> On 5 Mar 2020, at 12:07, Bruce Kellett <[email protected]> wrote: > > On Thu, Mar 5, 2020 at 9:59 PM Bruno Marchal <[email protected] > <mailto:[email protected]>> wrote: > On 5 Mar 2020, at 04:54, Bruce Kellett <[email protected] > <mailto:[email protected]>> wrote: >> On Thu, Mar 5, 2020 at 2:02 PM 'Brent Meeker' via Everything List >> <[email protected] <mailto:[email protected]>> >> wrote: >> On 3/4/2020 6:45 PM, Bruce Kellett wrote: >>> On Thu, Mar 5, 2020 at 1:34 PM 'Brent Meeker' via Everything List >>> <[email protected] >>> <mailto:[email protected]>> wrote: >>> On 3/4/2020 6:18 PM, Bruce Kellett wrote: >>>> >>>> But one cannot just assume the Born rule in this case -- one has to use >>>> the data to verify the probabilistic predictions. And the observers on the >>>> majority of branches will get data that disconfirms the Born rule. (For >>>> any value of the probability, the proportion of observers who get data >>>> consistent with this value decreases as N becomes large.) >>> >>> No, that's where I was disagreeing with you. If "consistent with" is >>> defined as being within some given fraction, the proportion increases as N >>> becomes large. If the probability of the an even is p and q=1-p then the >>> proportion of events in N trials within one std-deviation of p approaches >>> 1/e and N->oo and the width of the one std-deviation range goes down at >>> 1/sqrt(N). So the distribution of values over the ensemble of observers >>> becomes concentrated near the expected value, i.e. is consistent with that >>> value. >>> >>> >>> But what is the expected value? Does that not depend on the inferred >>> probabilities? The probability p is not a given -- it can only be inferred >>> from the observed data. And different observers will infer different values >>> of p. Then certainly, each observer will think that the distribution of >>> values over the 2^N observers will be concentrated near his inferred value >>> of p. The trouble is that that this is true whatever value of p the >>> observer infers -- i.e., for whatever branch of the ensemble he is on. >> >> Not if the branches are unequally weighted (or numbered), as Carroll seems >> to assume, and those weights (or numbers) define the probability of the >> branch in accordance with the Born rule. I'm not arguing that this doesn't >> have to be put in "by hand". I'm arguing it is a way of assigning measures >> to the multiple worlds so that even though all the results occur, almost all >> observers will find results close to the Born rule, i.e. that self-locating >> uncertainty will imply the right statistics. >> >> But the trouble is that Everett assumes that all outcomes occur on every >> trial. So all the branches occur with certainty — > > In the 3p view, but then the “self-locating” idea explains that QM predicts > that the observers abstained do not see the “other branches” (“they don’t > even feel the split”, as Everett argued correctly). > > > But each individual can test the probability predictions from the > first-person data obtained on his branch. And most will find that the Born > rule is disconfirmed if Everett is true.
That is not in Everett, and Graham, like Hartle and Gell'man explain why we must not compute or infer the probability. Either you add the Born Rule, or you use the Paulette Février calculus, or Gleason theorem, to get the probabilities. The coefficient of the terms, when squared, gives the relative probabilities on the consistent histories. That is rather clear also in the work of Griffith and Omnes. > >> 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. The quantum formalism imposes such distribution, although in a relative way (which is threatening the notion of worlds, but this is no problem with mechanism, as world are pure phenomenological construct made by the relative numbers in arithmetic). > 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. Yes, you need to go beyond Everett, and compute the probabilities on all computations, and abandon the notion of worlds, and eventually this asks to abandon physicalism. > > (Except that we can’t use the universal wave no more, but then we do recover > it in arithmetic, like it was necessary, so no problem at all, except > difficult mathematics …). > > > > >> That is to assume that the branches occur with the probabilities that they >> would have in a single-world scenario. To assume that branches have >> different weights is in direct contradiction to the basic postulates the the >> many-worlds approach. > > Since the paper by Graham, nobody count the worlds by the distinguishable > outcome, but use Gleason or Kochen, or other manner to attribute a weighting. > > And that is contradicted by the data. ? > >> It is not that one can "put in the weights by hand"; it is that any >> assignment of such weights contradicts that basis of the interpretation, >> which is that all branches occur with certainty. > > > They all occur with certainty, but the formalism explain why, from the first > person perspective, they all occur with relative weighted uncertainties. > > > That is false. How many times do I have to prove to you that this does not > work. ? Bruno > > Bruce > > There is only “relative states”, some sharable, some non sharable. > > Bruno > > -- > 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] > <mailto:[email protected]>. > To view this discussion on the web visit > https://groups.google.com/d/msgid/everything-list/CAFxXSLS6HP-mMTsb8ScFDbs1Eb3ayb5ea%2BPw%3D-Hq_Jv9Q%3DMj%3DA%40mail.gmail.com > > <https://groups.google.com/d/msgid/everything-list/CAFxXSLS6HP-mMTsb8ScFDbs1Eb3ayb5ea%2BPw%3D-Hq_Jv9Q%3DMj%3DA%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 on the web visit https://groups.google.com/d/msgid/everything-list/4FDFD8FB-7D69-416D-BBDF-E7EAB89A3794%40ulb.ac.be.
