N88
- 225
- 12
It seems to me that this answer leaves me with my original difficulty with Bell's workings. I agree that, given 4N objects prepared in the same state, the average of each of the 4 terms in RHS (9.33) would be unchanged when each is tested over N runs. For example, under EPRB, the first such RHS term would reduce to -a.b; and so on.PeterDonis said:They don't have to be the same objects; they just have to be taken from an ensemble of objects all prepared in the same state.
But (9.33) is based on the an appearing again in the second term on RHS; the bn appearing again in the third term on RHS; etc. So the problem (that I am wrestling with) goes back to eqn (9.32). There we see 4 pairs of terms created from 4 terms; I say that there should be 8 terms. For how is it possible to match the a in the first pair with the a result in the second? For in one run it may be +1, in the next -1.
This seems to be recognised by Isham in the 2nd paragraph: "The central realist assumption we are testing is that each particle has a definite value at all times in any direction of spin. We let an denote 2/h times the value of a⋅S possessed by particle 1 in the n'th element of the collection. Thus an = ±1 if a⋅S = ±h/2".
How can particle 1 in the n'th element be tested twice?
NB: As a local-realist, that "central realist assumption" has no place in my thinking: for surely it is false?