4 ms·
Violations of Boole's inequalities only happen when you do something bad with probabilities, you just need to be a little creative. Experimenters observe CHSH
by GistNoesis 5y ago
Violations of Boole's inequalities only happen when you do something bad with probabilities, you just need to be a little creative.
Experimenters observe CHSH > 2 because they are likely doing something bad with probabilities : the same bad things with probabilities you need to do to violate Boole's inequalities.
The measurement process of the pairs by the detectors implicitly do some additional filtering on the hidden_state. Typically when you define what your detection threshold for registering a blip is, or what you consider is a coincident pair by filtering based upon the time of measurement.
When you try to apply Boole's inequality to something like
P( anything | measurement1, measurement2, 0< hidden_state < 1),
P( anything | measurement1, 0< hidden_state < 1),
P( anything | measurement2, 0< hidden_state < 1)
The chain rule doesn't apply the same way. Because hidden_state is not a specific value but a range of values : The joint probability is not factoring as a product of conditionals anymore.
You have some incompatible sets of events and you can observe CHSH type violation for your Boole's inequalities.
- eigenket 5y ago> Violations of Boole's inequalities only happen when you do something bad with probabilities. I sort of agree with this, if by "something bad" you mean not obeying the axioms of probabilities. > Experimenters observe CHSH > 2 because they are likely doing something bad with probabilities. I disagree with this completely. Note that the probability distribution I gave just now (which is an honest probability distribution) violates CHSH. It obtains a CHSH score of 4, indeed. I am not doing anything "bad" with my probabilities to obtain this score, I'm just providing a probability distribution that does not obey Bell's additional locality assumption. I ask you specifically, what is "bad" about the distribution P(a,b| x,y) = 1/2 if ab = x xor y and zero otherwise which makes it obtain a CHSH score of 4?
- GistNoesis 5y agoI think you are missing the point I'm trying to make. You can violate CHSH by giving a non-local distribution (That's what happening with the distribution you provide). But there is more than one way to skin a cat. You can also violate them using the mathematical trick as I suggested above. This trick to replicate the result seems a lot more plausible than non-locality. > not obeying the axioms of probabilities. That's exactly what I blame the experimenters for. By ignoring that they are conditioning on the hidden variable, they are treating what is a "conditional probability" exactly the same way as it if was a "standard probability". Which is a big no-no from the math standpoint.
- eigenket 5y agoCould you give an example of exactly what you think they're doing wrong? I've read a few of these papers and I've not seen any such issues.
- GistNoesis 5y agoWhat convinced me of what is happening is, and of the pointlessness of Bell theorem is : Marian Kupczynski : Closing the Door on Quantum Nonlocality : https://philpapers.org/archive/KUPCTD.pdf https://philpapers.org/archive/KUPCTD.pdf