DrChinese
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Morbert said:1. I think I see the point of disagreement: We have four subsets HH, HV, VH, VV when no swap occurs. We have four subsets HH', HV', VH', VV' when a swap occurs (or at least HH' and VV'). I don't think it's the case that HH = HH' and VV = VV'
From Ma:
The operation "quarter wave plates off and then polarization measurement" will select a different four subsets from "quarter wave plates on and then polarization measurement"
2. As this is converging with the other thread I might give my responses there
1. Is HH = HH' and VV = VV' ?
Well of course this is true, answer YES. It has to be, this is physically dependent on the polarization characteristics of a beam splitter. No amount of interference or interaction between 2 photons overlapping in a beam splitter is going to change H to V (or vice versa). There is no evidence otherwise, and there is no theory to support that speculative idea. It is easily testable, although I have never seen such an experiment. (Of course, there are many experiments one might perform to confirm the predictions of QM that have never been performed.)
Next, you ask: Are the selected [4 fold] subsets the same? Answer is of course, NO, as the authors say in your quote. That's because the swap induces a change in the 1 & 4 pairs - they are correlated in a specific Bell state, when they weren't before.
2. And I answered your equivalent question in more detail over there in this post.