Le dim. 23 févr. 2025, 06:15, Brent Meeker <meekerbr...@gmail.com> a écrit :
> > > On 2/12/2025 11:01 PM, Bruce Kellett wrote: > > On Thu, Feb 13, 2025 at 5:57 PM Quentin Anciaux <allco...@gmail.com> > 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. >> > Are you postulating more that 2^N sequences so the there can be more than > one sequence of a given proportion of 1's and 0's with one observer each or > are you postulating more than one observer in a given sequence? The former > corresponds to branch counting which JKC pointed out doesn't work because > it would imply retroactive changes in amplitudes based on future > measurement decisions. > > >> Simply stating that each branch contains "one observer" >> > If a branch contains more that one observer they must still just observe > the same sequence. So they can't add to the weight of that sequence. > Brent, You’re treating "branches" as isolated, discrete units, but if the wavefunction remains a continuous superposition, then what we call "a branch" is just an approximation—a macroscopic coarse-graining of many micro-branches. Decoherence prevents interference between them, but it does not imply a strict one-to-one mapping between observer instances and branches. If more observer instances exist in a high-amplitude region of the wavefunction, then an observer randomly drawn from the total set of observers is overwhelmingly likely to experience a sequence in proportion to its measure, not because the sequence itself is somehow weighted, but because there are simply more instances of the observer experiencing it. This is not just an abstract claim—it follows directly from how measure works in probability. If you duplicate a computational process a million times and run it on different hardware, the subjective experience of that process does not exist in just one instance. Similarly, in MWI, if decoherence results in more observer instances in a high-measure region, then most self-locating observers will find themselves in those regions. So the issue isn’t that observers "add to the weight of a sequence"—the weight already exists in the wavefunction amplitudes. The observer simply finds themselves in a sequence that corresponds to their proportion in the overall measure. Quentin > 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 everything-list+unsubscr...@googlegroups.com. > 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 everything-list+unsubscr...@googlegroups.com. > To view this discussion visit > https://groups.google.com/d/msgid/everything-list/6c952989-edb7-401e-a108-5ceec91e4219%40gmail.com > <https://groups.google.com/d/msgid/everything-list/6c952989-edb7-401e-a108-5ceec91e4219%40gmail.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 everything-list+unsubscr...@googlegroups.com. To view this discussion visit https://groups.google.com/d/msgid/everything-list/CAMW2kAphCQt9hp710SwevWH-R5j-9gCmh5_xDQeVgeM7cpgV8w%40mail.gmail.com.
