Bruce, You keep treating the existence of all 2^N sequences as if it implied equal measure, but that conclusion is exactly what is under debate. Everett's framework does not equate existence with weight: amplitudes encode structural information about correlations and observer densities, and experiments confirm this via the Born rule.
Your argument sidesteps this entirely by assuming amplitudes are irrelevant, which is precisely the point you need to demonstrate, not assume. Quentin All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer) Le mer. 27 août 2025, 07:28, Bruce Kellett <[email protected]> a écrit : > On Wed, Aug 27, 2025 at 2:56 PM Quentin Anciaux <[email protected]> > wrote: > >> Bruce, >> >> Your reasoning is flawed because you conflate the existence of sequences >> with their measure. Yes, the set of 2^N sequences is the same regardless of >> a and b, but the amplitudes attached to them are not irrelevant >> bookkeeping. In Everett's framework, the squared amplitudes define the >> structure of the wavefunction and thus the density of observer-instances >> across sequences. >> > > The amplitudes play no role in the argument. Again, if you think that the > squared amplitudes give multiple observers per branch, then it is up to you > to demonstrate this mathematically. > > Setting a = b to argue that all sequences have equal weight and then >> generalizing that conclusion to arbitrary a and b is invalid. >> > > Prove that it is invalid! In the case of a = b, all sequences certainly > have equal weight. And the same sequences occur for any values of a and b. > > It ignores the very feature that makes different amplitudes significant: >> they change the relative contribution of each branch to future correlations >> and observed frequencies. You cannot dismiss this and still claim to be >> deriving anything about measure. >> > > You still seem to be thinking in terms of binomial distributions from N > trials with specific probabilities for success. > You have missed the significance of the fact that all outcomes occur on > each trial. > > If you want to argue that amplitudes have no bearing on measure, you need >> an independent justification for why quantum mechanics' core mathematical >> structure and its experimental validation via the Born rule should be >> discarded. Otherwise, you are assuming your conclusion. >> > > No, I am proving that the observed frequencies do not conform to the > expected Born probabilities. How do you think that the Born rule is > demonstrated in practice? > > Look, the argument is very simple. Given the 2^N binary sequences from N > trials, the number of zeros (ones) in each sequence varies, but the > amplitude of zero in the original wavefunction, a, remains the same. > Consequently, the expected Born probability is |a|^2 for every sequence. > But the proportion of success (zeros) differs between sequences, leading to > a contradiction. This argument is independent of the amplitudes for the > sequences, and the number of observers per sequence. These, and other > things, are just items that you have introduced as distractions. It is time > you actually considered the argument being made. > > 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/CAFxXSLRXquVqkS%3DL71C%3DBn6SRHRsv-HVPi_idhGSmTtrmUaduA%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLRXquVqkS%3DL71C%3DBn6SRHRsv-HVPi_idhGSmTtrmUaduA%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/CAMW2kAqgZ6iP-mHeiKD%2BZmJcRgVwbCs5VVdM%3DLfAz9ohsvmbGA%40mail.gmail.com.
