On Thu, Aug 28, 2025 at 3:52 PM Jesse Mazer <[email protected]> wrote:
> 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 > It is not clear how this applies to the case under discussion. Could you be more specific? 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/CAFxXSLTnvKm0pNqoDBevWW1a4qsgYnvY-mpUXMO%2BhbuvVPC51w%40mail.gmail.com.
