On Tue, Aug 26, 2025 at 10:30 PM Quentin Anciaux <[email protected]> wrote:
> Bruce,
>
> You keep assuming one observer per sequence and uniform sampling, but that
> assumption is yours, not Everett’s. In Everett’s framework, the relative
> weights aren’t arbitrary — they’re determined by the amplitudes in the
> wavefunction. By rejecting that, you’re refuting a simplified model of your
> own making, not MWI itself. If your argument truly applied to Everett’s
> theory, you should be able to show how it addresses the role of measure
> instead of ignoring it.
>
I think you need to actually work through the calculation/construction. One
observer per sequence is implied by Everett's theory. If you have an
initial state such as
a|0> + b|1>
and every possible result is realized in every trial: the first trial gives
you two branches, with results |0> and |1> respectively. The observer
splits so that there is a single observer on each branch. Then, the next
trial for the observer who saw |0> again causes a split, so we get two
further branches, 00 and 01, with a single observer on each branch.
Continuing this, we get branches 0000.., 0100..., 011,00... and so on, so
there are 2^N sequences (branches) after N repetitions. On each repetition,
the branch, and the associated observer again splits, so you get two new
branches, with a single observer on each. Consequently, a straightforward
Everettian construction proves that there is only one observer on each
branch, and there is no branch without a single observer.
If you disagree with this result, then it is up to you to give the
calculation that produces multiple observers on some branches.
Of course, each sequence in this process has an amplitude, coming from the
amplitudes in the initial state. For example, the sequence with u zeros has
an amplitude
a^u*b^{N-u}. But these are just amplitudes carried through. They do not
represent weights or probabilities until you impose Born's rule, and we are
not doing that in this calculation. The question of measure is answered by
the above construction: each sequence has the same measure.
It is important to realize that the same 2^N sequences are obtained
whatever the initial amplitude a, and b might be. Thus the set of sequences
does not come from the binomial theorem with a probability of success given
by |a|^2, because you get the same sequence for all values of a and b,
which is not the binomial theorem result. Equal measure over the sequences
is a result of the construction, not of some mathematical theorem related
to the binomial distribution.
Bruce
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