On Thu, Aug 28, 2025 at 4:03 AM Bruce Kellett <[email protected]> wrote:
> 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 were discussing a case of this form: "This is easily seen if one considers a wave function with a binary outcome, |0> and |1> for example. After N repeated trials, one has 2^N strings of possible outcome sequences. One can count the number of, say, ones in each possible outcome sequence." If we are interested in statistics for N trials, let's define a "supertrial" as a sequence of N trials of the individual measurement, and say that we are repeating many supertrials and recording the results of all the individual trials in each supertrial using some kind of physical memory (persistent 'pointer states'). Each supertrial has 2^N possible outcomes, and for a given supertrial outcome O (like up, down, up, up, up, down for N=6) you can define a measurement operator on the pointer states whose eigenvalues correspond to what the records would tell you about the fraction of supertrials where the outcome was O. If I'm understanding the result in those references correctly, then if one models the interaction between quantum system, measuring apparatus, and records using only the deterministic Schrodinger equation, without any collapse assumption or Born rule, one can show that in the limit as the number of supertrials goes to infinity, all the amplitude for the whole system including the records becomes concentrated on state vectors that are parallel to the eigenvector of the measurement operator with the eigenvalue that exactly matches the frequency of outcome O that would have been predicted if you *had* used the collapse assumption and Born rule for individual measurements. And this should be true even if the probability for up vs. down on individual measurements was not 50/50 given the experimental setup. Jesse -- You received this message because you are subscribed to the Google Groups "Everything List" group. 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