You are still assuming that each measurement results in a discrete split with exactly one observer per branch, which is an interpretation, not a derivation. Nothing in the Schrödinger equation forces branches to be discrete rather than continuously superposed structures with relative measure. Your reasoning assumes what it wants to prove: that branching is a countable process rather than a differentiation of an already superposed structure.
Publish it, what are you afraid of? Being proved wrong? Quentin Le mer. 26 févr. 2025, 11:04, Bruce Kellett <[email protected]> a écrit : > On Wed, Feb 26, 2025 at 7:08 PM Quentin Anciaux <[email protected]> > wrote: > >> Le mer. 26 févr. 2025, 01:53, Brent Meeker <[email protected]> a >> écrit : >> >>> On 2/25/2025 5:09 PM, Russell Standish wrote: >>> >>> On Wed, Feb 26, 2025 at 11:01:11AM +1100, Bruce Kellett wrote: >>> >>> On 2/24/2025 6:09 PM, Quentin Anciaux wrote: >>> >>> Your claim that "one observer per branch" follows directly from >>> unitary >>> evolution is an assumption, not a derivation. >>> >>> >>> No, it is a straightforward derivation from the formalism. If you don't >>> understand that, it is just further confirmation of the fact that you >>> understand very little about quantum mechanics. >>> >>> Not at all - it is an assumption you're making, and the nub of the entire >>> argument between you and Quentin. >>> >>> Do you understand Quentin's theory? ISTM it's just "Observer counting" >>> where the counts instantiate the Born rule ex hypothesi. It's branch >>> counting by another name. >>> >>> Brent >>> >> >> Brent, >> >> You’re still misrepresenting the argument. It’s not branch counting under >> another name, it’s about how measure determines observer frequencies. The >> issue is whether the number of observer instances scales with amplitude >> squared, not whether we simply count branches. If all branches were >> weighted equally, MWI would have been dead on arrival, because it wouldn’t >> match experiments. >> >> The claim that “one observer per branch” is a direct consequence of >> unitary evolution is just an assumption, it’s not something derived from >> the Schrödinger equation. >> > > It is derived from that, or the Schrodinger equation enhanced with unitary > evolution and the linearity of Hilbert space. > > Since you clearly don't get it. Let me spell it out in baby steps. > > We start from the wave function for some system, say |psi>. This is the > expanded in some basis like |psi> = a|0> + b|1>, where I have taken a two > dimensional space for clarity and convenience, although the argument is > easily expanded to an arbitrary number of independent basis states. > > We then measure this state (or subject it to some interaction). > |psi>|O>|E> where |O> is an observer, and |E> is the environment which can > include anything else that is relevant. Linear unitary evolution then > entangles both the observer and the environment with the object state: > > |psi>|O>|E> = (a|0> + b|1>)|O>|E> --> a|O sees zero>|E records > zero>|0> + b|O sees one>|E records one>|1>, > > One can readily see that there is one, and only one, copy of the observer > for each branch. Decoherence renders these branches approximately > orthogonal, and leads to the notion of independent worlds. The argument > can, of course, be readily generalized to a state with any number of basis > vectors. In no case, do we get more than one copy of the observer on any > branch, and there are no branches without a copy of the observer. > > All of this is just elementary linear unitary evolution, taught in general > quantum mechanics courses. If you want to deny this, you have to go to some > other theory which is incompatible with quantum mechanics. > > 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/CAFxXSLR8x8s7N6-gQmgSmAj_TD8MCi%3DB7nKTpbxkGWB1PUXEOA%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLR8x8s7N6-gQmgSmAj_TD8MCi%3DB7nKTpbxkGWB1PUXEOA%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/CAMW2kArJhmDFcHqzcW2%3DJJ_V2RN6nFgzbT%2BoA10z33BTv4VY5g%40mail.gmail.com.
