On Thu, Mar 5, 2020 at 3:23 PM 'Brent Meeker' via Everything List < everything-list@googlegroups.com> wrote:
> On 3/4/2020 7:54 PM, Bruce Kellett wrote: > > On Thu, Mar 5, 2020 at 2:02 PM 'Brent Meeker' via Everything List < > everything-list@googlegroups.com> 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 < >> everything-list@googlegroups.com> 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 -- there is no "weight" > that differentiates different branches. 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. 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. > > > All branches occur with certainty so long as their weight>0. Yes, Everett > simply assumed they all occur. Take a simple branch counting model. > Assume that at each trial a there are a 100 branches and a of them are |0> > and b are |1> and the values are independent of the prior values in the > sequence. So long as a and b > 0.1 every value, either |0> or |1> will > occur at every branching. But almost all observers, seeing only one > sequence thru the branches, will infer P(0)~|a|^2 and P(1)~|b|^2. > > Do you really disagree that there is a way to assign weights or > probabilities to the sequences that reproduces the same statistics as > repeating the N trials many times in one world? It's no more than saying > that one-world is an ergodic process. > I am saying that assigning weights or probabilities in Everett, by hand according to the Born rule, is incoherent. Consider a state, |psi> = a|0> + b|1>, and a branch such that the single-world probability by the Born rule is p = 0.001. (Such a branch can trivially be constructed, for example, with a^2 = 0.9 and b^2 = 0.1). Then according to Everett, this branch is one of the 2^N branches that must occur in N repeats of the experiment. But, by construction, the single world probability of this branch is p = 0.001. So if MWI is to reproduce the single-world probabilities, we have with certainty a branch with weight p = 0.001. Now this is not to say that we certainly have a branch with p = 0.001; it is, rather, the conjunction of two statements: (a) the branch probability is p = 0.001, and (b) the branch probability is p = 1.0. These two statements are incompatible, so any assignment of weights to Everettian branches is incoherent. 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 everything-list+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLS%3DLTwRWURF_Tmv-3ThXNGE7TqwGo927sCdnYW5cAw_dw%40mail.gmail.com.
