Awesome, and the less dense inner knowledge of thought to come together with some material sciences always fascinating
Ilsa Bartlett Institute for Rewiring the System http://ilsabartlett.wordpress.com http://www.google.com/profiles/ilsa.bartlett www.hotlux.com/angel <http://www.hotlux.com/angel.htm> "Don't ever get so big or important that you can not hear and listen to every other person." -John Coltrane On Tue, Aug 26, 2025, 5:14 PM Bruce Kellett <[email protected]> wrote: > On Wed, Aug 27, 2025 at 9:11 AM Bruce Kellett <[email protected]> > wrote: > >> 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 >> > > I can expand a little on the argument for equal weight for each sequence. > From the above, the sequence with u zeros has an amplitude a^ub^{N-u}. If > we take the case with equal probabilities, a = b = 1/sqrt(2), this > amplitude becomes 1/2^N. That is, in this case, the amplitude is > independent of u, the number of zeros in the sequence. In other words, all > sequences have the same amplitude, and hence the same weight or measure. > Now we know that the same set of sequences is obtained for all values of a > and b. So the result for a = b = 1/sqrt(2) is typical, and we can conclude > that all sequences have the same weight for all cases. Note again that > these sequences cover all possible binary sequences of length N, so we must > get the same set of sequences for any a and b. The sequences are obtained > by the construction, and, except in special cases like a = b, they are not > the same as sequences from the binomial distribution for different > probabilities of success. > > 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/CAFxXSLTJiaTBF9ONrzEXr5nX3m2Y%3DhWrFG0ZqcfecnPLDvEaTQ%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLTJiaTBF9ONrzEXr5nX3m2Y%3DhWrFG0ZqcfecnPLDvEaTQ%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]. 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