Bruce, You’re conflating two different levels: the combinatorial existence of 2^N sequences and their statistical weight. Yes, the Schrodinger equation produces all 2^N sequences regardless of amplitudes, but the amplitudes control the measure over those sequences. Changing |a|² and |b|² doesn’t alter which sequences exist, it alters how observers are distributed among them. That’s exactly what the Born rule captures: almost all self-locating observers end up in high-measure branches. Treating all sequences as equally weighted is equivalent to replacing quantum mechanics with flat branch counting, which is not what the formalism prescribes.
Quentin All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer) Le mar. 26 août 2025, 10:02, Bruce Kellett <[email protected]> a écrit : > On Tue, Aug 26, 2025 at 4:56 PM Quentin Anciaux <[email protected]> > wrote: > >> Bruce, >> >> The measure isn’t something I’m inventing, it’s implicit in the squared >> amplitudes of the wavefunction. The Schrödinger equation preserves the L² >> norm, and decoherence ensures that branches with extremely low amplitudes >> contribute negligibly to observer statistics. Ignoring that structure and >> treating all branches as equally weighted is not quantum mechanics, it’s >> just branch counting under a flat prior. >> > > The trouble with that idea is that the Schrodinger equation is insensitive > to the amplitudes. You can change the amplitudes on the wavefunction with > binary outcomes and you get exactly the same set of 2^N sequences. So > actually all the sequences have the same weight -- construction from > repeated trials with both outcomes realized on every trial ensures that all > 2^N sequences occur with equal weight. So there is no way in which there > are low weight or low probability sequences. Anyway, such an argument fails > because you can change the Born probability of a zero at will, by changing > the amplitudes in the original wave function. And the proportion of zeros > in any sequence is an estimate of the probability of obtaining a zero. > These proportions change over sequences, so most will give a probability > estimate that disagrees with the Born probability. > > 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 [email protected]. > To view this discussion visit > https://groups.google.com/d/msgid/everything-list/CAFxXSLTRTqsqc6SNUrp%3DKj-epnrqj9KGDRK8wb%2Bvpfu0-zyTEQ%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLTRTqsqc6SNUrp%3DKj-epnrqj9KGDRK8wb%2Bvpfu0-zyTEQ%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]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/CAMW2kAqTwJ6cmYTRZQjKrSxXTvMQ7F1xo58a1fjwLmQN6TpUKQ%40mail.gmail.com.
