Brent, The key point is that a branch isn’t a single, discrete unit—it’s a coarse-grained structure emerging from decoherence. The fact that a branch has "only one result" is an approximation because, in reality, the wavefunction remains a superposition of countless micro-branches. What we call "a branch" is just a region where interference is negligible, but within that, there are still subtler partitions based on amplitude.
Yes, observer instances are in orthogonal branches, but the partitioning isn’t uniform. The measure of a branch corresponds to the number of observer instances experiencing that outcome, not to the number of distinct sequences. This means that instead of "one observer per branch," there are exponentially more copies of an observer in high-amplitude branches than in low-amplitude ones. The difference is subtle but crucial: If you assume one observer per branch, you get branch counting and no Born rule. But if you recognize that branches are not discrete and observer instances scale with measure, you naturally recover Born probabilities. Quentin Le ven. 21 févr. 2025, 04:03, Brent Meeker <[email protected]> a écrit : > > > On 2/19/2025 11:16 PM, Quentin Anciaux wrote: > > > > Le jeu. 20 févr. 2025, 07:55, Brent Meeker <[email protected]> a > écrit : > >> What does it mean to have more than one observer in a branch? A branch >> forms because the result is orthogonal to the other different results. Of >> course any number of persons can observe it, which by construction means >> they are in that branch. Is that what you mean? >> >> Brent >> > > Brent, > > A branch isn't a single discrete entity—it's a region in the wavefunction > where decoherence prevents interference. The distinction matters because if > the wavefunction remains a superposition of infinite components, then what > we call a "branch" is just an approximate partition, not a fundamental unit. > > But it has only one result...right? That's what makes it a branch and not > two branches. > > > Saying "one observer per branch" assumes a sharp branching structure, but > if the wavefunction maintains an underlying continuous structure, then what > we experience as "branches" are really clusters of high-measure observer > instances. More than one observer in a branch means that, within that > region of the wavefunction, there are exponentially more instances of an > observer experiencing a high-amplitude outcome than a low-amplitude one. > > In other words, observer count isn't tied to branch count—it's tied to > measure. > > Which is the number of observers, each of which is in a branch orthogonal > to all other branches...right? > > Brent > > What matters isn’t how many "branches" exist but how many copies of an > observer exist in each, which is why most observers see outcomes aligning > with the Born rule. > > Quentin > > >> On 2/19/2025 10:48 PM, Quentin Anciaux wrote: >> >> If you assume one observer per branch, you assume what you want to prove: >> that measure plays no role. >> >> >> -- >> 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/fc277de3-baaa-4029-b71e-a4115d76e3fc%40gmail.com >> <https://groups.google.com/d/msgid/everything-list/fc277de3-baaa-4029-b71e-a4115d76e3fc%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 [email protected]. > To view this discussion visit > https://groups.google.com/d/msgid/everything-list/CAMW2kAo6crOnadBmhkHBm0F67Y%3D%2Be25%2BmNOXodPtOwjxXPmXsg%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAMW2kAo6crOnadBmhkHBm0F67Y%3D%2Be25%2BmNOXodPtOwjxXPmXsg%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/016cc873-002c-4f6b-a0d1-07dab24517be%40gmail.com > <https://groups.google.com/d/msgid/everything-list/016cc873-002c-4f6b-a0d1-07dab24517be%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 [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/CAMW2kAougfoZD5orpyPdqNaRQ%2BDsw%3DTi-GcFp2XTLXcdss%3D%3DFA%40mail.gmail.com.
