On 2/12/2025 11:01 PM, Bruce Kellett wrote:
On Thu, Feb 13, 2025 at 5:57 PM Quentin Anciaux <allco...@gmail.com>
wrote:
Bruce,
You argue that MWI predicts a uniform distribution of outcomes
because all sequences exist and each branch contains exactly one
observer. Since experiments follow the Born rule instead, you
claim MWI is falsified. But this assumes that measure has no
effect—something you have not proven.
The fact that 2^N sequences exist does not mean they all
contribute equally to an observer’s experience. That’s the core
issue. If measure determines how many copies of an observer exist
in different branches, then high-measure branches dominate
experience. This would naturally lead to Born-rule frequencies,
without contradicting experiment.
Are you postulating more that 2^N sequences so the there can be more
than one sequence of a given proportion of 1's and 0's with one observer
each or are you postulating more than one observer in a given sequence?
The former corresponds to branch counting which JKC pointed out doesn't
work because it would imply retroactive changes in amplitudes based on
future measurement decisions.
Simply stating that each branch contains "one observer"
If a branch contains more that one observer they must still just observe
the same sequence. So they can't add to the weight of that sequence.
and that measure is irrelevant does not prove MWI is falsified—it
assumes your conclusion. If you want to show MWI is incompatible
with experiment, you need more than just claiming that measure
plays no role; you need to justify why quantum experiments
consistently match despite your assertion that all sequences
should be equally likely.
The fact is that you get the same 2^N binary sequences from the binary
state |psi> = a|0> + b|1> whatever the values of a and b. My case is
proven.
Bruce
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