On Fri, Mar 6, 2020 at 5:22 PM 'Brent Meeker' via Everything List < [email protected]> wrote:
> On 3/5/2020 10:07 PM, Bruce Kellett wrote: > > On Fri, Mar 6, 2020 at 11:33 AM Bruce Kellett <[email protected]> > wrote: > >> 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: >>> >>> 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 >> > > To put this another way., if a=sqrt(2/3) and b=sqrt(1/3), then if an > observer is to conclude, from his data, that 0 is twice as likely as 1, he > must see approximately twice as many zeros as ones. This cannot be achieved > by simply multiplying the number of branches on a zero result. Multiplying > the number of branches does not change the data within each branch, > > > Sure it does. The observer is twice as likely to add on 0 branches to his > sequence of observations as to ad a 1 branch. So more observers will see > an excess of 0s over 1s. > The observer does not get to add branches to his sequence at will. Whether more observers see an excess of zeros or not does not affect what each individual observer sees. so observers will obtain exactly the same statistics as they would for > a=b=1/sqrt(2). As I have repeatedly said, the data on each (and every) > branch is independent of the weights or coefficients. This is a trivial > consequence of having every result occur on every trial. Even if zero has > weight 0.99, and one has weight 0.01, at each fork there is still one > branch corresponding to zero, and one branch corresponding to one. > > > That was Everett's original idea. But if at each trial there are 99 forks > with |0> and 1 fork with |1> then there will be many observers who have > observed only |0>'s after say 20 trials and few or none who will have > observed only |1>'s . > But it is not a question of how many observers see a particular string: the issue is what each observer sees from his own data. Since this is a deviation from Everett's relative state idea, you have departed from the Schrodinger equation, and have not really replaced it with a viable dynamical equation that will multiply branches in the required way. Multiplying the number of zero branches at each fork does not change the > statistics within individual branches. > > > Yes it does. > Think again -- that is just an absurd comment. Every time a zero occurs in a sequence, another identical sequence is added. That does not change anything within the sequence. There are, after all, only 2^N possible binary bit strings of length N. Whatever the observed sequence up to given trial, the observer is more > likely to add a |0> to his sequence on the next trial if there are more > zero branches. > As above -- the observer does not get to add anything to his sequence -- it is data that he is given. Actually, in the 2:1 ratio of zero branches to one branches, one will end up with 3^N branches in total (since each duplicated zero could be coded as 2, giving 3 branches to be added at each fork). And there is a separate observer for each branch. It is what these observers can infer from their data that is important -- not how many of them there are. And it is the data from within his branch that the physicist must use to > test the theory. Even if he is selected at random from some population > where the number of branches is proportional to the weights, he still has > only the data from within a single branch against which to test the theory. > Multiplying branches is as irrelevant as imposing branch weights. > > That is where I think the attempt to force the Born rule on to Everett > must inevitable fail -- there is no way that one can arrange fork dynamics > so that there will always be twice as many zeros as ones along each branch > (for the case a^2=2/3, b^2=1/3). > > > Not along each observed sequence. But there will be many more sequences > with twice as many zeros than sequences with other proportions. > > Yes, because rather than 2^N sequences, we will have many duplicates, so > that there are 3^N sequences (and observers). But the first-person > experience is still limited to the data within a sequence, not over > sequences. > In the full set of all 2^N branches there will, of course, be branches in > which this is the case. But that is just because when every possible bit > string is included, that possibility will also occur. The problem is that > the proportion of branches for which this is the case becomes small as N > increases. > > But not the proportion of branches which are within a fixed deviation from > 2:1. That proportion will increase with N. > I disagree. I can see that I'm going to have to write a program to produce and example > for you. > That might be interesting. I will attempt to do the calculation for the 2:1 ratio. 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]. 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