Bruce, Listing 2^N sequences does not prove that measure is irrelevant. Yes, all sequences exist, but that does not mean they all contribute equally to an observer’s experience. Your argument assumes that each sequence corresponds to exactly one branch with exactly one observer, but nothing in unitary evolution enforces such a strict partitioning.
Branches are not necessarily discrete, isolated entities—they could always remain in a superposition of infinite sub-branches, with measure determining the relative weight of each. If measure partitions reality rather than simply labeling sequences, then high-measure branches dominate experience, naturally leading to the Born rule. Your argument assumes its conclusion: that each sequence corresponds to one and only one branch, with no role for measure. But you haven’t proven this—just asserted it. Quentin Le jeu. 13 févr. 2025, 08:01, Bruce Kellett <[email protected]> a écrit : > On Thu, Feb 13, 2025 at 5:57 PM Quentin Anciaux <[email protected]> > wrote: > >> Bruce, >> >> You argue that MWI predicts a uniform distribution of outcomes because >> all sequences exist and each branch contains exactly one observer. Since >> experiments follow the Born rule instead, you claim MWI is falsified. But >> this assumes that measure has no effect—something you have not proven. >> >> The fact that 2^N sequences exist does not mean they all contribute >> equally to an observer’s experience. That’s the core issue. If measure >> determines how many copies of an observer exist in different branches, then >> high-measure branches dominate experience. This would naturally lead to >> Born-rule frequencies, without contradicting experiment. >> >> Simply stating that each branch contains "one observer" and that measure >> is irrelevant does not prove MWI is falsified—it assumes your conclusion. >> If you want to show MWI is incompatible with experiment, you need more than >> just claiming that measure plays no role; you need to justify why quantum >> experiments consistently match despite your assertion that all sequences >> should be equally likely. >> > > The fact is that you get the same 2^N binary sequences from the binary > state |psi> = a|0> + b|1> whatever the values of a and b. My case is proven. > > 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 [email protected]. > To view this discussion visit > https://groups.google.com/d/msgid/everything-list/CAFxXSLQGGNQZnr8JkO5h8fGVAETWsk2fheYpH8ruSr8N7GNXhQ%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLQGGNQZnr8JkO5h8fGVAETWsk2fheYpH8ruSr8N7GNXhQ%40mail.gmail.com?utm_medium=email&utm_source=footer> > . > -- 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 [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/CAMW2kArhg%3DNcAsuxFhK8cGcck_WHbCzUYTQz5bwaSpF%2BKT-6mA%40mail.gmail.com.
