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.

Simply stating that each branch contains "one observer" 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.

Quentin

Le mer. 12 févr. 2025, 23:44, Bruce Kellett <[email protected]> a
écrit :

> On Thu, Feb 13, 2025 at 9:19 AM Quentin Anciaux <[email protected]>
> wrote:
>
>> Bruce,
>>
>> You insist that no matter what is added to MWI, it cannot recover the
>> Born rule. But that’s not an argument, it’s a claim.
>>
>
> It is the conclusion to an argument.
>
> The fact that multiple approaches attempt to derive it suggests the issue
>> is far from settled. Dismissing them without engaging with their reasoning
>> does not refute them. (I still have to read the links from Brent)
>>
>
> If you show that something cannot work, it is no longer necessary to show
> why every individual attempt fails. Have you never heard of an
> impossibility theorem?
>
>
> You also say you don’t know what it would mean to "count sequences" yet
>> your argument relies on using observed frequency in a single binary
>> sequence to estimate probability. But this only works if all sequences
>> contribute equally to experience, which is precisely the assumption you
>> claim not to make.
>>
>
> Such an idea makes no sense, and it plays no role in my argument. You get
> an estimate of the probability of getting a zero by counting the number of
> zeros in your binary string. Statistics 101.
>
> If every observer in MWI experiences a sequence independent of its
>> amplitude, then we should observe a uniform distribution of outcomes rather
>> than Born-rule statistics.
>>
>
> Rubbish. You clearly don't know what you are talking about. There is an
> independent copy of the original observer on every branch -- on every
> sequence.
>
> That does not happen, which suggests that something in the structure of
>> MWI suppresses low-measure branches from dominating experience. (See
>> lottery example with multiple printing of same number)
>>
>> You ask why the same 2^N sequences appear regardless of the initial
>> amplitudes. That’s expected, unitary evolution does not prevent sequences
>> from existing, all do exists exhypothesi.  But the real question is whether
>> all sequences contribute equally to observer experience.
>>
>
> There is an independent copy of the observer observing every sequence.
>
> If you believe MWI fundamentally cannot account for the Born rule, you
>> need more than dismissals. Naive sequence (aka branch) counting is not
>> correct to infer probability without measure of such sequence in the set.
>> Your argument that measure has no effect remains unsupported imo.
>>
>
>  No-one here is doing branch counting (except, perhaps you!)
>  And you really do need to do the math to see that measure (or the weight
> of each branch) has no effect. The fact that you do not understand that the
> N trials give the same 2^N sequences, whatever the amplitudes, might
> explain why you think that my claim is unsupported.
>
> Bruce
>
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