On 3/5/2020 10:07 PM, Bruce Kellett wrote:
On Fri, Mar 6, 2020 at 11:33 AM Bruce Kellett <[email protected]
<mailto:[email protected]>> wrote:
On Fri, Mar 6, 2020 at 11:08 AM 'Brent Meeker' via Everything List
<[email protected]
<mailto:[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.
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 .
Multiplying the number of zero branches at each fork does not change
the statistics within individual branches.
Yes it does. 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.
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.
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 can see that I'm going to have to write a program to produce and
example for you.
Brent
Consequently, the majority of observers will conclude that the Born
rule is disconfirmed. This is not in accordance with observation, so
Everett fails as a scientific theory -- it cannot account for our
observation of probabilistic results.
Bruce
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