Brent, The issue is precisely that if you start with only unitary evolution and no additional assumptions, you don’t get probabilities at all—just a deterministic wavefunction. That’s why the Born rule must be explained rather than assumed. The problem is not that MWI contradicts the Born rule, but that it needs to derive it without assuming it.
Your argument boils down to saying, "The Born rule is empirically confirmed, so MWI must explicitly postulate it." But any interpretation of QM, including single-world ones, requires some justification for why the squared amplitude determines probability. If MWI needs additional reasoning to get there, that’s an open question, not a refutation. You say that different values of a and b still produce the same sequences, but What you’re missing is that in MWI, observer instances are not evenly distributed across all sequences—this is fundamental to Everett’s relative state interpretation. If all sequences contributed equally, there would be no need for measure at all, and MWI would have been dismissed from the start. Everett’s entire motivation was to account for probability within a deterministic framework, meaning your argument misrepresents what MWI actually proposes. The number of observer instances experiencing each sequence scales with the amplitude squared, so the observed frequencies match the Born rule. This is not an arbitrary postulate—it follows from the way measure is assigned in the wavefunction. If you believe this is just "counting branches," you are assuming that all branches have equal weight, which is exactly what’s in question. The fact that small N allows some spread in observed averages is true in any probabilistic system—classical or quantum—but that does not mean the amplitudes are irrelevant to observer statistics. Ultimately, the challenge is not proving the Born rule holds empirically—we know it does. The question is whether it emerges naturally from unitary evolution or must be separately assumed. Simply asserting that MWI "doesn’t work without the Born rule" is not an argument—it’s a restatement of the problem. Quentin Le lun. 24 févr. 2025, 21:43, Brent Meeker <meekerbr...@gmail.com> a écrit : > > > On 2/24/2025 3:07 AM, Quentin Anciaux wrote: > > 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. > > Which means you need some different, separate theory about how "observer > distribution" is determined in a way that depends on the 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. > > Not necessarily a "vast majority". For a short sequence, small N, there > should be a lot of spread in the observed average. Only for large N does > the distribution become sharply peaked. "Follow the Born rule" is a little > ambiguous since it's probabilitistic. "Consistent with the Born rule" is > better. > > You are simply assuming that observer distribution is uniform across > sequences, which contradicts the entire structure of quantum mechanics. > > You mean it doesn't comport with the Born rule...and the Born rule is > empirically confirmed...which is the whole point. The Born rule must be > invoked to assign numbers of observers, aka probabilities, to the 2^N > sequences. > > > 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, > > It doesn't contradict it unless you take it as the whole of the theory, > without the Born rule. If you don't include the Born rule, or an > equivalent postulate, then you are left with nothing but counting the > branches, which doesn't work for a=/=b. > > Brent > > > 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. > To view this discussion visit > https://groups.google.com/d/msgid/everything-list/CAMW2kArvTk%2BhCcOF8sdqi_oKp%3DtpKbherpGg5D20vFVrUevdpw%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAMW2kArvTk%2BhCcOF8sdqi_oKp%3DtpKbherpGg5D20vFVrUevdpw%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. > To view this discussion visit > https://groups.google.com/d/msgid/everything-list/82f98fcd-9a2c-458f-aa0b-e70d4327c133%40gmail.com > <https://groups.google.com/d/msgid/everything-list/82f98fcd-9a2c-458f-aa0b-e70d4327c133%40gmail.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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