Bruce, Your argument still assumes what it wants to prove. Yes, the 2^N sequences exist independently of the amplitudes, but the claim that they all contribute equally to probability remains an implicit assumption in your reasoning. The fact that the sequences themselves don’t depend on a and b does not mean that their statistical weights—how often an observer finds themselves in each sequence—are independent of amplitudes.
Your core mistake is treating the observed sequences as primary while ignoring the fact that the number of observer instances experiencing each sequence is what matters. The amplitudes determine the measure of each branch, meaning the vast majority of observers end up in sequences that follow the Born rule. You are simply assuming that observer distribution is uniform across sequences, which contradicts the entire structure of quantum mechanics. You keep insisting that different values of a and b still produce the same sequences. Yes, the sequences are the same in a mathematical sense, but that does not mean they are observed with equal frequency. This is like saying that rolling a biased die produces the same possible numbers as a fair die and then concluding that all outcomes are equally likely. The actual distribution of observed frequencies follows the amplitudes squared, as the Born rule predicts. If you believe MWI contradicts the Born rule, publish a formal proof. Otherwise, repeating the same flawed argument does not make it more correct. Quentin Le lun. 24 févr. 2025, 11:49, Bruce Kellett <bhkellet...@gmail.com> a écrit : > On Mon, Feb 24, 2025 at 5:31 PM Quentin Anciaux <allco...@gmail.com> > wrote: > >> Bruce, >> >> Your assertion is absurd. The existence of 2^N sequences is not in >> question—it's a direct prediction of MWI. The issue is whether all >> sequences contribute equally to probability, which they demonstrably do >> not, as experiments confirm the Born rule, not a uniform distribution. >> > > You are persisting with this idea that my argument is that all the 2^N > "sequences contribute equally to probability". But I have never made any > such claim, and such an idea is completely beside the point. The first > important point comes after I mention the 2^N binary sequences. I point out > that these come from the binary state a|0> + b|1>, but the sequences > themselves are independent of the amplitudes a and b. This means that if > you repeat N trials with different amplitudes a and b, you will get exactly > the same 2^N sequences. The point of this is that the amplitudes themselves > play no role in the formation of the experimentally observed binary > sequences. > > You must remember that the sequences of UP and DOWN (or zero and one in my > coding) are the data that any experimentalist measuring the spin > projections of N spin-half atoms will get. They are the data he/she will > work with, and that data is independent of the amplitudes. > > The second point is that from that experimental data, the experimentalist > can get an estimate of the probability of obtaining a zero. Say there are r > zeros in someone's observed sequence. Then a good estimate of the > probability of getting a zero will be p = r/N. If one compares this with > the Born rule prediction, which is |a|^2, then in the majority of cases the > experimentalist will find the Born rule to be disconfirmed: his estimate p > = r/N will not equal |a|^2. Since the same applies for any sequence in the > set of 2^N: the value of r for different experimentalists, will range from > zero to one. But all have the same Born rule probability, |a|^2. This is > what happens when the data from such an experiment is used for theory > confirmation. MWI is not confirmed, whereas the single world model is > confirmed on every occasion. > > The problem is made more acute if you consider that we can take completely > different values for the amplitudes a and b, but we will get the same > sequences, and the same set of values of p = r/N, even though the Born > probabilities |a|^2 are wildly different. Of course, there is a value of r > in the set of sequences that agrees with the Born probability, simply > because the set of sequences covers all possibilities, including the set of > zeros and ones that would be obtained by a single experimentalist in a > single world scenario. In the single world, the Born rule is confirmed in > every case, Though in the many-world case, the majority of experimentalists > will find that the Born rule is violated. > > 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/CAFxXSLRZQaOjFpJV1Bsfq4qHcz3B79QOS8x45mJ6%2BiCfDm%2BRng%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLRZQaOjFpJV1Bsfq4qHcz3B79QOS8x45mJ6%2BiCfDm%2BRng%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 everything-list+unsubscr...@googlegroups.com. 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