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Gordon Watson
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This thread is intended to focus on arithmetic issues, mainly raised by JesseM, vanesch, DrChinese, ThomasT, in the context of: "What's wrong with this local realistic counter-example to Bell's theorem?" https://www.physicsforums.com/showthread.php?t=475076 :
For newcomers to the issues, I am committed to reply to each and every question that arises in the above context. So further arithmetic questions, arising from that context, will also be answered here.
Thus, to be clear, this thread focusses on the arithmetic associated with quantum-entangled spin-1/2 particles in the EPR-Bohm experiment (Bohm 1951) and its analysis in Bell (1964).
(If the need arises, I'd propose to answer spin-1 questions in another thread -- to avoid arithmetic confusion.)
PS: I will attempt to answer each question in a separate Post, to facilitate follow-up questions. And I would very much appreciate ONE question per Post -- because it's difficult to reply to long Posts on my small-screen computer. Thanks!
.........
The main terms and equations are defined in PDF2 [post=3191024]attached here[/post], with partial-errata in https://www.physicsforums.com/showpost.php?p=3191819&postcount=86
.........
The main equations that we'll be discussing are similar to those which follow -- "etc." to be understood after each equation.
The OP-style identification is intended to lead back to this OP; the identifier on the Right is that found in PDF2 [post=3191024]attached here[/post], Appendix A:
From PDF2 ([post=3191024] attached here[/post]), Appendix A:
(OP-A) P(a+|a) = P(a–|a) =1/2. (A4).
(OP-B) P(ab++|ab) = P(ab++|a)/2 + P(ab++|b)/2. (A0a).
(OP-C) P(ab++|abc) = P(ab++|a)/3 + P(ab++|b)/3 + P(ab++|c)/3
= 2P(ab++|ab)/3 + P(ab++|c)/3. (A0b).
(OP-D) ∴ P(ab++|ab) = 3P(ab++|abc)/2 – P(ab++|c)/2.
From PDF2 ([post=3191024]attached here[/post]), Table A3.c, eqn. (A3.3a):
(OP-E) P(ab++|ab) = 3P(ab++|abc)/2 – (Cac.Sbc/2 + Sac.Cbc/2)/2.
From PDF2 ([post=3191024]attached here[/post]), Table 1:
(OP-F) P(ab++|abc) = P3 + P4 = Sab/3 + (Cac.Sbc/2 + Sac.Cbc/2)/3.
From (OP-E) and (OP-F):
(OP-G) P(ab++|ab) = Sab/2; (A0c); and PDF2, Table 2;
in full accord with QM.
.......
My next Posts will be replies to arithmetic questions from JesseM and vanesch.
This thread is intended to focus on arithmetic issues, mainly raised by JesseM, vanesch, DrChinese, ThomasT, in the context of: "What's wrong with this local realistic counter-example to Bell's theorem?" https://www.physicsforums.com/showthread.php?t=475076 :
For newcomers to the issues, I am committed to reply to each and every question that arises in the above context. So further arithmetic questions, arising from that context, will also be answered here.
Thus, to be clear, this thread focusses on the arithmetic associated with quantum-entangled spin-1/2 particles in the EPR-Bohm experiment (Bohm 1951) and its analysis in Bell (1964).
(If the need arises, I'd propose to answer spin-1 questions in another thread -- to avoid arithmetic confusion.)
PS: I will attempt to answer each question in a separate Post, to facilitate follow-up questions. And I would very much appreciate ONE question per Post -- because it's difficult to reply to long Posts on my small-screen computer. Thanks!
.........
The main terms and equations are defined in PDF2 [post=3191024]attached here[/post], with partial-errata in https://www.physicsforums.com/showpost.php?p=3191819&postcount=86
.........
The main equations that we'll be discussing are similar to those which follow -- "etc." to be understood after each equation.
The OP-style identification is intended to lead back to this OP; the identifier on the Right is that found in PDF2 [post=3191024]attached here[/post], Appendix A:
From PDF2 ([post=3191024] attached here[/post]), Appendix A:
(OP-A) P(a+|a) = P(a–|a) =1/2. (A4).
(OP-B) P(ab++|ab) = P(ab++|a)/2 + P(ab++|b)/2. (A0a).
(OP-C) P(ab++|abc) = P(ab++|a)/3 + P(ab++|b)/3 + P(ab++|c)/3
= 2P(ab++|ab)/3 + P(ab++|c)/3. (A0b).
(OP-D) ∴ P(ab++|ab) = 3P(ab++|abc)/2 – P(ab++|c)/2.
From PDF2 ([post=3191024]attached here[/post]), Table A3.c, eqn. (A3.3a):
(OP-E) P(ab++|ab) = 3P(ab++|abc)/2 – (Cac.Sbc/2 + Sac.Cbc/2)/2.
From PDF2 ([post=3191024]attached here[/post]), Table 1:
(OP-F) P(ab++|abc) = P3 + P4 = Sab/3 + (Cac.Sbc/2 + Sac.Cbc/2)/3.
From (OP-E) and (OP-F):
(OP-G) P(ab++|ab) = Sab/2; (A0c); and PDF2, Table 2;
in full accord with QM.
.......
My next Posts will be replies to arithmetic questions from JesseM and vanesch.
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