Chaos is sensitivity to initial conditions. Could the randomness in quantum mechanics simply be a manifestation of chaos? The initial conditions would be both the initial state of a particle, and perhaps initial conditions of the measurement device at the time of measurement, and in the fields present at the particle location at the time of measurement. Roger Penrose believes gravity plays a role in the collapse of the wavefunction, so gravity could be included here (in factoring in of initial conditions).
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2$\begingroup$ Bell's inequalities, which have been experimentally tested show that there is no local hidden variable theory that can reproduce the results of quantum mechanics (modulo some small loopholes). What this means is that a classical theory in the normal sense simply cannot reproduce quantum mechanics. Whether or not the classical theory is chaotic is irrelevant. There has to be something strange going on. The mainstream view is that the "something" is that the world is quantum, and not deterministic. $\endgroup$– AndrewCommented Jul 1, 2022 at 20:50
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1$\begingroup$ I recognize there is a mainstream view. However, a few prominent physicists believe things are not inherently random at the core. I am simply asking for what is possible given current experimental observations. $\endgroup$– Joseph Robert JepsonCommented Jul 1, 2022 at 20:53
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2$\begingroup$ Chaos is not a solution to the problem, because chaos is a feature of classical mechanics. Classical mechanics is not able to reproduce the results of Bell's inequality tests. To explain observations while avoiding standard quantum mechanics, you have to change classical mechanics in a deep way, such as making it nonlocal (which is hard to jive with special relativity because in special relativity nonlocality implies acausal transfer of information, meaning you can kill your grandfather), or by appealing to superdeterminism. $\endgroup$– AndrewCommented Jul 1, 2022 at 20:57
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1$\begingroup$ I do like your above comment, but I guess I would like some more specific rigorous reasons for that. Also appreciate the link. $\endgroup$– Joseph Robert JepsonCommented Jul 1, 2022 at 21:00
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1$\begingroup$ Bell's inequalities are the rigorous statement. Both the mathematical proof and experimental tests are described in the wikipedia article about Bell's inequalities I linked above. $\endgroup$– AndrewCommented Jul 1, 2022 at 21:01
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2 Answers
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No. Even if one can reproduce exactly the initial state and the measurements are ideal (thus even if the initial conditions are exactly identical) the outcomes can be randomly different.
Thus, even if multiple copies of a system are exactly prepared so the initial state is the spin-up state along $+\hat z$ $\vert +;z\rangle$ (exactly same initial conditions on all copies) and the copies are fed in identical copies of a perfect apparatus designed to measure the $\hat y$ component of spin, the outcome of any one measurement would have 50/50 chance of being either $\vert +;y\rangle$ or $\vert -;y\rangle$.
answered Jul 2, 2022 at 2:31
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$\begingroup$ @CarlosGauss Not sure I follow. There is no interpretation here; the interpretation comes after, in explaining the (possibly apparent) randomness, not in the factual observation that the outcomes are 50/50. $\endgroup$ Commented Jul 2, 2022 at 3:08
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$\begingroup$ @CarlosGauss I am not suggesting an explanation to this randomness (real or apparent), just claiming it is random inasmuch as identical preparation and perfect measurement do not lead to identical outcomes (hence independence on initial conditions). $\endgroup$ Commented Jul 2, 2022 at 9:53
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$\begingroup$ @CarlosGauss I edited my answer to hopefully remove some confusion. $\endgroup$ Commented Jul 2, 2022 at 10:02
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$\begingroup$ I see, thanks, I am sorry about the confusion, I erased my comment $\endgroup$– user338734Commented Jul 2, 2022 at 11:28
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1$\begingroup$ well you are incorrect. Randomness in quantum mechanics is not something that is tied to the measuring device or the initial conditions. Calculations where you assume these are perfect still yield random distributions. $\endgroup$ Commented Jul 15, 2022 at 15:30
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My two cents of the euro, you ask:
Is randomness in the collapse of the wavefunction in quantum mechanics simply a manifestation of chaos, and not inherently random at all?
Do an experiment with a dice.After 1000 throws there is a distribution with the number of times the six faces came up. If it you get a flat probability distribution you decide on true randomness of classical mechanics probable motions on the dice. If you see a biased distribution, you attribute it to a weighted dice.
The experiments at the quantum level do not show, cannot be modeled with classical mechanics statistical distributions. We attribute the bias to quantum mechanics which by now is an elaborate mathematical model , that fits the distributions and, important , predicts correctly new ones. And the collapse is not random in the theory, it is biased with the wavefunction distribution ($Ψ^*Ψ$)
My point is that the "collapse" is biased, not random according to mainstream theory. Chaos would produce random distributions, but the data, crossections and decays directly connected to the probability distribution are not random.
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$\begingroup$ Is it true that chaos cannot produced biased distributions?? If you could add more explanation to your answer that would be helpful. $\endgroup$ Commented Jul 11, 2022 at 18:35
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$\begingroup$ a specialist in chaos should answer that. It is a different question., $\endgroup$– anna vCommented Jul 11, 2022 at 19:10