On Thu, Aug 28, 2025 at 12:49 AM Bruce Kellett <[email protected]> wrote:
> On Thu, Aug 28, 2025 at 2:30 PM Brent Meeker <[email protected]> > wrote: > >> On 8/27/2025 5:07 PM, Bruce Kellett wrote: >> >> On Thu, Aug 28, 2025 at 5:16 AM Brent Meeker <[email protected]> >> wrote: >> >>> I think some specificity would help this debate. Suppose N=6, so there >>> are 64 different sequences in 64 different worlds. The number of observers >>> is irrelevants; we can suppose the results are recorded mechanically in >>> each world. Further suppose that a=b so there is no question of whether >>> amplitudes are being respected. Then in one of the worlds we have 011000. >>> Per the Born rule its probability is 0.2344. In MWI it is 1/64=0.0156. >>> The difference arises because the observers applying the Born rule looks at >>> it as an instance of 2 out of 6 successes. >>> >> >> The trouble with this is that you are treating this as an instance of >> Bernoulli trials with probability p= 0.5. When every outcome occurs with >> every trial we no longer have a Binomial distribution The Binomial >> distribution assumes that you have x successes out of N trials. In the >> Everettian case you have one success on every trial. >> >> So your probability above for 2 successes applies to Bernoulli trials >> with one as the 'success'. The thing is that the probability of getting a >> zero is also 1/2, so we also have four successes out of six trials in your >> example. The Binomial probability for this result is also 0.2344. Actually, >> if we regard this experiment as a test of the Born rule, we have four zeros >> in 6 trials, which gives an estimate of the probability as 4/6 = 0.667, or >> as two ones in 6 trials which gives an estimate of the probability as 2/6 = >> 0.333, neither estimate is equivalent to |a|^2 = 0.5. The difference >> becomes more pronounced as N increases. The problem with your analysis is >> that you are assuming a binomial distribution. and we do not have any such >> distribution. >> >> That's how an experimenter will compute the probability of 2 successes in >> 6 trials given p=0.5. Yes, the probability estimate is 0.33 given 2 in >> 6. The probability of the result given p is not generally the same as the >> estimate of p given the result. The former is binomial distributed. The >> estimates of p aren't part of a distribution since they don't add up to 1. >> > > You are still assuming that the results follow a binomial distribution. > This is not the case. As I said, in the Everett situation, when every > outcome is realized in every trial, you have one success in every trial, > and the resulting distribution is not binomial. The normal approximation > for large N, as Jesse seems to assume, simply does not hold, since the > distribution is not binomial. > I don't assume anything myself, I point to some sources by physicists below indicating there is a mathematical derivation of the fact that if physical records ('pointer states') of a series of N trials are kept, and you define a measurement operator that only tells you the fraction of those recorded trials with some result, then one can show that the deterministic equation of wavefunction evolution (without collapse assumption/Born rule) implies that the quantum state of the records approaches the "correct" eigenstate of this measurement operator (the one that matches what would be predicted using the Born rule) in the limit as N approaches infinity. https://books.google.com/books?id=_HgF3wfADJIC&lpg=PP1&pg=PA238 https://www.academia.edu/6975159/Quantum_dispositions_and_the_notion_of_measurement > > 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/CAFxXSLQ_N1_wYg7aQBMabdigBEnR4uRJuArSC0Kiv26brfzDOg%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLQ_N1_wYg7aQBMabdigBEnR4uRJuArSC0Kiv26brfzDOg%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/CAPCWU3JbBM4rSehGVwE0Hj5kY%3DWP9yshKiyouKFd%3Dx0QkA9nFg%40mail.gmail.com.
