On 3/5/2020 3:44 PM, Bruce Kellett wrote:
OR postulate that the splits are into many copies so that the branch count gives the Born statistics.That has possibilities, but I think it cannot work either. After all, each observer just sees a sequence of results -- he is unaware of other branches or sequences, so does not know how many branches are the same as his. The 1p/3p distinction comes into play again. Any attempt to make multiple branches reproduce probabilities necessarily confuses this distinction. You have to think in terms of what data an observer actually obtains. Thinking about what happens in the "other worlds" is illegitimate.Consider the many copies case as an ensemble and it will reproduce the Born statistics even though it is deterministic. This is easy to see because every sequence a single observer has seen is the result of a random choice at the split of which path you call "that observer".But the weights do not influence that split, so the observer cannot see the weights.
Not weights, multiple branches. Brent -- 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/c0b51f3b-d106-0f21-1822-df6d5088658c%40verizon.net.
