Bruce, Yes, every possible experience is lived by some version of me in MWI, but that does not mean all experiences are equally likely or subjectively equivalent. The measure of a branch determines how many copies of me experience a given outcome. In practice, my conscious experience will overwhelmingly be shaped by the branches with higher measure, not by the rare and improbable ones.
For example, if a quantum event has a 1% probability, then there will be branches where I observe it, but they will be exponentially fewer than those where I do not. The measure is not just an abstract number—it reflects the relative weight of different outcomes in the wavefunction. This is why, as an observer, I will almost always see frequencies matching the Born rule, because the majority of my copies exist in branches where this distribution holds. Your argument assumes that since all branches exist, they must be equiprobable, but this ignores the fact that measure determines how many copies of an observer exist in each branch. In a lottery, every ticket exists, but some are printed in larger quantities. Saying "all branches exist, so they must be equal" is as flawed as saying "all lottery tickets exist, so all should win equally." Ultimately, my conscious experience is not determined by the mere existence of branches, but by the relative number of copies of me in each. Low-measure branches do exist, but they are not representative of my experience. This is why MWI naturally leads to the Born probabilities, without assuming collapse or introducing an arbitrary rule. Your reasoning collapses probability into mere branch-counting, but probability is about where observers actually find themselves, not about an abstract collection of sequences. Quentin Le lun. 10 févr. 2025, 04:14, Bruce Kellett <bhkellet...@gmail.com> a écrit : > On Mon, Feb 10, 2025 at 1:47 PM Russell Standish <li...@hpcoders.com.au> > wrote: > >> On Mon, Feb 10, 2025 at 11:35:11AM +1100, Bruce Kellett wrote: >> > On Mon, Feb 10, 2025 at 10:52 AM Russell Standish < >> li...@hpcoders.com.au> >> > wrote: >> > >> > >> > > On Mon, Feb 10, 2025 at 09:25:57AM +1100, Bruce Kellett wrote: >> > > > On Mon, Feb 10, 2025 at 8:49 AM Russell Standish >> > > wrote: >> > > > >> > > > On Thu, Feb 06, 2025 at 11:38:52AM +1100, Bruce Kellett >> wrote: >> > > > > >> > > > > Many worlds theory does not have any comparable way >> of relating probabilities >> > > > > to the properties of the wave function. In fact, if al >> l possibilities are >> > > > > realized on every trial, the majority of observers >> will get results that >> > > > > contradict the Born probabilities. >> > > > > >> > > > >> > > > I'm not sure what you mean by "contradict", but the >> majority of >> > > > observers will get results that lie within one standard >> deviation of >> > > > the expected value (ie mean) according to the >> distribution of Born >> > > > probabilities. If this is what you mean by >> "contradict", then you are >> > > > trivially correct, but uninteresting. If you mean the >> above statement >> > > > is false according to the MWI, then I'd like to know >> why. It sure >> > > > doesn't seem so to me. >> > > > >> > > > >> > > > It does depend on what value you take for N, the number of >> trials. In the limit >> > > > of very large N, the law of large numbers does give the >> result you suggest. But >> > > > for intermediate values of N, MWI says that there will >> always be branches for >> > > > which the ratio of successes to N falls outside any >> reasonable error bound on >> > > > the expected Born value. >> > > > >> > > > This problem has been noted by others, and when asked about >> it, Carroll simply >> > > > dismissed the poor suckers that get results that invalidate >> the Born Rule as >> > > > just poor unlucky suckers. Sure, in a single world system, >> there is always a >> > > > small probability that you will get anomalous results. But >> that is always a >> > > > small probability. Whereas, in MWI, there are always such >> branches with >> > > > anomalous results, even for large N. The difference is >> important. >> > > > >> > > >> > > Yes, but the proportion of "poor unlucky suckers" in the set >> of all >> > > observers becomes vanishingly small as the number of >> observers tend to >> > > infinity. >> > > >> > > >> > > The number of trials does not have to tend to infinity. That is >> just the >> > > frequentist mistake. >> > > >> > >> > I wasn't talking about the number of trials, but the number of >> > observers. That is either astronomically large or an actual >> infinity. >> > >> > >> > There are only ever 2^N branches under consideration, so only ever 2^N >> observers. >> >> These 2^N observers will be unevenly weighted. If you want to do even >> weighting by some sort of symmetry argument, you will need to >> subdivide more finely. >> > > And where do these weights come from? If you do the construction, all 2^N > branches arise in the same way, so they have the same weight. No > indifference principle need be applied. But in any case, whatever weights > you might imagine these sequences of binary outcomes to have, the results > are the same. For large N, the vast majority of observers will find a 50/50 > split of zeros and ones, (within a standard deviation). So regardless of > the "weights", most will disagree with the Born result, which depends on > the coefficients in the original wave function: > > |psi> = a|0> + b|1>. > > > Additional branches due to decoherence can be disregarded for these >> purposes since >> > all such observers only duplicate some that have already been counted. >> > >> > >> > > As JC says, we don't know if the number of observers is >> countably >> > > infinite (which would be my guess), uncountably infinite or >> just plain >> > > astronomically large. In any case, the proportion of >> observers seeing >> > > results outside of one standard deviation is of measure zero >> for >> > > practical purposes. If that is not the case, please explain. >> > > >> > > >> > > The number of anomalous results in MWI is not of measure zero in >> any realistic case. >> > > >> > >> > I'm trying to see why you say that. >> > >> > >> > Read Kent. >> >> Sure - I will get to it eventually. And probably have my own issues with >> that. >> >> In the meantime, you will need to excuse my incredulity in your >> statements. >> >> > >> > >> > > > The other point is that the set of branches obtained in >> Everettian many worlds >> > > > is independent of the amplitudes, or the Born probabilities >> for each outcome, >> > > > so observations on any one branch cannot be used as >> evidence, either for or against the theory. >> > > > >> > > >> > > We've had this discussion before. They're not independent, >> because the >> > > preparation of the experiment that defines the Born >> probabilities >> > > filters the set of allowed branches from which we sample the >> > > measurements. >> > > >> > > >> > > I don't know what this means. >> > > >> > >> > In preparing the experiment, you are already filtering out the >> > observers who choose to observe something different. And that >> > definitely changes the set of worlds, or branches under >> > consideration. So you cannot say (as you did) "the set of branches >> > obtained in Everettian many worlds is independent of the >> > amplitudes". Whether the set of branches changes in precisely the >> way >> > to recover the Born rule is a different question, of course, and >> > obviously rather hard to prove. >> > >> > >> > There is no such selective state preparation. Nobody gets filtered out >> in this >> > way. You are just making things up. >> > >> >> The way you brought it up earlier was to assume two SG apparatuses >> where observers were free to rotate one of the apparatus by angle θ >> with respect to the other. You inappropriately applied an indifference >> principle to assign a uniform weight to all branches. >> > > As I have said. I did not apply the indifference principle because the > sequences would have equal weight simply by construction. Besides, the > "weights" assigned to these sequences are essentially irrelevant. > > Yes - the decision to set the apparatus at angle θ filters the set of >> observers to those who make the same choice. And that filtering makes >> a difference in this case. >> > > What utter nonsense. I used the case of a beam of spin-half particles > polarized along the x-axis so that I could then observe these with a > rotatable S-G magnet in order to get a range of values for the amplitudes a > and b in the wave function |psi> above. There is no filtering. You have > just not understood what is going on. > > I'll read Kent's paper to see if he makes the same error. >> > > Kent does not use this particular example, although he does point out that > the "weights" assigned to each sequence in the 2^N possible sequences are > irrelevant. > > 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/CAFxXSLRbPrij2JN1%3DugeBTQou9Kwvvd%2Bk%3Dd8pxFbVH19SX57DQ%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLRbPrij2JN1%3DugeBTQou9Kwvvd%2Bk%3Dd8pxFbVH19SX57DQ%40mail.gmail.com?utm_medium=email&utm_source=footer> > . > -- You received this message because you are subscribed to the Google Groups "Everything List" group. 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