Quantum physics. I don’t see why it would be indistinguishable from 50%.
Agree that there will be some decoherence. My guess is decoherence would mostly leave particle position at this scale intact, and if it becomes a huge factor, I would want the question to be settled on the basis being able to predict which side has higher irreducible uncertainty (i.e. which side had higher amplitude, if I am using that concept correctly).
Citing https://arxiv.org/abs/cond-mat/9403051: “Furthermore if a quantum system does possess this property (whatever it may be), then we might hope that the inherent uncertainties in quantum mechanics lead to a thermal distribution for the momentum of a single atom, even if we always start with exactly the same initial state, and make the measurement at exactly the same time.”
Then the author proceed to demonstrate that it is indeed the case. I guess it partially answers the question: quantum state thermalises and you’ll get classical thermal distribution of measurement results of at least some measurements even when measuring the system in the same quantum state.
The less initial uncertainty in energy the faster the system thermalises. That is to slow quantum thermalisation down you need to initialize the system with atoms in highly localized positions, but then you can’t know their exact velocities and can’t predict classical evolution.
Decoherence (or any other interpretation of QM) will definitely lead to a pretty uniform distribution over this sort of time scale. Just as in the classical case, the underlying dynamics is extremely unstable within the bounds of conservation laws, with the additional problem that the final state for any given perturbation is a distribution instead of a single measurement.
If there is any actual asymmetry in the setup (e.g. one side of the box was 0.001 K warmer than the other, or the volumes of each side were 10^-9 m^3 different), you will probably get a very lopsided distribution for an observation of which side has more molecules regardless of initial perturbation.
If the setup is actually perfectly symmetric though (which seems fitting with the other idealizations in the scenario), the resulting distribution of outcomes will be 50:50, essentially independent of the initial state within the parameters given.
Quantum physics. I don’t see why it would be indistinguishable from 50%.
Agree that there will be some decoherence. My guess is decoherence would mostly leave particle position at this scale intact, and if it becomes a huge factor, I would want the question to be settled on the basis being able to predict which side has higher irreducible uncertainty (i.e. which side had higher amplitude, if I am using that concept correctly).
Citing https://arxiv.org/abs/cond-mat/9403051: “Furthermore if a quantum system does possess this property (whatever it may be), then we might hope that the inherent uncertainties in quantum mechanics lead to a thermal distribution for the momentum of a single atom, even if we always start with exactly the same initial state, and make the measurement at exactly the same time.”
Then the author proceed to demonstrate that it is indeed the case. I guess it partially answers the question: quantum state thermalises and you’ll get classical thermal distribution of measurement results of at least some measurements even when measuring the system in the same quantum state.
The less initial uncertainty in energy the faster the system thermalises. That is to slow quantum thermalisation down you need to initialize the system with atoms in highly localized positions, but then you can’t know their exact velocities and can’t predict classical evolution.
Decoherence (or any other interpretation of QM) will definitely lead to a pretty uniform distribution over this sort of time scale. Just as in the classical case, the underlying dynamics is extremely unstable within the bounds of conservation laws, with the additional problem that the final state for any given perturbation is a distribution instead of a single measurement.
If there is any actual asymmetry in the setup (e.g. one side of the box was 0.001 K warmer than the other, or the volumes of each side were 10^-9 m^3 different), you will probably get a very lopsided distribution for an observation of which side has more molecules regardless of initial perturbation.
If the setup is actually perfectly symmetric though (which seems fitting with the other idealizations in the scenario), the resulting distribution of outcomes will be 50:50, essentially independent of the initial state within the parameters given.