Bruce, You keep repeating the same circular reasoning: you assume one observer per branch, then conclude uniform sampling by observers, then use that to claim equal measure. But Everett’s relative-state formulation does not require discrete worlds or a uniform observer distribution — that is your interpretation, not a derivation.
Carrying amplitudes through unitarity without giving them any role is precisely the point of contention. Ignoring them does not make their influence vanish; it only means you are not engaging with the core question. If you assert that your construction proves all branches have equal measure, then you are assuming the conclusion you're trying to establish. If you are certain this invalidates any amplitude-based measure and thus the Born rule, the proper way forward is still the same: publish the derivation and let it stand under peer review. Repeating it here without addressing counterarguments doesn't make it more correct, just more dogmatic. Quentin All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer) Le mer. 27 août 2025, 01:11, Bruce Kellett <[email protected]> a écrit : > On Tue, Aug 26, 2025 at 10:30 PM Quentin Anciaux <[email protected]> > wrote: > >> Bruce, >> >> You keep assuming one observer per sequence and uniform sampling, but >> that assumption is yours, not Everett’s. In Everett’s framework, the >> relative weights aren’t arbitrary — they’re determined by the amplitudes in >> the wavefunction. By rejecting that, you’re refuting a simplified model of >> your own making, not MWI itself. If your argument truly applied to >> Everett’s theory, you should be able to show how it addresses the role of >> measure instead of ignoring it. >> > > I think you need to actually work through the calculation/construction. > One observer per sequence is implied by Everett's theory. If you have an > initial state such as > a|0> + b|1> > and every possible result is realized in every trial: the first trial > gives you two branches, with results |0> and |1> respectively. The observer > splits so that there is a single observer on each branch. Then, the next > trial for the observer who saw |0> again causes a split, so we get two > further branches, 00 and 01, with a single observer on each branch. > Continuing this, we get branches 0000.., 0100..., 011,00... and so on, so > there are 2^N sequences (branches) after N repetitions. On each repetition, > the branch, and the associated observer again splits, so you get two new > branches, with a single observer on each. Consequently, a straightforward > Everettian construction proves that there is only one observer on each > branch, and there is no branch without a single observer. > > If you disagree with this result, then it is up to you to give the > calculation that produces multiple observers on some branches. > > Of course, each sequence in this process has an amplitude, coming from the > amplitudes in the initial state. For example, the sequence with u zeros has > an amplitude > a^u*b^{N-u}. But these are just amplitudes carried through. They do not > represent weights or probabilities until you impose Born's rule, and we are > not doing that in this calculation. The question of measure is answered by > the above construction: each sequence has the same measure. > > It is important to realize that the same 2^N sequences are obtained > whatever the initial amplitude a, and b might be. Thus the set of sequences > does not come from the binomial theorem with a probability of success given > by |a|^2, because you get the same sequence for all values of a and b, > which is not the binomial theorem result. Equal measure over the sequences > is a result of the construction, not of some mathematical theorem related > to the binomial distribution. > > 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/CAFxXSLRPTdg1tp4hTamZMEBD4hzy_w5ysHDOPfpeqTN02NOLeg%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLRPTdg1tp4hTamZMEBD4hzy_w5ysHDOPfpeqTN02NOLeg%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/CAMW2kAonM6sRRnMKTUEOpWEJYWB5eXVAC04wQYBit%3DYdyXnuNQ%40mail.gmail.com.
