On 2/19/2025 11:16 PM, Quentin Anciaux wrote:
Le jeu. 20 févr. 2025, 07:55, Brent Meeker <[email protected]> a
écrit :
What does it mean to have more than one observer in a branch? A
branch forms because the result is orthogonal to the other
different results. Of course any number of persons can observe
it, which by construction means they are in that branch. Is that
what you mean?
Brent
Brent,
A branch isn't a single discrete entity—it's a region in the
wavefunction where decoherence prevents interference. The distinction
matters because if the wavefunction remains a superposition of
infinite components, then what we call a "branch" is just an
approximate partition, not a fundamental unit.
But it has only one result...right? That's what makes it a branch and
not two branches.
Saying "one observer per branch" assumes a sharp branching structure,
but if the wavefunction maintains an underlying continuous structure,
then what we experience as "branches" are really clusters of
high-measure observer instances. More than one observer in a branch
means that, within that region of the wavefunction, there are
exponentially more instances of an observer experiencing a
high-amplitude outcome than a low-amplitude one.
In other words, observer count isn't tied to branch count—it's tied to
measure.
Which is the number of observers, each of which is in a branch
orthogonal to all other branches...right?
Brent
What matters isn’t how many "branches" exist but how many copies of an
observer exist in each, which is why most observers see outcomes
aligning with the Born rule.
Quentin
On 2/19/2025 10:48 PM, Quentin Anciaux wrote:
If you assume one observer per branch, you assume what you want
to prove: that measure plays no role.
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