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by jlokier 2091 days ago
> That increases the information content, from O(N) for N particles to (worst case) O(2^N).

Classical (model) molecules have infinite bits of information: Their motion parameters are "analogue", which you can think of as real numbers, or infinite precision numbers.

For a digital computer, that takes infinite bits unless you know of a constraint upon them.

You can't call that O(N). And you can't say they have mass proportional to that kind of information either, because that would be infinite mass.

The quantum molecules are in a bounded box. Individual eigenstates are constrained by quantum mechanics in a box into bound states, which are countably enumerable as integers. If you have an upper bound on the energy of the entire box contents, there's a maximum integer required, therefore finite bits to encode an eigenstate.

The coefficients associated to each eigenstate in the general wavefunction are complex (and therefore infinite bits), while subject to various constraints, but they are unobservable. Observations select among eigenstates, each of which is represented in finite bits.

So is it infinite (like the classical model) or finite?

But observation is meaningless in the "special ball is a quantum computer" model. How much information does the "special ball" need from its environment, if it's allowed to entangle with that environment, to outsmart the quantized chaos around itself? Qubits linked to physical measurements and actions are full of paradoxes arising from the mathematics, which makes thought experiments useful. Where does the computation even take place, given that entanglement makes qubits non-local? In the special ball, or in all the balls it's entangled with, affecting them all subtly? I don't think this information question is simple enough to hand-wave as O(N) or O(2^N).