Hidden Variables
Bell's theorem - for or against Hidden Variables? - Printable Version

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RE: Bell's theorem - for or against Hidden Variables? - secur - 07-16-2016

Although I haven't been too impressed by some of the arguments made, I have been impressed by the good nature of the debate. It's been mentioned that on some other boards the discussion has been acrimonious. I don't know about that; haven't read those threads; but here there's been no such problem, so far. I'd like to thank everyone involved for keeping it that way!


RE: Bell's theorem - for or against Hidden Variables? - gill1109 - 07-16-2016

(07-15-2016, 08:20 PM)FrediFizzx Wrote: LOL!  The QRC is "rigged" by the same thing we were discussing here.  That it is impossible for a quantum experiment to test the quadruple in Bell-CHSH.  Only a fool would try to play the QRC game.

I recall that you made an attempt to win a similar challenge two years ago. I'm glad you learnt from the experience.

Sascha Vongehr's Quantum Randi Challenge https://arxiv.org/abs/1207.5294 is a didactic tool. Basically the challenge is to program a simulation of a local hidden variables model, which systematically and substantially violates Bell's inequality, no cheating allowed. The challenge is rigged in the sense that this is, of course, impossible.


RE: Bell's theorem - for or against Hidden Variables? - Thomas Ray - 07-16-2016

(07-16-2016, 11:32 AM)gill1109 Wrote:
(07-15-2016, 08:20 PM)FrediFizzx Wrote: LOL!  The QRC is "rigged" by the same thing we were discussing here.  That it is impossible for a quantum experiment to test the quadruple in Bell-CHSH.  Only a fool would try to play the QRC game.

I recall that you made an attempt to win a similar challenge two years ago. I'm glad you learnt from the experience.

Sascha Vongehr's Quantum Randi Challenge https://arxiv.org/abs/1207.5294 is a didactic tool. Basically the challenge is to program a simulation of a local hidden variables model, which systematically and substantially violates Bell's inequality, no cheating allowed. The challenge is rigged in the sense that this is, of course, impossible.

If the intent is to simulate local hidden variables, what is hidden?


RE: Bell's theorem - for or against Hidden Variables? - Thomas Ray - 07-16-2016

Let me put it in terms of the question you ignore, Richard:

Can (the value of) h-bar be degenerate in a consistent quantum theory?

The answer should make evident that a "dictionary" approach to the problem is just naive.  If consensus is all we need as a "didactic tool", then we have no need of hidden variables, because we simply assume a principle that forbids them.  That's not science, nor even good philosophy.

Suppose the combination to a locked safe is hidden inside.


RE: Bell's theorem - for or against Hidden Variables? - FrediFizzx - 07-16-2016

(07-16-2016, 01:26 PM)Thomas Ray Wrote: Suppose the combination to a locked safe is hidden inside.

Kobayashi Maru -- Kirk would blow the safe up. Big Grin

"...no cheating allowed" LOL! But it is OK for QRC to cheat. Not! It is impossible for a quantum experiment to test E(a, c) - E(b, a) - E(b, c) for Bell's original inequality just like it is impossible for the quadruple in Bell-CHSH. So how do they get around that for QM? They cheat.


RE: Bell's theorem - for or against Hidden Variables? - Thomas Ray - 07-16-2016

(07-16-2016, 09:49 PM)FrediFizzx Wrote:
(07-16-2016, 01:26 PM)Thomas Ray Wrote: Suppose the combination to a locked safe is hidden inside.

Kobayashi Maru  -- Kirk would blow the safe up. Big Grin

"...no cheating allowed"  LOL!  But it is OK for QRC to cheat.  Not!  It is impossible for a quantum experiment to test E(a, c) - E(b, a) - E(b, c) for Bell's original inequality just like it is impossible for the quadruple in Bell-CHSH.  So how do they get around that for QM?  They cheat.

Indeed, Fred.  Or they just say, no one has ever opened this safe, and never will.  Case closed.


RE: Bell's theorem - for or against Hidden Variables? - gill1109 - 07-17-2016

(07-15-2016, 05:42 PM)secur Wrote: gill1109: In a certain context, local realism implies Bell's inequality. Therefore, if in this context we observe that Bell's inequality is violated, we may deduce that local realism is not true.

I agree.

If A then B.
But, not B.
Therefore, not A !

I'm not quite sure who's more illogical here. It's illogical to deny proof by contradiction - a fundamental tenet of scientific reasoning -, as FreddiFizzx and others are doing. But it might be even more illogical to waste your time arguing about it! Wink
You're absolutely right, Secur.


RE: Bell's theorem - for or against Hidden Variables? - Thomas Ray - 07-17-2016

(07-17-2016, 09:43 AM)gill1109 Wrote:
(07-15-2016, 05:42 PM)secur Wrote: gill1109: In a certain context, local realism implies Bell's inequality. Therefore, if in this context we observe that Bell's inequality is violated, we may deduce that local realism is not true.

I agree.

If A then B.
But, not B.
Therefore, not A !

I'm not quite sure who's more illogical here. It's illogical to deny proof by contradiction - a fundamental tenet of scientific reasoning -, as FreddiFizzx and others are doing. But it might be even more illogical to waste your time arguing about it! Wink
You're absolutely right, Secur.

As I said, there's no interest in opening the safe.  It's "illogical."  However, logic is only as good as its premises.

There are two questions I have put to you, Richard, that you have declined to answer:

1.  What is your measure space?
2.  Can Planck's constant go to zero?

Quantitative answers to these questions change the premises, and the logic of Bell's theorem falls apart.  Fundamental physics needs a good safecracker, like Feynman.


RE: Bell's theorem - for or against Hidden Variables? - Thomas Ray - 07-19-2016

Well, it's not surprising to be greeted with silence -- because it's been long known that if Planck's constant were zero, the world would be classical. The value of h-bar is an empirical measurement, with no compelling theory behind it.

Seems the temple of Bell has a crumbling foundation. Its logic can't hold against what we now know of topology.


RE: Bell's theorem - for or against Hidden Variables? - Schmelzer - 07-19-2016

Simply nobody understands how some theoretical considerations about \(\hbar\) going to zero have any relation to Bell's theorem. Of course, with \(\hbar\) as zero the Bell experiment itself becomes meaningless. The point being?