3 ms·

> Hint : Listing the assumptions doesn't work Why not? I could see that it might not if you are not clear about your assumptions

GistNoesis 2y ago
It's circular reasoning, hidden in the definition of your assumptions. By defining not clearly what a measurement is and observations are. You must let the cat step out of the box your definitions put you in. You have infinite freedom in your choices of definitions, listing assumptions is creating a false dichotomy. Especially when doing so conclude to exclude the most probable assumption : Locality. Preserve locality, and find another self consistent theory which define properly what according to it a measurement is, an not take measurement and observations as axioms.
- eigenket 2y ago
  Will you grant me that it is at least possible to derive Bell's inequality by listing out a complete set of assumptions (including assumptions that define what a measurement is and what observations are)? Of course you personally may disagree with some of these axioms (indeed, if you take Bell's theorem seriously you must), but certainly it is possible to list them, and thereby derive Bell's inequality?
  - GistNoesis 2y ago
    Bell's theorem is a theorem. If hypothesis applies conclusion must follow. That's math. Everything is fine with it (They are a reformulation of "Bonferroni inequalities" or "Boole's_inequality" by the way). You've got to reframe the problem so that Bell's theorem doesn't apply. When you build your theory, if you manage to define what a measurement is, so that you don't satisfy the hypothesis of the Bell's theorem, you get to avoid having to have its conclusions. One of Bell's theorem implied hypothesis is that measurements/observations are probabilities, so by defining measurement instead as a conditional probability, you get to avoid being subjected to Bell's inequalities. It's inductive reasoning, you don't get truth you only get self consistency, and a model that looks much nicer than QM.
    - eigenket 2y ago
      > You've got to reframe the problem so that Bell's theorem doesn't apply. When you build your theory, if you manage to define what a measurement is, so that you don't satisfy the hypothesis of the Bell's theorem, you get to avoid having to have its conclusions. This (in my opinion) a bad way of explaining how the standard reasoning goes. We start with a list of assumptions, we prove this inequality which it turns out is not satisfied, we reject (at least) one of our assumptions. These is no crackpottery here, this is the norm. > by defining measurement instead as a conditional probability This sounds like it probably doesn't get you anywhere, but I'll bite - what are we conditioning on? In the standard formulation of Bell's theorem they are conditional on the "hidden variable" we are assuming exists, as well as any relevant measurement settings but it sounds like you're imagining something wilder than that.
      - GistNoesis 2y ago
        >what are we conditioning on? The local hidden state, but you don't get to set it from inside the universe when you do an experiment (this local hidden state is unobservable). From inside the universe based on this hidden state, everything behave classically, pseudo-randomly based on the local hidden state. But because you don't get to set the local hidden state during your experiment if you want to calculate the probabilities, you have to integrate over the possible values of the unknown hidden state, and this allows you to recover the strange looking quantum correlations. Doing repeated experiment inside a universe mean picking a different initial local hidden state (because it's unobservable). [Spoiler ahead] The original idea is not from me, if you want the nitty gritty details, look at the work of Marian Kupczynski (Closing the Door on Quantum Nonlocality https://philarchive.org/archive/KUPCTDv1 https://philarchive.org/archive/KUPCTDv1 ). Or his more recent works. I have made a straight forward implementation (3 years ago) of it to convince myself with a Monte Carlo simulation : https://gist.github.com/unrealwill/2a48ea0926deac4011d26842627b69c9 https://gist.github.com/unrealwill/2a48ea0926deac4011d268426... [End Spoiler]
        eigenket 2y ago
        Everything up to the [spoiler ahead] in this comment is (as far as I can tell) exactly how things work in standard formulations of Bell's inequality. There's nothing weird or crackpot there. Your numerical code is impossible for me to read without some basic idea of what you're trying to show, but I'd like to point out that numpy has functions like np.radians, and np.deg2rad to convert from degrees to radians, you don't have to make your own.
- wallfacer120 2y ago
  [dead]

GistNoesis 2y ago

It's circular reasoning, hidden in the definition of your assumptions. By defining not clearly what a measurement is and observations are. You must let the cat step out of the box your definitions put you in. You have infinite freedom in your choices of definitions, listing assumptions is creating a false dichotomy. Especially when doing so conclude to exclude the most probable assumption : Locality. Preserve locality, and find another self consistent theory which define properly what according to it a measurement is, an not take measurement and observations as axioms.

eigenket 2y ago
Will you grant me that it is at least possible to derive Bell's inequality by listing out a complete set of assumptions (including assumptions that define what a measurement is and what observations are)? Of course you personally may disagree with some of these axioms (indeed, if you take Bell's theorem seriously you must), but certainly it is possible to list them, and thereby derive Bell's inequality?
- GistNoesis 2y ago
  Bell's theorem is a theorem. If hypothesis applies conclusion must follow. That's math. Everything is fine with it (They are a reformulation of "Bonferroni inequalities" or "Boole's_inequality" by the way). You've got to reframe the problem so that Bell's theorem doesn't apply. When you build your theory, if you manage to define what a measurement is, so that you don't satisfy the hypothesis of the Bell's theorem, you get to avoid having to have its conclusions. One of Bell's theorem implied hypothesis is that measurements/observations are probabilities, so by defining measurement instead as a conditional probability, you get to avoid being subjected to Bell's inequalities. It's inductive reasoning, you don't get truth you only get self consistency, and a model that looks much nicer than QM.
  - eigenket 2y ago
    > You've got to reframe the problem so that Bell's theorem doesn't apply. When you build your theory, if you manage to define what a measurement is, so that you don't satisfy the hypothesis of the Bell's theorem, you get to avoid having to have its conclusions. This (in my opinion) a bad way of explaining how the standard reasoning goes. We start with a list of assumptions, we prove this inequality which it turns out is not satisfied, we reject (at least) one of our assumptions. These is no crackpottery here, this is the norm. > by defining measurement instead as a conditional probability This sounds like it probably doesn't get you anywhere, but I'll bite - what are we conditioning on? In the standard formulation of Bell's theorem they are conditional on the "hidden variable" we are assuming exists, as well as any relevant measurement settings but it sounds like you're imagining something wilder than that.
    - GistNoesis 2y ago
      >what are we conditioning on? The local hidden state, but you don't get to set it from inside the universe when you do an experiment (this local hidden state is unobservable). From inside the universe based on this hidden state, everything behave classically, pseudo-randomly based on the local hidden state. But because you don't get to set the local hidden state during your experiment if you want to calculate the probabilities, you have to integrate over the possible values of the unknown hidden state, and this allows you to recover the strange looking quantum correlations. Doing repeated experiment inside a universe mean picking a different initial local hidden state (because it's unobservable). [Spoiler ahead] The original idea is not from me, if you want the nitty gritty details, look at the work of Marian Kupczynski (Closing the Door on Quantum Nonlocality https://philarchive.org/archive/KUPCTDv1 https://philarchive.org/archive/KUPCTDv1 ). Or his more recent works. I have made a straight forward implementation (3 years ago) of it to convince myself with a Monte Carlo simulation : https://gist.github.com/unrealwill/2a48ea0926deac4011d26842627b69c9 https://gist.github.com/unrealwill/2a48ea0926deac4011d268426... [End Spoiler]
      - eigenket 2y ago
        Everything up to the [spoiler ahead] in this comment is (as far as I can tell) exactly how things work in standard formulations of Bell's inequality. There's nothing weird or crackpot there. Your numerical code is impossible for me to read without some basic idea of what you're trying to show, but I'd like to point out that numpy has functions like np.radians, and np.deg2rad to convert from degrees to radians, you don't have to make your own.
wallfacer120 2y ago
[dead]