Your argument suggests that since measure is not directly observable, it cannot influence what we experience. But this is incorrect for the same reason that probability distributions matter in classical systems:
You cannot observe probability itself, only its consequences over many trials. In a biased coin flip (90% heads, 10% tails), every sequence of flips exists in MWI. But most copies of an observer will find themselves in sequences where heads appear 90% of the time. The fact that all sequences exist does not mean they contribute equally to an observer’s experience. You experience the weight of measure indirectly—just like in classical probability, where we experience frequencies that match probabilities over many trials. Le lun. 10 févr. 2025, 23:32, Quentin Anciaux <[email protected]> a écrit : > > > Le lun. 10 févr. 2025, 23:21, Bruce Kellett <[email protected]> a > écrit : > >> On Mon, Feb 10, 2025 at 11:09 PM Quentin Anciaux <[email protected]> >> wrote: >> >>> Bruce, >>> >>> Your argument assumes that all measurement sequences are equally likely, >>> which is false in MWI. The issue is not about which sequences exist (they >>> all do) but about how measure is distributed among them. The Born rule does >>> not emerge from simple branch counting—it emerges from the relative measure >>> assigned to each branch. >>> >> >> You really are obsessed with the idea that I am assuming that all >> measurement sequences are equally likely. It does not matter how many times >> I deny this, and point out how my argument does not depend in any way on >> such an assumption, you keep insisting that that is my error. I think you >> should pay more attention to what I am saying and not so much to your own >> prejudices. >> >> Each of the binary sequences that result from N trials of measurements on >> a 2-component system will exist independently of the original amplitudes. >> For example, the sequence with r zeros, and (N - r) ones, will have a >> coefficient a^r b^(N-r). You are interpreting this as a weight or >> probability without any evidence for such an interpretation. If you impose >> the Born rule, it is the Born probability of that sequence. But we have not >> imposed the Born rule, so as far as I am concerned it is just a number. And >> this number is the same for that sequence whenever it occurs. The point is >> that I simply count the zeros (and/or ones) in each sequence. This gives an >> estimate of the probability of getting a zero (or one in that sequence). >> That estimate is p = r/N. Now that probability estimate is the same for >> every occurrence of that sequence. In particular, the probability estimate >> is independent of the Born probability from the initial state, which is >> simply a^2. >> >> The problem here is that we get all possible values of the probability >> estimate p = r/N from the set of 2^ binary sequences that arise from every >> set of N trials. This should give rise to concern, because only very few of >> these probability estimates are going to agree with the Born probability >> a^2. You cannot, at this stage, use the amplitudes of each sequence to >> downweight anomalous results because the Born rule is not available to you >> from the Schrodinger equation. >> >> The problem is multiplied when you consider that the amplitudes in the >> original state |psi> = a|0> + b|1> are arbitrary, so the true Born >> probabilities can take on any value between 0 and 1. This arbitrariness is >> not reflected in the set of 2^N binary sequences that you obtain in any >> experiment with N trials because you get the same set for any value of the >> original amplitudes >> >> >> You claim that in large N trials, most sequences will have an equal >>> number of zeros and ones, implying that the estimated probability will tend >>> toward 0.5. But this ignores that the wavefunction does not generate >>> sequences with uniform measure. The amplitude of each sequence is >>> determined by the product of individual amplitudes along the sequence, and >>> when you apply the Born rule iteratively, high-measure sequences dominate >>> the observer’s experience. >>> >>> Your mistake is treating measurement as though every sequence has equal >>> likelihood, which contradicts the actual evolution of the wavefunction. >>> Yes, there are 2^N branches, but those branches do not carry equal measure. >>> The vast majority of measure is concentrated in the sequences that match >>> the Born distribution, meaning that nearly all observers find themselves in >>> worlds where outcomes obey the expected frequencies. >>> >>> This is not speculation; it follows directly from the structure of the >>> wavefunction. The weight of a branch is not just a number—it represents the >>> relative frequency with which observers find themselves in different >>> sequences. The fact that a branch exists does not mean it has equal >>> relevance to an observer's experience. >>> >>> Your logic would apply if MWI simply stated that all sequences exist and >>> are equally likely. But that is not what MWI says. It says that the measure >>> of a branch determines the number of observer instances that experience >>> that branch. The overwhelming majority of those instances will observe the >>> Born rule, not because of "branch counting," but because high-measure >>> sequences contain exponentially more copies of any given observer. >>> >>> If your argument were correct, QM would be falsified every time we ran >>> an experiment, because we would never observe Born-rule statistics. >>> >> >> That is the point I am making. MWI is disconfirmed by every experiment. >> QM remains intact, it is your many worlds interpretation that fails. >> >> >> Yet every experiment confirms the Born rule, which means your assumption >>> that "all sequences contribute equally" is demonstrably false. >>> >> >> Since I do not make that assumption, your conclusion is wrong. >> >> You are ignoring that measure, not count, determines what observers >>> experience. >>> >> >> When you do an experiment measuring the spin projection of some >> 2-component state, all that you record is a sequence of zeros and ones, >> with r zeros and (N - r) ones. You do not ever see the amplitude of that >> sequence. It has no effect on what you measure, so claiming that it can up- >> or down-weight your results is absurd. >> >> Bruce >> > > Your argument is based on treating the measurement process as merely > counting sequences of zeros and ones, while dismissing the amplitudes as > “just numbers.” But this ignores that the wavefunction governs the > evolution of the system, and the amplitudes are not arbitrary labels—they > encode the structure of reality. The Schrodinger equation evolves the > system deterministically, and when measurement occurs, the measure of each > branch determines how many observer instances find themselves in it. > > You claim that the amplitude of a sequence does not affect what is > measured, yet this is exactly what determines how many observers experience > a given sequence. The claim that “you do not ever see the amplitude” misses > the point: you do not directly observe measure, but you observe its > consequences. The reason we see Born-rule statistics is that the measure > dictates the relative number of observers experiencing different sequences. > > Quentin > >> -- >> 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/CAFxXSLS518CcE8Wuj3FHe_66hnGN-KZFb%2BwOJ7-cHC_geo1N_w%40mail.gmail.com >> <https://groups.google.com/d/msgid/everything-list/CAFxXSLS518CcE8Wuj3FHe_66hnGN-KZFb%2BwOJ7-cHC_geo1N_w%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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