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 <[email protected]> a écrit : > On Wed, Feb 26, 2025 at 10:26 PM Quentin Anciaux <[email protected]> > 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 <[email protected]> a >> écrit : >> >>> On Wed, Feb 26, 2025 at 7:08 PM Quentin Anciaux <[email protected]> >>> 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 [email protected]. > 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 [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/CAMW2kApAiWuKE5OEtj5V-AROpJ0davH7HZ0LRXvby7TE-OvWgQ%40mail.gmail.com.
