made many times is that the reason for the violation is due to an erroneous substitution of actual results from two separate systems
gill1109 wrote:minkwe wrote:The argument I have made many times is that the reason for the violation is due to an erroneous substitution of actual results from two separate systems, into an expression derived using actual & counter-factual results from a single system.
There is no erroneous substitution of actual results from two separate systems.
minkwe wrote:gill1109 wrote:minkwe wrote:The argument I have made many times is that the reason for the violation is due to an erroneous substitution of actual results from two separate systems, into an expression derived using actual & counter-factual results from a single system.
There is no erroneous substitution of actual results from two separate systems.
Seriously Richard?
Astounding!
Do you want me to show you where it is done in your papers and in the literature by ALL experimentalists who have done Bell-type experiments?
gill1109 wrote:As to the literature of all experimentalists, I am afraid that they are often pretty fuzzy about the logic of what they are doing. They just see the chance of getting famous by doing a difficult experiment (it is difficult, you know!) which some theoretician has said is important. They don't understand the theoretician's reasoning.
I'm afraid the literature is a big big mess. A lot of people are very confused, including, I'm afraid to say, many famous experimentalists. I can show you some truly incredible nonsense, if you are interested. But only after you have done my experiment. It might teach you something useful, something you don't seem to know about yet.
minkwe wrote:gill1109 wrote:As to the literature of all experimentalists, I am afraid that they are often pretty fuzzy about the logic of what they are doing. They just see the chance of getting famous by doing a difficult experiment (it is difficult, you know!) which some theoretician has said is important. They don't understand the theoretician's reasoning.
I'm afraid the literature is a big big mess. A lot of people are very confused, including, I'm afraid to say, many famous experimentalists. I can show you some truly incredible nonsense, if you are interested. But only after you have done my experiment. It might teach you something useful, something you don't seem to know about yet.
Richard, I'm talking about the same experimentalists you have cited in your papers, whose claims you have repeated in your papers, and with whom you have jointly written papers. Are you now throwing them under the bus? You did not say if you wanted me to show you were you did infact make the substitution in your papers. Do you want me to show it, or are you ready to retract the claim that nobody is doing that.
You still haven't mentioned where my argument above is wrong. Can it not be expressed in English?
I am not answering any more questions from you till you have (a) done my experiment, (b) translated the code for determine the bet outcome into Python.
There is a lot of work to do. Because of your lack of cooperation I have to learn Python. Oh well...
Ben6993 wrote:Michel, I have switched to this new thread for my reply.
I have worked through Case 1 and 2 and Experiments I and II with the biased coin. In Case 2, for five throws [using SIGN(RAND()-0.375) for each throw in Excel] I obtained <A> = 1 and <B> = 0.6. So <A> + <B> = 1.6 which still does not exceed 2. As to be expected, because neither <A> nor <B> can exceed 1.
I had assumed that the four pairs of angles produced outcomes based on four different sets of balls, so that no counterfactuals were needed, and no ball was measured twice.
(Although, as these are macroscopic entities it would not matter if the same ball was measured many times at many different angles? A measurement of a macroscopic object cannot be wrong like a counterfactual calculation?)
I do not see a 'counterfactual' problem if all correlations are based on measurements, so the four sets are all treated equally.
import numpy
N = 100000 # number of tosses
t = 0.375 # Tail bias
print "Case 1: Single coin toss (Experiment I)"
A = numpy.random.choice([-1, 1], p=(t, 1-t), size=N)
B = -A
A_B = A + B
print "%0.2f <= E(A) + E(B) <= %0.2f" % (A_B.min(), A_B.max())
print "<A> + <B> = %0.2f + %0.2f = %0.2f" % (A.mean(), B.mean(), A_B.mean())
print
print "Case 2: Two coin tosses (Experiment II)"
A = numpy.random.choice([-1, 1], p=(t, 1-t), size=N)
B = numpy.random.choice([-1, 1], p=(t, 1-t), size=N)
A_B = A + B
print "%0.2f <= E(A) + E(B) <= %0.2f" % (A_B.min(), A_B.max())
print "<A> + <B> = %0.2f + %0.2f = %0.2f" % (A.mean(), B.mean(), A_B.mean())
print
print "Case 3: Re-use just one data stream from Experiment II + its counterfactual"
C = - A
A_B = A + C
print "%0.2f <= E(A) + E(B) <= %0.2f" % (A_B.min(), A_B.max())
print "<A> + <B> = %0.2f + %0.2f = %0.2f" % (A.mean(), C.mean(), A_B.mean())
print
print "Case 3: Re-use the other data stream from Experiment II + its counterfactual"
D = - B
A_B = D + B
print "%0.2f <= E(A) + E(B) <= %0.2f" % (A_B.min(), A_B.max())
print "<A> + <B> = %0.2f + %0.2f = %0.2f" % (B.mean(), D.mean(), A_B.mean())
Case 1: Single coin toss (Experiment I)
0.00 <= E(A) + E(B) <= 0.00
<A> + <B> = 0.25 + -0.25 = 0.00
Case 2: Two coin tosses (Experiment II)
-2.00 <= E(A) + E(B) <= 2.00
<A> + <B> = 0.25 + 0.25 = 0.50
Case 3: Re-use just one data stream from Experiment II + its counterfactual
0.00 <= E(A) + E(B) <= 0.00
<A> + <B> = 0.25 + -0.25 = 0.00
Case 3: Re-use the other data stream from Experiment II + its counterfactual
0.00 <= E(A) + E(B) <= 0.00
<A> + <B> = 0.25 + -0.25 = 0.00
gill1109 wrote:I furthermore suggest we erase the words "proof", "inequality", "bound", and "violation" from the vocabulary of experimental physics. They are superfluous and apparently cause a lot of confusion.
More precisely: the words can be avoided; but if they are used, they need very careful qualification. EG "statistical proof" might be better than "proof", but still one needs to explain what is meant by "statistical proof" ... apparently a lot of people have never even thought about the concept.
Ben6993 wrote:Michel, I have done more calculations following your reply.
CHSH statistic = E(a, b) − E(a, b′) + E(a′, b) + E(a′ b′). The settings a, a′, b and b′ are 0°, 90°, 45° and 135°, respectively
For the trivial case of a CHSH run of one pair of particles, CHSH = AB - AB' + A'B + A'B'
If A and B are measurement outcomes, and A' (90°) and B'(135°) are counterfactual estimates based on A (0°) and B(45°) such that A' = -A and B' = -B.
Then
CHSH = -2 when A = 1 and B = -1
CHSH = -2 when A = -1 and B = 1
CHSH = 2 when A = 1 and B = 1
CHSH = 2 when A = -1 and B = -1.
These are the only four possible outcomes
.
But if A, A', B and B' are all different and independent particles, the possible outcomes are greater and can be any one of 16 combinations ranging from A = B = A' = B' = 1 through to A = B = A' = B' = -1. In each case the CHSH statistic is again either 2 or -2.
import itertools
[''.join(v) for v in itertools.product('+-', repeat=8)]
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