On Tue, Feb 18, 2025 at 10:11 PM Quentin Anciaux <[email protected]> wrote:
> Bruce,
>
> You’re misrepresenting my example. There are still only two possible
> outcomes per trial—seeing 0 or seeing 1. The key difference is that the
> number of observers in each outcome is not equal, which is precisely what
> happens in MWI due to the wavefunction amplitudes.
>
> Your argument relies on counting sequences alone, assuming that all 2^N
> sequences contribute equally to probability estimates. But this is exactly
> the point of contention. In MWI, amplitude determines the number of
> observer instances experiencing each sequence, just as in my thought
> experiment.
>
> Your claim that "each observer sees only one set of data" is trivial—it’s
> true in any probability framework. The issue is why most observers will see
> outcomes that align with the Born rule rather than a uniform 50/50 split.
>
> You state that after N trials, we get all 2^N sequences, regardless of
> amplitudes. But MWI does not say that each sequence has the same
> measure—only that each exists. The assumption that each sequence carries
> equal weight is what you need to prove, not assume.
>
> Your objection about my example "having ten outcomes" misrepresents the
> analogy. There are still only two possible outcomes per trial—the
> difference is that some outcomes have exponentially more observers
> experiencing them. This is exactly how amplitude influences measure in MWI.
>
> Your assertion that "one and only one copy of each outcome is produced"
> contradicts MWI’s entire framework. Unitarity doesn’t create a single copy
> per outcome—it preserves the entire superposition, which includes weighted
> branches.
>
> The fact that amplitudes affect observer frequencies is exactly what leads
> to Born-rule probabilities in MWI, rather than a naive uniform distribution.
>
> I'm afraid the sophistry is on your end, there are only two outcomes in my
> observer duplication outcome, seeing 0 or seeing 1.
>
> Your reasoning contains a categorical error: you misinterpret each copy as
> a "distinct outcome," whereas in reality, all copies in the same state see
> the same result. This does not create 10 outcomes but only two, with an
> asymmetric distribution of observers.
>
> Your reasoning is essentially saying "everything that exists must count
> equally," which is a circular assumption and has never been a property of
> quantum mechanics.
>
> So no, there are only two possible outcomes in my example, and your
> objection is invalid.
>
Starting from the two-component state that I have been talking about:
|psi> = a|0> + b|1>
and using only unitary evolution and the Schrodinger equation, I invite you
to prove mathematically the various claims that you have been making.
Bruce
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