6 ms·

> You usually show a pupil the problem with classical probabilities, and show that you can't violate Bell's Inequalities, then you show that Quantum Mechanics m

> You usually show a pupil the problem with classical probabilities, and show that you can't violate Bell's Inequalities, then you show that Quantum Mechanics managed to replicated the observed probabilities using a non-local way, and therefore you conclude that the world is non-local. If you do this you're doing a bad job at being a teacher. The way the argument should go is you start with a list of assumptions (of which locality is one), derive Bell's inequality from them, and determine that as Bell's inequality seems to be false in real experiments at least one of your assumptions was wrong. Then you can talk about quantum mechanics and explain which of these assumption are broken in quantum mechanics. If you have time you can have fun talking about different interpretations of quantum mechanics because (e.g.) Everettian Many Worlds is completely local, but still produces predictions matching quantum mechanics (and therefore breaks Bell's inequality).

GistNoesis 2y ago
>If you do this you're doing a bad job at being a teacher. If this wasn't sufficiently clear, I am not a teacher, I am a crackpot physicist. Hint : Listing the assumptions doesn't work. This is what I call the three-card monte argument. The ball is not under one of the three goblets, the ball is in the sleeve of the magician.
- eigenket 2y ago
  > 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.
meroes 2y ago
MWI is not local according to all the big names I read; Lev Vaidman, Tim Maudlin, and Im pretty sure David Wallace too.
- eigenket 2y ago
  You can probably define locality in a way that MWI is nonlocal, but you can also definitely define it in a way such that MWI is local. For me the most important thing about nonlocality is the lack of any "action at a distance", MWI satisfies this, but if you make more stringent demands it might not satisfy those.

GistNoesis 2y ago

>If you do this you're doing a bad job at being a teacher. If this wasn't sufficiently clear, I am not a teacher, I am a crackpot physicist. Hint : Listing the assumptions doesn't work. This is what I call the three-card monte argument. The ball is not under one of the three goblets, the ball is in the sleeve of the magician.

eigenket 2y ago
> 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.

meroes 2y ago

MWI is not local according to all the big names I read; Lev Vaidman, Tim Maudlin, and Im pretty sure David Wallace too.

eigenket 2y ago
You can probably define locality in a way that MWI is nonlocal, but you can also definitely define it in a way such that MWI is local. For me the most important thing about nonlocality is the lack of any "action at a distance", MWI satisfies this, but if you make more stringent demands it might not satisfy those.