On Thu, Mar 5, 2020 at 3:23 PM 'Brent Meeker' via Everything List < [email protected]> 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 < > [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]> 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 [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLS%3DLTwRWURF_Tmv-3ThXNGE7TqwGo927sCdnYW5cAw_dw%40mail.gmail.com.
