[a1369209993]

> an immediate objection would be "but we can't know everything to that much detail".

No, the immediate objection would be "but it's physically impossible to know everything to that much detail, because you don't have enough bits of storage[0], and also because of Heisenberg's uncertainty principle". (Both objections are suffient on their own to make this not work except possibly for a homogenous spherical molecule-shaped unphysically-light and -compact hypercomputer in a frictionless vacuum.)

0: That is, it's physically impossible to pack enough bits to describe a cloud of gas onto a storage medium massing significantly less than the entire cloud of gas.

[jlokier]

> because you don't have enough bits of storage

Inside the computer. That's what makes it a computational reducibility question and not a measurement information-availability objection.

(Also: For a twist, assume you have a quantum computer and they are quantum balls.)

> Heisenberg's uncertainty principle

This raises questions, certainly, but the answers aren't obvious when talking about repeated interactions with the many particles. In the box model, the balls are inevitably entangled with each other at the position-momentum level due to their collisions, even if that entanglement is undetectable in an analogous way to how their motions appear "random" classically.

Heisenberg does not apply to each ball independently when they are entangled. In this box model, as your little computer/mind/demon accumulates information-in-principle from many interactions, in addition to classical information it couples to that entangled state, and the independence of Heisenberg limits dissolves because they aren't really independent.

(Also: Once you invoke Heisenberg, you've also invoked quantum particles in a box self-interfering. In a box that reduces the amount of information you need to represent a single particle's state to an integer, bounded if the energy is bounded. I'm not sure if that also applies to multiple particles interacting chaotically.)

> except possibly for a homogenous spherical molecule in a frictionless vacuum.

Well, the model actually is about homogeneous spherical molecules, and vacuum at the molecular level is frictionless, so that's ok :-)

[mnl]

Being omniscient this special molecule can tell which identical molecule is which in the gas, that's some interesting physics.

Of course it can also predict the final states of collisions between other molecules. Even when that information just doesn't exist in the perfect a priori knowledge about the system, which is something that if this special molecule could obtain somehow and then store for later use should violate half a dozen theorems or so.

Really this could make some sense if we were talking about an ideal gas of classical particles that obey deterministic mechanics, but then not even the special molecule would be able to determine the initial conditions with sufficient precision to make useful predictions, beyond a short path and a few collisions in the system.

[jlokier]

> it can also predict the final states of collisions between other molecules. Even when that information just doesn't exist in the perfect a priori knowledge

Obviously the special molecule is a thought experiment to illustrate an idea under certain simplifying assumptions and extreme parameters, to understand the consequences. Nobody expects to make one.

And you are doing a proper job of arguing why it cannot work, as you are supposed to with a thought experiment.

But... it's not correct to reason that "it can't predict" when the "information just doesn't exist [...] a priori".

If the special molecule senses, computes and reacts entirely in the quantum realm itself, then its processing will be entangled with those other molecules.

Despite the absence of a priori final states, the special molecule is, in principle, able to select an entangled reaction to those final states anyway.

It's a bit like saying "I don't know if particle X will move to A or B later (and particle X hasn't decided either), but I can prepare myself into a state where if X moves to A then I will already be at A', and if X moves to B then I will already be at B'".

And if being at A' when X moves to A, or B' when X moves to B, means that X can't actually move to A or B, that entangled reaction will affect X so the question of A or B doesn't even arise in the first place.

[a1369209993]

> > because you don't have enough bits of storage

Edited to clairify, thanks.

> the particles are inevitably entangled with each other

That increases the information content, from O(N) for N particles to (worst case) O(2^N). Using a quantum computer at best reduces that back down to O(N) qubits.

> the model actually is about homogeneous spherical molecules

I don't think electron orbitals are spherical enough for that, much less nuclei or polyatomic molecules, but edited anyway.

[jlokier]

> 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).