On Thu, Mar 24, 2022 at 12:34 PM Brent Meeker <meekerbr...@gmail.com> wrote:

> On 3/22/2022 11:05 PM, Bruce Kellett wrote: > > On Wed, Mar 23, 2022 at 4:01 PM Brent Meeker <meekerbr...@gmail.com> > wrote: > >> On 3/22/2022 7:55 PM, Bruce Kellett wrote: >> >> >> Actually, that is where I started. I assumed that Alice and Bob were both >> able to collect results from N trials before they met. Then there are 2^N >> copies of each experimenter, and a potential (2^N)^(2^N) pairs when they >> meet. The trouble to be explained is that there are actually only 2^N pairs >> in a real experiment, each with inequality-violating correlations. What has >> happened to all the extra pairings that MWI must produce? (Most of which >> have correlations violating the quantum predictions.) >> >> >> Don't you mean (2^N)x(2^N). >> > > Probably. Both Alice and Bob split into 2^N copies for binary results. So > each of the 2^N Alice's splits into 2^N further copies, one for each copy > of Bob. > > > Why would Alice split again for each copy of Bob. There's no QM involved. > There is. That is what MWI predicts: for every splitting of Bob, the whole world, including copies of Alice, also splits. That is certainly the case if both Alice and Bob measure independent particles that they prepare themselves. The question is, "Why are there fewer joint branches when they measure entangled particles?" If you consider that all (2^N)*(2^N) branches with Alice-Bob pairings exist in the entangled case as well as the independent case, as MWI would predict if the measurements are truly local and independent, then a large number of the couples are going to find correlations that violate the quantum predictions. The fact that all experiments have found correlations that agree with QM is then hard to explain. Each Alice and each Bob has seen N events and has recorded them as a binary > number of N bits in their notebook. They are quasi-classical observers and > they're going to meet in the future. If there are not (2^N)x(2^N) meetings > there must be some quantum magic that it is eliminating or otherwise > weighting them. > That is the point that I am making. There must be some fairy dust or other magic at play if the MWI predictions are to agree with QM. As I have said before, if some of the branches have zero weight or zero probability, then such branches can never have been created. There is no mechanism which could make them vanish only when Alice and Bob meet. In many of the Everettian apologies, capital is made of the fact that the Alice-Bob meeting is also an interaction. But that interaction is not (or need not be) of the sort that could remove unwanted branches. For example, if Alice and Bob exchange results by email, there is no interaction that could possibly eliminate branches, or change data in their notebooks so that everything worked out. Such notions are in the realm of fantasy and magic. Bruce -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLTbQz5wzXfEZ563Eu_BS2%3DhH%2BUU_XYEm9KHijyOPCnxmQ%40mail.gmail.com.