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.
