4 ms·

Strong assertions.. I think that you'd better study the EPR paradox (E as in Einstein), Bell's inequality https://en.wikipedia.org/wiki/Bell%27s_theorem https:/

by renox 6y ago

Strong assertions.. I think that you'd better study the EPR paradox (E as in Einstein), Bell's inequality https://en.wikipedia.org/wiki/Bell%27s_theorem https://en.wikipedia.org/wiki/Bell%27s_theorem and the Alan Aspect (and other) experiments. To summarize: reality isn't "local" even though we cannot send information faster than light..

eigenket 6y ago
There are (at least) two local, deterministic interpretations of quantum mechanics (many worlds and superdeterminism). You are correct that Bell-type experiments put very strong constraints on such theories. If your theory is local it is going to be very weird.
- l33tman 6y ago
  This is not meant as a snarky comment - but I would be happy to hear how many worlds is a local theory (as I guess it isn't a fully operational theory in the sense that QM+Born rule is). Even if you assume the independent, local evolution of all "branches" of Psi, at some point you need to explain interference (as that is an observed phenomena in our real world) and the comparison of different swaths of Psi's is not a local operation. How is this handled in that interpretation?
  - eigenket 6y ago
    By local I mean local in the sense of special relativity. Stuff at a point in spacetime only affects stuff at that point and in its causal future. If you want to affect something somewhere else then you have to fire a light particle or whatever there. Interference between different "branches" of the wavefunction is local if you're comparing the value of the two branches at the same point(s). Edit: I should emphasise that interference is a completely local (in the sense I use it above) process in every interpretation of quantum mechanics I am aware of (even Copenhagen) generally the thing that makes quantum mechanics non-local in various interpretations is measurement.
    - l33tman 6y ago
      Yeah I guess we're thinking similarly in that case :) It IS annoying though that it seems like the "last step" to produce measurable results requires a global integration though even though all the field interactions are local. The path integral QFT approach even makes this explicitly manifest, and modern QFT is the gold standard of reality-matching-predictions no matter the field.. Oh BTW re your edit: I'm not sure what "local interference" would mean. If interference effects could be solved locally, there would be no entanglement, no violated Bell's theorem or EPR experiments etc.
      - eigenket 6y ago
        I think maybe we're talking about different things? Could you be more precise about the sort of interference effects you're talking about? I generally wouldn't call entanglement an "interference effect". When I think about interference my mind goes to the patterns emerging from double slit experiments. Consider a quantum teleportation experiment, I start with N (a|0> + b|1>) ( |00> + |11> ) where N fixes the normalisation I do a measurement on qubits 0 and 1 in the basis ( |00> + |11> ), ( |00> - |11> ), ( |10> + |01> ), ( |10> - |10> ) and observe the outcome associated with ( |00> + |11> ), so now my state is N ( |00> + |11> ) (a|0> + b|1>) Now we collect our grant money because we've teleported the state to the other end. I don't see where the interference effect(s) were, from my perspective everything looked local and nice except for the impact of measurement, which is nonlocal in the "standard" picture and unexplained in general. I think my perspective on interference comes from the fact that water waves and sound waves (for example) exhibit interference in pretty much the same way that wavefunctions do, but we can't break Bell inequalities with them since we don't have the same weirdness around measurement.
- renox 6y ago
  As these 'interpretations' provide no new prediction, I think that Occam's razor apply here..
  - eigenket 6y ago
    Occam's razor is not trivial to apply when everyone claims that their own interpretation is the simplest.

by renox 6y ago

eigenket 6y ago

There are (at least) two local, deterministic interpretations of quantum mechanics (many worlds and superdeterminism). You are correct that Bell-type experiments put very strong constraints on such theories. If your theory is local it is going to be very weird.

l33tman 6y ago
This is not meant as a snarky comment - but I would be happy to hear how many worlds is a local theory (as I guess it isn't a fully operational theory in the sense that QM+Born rule is). Even if you assume the independent, local evolution of all "branches" of Psi, at some point you need to explain interference (as that is an observed phenomena in our real world) and the comparison of different swaths of Psi's is not a local operation. How is this handled in that interpretation?
- eigenket 6y ago
  By local I mean local in the sense of special relativity. Stuff at a point in spacetime only affects stuff at that point and in its causal future. If you want to affect something somewhere else then you have to fire a light particle or whatever there. Interference between different "branches" of the wavefunction is local if you're comparing the value of the two branches at the same point(s). Edit: I should emphasise that interference is a completely local (in the sense I use it above) process in every interpretation of quantum mechanics I am aware of (even Copenhagen) generally the thing that makes quantum mechanics non-local in various interpretations is measurement.
  - l33tman 6y ago
    Yeah I guess we're thinking similarly in that case :) It IS annoying though that it seems like the "last step" to produce measurable results requires a global integration though even though all the field interactions are local. The path integral QFT approach even makes this explicitly manifest, and modern QFT is the gold standard of reality-matching-predictions no matter the field.. Oh BTW re your edit: I'm not sure what "local interference" would mean. If interference effects could be solved locally, there would be no entanglement, no violated Bell's theorem or EPR experiments etc.
    - eigenket 6y ago
      I think maybe we're talking about different things? Could you be more precise about the sort of interference effects you're talking about? I generally wouldn't call entanglement an "interference effect". When I think about interference my mind goes to the patterns emerging from double slit experiments. Consider a quantum teleportation experiment, I start with N (a|0> + b|1>) ( |00> + |11> ) where N fixes the normalisation I do a measurement on qubits 0 and 1 in the basis ( |00> + |11> ), ( |00> - |11> ), ( |10> + |01> ), ( |10> - |10> ) and observe the outcome associated with ( |00> + |11> ), so now my state is N ( |00> + |11> ) (a|0> + b|1>) Now we collect our grant money because we've teleported the state to the other end. I don't see where the interference effect(s) were, from my perspective everything looked local and nice except for the impact of measurement, which is nonlocal in the "standard" picture and unexplained in general. I think my perspective on interference comes from the fact that water waves and sound waves (for example) exhibit interference in pretty much the same way that wavefunctions do, but we can't break Bell inequalities with them since we don't have the same weirdness around measurement.
renox 6y ago
As these 'interpretations' provide no new prediction, I think that Occam's razor apply here..
- eigenket 6y ago
  Occam's razor is not trivial to apply when everyone claims that their own interpretation is the simplest.