As I understand it this result isn't based on Born probability calculations, it's just based on looking at a measurement operator whose eigenvectors/eigenvalues correspond to adding the results of records (pointer states) of multiple trials and dividing by the total number of trials to get a fraction. For example if you have a measuring device designed to measure the spin on the z-axis on five successive trails and record the results, then if a detailed measurement of the records (using the assumption of collapse and the Born rule) gave the results "up, up, down, up, down", then in that case after collapse the state vector of the records would be parallel to the eigenvector of this new measurement operator with the eigenvalue "3/5 up". But if the experiment you're repeating is supposed to give probabilities 1/3 up and 2/3 down on each trial using the Born rule, you can apparently show that if we consider the limit as the number of trials approaches infinity, and just use the deterministic Schrodinger equation for wavefunction evolution *without* ever directly invoking the Born rule or state vector collapse, then all the amplitude of the records becomes concentrated on state vectors parallel to the eigenvector with eigenvalue "1/3 up", and orthogonal to all the other eigenvectors of this measurement operator. But since the collapse assumption was never invoked, the result for every *individual* trial can in general still be in a superposition of the spin eigenvectors for that the record of that trial (eg the record of trial #3 is in some superposition of spin-up and spin-down), it's only the sum-divided-by-number-of-trials that approaches a definite answer.
On Wed, Aug 27, 2025 at 7:31 PM Brent Meeker <meekerbr...@gmail.com> wrote: > Yes, given a device that calculates the Born probability, almost all > worlds will agree on the probability. > > Brent > > On 8/27/2025 3:17 PM, Jesse Mazer wrote: > > Also, even without invoking the Born rule there is a result that if you > consider a "pointer state" that records the relative fractions of different > possible measurement results in a *series* of N systems prepared in the > same initial state, and consider the limit as N approaches infinity, in > this limit all the amplitude gets concentrated on the pointer state with > the fractions that correspond to the probabilities for individual > measurements predicted by the Born rule--see David Z Albert's comments at > https://books.google.com/books?id=_HgF3wfADJIC&lpg=PP1&pg=PA238 and the > paper discussing Mittelstaedt’s theorem at > https://www.academia.edu/6975159/Quantum_dispositions_and_the_notion_of_measurement > > Jesse > > On Wed, Aug 27, 2025 at 2:39 AM Quentin Anciaux <allco...@gmail.com> > wrote: > >> Bruce, >> >> Everett’s original formulation describes a universal wavefunction >> evolving unitarily, not discrete worlds with one observer per branch. Your >> argument assumes this mapping, but it is an interpretative choice, not a >> result derived from the Schrödinger equation. >> >> Also, your claim that all 2^N sequences have equal measure only holds if >> amplitudes are treated as irrelevant. In standard quantum mechanics, >> amplitudes directly determine observed frequencies via the Born rule, which >> has strong experimental support. Ignoring amplitudes means you are no >> longer analyzing Everett’s framework but a different model where the Born >> rule indeed fails. >> >> To refute Everett with Born included, you would need to show that even >> when squared amplitudes define a natural measure, the predicted observed >> frequencies still fail. Assuming uniform sampling over sequences does not >> establish that. >> >> This is why your derivation is not accepted: it relies on a hidden >> premise, one observer per branch with uniform sampling, which is not part >> of Everettian quantum mechanics. >> >> Quentin >> >> All those moments will be lost in time, like tears in rain. (Roy >> Batty/Rutger Hauer) >> >> Le mer. 27 août 2025, 07:32, Bruce Kellett <bhkellet...@gmail.com> a >> écrit : >> >>> On Wed, Aug 27, 2025 at 3:26 PM Quentin Anciaux <allco...@gmail.com> >>> wrote: >>> >>>> Bruce, >>>> >>>> If your derivation is as solid as you claim, then a skeptical referee >>>> is exactly who you should want to convince. Repeating the same argument >>>> here without engaging with the role of amplitudes will not make it any >>>> stronger. You cannot dismiss amplitudes entirely and then claim to have >>>> explained why measure must be uniform, that is circular. >>>> >>>> If you truly believe your reasoning refutes the Born rule within >>>> Everett’s framework, then publishing it is the only way to settle the >>>> matter. Otherwise, endlessly asserting it here looks less like confidence >>>> and more like avoidance. >>>> >>>> Your entire argument hinges on assuming uniform observer sampling by >>>> postulating one observer per branch. >>>> >>> >>> The argument does not depend on this. This shows nothing more than that >>> you have not understood the argument. >>> >>> But that is precisely the point under debate, not a derived result. If >>>> you ignore the role of amplitudes in defining the structure of the >>>> wavefunction, you're not engaging with Everett's formulation, only with >>>> your own simplified model. >>>> >>>> Until you demonstrate why amplitudes should be irrelevant within >>>> unitary evolution, claiming equal weights is just assuming your conclusion. >>>> >>> >>> I think, rather, that you should show how the argument I have made >>> depends on amplitudes when it clearly does not. It depends merely on the >>> proportion of zero outcomes in each sequence. And that does not depend on >>> the amplitudes. >>> >>> 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/CAFxXSLTqmwjWPL45KfJwEJRqr5_VOZETJZKZaCE3tZamgVBXbg%40mail.gmail.com >>> <https://groups.google.com/d/msgid/everything-list/CAFxXSLTqmwjWPL45KfJwEJRqr5_VOZETJZKZaCE3tZamgVBXbg%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/CAMW2kAp6jiQmTmu%3D%2Bd1p0XFf1axT6%2BBSp4EMbna1ZJ5%2BvDp4jQ%40mail.gmail.com >> <https://groups.google.com/d/msgid/everything-list/CAMW2kAp6jiQmTmu%3D%2Bd1p0XFf1axT6%2BBSp4EMbna1ZJ5%2BvDp4jQ%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/CAPCWU3LDe3qoLj_Kp1M9oxt1OR%3DXj2aGRVZuZwOr_t7b3Y6jWg%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAPCWU3LDe3qoLj_Kp1M9oxt1OR%3DXj2aGRVZuZwOr_t7b3Y6jWg%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/75aebb9c-106d-4f19-8236-a667b345681b%40gmail.com > <https://groups.google.com/d/msgid/everything-list/75aebb9c-106d-4f19-8236-a667b345681b%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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