Bruce, Your response assumes that unitary evolution inherently produces "one observer per branch" in a discrete way, but that’s not what follows from the wavefunction’s continuous structure. Everett’s relative state formulation does not propose discrete worlds but rather an evolving superposition where decoherence prevents interference. The fact that we describe macroscopic branches as "splitting" is a convenient approximation, not a fundamental aspect of the theory.
The key point you keep ignoring is that amplitudes are not just "carried along" without meaning—they define the structure of the wavefunction, and decoherence prevents low-amplitude branches from significantly contributing to observer experiences. Your claim that "one observer per branch" follows directly from unitary evolution is an assumption, not a derivation. If you insist that unitary evolution cannot produce probability weights, then your argument applies equally to any interpretation of quantum mechanics. The Born rule is a fact of experiment, and any valid interpretation must explain it. If you believe MWI cannot do so, you must show why—not just assert that it "hasn’t been done" while dismissing attempts to derive it. Assume your pride, publish and get the glory. Quentin Le lun. 24 févr. 2025, 23:29, Bruce Kellett <bhkellet...@gmail.com> a écrit : > On Tue, Feb 25, 2025 at 9:15 AM Quentin Anciaux <allco...@gmail.com> > wrote: > >> >> You keep assuming that one branch corresponds to one observer, but that’s >> simply not what MWI proposes. >> > > That is exactly what Everett (MWI) proposes. Have you not in the least > understood what I have been saying. On each measurement of the binary > wavefunction, the observer splits, so that there is one copy on each > branch. Continuing this process N times leads to the 2^N branches with one > observer on each sequence. There is only one observer per sequence > (branch), because that is what the unitary splitting process gives you. > > > Everett’s framework is about relative states, not discrete worlds with >> single, isolated observers. If you treat branches as coarse-grained >> partitions of an underlying continuous wavefunction, then observer >> instances scale with amplitude, >> > > But that is not what unitary evolution and the Schrodinger equation say. > It is just pure fantasy on your part. I can ask, where do these additional > observers on each branch come from? They are not included in any of the > mathematics of unitary evolution. > > and that’s what leads to the Born rule. The fact that all 2^N sequences >> exist doesn’t mean they contain the same number of observer copies. If you >> disagree, you need to justify why unitary evolution should produce equal >> weighting when the amplitudes explicitly define the structure of the >> wavefunction. >> > > That is what has been done. Unitary evolution produces one observer for > each branch, because the observers are copies that arise from unitary > splits according to the Schrodinger equation. At this stage, there are no > such things as branch weights, because they have not been defined. I am > just following unitary evolution and, although the amplitudes get carried > along, they have no particular meaning or significance until some such is > imposed from outside. > > 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/CAFxXSLRZL_k3t%2B%2Bo2BLjMi3ZS2eywVTbieEaaB7NxKqSZk98JQ%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLRZL_k3t%2B%2Bo2BLjMi3ZS2eywVTbieEaaB7NxKqSZk98JQ%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/CAMW2kAp-H6FzCtCVEaBg4BpByXTVC5BMnMJE%2Bd7k_kVGpLdJ7Q%40mail.gmail.com.
