minkwe wrote:Here is the question: Please demonstrate that QM violates Bell's inequality
by providing , the QM predictions for the terms
minkwe wrote:Here is the question: Please demonstrate that QM violates Bell's inequality
by providing , the QM predictions for the terms
minkwe wrote:So Guest and Heinera are demonstrating selective amnesia or something similar. I will advice them to go read the first page of this thread and then answer the question I asked.minkwe wrote:Here is the question: Please demonstrate that QM violates Bell's inequality
by providing , the QM predictions for the terms
Heinera wrote:I already told you in another thread: This expression only makes sense for LHV-models, and not for nonlocal models where the outcome depend on settings in both wings.
Heinera wrote:Nor is this the inequality we are testing against in experiments.
heinera wrote:Since the expectation of a random subset of values from a population is the same as the population expectation, we can replace the inequality with
This is the inequality we are testing against.
minkwe wrote:Bell's inequality is the following:
,
which makes use of the three terms all defined for the same set of outcomes .
The terms from experiments are each calculated from 3 separate sets of outcomes: yielding the three terms . As I have argued in the other thread, these terms do not have the same values as those in the original inequality, therefore we are not allowed to use the QM predictions for these terms in Bell's inequality as is customarily done to demonstrate violation by QM or Experiments. In other words, the common procedure of calculating
,
from QM and experimental data to proclaim violation of Bell's inequality is a farce. One other thing I argued in the other threads is that, we could be allowed to use those terms if and only if we can demonstrate that they are statistically equivalent through rearrangements. This ensures that must hold. Only under this condition are we allowed to use the QM prediction, and experimental results in Bell's inequality, and only under this condition do the original inequalities, derived on a single set, apply to data measured on disjoint sets. As I argued in the other thread, given 3 statistically independent sets of N outcome pairs, it is not possible to do such rearrangement in order to demonstrate the above equivalences. To summarize why, note that we could perform row permutations of the set of outcomes so that the column matches the column of the set. Then we could do row permutations of the set of outcomes so that the column matches the column of the set. One more rearrangement and we are home, since only the and pairs have not been matched. But, since only row permutations are allowed, any rearrangements to make match will undo the match between and . Therefore, it is not possible for 3 statistically independent sets of outcome pairs to satisfy simultaneously. Therefore the inequality does not apply, and the QM predictions for can not possibly be the same as the QM prediction for . Therefore it is not surprising that QM or even LHV theories would have different predictions for those terms.
minkwe wrote:heinera wrote:Since the expectation of a random subset of values from a population is the same as the population expectation, we can replace the inequality with
This is the inequality we are testing against.
No! you cannot do that magicians bait-and-switch trick. Please provide a detailed step-by-step derivation. The terms in Bell's inequality do not represent statistically independent outcomes of a population. You can not replace them with values from random statistically independent outcomes with or without LHV. This is basic statistics and has nothing do with any physics.
Heinera wrote:minkwe wrote:heinera wrote:Since the expectation of a random subset of values from a population is the same as the population expectation, we can replace the inequality with
This is the inequality we are testing against.
No! you cannot do that magicians bait-and-switch trick. Please provide a detailed step-by-step derivation. The terms in Bell's inequality do not represent statistically independent outcomes of a population. You can not replace them with values from random statistically independent outcomes with or without LHV. This is basic statistics and has nothing do with any physics.
A random sampla drawn from a population with expectation also has expectation . If I have a large set with mean , and randomly divide it into 3 or 4 subsets, each subset vil have a mean very close to . This is elementary probability theory. It is you who are invoking magic by believing that these means will miraculously be very different.
minkwe wrote:Heinera wrote:minkwe wrote:Is it too hard for you to see that the terms from your bogus inequality are statistically independent,
Heinera wrote:minkwe wrote:Heinera wrote:
They are no more or less statistically independent than the four subsets i described in my previous post. Only a fool would believe the four subsets would have very different means.
minkwe wrote:huh? Have you even read the argument? From the first post, repeated in my previous post to you? Only an idiot will continue to make arguments which are obviously false, without making an effort to read the post which demonstrates this fact.
minkwe wrote:Now where is the evidence that experiments violate Bell's inequality???
Mikko wrote:minkwe wrote:Now where is the evidence that experiments violate Bell's inequality???
It is in the reports of those experiments. The reports tell what was done and what happened.
However, the topic of this discussion is not experiments. It is quantum mechanics instead. More specifically, whether quantum mechanics predicts violation of Bell's inequalities. Apparently none of those who might know is motivated to reveal it here. Elsewhere we may read that it does. Easy to check with a simulation.
Heinera wrote:I already told you in another thread: This expression only makes sense for LHV-models, and not for nonlocal models where the outcome depend on settings in both wings. Nor is this the inequality we are testing against in experiments. Since the expectation of a random subset of values from a population is the same as the population expectation, we can replace the inequality with
This is the inequality we are testing against. Since the first inequality only makes sense for LHV-models, the second inequality only applies to LHV-models as well.
FrediFizzx wrote:Heinera wrote:I already told you in another thread: This expression only makes sense for LHV-models, and not for nonlocal models where the outcome depend on settings in both wings. Nor is this the inequality we are testing against in experiments. Since the expectation of a random subset of values from a population is the same as the population expectation, we can replace the inequality with
This is the inequality we are testing against. Since the first inequality only makes sense for LHV-models, the second inequality only applies to LHV-models as well.
Nope. This is the inequality that you are testing against.
Since the terms are independent.
Heinera wrote:They are all expectations of disjoint subsets drawn randomly from a common population ...
Heinera wrote:For an LHV-model, the expectations are not independent.
Joy Christian wrote:Heinera wrote:They are all expectations of disjoint subsets drawn randomly from a common population ...
A complete and utter hogwash.
Heinera wrote:Joy Christian wrote:Heinera wrote:They are all expectations of disjoint subsets drawn randomly from a common population ...
A complete and utter hogwash.
Because?
Heinera wrote:In , the sets of j,k, and l are random disjoint subsets of the set of i in .. Thus the terms will have approximately the same values in both inequalities for large N. This is truly elementary probability theory.
minkwe wrote:Bell's inequality is the following:
,
which makes use of the three terms all defined for the same set of outcomes .
The terms from experiments are each calculated from 3 separate sets of outcomes: yielding the three terms . As I have argued in the other thread, these terms do not have the same values as those in the original inequality, therefore we are not allowed to use the QM predictions for these terms in Bell's inequality as is customarily done to demonstrate violation by QM or Experiments. In other words, the common procedure of calculating
,
from QM and experimental data to proclaim violation of Bell's inequality is a farce. One other thing I argued in the other threads is that, we could be allowed to use those terms if and only if we can demonstrate that they are statistically equivalent through rearrangements. This ensures that must hold. Only under this condition are we allowed to use the QM prediction, and experimental results in Bell's inequality, and only under this condition do the original inequalities, derived on a single set, apply to data measured on disjoint sets. As I argued in the other thread, given 3 statistically independent sets of N outcome pairs, it is not possible to do such rearrangement in order to demonstrate the above equivalences. To summarize why, note that we could perform row permutations of the set of outcomes so that the column matches the column of the set. Then we could do row permutations of the set of outcomes so that the column matches the column of the set. One more rearrangement and we are home, since only the and pairs have not been matched. But, since only row permutations are allowed, any rearrangements to make match will undo the match between and . Therefore, it is not possible for 3 statistically independent sets of outcome pairs to satisfy simultaneously. Therefore the inequality does not apply, and the QM predictions for can not possibly be the same as the QM prediction for . Therefore it is not surprising that QM or even LHV theories would have different predictions for those terms.
Heinera wrote:Joy Christian wrote:Heinera wrote:They are all expectations of disjoint subsets drawn randomly from a common population ...
A complete and utter hogwash.
Because?
minkwe wrote:Heinera wrote:In , the sets of j,k, and l are random disjoint subsets of the set of i in .. Thus the terms will have approximately the same values in both inequalities for large N. This is truly elementary probability theory.
Like I thought, you have no answer to the very clear arguments which I've provided since the very first post of this thread, debunking this nonsense. You think because you append "This is truly elementary probability theory" to a statement that has absolutely nothing to do with probability theory, it changes the fact that it is junk? You simply pretend the argument does not exist. Here it is again, for your enjoyment:
Heinera wrote:minkwe wrote:Heinera wrote:In , the sets of j,k, and l are random disjoint subsets of the set of i in .. Thus the terms will have approximately the same values in both inequalities for large N. This is truly elementary probability theory.
Like I thought, you have no answer to the very clear arguments which I've provided since the very first post of this thread, debunking this nonsense. You think because you append "This is truly elementary probability theory" to a statement that has absolutely nothing to do with probability theory, it changes the fact that it is junk? You simply pretend the argument does not exist. Here it is again, for your enjoyment:
Your "clear arguments" was thoroughly debunked by Jochen. No need to repeat that (use the search function in this forum instead) If you think statistics and probability theory is nonsense, or that it does not apply in this case, then so be it. Just stay away from the casinos, or you'll soon find yourself to be a very poor man.
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