Bruce, I’ve already provided multiple counterexamples, but you keep repeating the same assertions as if that’s a substitute for addressing them. If you’re so convinced of your "proof," publish it instead of endlessly posturing here. Otherwise, this is just noise.
Quentin Le mer. 26 févr. 2025, 22:08, Bruce Kellett <bhkellet...@gmail.com> a écrit : > On Wed, Feb 26, 2025 at 10:26 PM Quentin Anciaux <allco...@gmail.com> > wrote: > >> 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. >> > > I find it interesting that you haven't even attempted to answer the > detailed argument that I made (below). That, to me, suggests that you do > not have any coherent response to offer. So you just repeat your normal > smoke-and-mirrors trick and hope that I will be diverted away from the main > points. I think you need to up your game if you want to make any progress > here. > > Bruce > > >> Publish it, what are you afraid of? Being proved wrong? >> >> Quentin >> >> Le mer. 26 févr. 2025, 11:04, Bruce Kellett <bhkellet...@gmail.com> a >> écrit : >> >>> On Wed, Feb 26, 2025 at 7:08 PM Quentin Anciaux <allco...@gmail.com> >>> wrote: >>> >>>> >>>> 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 everything-list+unsubscr...@googlegroups.com. > To view this discussion visit > https://groups.google.com/d/msgid/everything-list/CAFxXSLQGreXJZ01ukTV5PJSnM9g8QpRdhTmcog3Hz1rHoA%2BkHw%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLQGreXJZ01ukTV5PJSnM9g8QpRdhTmcog3Hz1rHoA%2BkHw%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/CAMW2kApAiWuKE5OEtj5V-AROpJ0davH7HZ0LRXvby7TE-OvWgQ%40mail.gmail.com.
