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06132016, 08:04 AM
(This post was last modified: 06132016, 08:04 AM by Schmelzer.)
(06132016, 07:45 AM)FrediFizzx Wrote: Sure the evaluation doesn't require anything from \(S^3\) but how those +1 and 1 outcomes happen does involve \(S^3\). Why are you mixing that up?
The argument is about Einsteincausal realistic explanations, and these are shown to be of the form
\[ E(a,b) = \int A(a,\lambda)B(b,\lambda) \rho(\lambda)d \lambda\]
with \(A(a,\lambda)\) and \(B(b,\lambda) \in \{1,1\}\), because these A, B are the measurement results and the a, b are the parameters of the experiment chosen by the experimenters. So, these are not part of some hidden reality, they are part of some observable reality. The hidden part is the \(\lambda\) and the particular choice of the functions \(A(a,\lambda)\) and \(B(b,\lambda)\), but not that they have values in +1,1.
(06132016, 07:45 AM)FrediFizzx Wrote: And from the classical optical experiment, you can see that +/1 outcomes are not required for an experiment anyways. So we do have an LHV model that produces the same prediction as QM for the EPRBohm scenario. I really think you should be asking other questions.
I can have a look out of the window and observe if it is raining now, this will have only nonnegative results (how much raining). And I can ask you other questions, like if you have had a nice day. But this would be irrelevant. The experiment proposed by Bell's theorem has outcomes +1 or 1, and this experiment seems unexplainable by realistic Einsteincausal theories. So, if I want to know if realistic Einsteincausal theories are viable, I have to think about this particular problem, with outcomes +1 or 1.
(06132016, 07:45 AM)FrediFizzx Wrote: It is abundantly clear that the classical experiment that exceeds the CHSH bound of 2 is using classical fields. Perhaps you need to study it more thoroughly?
It is not clear. I have given the quote which shows that this is not clear at all. And, anyway, it is irrelevant.
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06132016, 09:14 AM
(This post was last modified: 06132016, 12:16 PM by gill1109.)
FreddiFizzx is right, that there is nothing wrong with Christian's model. It is a legitimate LHV model of the kind covered by Bell's theorem.
The definitions of \(A(a, \lambda)\) and \(B(b, \lambda)\) in equations (1) and (2) of http://arxiv.org/pdf/1103.1879v2.pdf do involve \(S^3\); indeed, according to their definitions, they are both elements of \(S^3\) (the set of unit length quaternions). But it turns out, by simple geometric algebra computations (provided one can correctly guess the definitions of D and L), that \(A(a, \lambda) = \lambda\) and \(B(b, \lambda) =  \lambda\), as also stated implicitly in (1) and (2). Since \(\lambda\) = +/ 1 (completely at random) the outcomes are +/1.
The correlation is defined in equation (4) by a conventional formula. We find immediately that E(a, b) = 1. No violation of Bell's inequality.
There seem to be problems with Christian's computation of E(a, b).
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(06132016, 09:14 AM)gill1109 Wrote: FreddiFizzx is right, that there is nothing wrong with Christian's model. It is a legitimate LHV model of the kind covered by Bell's theorem.
The definitions of \(A(a, \lambda)\) and \(B(b, \lambda)\) in equations (1) and (2) of http://arxiv.org/pdf/1103.1879v2.pdf do involve \(S^3\); indeed, according to their definitions, they are both elements of \(S^3\) (the set of unit length quaternions). But it turns out, by simple geometric algebra computations (provided one can correctly guess the definitions of D and L), that \(A(a, \lambda) = \lambda\) and \(B(b, \lambda) =  \lambda\), as also stated implicitly in (1) and (2). Since \(\lambda\) = +/ 1 (completely at random) the outcomes are +/1.
The correlation is defined in equation (4) by a conventional formula. We find immediately that E(a, b) = 1. No violation of Bell's inequality.
There seem to be problems with Christian's computation of E(a, b).
Nope. The correlation has to be calculated as done in eq. (59) in order to have the effects of \(S^3\) on producing the outcomes. If you don't do the correlation calculation that way then you are just doing \(R^3\). It is a very simple thing. But I guess you two will never understand that.
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(06132016, 04:40 PM)FrediFizzx Wrote: (06132016, 09:14 AM)gill1109 Wrote: FreddiFizzx is right, that there is nothing wrong with Christian's model. ...
....The correlation has to be calculated as done in eq. (59) in order to have the effects of \(S^3\) on producing the outcomes. ...
The only thing which is wrong is that it cannot work. Of course, one can include \(\{1,+1\}\) into the Monster group or whatever, but this will not have any effect. You have to do something wrong to have an effect of the inclusion of \(\{1,+1\}\) into the Monster group or whatever.
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(06132016, 05:58 PM)Schmelzer Wrote: (06132016, 04:40 PM)FrediFizzx Wrote: ....The correlation has to be calculated as done in eq. (59) in order to have the effects of \(S^3\) on producing the outcomes. ...
The only thing which is wrong is that it cannot work. Of course, one can include \(\{1,+1\}\) into the Monster group or whatever, but this will not have any effect. You have to do something wrong to have an effect of the inclusion of \(\{1,+1\}\) into the Monster group or whatever.
Nonsense. Once you have the result of a.b you can backtrack to the +/ 1 outcomes. IOW, the outcomes must have been such to produce a.b in \(S^3\) in the first place.
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(06132016, 06:25 PM)FrediFizzx Wrote: (06132016, 05:58 PM)Schmelzer Wrote: (06132016, 04:40 PM)FrediFizzx Wrote: ....The correlation has to be calculated as done in eq. (59) in order to have the effects of \(S^3\) on producing the outcomes. ...
The only thing which is wrong is that it cannot work. Of course, one can include \(\{1,+1\}\) into the Monster group or whatever, but this will not have any effect. You have to do something wrong to have an effect of the inclusion of \(\{1,+1\}\) into the Monster group or whatever.
Nonsense. Once you have the result of a.b you can backtrack to the +/ 1 outcomes. IOW, the outcomes must have been such to produce a.b in \(S^3\) in the first place.
But how to do this with the \(A(a,\lambda), B(b,\lambda)\in \{1,1\}\)? Nor including \(\{1,1\}\) into \(S^3\), nor into the Monster group changes anything about the rules how to multiply +1 with 1.
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(06132016, 06:32 PM)Schmelzer Wrote: (06132016, 06:25 PM)FrediFizzx Wrote: (06132016, 05:58 PM)Schmelzer Wrote: (06132016, 04:40 PM)FrediFizzx Wrote: ....The correlation has to be calculated as done in eq. (59) in order to have the effects of \(S^3\) on producing the outcomes. ...
The only thing which is wrong is that it cannot work. Of course, one can include \(\{1,+1\}\) into the Monster group or whatever, but this will not have any effect. You have to do something wrong to have an effect of the inclusion of \(\{1,+1\}\) into the Monster group or whatever.
Nonsense. Once you have the result of a.b you can backtrack to the +/ 1 outcomes. IOW, the outcomes must have been such to produce a.b in \(S^3\) in the first place.
But how to do this with the \(A(a,\lambda), B(b,\lambda)\in \{1,1\}\)? Nor including \(\{1,1\}\) into \(S^3\), nor into the Monster group changes anything about the rules how to multiply +1 with 1.
??? I just told you how to do it if you want to do the correlation calculation in \(R^3\).
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(06132016, 07:14 PM)FrediFizzx Wrote: ??? I just told you how to do it if you want to do the correlation calculation in \(R^3\).
I'm not guided by wishful thinking, so it does not matter for me what one has to do if one wants to reach some result one likes. I care about what is possible if \(A(a,\lambda), B(b,\lambda)\in \{1,1\}\), and in this case I have a simple, elementary proof which shows me that Bell's inequality will be the result.
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(06132016, 07:31 PM)Schmelzer Wrote: (06132016, 07:14 PM)FrediFizzx Wrote: ??? I just told you how to do it if you want to do the correlation calculation in \(R^3\).
I'm not guided by wishful thinking, so it does not matter for me what one has to do if one wants to reach some result one likes. I care about what is possible if \(A(a,\lambda), B(b,\lambda)\in \{1,1\}\), and in this case I have a simple, elementary proof which shows me that Bell's inequality will be the result.
The only reason you think it is "wishful thinking" is because your prejudice is so high it is blocking you from seeing the truth.
Let me be more mathematically specific. It really is quite simple. You do the correlation calculation in \(S^3\) and get the result of \(a \cdot b\). Now after backtracking, any set of +/1 outcomes that satisfy that result will do. Ya take that set and do the correlation calculation in \(R^3\) like they do for the experiments. You will find that the result is \(a \cdot b\). So simple.
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(06132016, 07:44 PM)FrediFizzx Wrote: Let me be more mathematically specific. It really is quite simple. You do the correlation calculation in \(S^3\) and get the result of \(a \cdot b\). Now after backtracking, any set of +/1 outcomes that satisfy that result will do. Ya take that set and do the correlation calculation in \(R^3\) like they do for the experiments. You will find that the result is \(a \cdot b\). So simple.
What "backtracking"? Or you have welldefined functions \(A(a,\lambda), B(b,\lambda)\) with values in \(\{1,1\}\), together with a probability distribution \(\rho(\lambda)d\lambda\), or you don't.
