- #36
SpectraCat
Science Advisor
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DrChinese said:And yet: you don't see anything weird about your prediction that if Charlie performs the BSM after A & D are detected, then there is no entanglement. Because that prediction seems weird to me. To me it implies (and would confirm) the absolute existence of a uniform time frame. Something which as far as I know has never been observed.
I am also fairly convinced by DeMystifier's arguments here, for the following reason.
When Alice and Bob measure A & D, then that destroys the entanglement between A&B, and between C & D. Thus, when Charlie performs his operation, he is working with photons that have already undergone decoherence, so there is nowhere to transfer the entanglement to. Sure, B & C become entangled if Charlie makes a BSM, but that no longer has any ramifications for A & D, because they are no longer entangled with B & C, respectively, when Charlie makes the BSM (or not).
So, based on this, I think there will be no violation of a Bell's inequality for A & D based on Charlie's choice of whether or not to conduct a BSM on B & C. But I really want to see the results of the experiment!