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I tried to understand the Bohm interpretation and this is what picture appeared to me. Please tell me if I understood something incorrectly.

  • All particles have definite positions and follow deterministic rules of dynamics

  • Any future configuration of an isolated subsystem is only dependent on initial conditions

  • Even slight differences in initial conditions may result in huge differences in the result.

The problem is that those initial conditions are inherently unknown. This is fundamental: even if an observer manages to measure the whole Bohm state of the entire universe, he still would not know the Bohm state of himself.

This is like making predictions about the future states of a three-body system based on Newtonian mechanics with initial coordinates known only with finite precision. Due to apparent chaoticity of the solution the possible results may be dramatically different.

Correct me if I am wrong.

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closed as off-topic by DanielSank, abc, John Rennie, JamalS, Rob Jeffries Jan 12 at 12:47

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This is correct. I don't see a question however. It's a little worse than what you say--- the Bohm particle positions are inherently unobservable because all you can see is the Everett branch they select to make real, and that's not enough information to determine anything at all about the positions, beyond a vague probability distribution as $|\psi|^2$. – Ron Maimon Jul 22 '12 at 9:54
As I understand there are no Everett branches in Bohm interpretation. – Anixx Jul 22 '12 at 12:18
There is a full QM wavefunction in Bohm, so there are branches. The branches are always there in QM, the only question is whether you consider the unobserved branch "real" or "unreal" (which is positivistically meaningless). The only difference in Bohm is that one of the branches has particles running around, this is the real branch, and the other branches are empty, so they are unreal. So real/unreal is determined dynamically. But they are still there in the formalism to reproduce the interference from far-away branches recohering. You can see this by Bohmian simulation of Shor's algorithm. – Ron Maimon Jul 22 '12 at 19:32
Where would I find a published account of the Bohmian simulation of Shor's algorithm? That sounds interesting. – Francis Davey Feb 3 '13 at 12:23
This question appears to be off-topic because it isn't a question. – DanielSank Jan 12 at 9:22

1 Answer 1

I think you are correct, but your post is a bit vague, so maybe not. therefore I'll try to be more explicit.

In general, the initial state needs to also include the spin. For instance for a spin 1/2 particle you need to specify a point on the unit sphere, e.g. $(x,y,z)$ with $x^2+y^2=z^2=1$ (or alternatively a unit oriented plane, or anything else with the right dynamical degrees of freedom). Then the positions of the particles, as well as the wave (in configuration space) as well as the spin vectors/planes all need to be specified and then they co-evolve. And you don't know the positions of the particles, and you're not going to know, but what do you expect? All you know is that the universe is what is has to be to be consistent with the observations already made and the data already acquired, no more and no less. Since multiple setups are consistent with our finite measurements we aren't going to know enough detail to be absolutely sure of most things.

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