- #1
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Bell (1964) http://cds.cern.ch/record/111654/files/vol1p195-200_001.pdf has 3 unnumbered equations following his equation (14). Let them be (14a)-(14c). Bell then uses his equation (1) to move from (14a)-(14b). It seems to me that he uses this:
[A(b,λ)]2 = 1. (X)
Now Bell (page 195) does not mind whether λ is continuous or discrete. So if we let λ be discrete, Bell needs this:
[A(b,λi)][A(b,λj)] = 1; (Y)
because each λ is drawn from a different run of the experiment (N times with λi from the test with the detectors set at a and b; N times with λj from the other test with the detectors set at a and c). So (with i = 1, 2, ..., N; j = N+1, N+2, …, 2N), is Bell assuming that he has a set of particles that he can test twice, and in the same order, in each test? So then λi = λj every time?
And is such an assumption in keeping with EPR and EPRB, the paper and the experiment that he is studying?
Because if λi ≠ λj:
[A(b,λi)][A(b,λj)] = ± 1; (Z)
and then his (14a) ≠ (14b).
Thank you.
Edited to fix brackets.
[A(b,λ)]2 = 1. (X)
Now Bell (page 195) does not mind whether λ is continuous or discrete. So if we let λ be discrete, Bell needs this:
[A(b,λi)][A(b,λj)] = 1; (Y)
because each λ is drawn from a different run of the experiment (N times with λi from the test with the detectors set at a and b; N times with λj from the other test with the detectors set at a and c). So (with i = 1, 2, ..., N; j = N+1, N+2, …, 2N), is Bell assuming that he has a set of particles that he can test twice, and in the same order, in each test? So then λi = λj every time?
And is such an assumption in keeping with EPR and EPRB, the paper and the experiment that he is studying?
Because if λi ≠ λj:
[A(b,λi)][A(b,λj)] = ± 1; (Z)
and then his (14a) ≠ (14b).
Thank you.
Edited to fix brackets.