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why do atoms not collapse on themselves. Doesnt this problem rule hidden variables as invalid as the heisenburg uncertainty is the solution to the problem because it says electrons exist in a probability cloud and is in many places at once, not at a single location. If hidden variables are right the electron has a well defined position orbiting the nucleus and where does the energy come from? Doesnt this mean if HV are right it would collapse in again. whats the answer?

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Comment to the question(v1): The title question is currently a possible duplicate of physics.stackexchange.com/q/9415/2451 and links therein. The main text contains another question about hidden variables. It would be good if OP (or somebody else?) could make the title reflect the main purpose of the question. –  Qmechanic Jan 23 '13 at 11:34
Depends what you mean by hidden variable theory. The paradigmatic example of a hidden varialbe theory that works (in that it exactly reproduces the predictions of ordinary QM) is Bohmian mechanics, which explains all of nonclassical behaviour through a non-local "quantum potential" that enters into the equation of motion for particles (a slight generalisation of $F=ma$). If you mean a local hidden variable theory then these are already ruled out by entanglement experiments. –  Michael Brown Jan 23 '13 at 11:43
Michael, others would disagree with you on that. If we assume they are valid option, are they ruled out by the fact atoms would collapse in on themselves because having a well defined postion at all times would mean they collapse in? –  lee hudson Jan 23 '13 at 11:50
and what does this mean for observed atoms in other interpretations?. observed atoms have there wavefunction collapse so that they have a single definite location and the observed electron would have a single definite orbit. in this case why wouldnt the atom collapse in on itself? –  lee hudson Jan 23 '13 at 12:06
@leehudson "having a well defined postion at all times would mean they collapse in" - No. It depends on the dynamics. Bohmian mechanics is a counter-example - a hidden variable theory where electrons have definite positions but atoms are stable. It achieves this by modifying the laws of motion in a non-local way. Only non-local hidden variable theories have a chance of working. Local ones are ruled out by other experiments. There really isn't any debate about this, except for semantic loopholes. –  Michael Brown Jan 23 '13 at 12:31

2 Answers 2

The fact that atoms exist does not rule out a hidden variable formulation (though for the record such formulations are ruled out by other experiments).

For example the electron wavefunction of a hydrogen atom gives us the probability of finding the electron at any particular point. However the electron could still have a definite hidden position, which would of course be a function of time. As an analogy consider a rubber ball bouncing around at very high velocity in a closed box. We can define a function that gives us the probability of finding the tennis ball at any particular point in the box, but the tennis ball still has a well defined position.

The point is that in the macroscopic case of the tennis ball there is a well defined position, but for the electron in the hydrogen atom the position is simply not defined until we interact with the atom in some way that localises the electron. The position we get is the position at which the electron interacts, not some hypothetical position where it was located immediately prior to the interaction.

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john and others, so electrons can have a definite single location at all times, orbit a nucleus like this and not collapse in on the nucleus according to HV theories? –  lee hudson Jan 23 '13 at 13:48

Sometimes some atoms collapse. Take a positronium as an example. It collapses for sure. Other atoms can collapse if a neutrino hits them. There is a capture of electron and conversion of an electron, proton, and neutrino into a neutron. It happens rarely, so we see atoms as stable ones.

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thanks. when we observe and measure the electron also the wavefunction collapses and we get a single electron in a definite state going around the nucleus? - is that right? –  lee hudson Jan 23 '13 at 13:58
@leehudson: Right. But I do not like the term "wave function collapse" because it does not collapse. Instead, in one measurement we retrieve one bit of information about the wave function. The latter needs many bits to be fully described, just like a photo needs many pixels to be meaningful. –  Vladimir Kalitvianski Jan 23 '13 at 14:01

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