Timeline for Does Bell's theorem exclude local hidden variables as explanation for radioactive decay?
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| Aug 12, 2015 at 14:56 | comment | added | Morty | dmckee: Yes everything is quantum but that doesn't mean that any seemingly indeterministic thing is due to quantum. Take the weather - it is very hard to predict but I think most people would agree that is not due to the potential indeterminism of the quantum but due to "hidden variables" and the chaotic nature of the system. And I don't think people would say Bell's theorem precludes such hidden variables. So my question is, could nuclear decay be the same thing... i.e. deterministic at the core? | |
| Aug 11, 2015 at 16:12 | comment | added | dmckee --- ex-moderator kitten | Everything is quantum at the root of it, and nuclear decay happens on the sub atomic scale, so it is quantum all the time. | |
| Aug 11, 2015 at 14:39 | comment | added | Morty | OK my question was more if the seeming indeterminism in radioactive decay is even known to be due to quantum. You could imagine that all the nuclei of a given isotope on earth is currently in a random state of a completely deterministic (maybe non-quantum, mechanical-like) system and chaotically, but deterministically, changes from state to state and wille ventually reach a "decay state". If that is the case, then Bell's theorem is irrelevant with respect to radioactive decay. | |
| Aug 10, 2015 at 20:37 | history | answered | dmckee --- ex-moderator kitten | CC BY-SA 3.0 |