On Fri, Mar 6, 2020 at 11:08 AM 'Brent Meeker' via Everything List < [email protected]> wrote:
> On 3/5/2020 3:33 PM, Bruce Kellett wrote: > > On Fri, Mar 6, 2020 at 10:18 AM 'Brent Meeker' via Everything List < > [email protected]> wrote: > >> On 3/5/2020 2:01 PM, Bruce Kellett wrote: >> >> On Fri, Mar 6, 2020 at 8:17 AM 'Brent Meeker' via Everything List < >> [email protected]> wrote: >> >>> On 3/5/2020 3:07 AM, Bruce Kellett wrote: >>> >>> there is no "weight" that differentiates different branches. >>>> >>>> >>>> Then the Born rule is false, and the whole of QM is false. >>>> >>> >>> No, QM is not false. It is only Everett that is disconfirmed by >>> experiment. >>> >>> Everett + mechanism + Gleason do solve the core of the problem. >>>> >>> >>> No. As discussed with Brent, the Born rule cannot be derived within the >>> framework of Everettian QM. Gleason's theorem is useful only if you have a >>> prior proof of the existence of a probability distribution. And you cannot >>> achieve that within the Everettian context. Even postulating the Born rule >>> ad hoc and imposing it by hand does not solve the problems with Everettian >>> QM. >>> >>> What needs to be derived or postulated is a probability measure on >>> Everett's multiple worlds. I agree that it can't be derived. But I don't >>> see that it can't be postulated that at each split the branches are given a >>> weight (or a multiplicity) so that over the ensemble of branches the Born >>> rule is statistically supported, i.e. almost all sequences will satisfy the >>> Born rule in the limit of long sequences. >>> >> >> Unfortunately, that does not work. Linearity means that any weight that >> you assign to particular result remains outside the strings, so data within >> each string are independent of any such assigned weights. The weights would >> not, therefore, show up in any experimental results. The weights can only >> work in a single-world version of the model. >> >> >> True. But the multiplicity still works. >> > > No, it doesn't. Just think about what each observer sees from within his > branch. > > > It's what an observer has seen when he calculates the statistics after N > trials. If a>b there will be proportionately more observers who saw more > 0s than those who saw more 1s. Suppose that a2=2/3 and b2=1/3. Then at > each measurement split there will be two observers who see 0 and one who > sees 1. So after N trials there will be N^3 observers and most of them > will have seen approximately twice as many 0s as 1s. > >From within any branch the observer is unaware of other branches, so he cannot see these weights. His statistics will depend only on the results on his branch. In order for multiple branches to count as probabilities, you have to appeal to some Self-Selecting-Assumption (SSA) in the 3p sense: you have to consider that the observer self-selects at randem from the set of all observers. Then, since there are more branches according to the weights, the probability that the randomly selected observer will see a branch that is multiplied over the ensemble will depend on the number of branches with that exact sequence. But this is not how it works in practise, because each observer can ever only see data within his branch, even if that observer is selected at random from among all observers, he will calculate statistics that are independent of any branch weights. 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 on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLSRNgiqwhz9bBBHtBa0U_%3Do-s5MNs6aVjguXRNByqunyw%40mail.gmail.com.
