Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

It seems to be common consensus that the world is non-deterministic and this is proved by Bell's theorem.

But even though Bell's experiments proved that the theory of quantum mechanics work, How does it prove the non-existent of local hidden variables?

Isn't it possible that there are hidden variables at work, and the results that were derived from these hidden variables coincide with the predictions of quantum mechanics?

share|improve this question
2  
You're mixing up two different ideas. The evolution of the wavefunction is completely deterministic. It is just the apparent collapse of the wavefunction to give physical observables that appears to be non-deterministic. I say "appears" because decoherence explains even the collapse as a deterministic interaction of the wavefunction with a poorly characterised environment. NB the determinism is restricted to the wavefunction. Local hidden variables are neither required nor involved. –  John Rennie Jan 1 '13 at 14:33
    
@JohnRennie, are you saying that Bell's theorem does not rule out the possibility of local hidden variables? –  Pacerier Jan 2 '13 at 12:59
    
Here is a Search for hidden variable questions on Phys.SE. –  Qmechanic Feb 25 '13 at 20:35

3 Answers 3

The term "local hidden variables" is a poor expression, and I haven't found the term used by John Bell anywhere. Bell showed that any theory with pre-existing spin cannot produce the right correlations predicted by quantum mechanics. Almost nobody I talk to seems to get this. Tim Maudlin is one. I highly suggest reading section 3 of this paper. It really explains Bell's theorem, and you won't walk around as so many do with the belief that "local hidden variables" is what Bell ruled out. He ruled out pre-existing properties (specifically spin) in any theory, local or not.

http://www.bslps.be/meaningWF.pdf

share|improve this answer
    
This is obviously wrong, or at least misleading. The wavefunction is a pre-existing property of the system! It is just not a local property, it lives in configuration space. "Local beables" was the phrase used by Bell and I agree it is better than "hidden variables" (after all, the wavefunction is a hidden variable in a sense), but the word local is rather key. –  Mark Mitchison Feb 27 '13 at 23:37

This is a very specific question. Bell's theorem rules nothing out or in. Bell made the assumption that hidden variables existed, and using simple statistical arguments he derived a set of inequalities. If hidden variables existed they should make a measurable contribution to the correlatiions of spins. Therefore, if the measured correlations satisfied Bell's inequality it would support the existence of hidden variables. But if the inequality is violated then the assumption about the existence of hidden variable is false and quantum mechanical predictions are correct. This is similar to the proof by "reductio ad absurdum" in geometry or pure mathematics. All experiments that have checked Bell's inequality so far, have shown that the experimental data violate them. Thus hidden variables do not find support by experiment!

share|improve this answer
    
Good answer, and I gave you +1. However, your answer is practically the same as mine. It is true that Bell showed that that the correlation isn't a result of pre-existing properties. In other words, let's say someone wanted to explain how the particles "know what to do". Let's say that this person says, "They discuss how they are going to behave in advance". Bell showed that this explanation can't work. What happens is that they make a decision at some time, and somehow their decisions are transmitted instantly to the other photon, e.g. spooky action at a distance. –  user7348 Jan 23 '13 at 1:02

Bell's theorems indeed rule out simple theories where hidden variables obey local equations. However, no matter how you reason, it's always at some point where you need another assumption. In its simplest form, it is the assumption that two observers, Bob an Alice, have the "free will" to choose along which axis they will measure the spin of a particle (photon, electron, or something else). Well, one could object that in a deterministic theory they have no such free will; their decisions were made in the far past.

But that does not invalidate Bell, because now you can say: Bell's theorem would imply that entangled photons emitted by a physical source are correlated non locally in an unnatural way with the nerves in Bob's and Alice's brains long before they made their decisions. That's called "conspiracy". So now the assumption is: there can't be conspiracy. Can't there? Spacelike non-local correlations in physical states are common in the physical world. In fact, in quantum field theory it's the propagators of all physical particles that describe correlations, and they do not vanish far outside the light cone. But the kind of conspiracy quantum systems seem to display (when described in terms of "hidden variables") looks disgusting to many researchers. So it is usually dismissed. Is "disgusting" a sound mathematical argument? You decide ...

share|improve this answer
    
@ G. 't Hooft: Why not simply accept non-locality instead? It seems like a much more plausible way to explain things. Don't be afraid of spooks! –  user7348 Feb 27 '13 at 23:26
1  
@G. 't Hooft Might you have a(n implied) response (somewhere) to the comment of Zeilinger, which doesn't appeal to (im)plausibility and regards the validity of science: "[W]e always implicitly assume the freedom of the experimentalist... This fundamental assumption is essential to doing science. If this were not true, then, I suggest, it would make no sense at all to ask nature questions in an experiment, since then nature could determine what our questions are, and that could guide our questions such that we arrive at a false picture of nature." –  Glen The Udderboat Feb 28 '13 at 0:39
1  
@user7348 : Non-locality would generate serious trouble with special relativity and causality. And I don't need it. That's why I don't introduce it. Non-local correlations is not the same as non-locality in the equations of motion. In QFT, the equations of motion are local but the vacuum-correlations are not. This is because the vacuum is a special solution of the e.o.m. –  G. 't Hooft Mar 1 '13 at 11:26
2  
@Gugg: I just don't agree with the Zeilinger quote. Determinism indeed implies that the experimenter's decisions, and questions, are generated by physical forces themselves, so his attitude would dismiss determinism categorically, and I am not ready to go that far. And my bottom line remains to be a simple one: I now have models telling me what might happen, and what they say does not disturb me. Important: I still keep causality intact. –  G. 't Hooft Mar 1 '13 at 11:26
    
@ G. 't Hooft: You don't think it's possible to have a non-local theory of relativity? I mean Einsteinian relativity, not the relativity of Lorentz where there is a hidden preferred frame. So, can we have non-locality and preserve Einstein's insights? You seem to say this is impossible. I'm not so sure. –  user7348 Mar 1 '13 at 15:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.