On 3/5/2020 3:33 PM, Bruce Kellett wrote:
On Fri, Mar 6, 2020 at 10:18 AM 'Brent Meeker' via Everything List
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<mailto:[email protected]>> wrote:
On 3/5/2020 2:01 PM, Bruce Kellett wrote:
On Fri, Mar 6, 2020 at 8:17 AM 'Brent Meeker' via Everything List
<[email protected]
<mailto:[email protected]>> wrote:
On 3/5/2020 3:07 AM, Bruce Kellett wrote:
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. 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.
What needs to be derived or postulated is a probability
measure on Everett's multiple worlds. I agree that it can't
be derived. But I don't see that it can't be postulated that
at each split the branches are given a weight (or a
multiplicity) so that over the ensemble of branches the Born
rule is statistically supported, i.e. almost all sequences
will satisfy the Born rule in the limit of long sequences.
Unfortunately, that does not work. Linearity means that any
weight that you assign to particular result remains outside the
strings, so data within each string are independent of any such
assigned weights. The weights would not, therefore, show up in
any experimental results. The weights can only work in a
single-world version of the model.
True. But the multiplicity still works.
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.
Brent
Bruce
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