Inspired by this question, I wonder if classical (chaotic) double pendulums are actually in a quantum superposition when not observed.

Chaotic physical systems are systems where slightly different initial positions eventually produce substantially different outcomes after some time.

For example, after enough time, even the smallest differences of starting positions of a double-pendulum will end up producing very different outcomes. Now (loosely speaking) even a massive system will have some quantum uncertainty to its center-of-mass position (yes, a pendulum really is a more complicated system but I suspect there is some level of truth to this...there surely is some quantum uncertainty to a massive quantum system's center-of-mass, right?). So, a massive system is therefore in a superposition of different tiny changes in positions (as long as these tiny positions are not "measured").

If you consider the classical evolution of each of these superpositions (of tiny changes in position of the massive quantum system) - they evolve chaotically, and eventually there will be a time when the outcomes are completely different. Yet, if the pendulum is completely isolated - wouldn't it remain in a superposition until measured? This would suggest that you could have "Schrodinger's Pendulums" where the macroscopic outcome is substantially different upon measurement.

Is there any sense to this idea? I suspect the responses will insist that decoherence ruins the fun, but I don't exactly get the intuition for why this can be argued for in general. If you think of a pendulum as a gigantic mass of particles all with harmonic oscillator potentials all connected to their nearest neighbor - then how does this system's positional uncertainty decohere without being explicitly measured by an observer?

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    $\begingroup$ As you suggested, this is like Schrödinger's cat (but your version is better, because it's more civilized). The same concepts apply in both cases. If you're not so sure that the same concepts apply in both cases, then can you clarify what you think is the key difference? $\endgroup$ Aug 11 '21 at 1:11
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    $\begingroup$ @ChiralAnomaly, uh, well in the case of Schrodinger's cat there are some arguments that you're not really going to be able to prepare a superposition because of "decoherence." I've never quite understood the precise argument there - but thinking about it more - I think you're probably right that it might just be the exact same thing. Although this does seem to suggest that probably most classical systems (when not observed) are actually in quantum superpositions until you measure them (since most things in nature some some degree of butterfly effect). $\endgroup$ Aug 11 '21 at 1:58
  • $\begingroup$ I don't think the argument is precise, which might explain why you don't feel like you've understood it. :) Decoherence doesn't prevent such superpositions on paper, it just prevents any feasible measurement from revealing the existence of such a superposition -- which is the same as saying that we can't prepare such a superposition in the lab in any feasibly-verifiable way. So whether or not such superpositions exist physically might be untestable, which might make your question kind of hard to answer definitively. $\endgroup$ Aug 11 '21 at 3:24
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    $\begingroup$ With Schrödingers Cat there are only two states because the radioactive atom is in a superposition of two states. With the double pendulum there were infinitely many states in superposition (if the answer to your question is yes). Great question but I have no clue how to tackle it. I think that for an idealised pendulum (two point masses, constraints for their distance, gravitational potential) it is maybe not even too difficult to write down the Schrödinger Equation. $\endgroup$ Aug 11 '21 at 5:15
  • $\begingroup$ I come from the other question and I am finding this article interesting (it deals with a chaotic asteroid instead of a double pendulum, however). Maybe it helps to understand the role of decoherence a bit more. $\endgroup$ Aug 12 '21 at 12:48

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