3 ms·

Speaking as a lay person, I think the difference might be that it's specifically about local hidden variables. If two particles are coupled, there's no per-part

Speaking as a lay person, I think the difference might be that it's specifically about local hidden variables. If two particles are coupled, there's no per-particle hidden variable?

6gvONxR4sf7o 5y ago
edit-too-late-to-edit: I remembered something that makes me think it's actually more complicated. If I write 0 and 1 on two different pieces of paper, then flip a coin to decide which paper to give you, we have "entangled" unknowns. When I reveal my paper, we instantly know what's on yours. The joint distribution can't be described per-particle, but we don't consider it spooky. So I think there's something more to it.
- BeKindAndLearn 5y ago
  Because your example does not contain superpositions. With entanglement the pieces of paper are both 0 and 1 at the same time until they are observed - and only then is it that the superposition of both particles become resolved.
  - 6gvONxR4sf7o 5y ago
    Tomayto - tomahto, right? Physically speaking, what’s the difference between an unresolved superposition and a different kind of unresolved random variable?
choeger 5y ago
Seriously, what's the difference between a non-local variable and a deterministic function that yields a pseudo-random value? From what I understand, we have some function f that can be evaluated by the measurement and we have another, entangled function g, that can also be evaluated but will yield the opposite of f. Now instead of assuming that f is truly random and somehow communicates its value instantaneously to g when evaluated, we could also assume that f and g contain a copy of the same pseudo-random number generator and the same seed. In both cases the interpretation of the model is weird, of course. But I don't see a fundamental difference here that wouldn't allow for hidden variables. It's just that the hidden variables would be have very non-trivial domains.

Speaking as a lay person, I think the difference might be that it's specifically about local hidden variables. If two particles are coupled, there's no per-particle hidden variable?

6gvONxR4sf7o 5y ago

edit-too-late-to-edit: I remembered something that makes me think it's actually more complicated. If I write 0 and 1 on two different pieces of paper, then flip a coin to decide which paper to give you, we have "entangled" unknowns. When I reveal my paper, we instantly know what's on yours. The joint distribution can't be described per-particle, but we don't consider it spooky. So I think there's something more to it.

BeKindAndLearn 5y ago
Because your example does not contain superpositions. With entanglement the pieces of paper are both 0 and 1 at the same time until they are observed - and only then is it that the superposition of both particles become resolved.
- 6gvONxR4sf7o 5y ago
  Tomayto - tomahto, right? Physically speaking, what’s the difference between an unresolved superposition and a different kind of unresolved random variable?

choeger 5y ago

Seriously, what's the difference between a non-local variable and a deterministic function that yields a pseudo-random value? From what I understand, we have some function f that can be evaluated by the measurement and we have another, entangled function g, that can also be evaluated but will yield the opposite of f. Now instead of assuming that f is truly random and somehow communicates its value instantaneously to g when evaluated, we could also assume that f and g contain a copy of the same pseudo-random number generator and the same seed. In both cases the interpretation of the model is weird, of course. But I don't see a fundamental difference here that wouldn't allow for hidden variables. It's just that the hidden variables would be have very non-trivial domains.