On 2/24/2025 3:07 AM, Quentin Anciaux wrote:
Bruce,
Your argument still assumes what it wants to prove. Yes, the 2^N
sequences exist independently of the amplitudes, but the claim that
they all contribute equally to probability remains an implicit
assumption in your reasoning. The fact that the sequences themselves
don’t depend on a and b does not mean that their statistical
weights—how often an observer finds themselves in each sequence—are
independent of amplitudes.
Which means you need some different, separate theory about how "observer
distribution" is determined in a way that depends on the amplitudes.
Your core mistake is treating the observed sequences as primary while
ignoring the fact that the number of observer instances experiencing
each sequence is what matters. The amplitudes determine the measure of
each branch, meaning the vast majority of observers end up in
sequences that follow the Born rule.
Not necessarily a "vast majority". For a short sequence, small N, there
should be a lot of spread in the observed average. Only for large N
does the distribution become sharply peaked. "Follow the Born rule" is
a little ambiguous since it's probabilitistic. "Consistent with the Born
rule" is better.
You are simply assuming that observer distribution is uniform across
sequences, which contradicts the entire structure of quantum mechanics.
You mean it doesn't comport with the Born rule...and the Born rule is
empirically confirmed...which is the whole point. The Born rule must be
invoked to assign numbers of observers, aka probabilities, to the 2^N
sequences.
You keep insisting that different values of a and b still produce the
same sequences. Yes, the sequences are the same in a mathematical
sense, but that does not mean they are observed with equal frequency.
This is like saying that rolling a biased die produces the same
possible numbers as a fair die and then concluding that all outcomes
are equally likely. The actual distribution of observed frequencies
follows the amplitudes squared, as the Born rule predicts.
If you believe MWI contradicts the Born rule,
It doesn't contradict it unless you take it as the whole of the theory,
without the Born rule. If you don't include the Born rule, or an
equivalent postulate, then you are left with nothing but counting the
branches, which doesn't work for a=/=b.
Brent
publish a formal proof. Otherwise, repeating the same flawed argument
does not make it more correct.
Quentin
Le lun. 24 févr. 2025, 11:49, Bruce Kellett <bhkellet...@gmail.com> a
écrit :
On Mon, Feb 24, 2025 at 5:31 PM Quentin Anciaux
<allco...@gmail.com> wrote:
Bruce,
Your assertion is absurd. The existence of 2^N sequences is
not in question—it's a direct prediction of MWI. The issue is
whether all sequences contribute equally to probability, which
they demonstrably do not, as experiments confirm the Born
rule, not a uniform distribution.
You are persisting with this idea that my argument is that all the
2^N "sequences contribute equally to probability". But I have
never made any such claim, and such an idea is completely beside
the point. The first important point comes after I mention the 2^N
binary sequences. I point out that these come from the binary
state a|0> + b|1>, but the sequences themselves are independent of
the amplitudes a and b. This means that if you repeat N trials
with different amplitudes a and b, you will get exactly the same
2^N sequences. The point of this is that the amplitudes themselves
play no role in the formation of the experimentally observed
binary sequences.
You must remember that the sequences of UP and DOWN (or zero and
one in my coding) are the data that any experimentalist measuring
the spin projections of N spin-half atoms will get. They are the
data he/she will work with, and that data is independent of the
amplitudes.
The second point is that from that experimental data, the
experimentalist can get an estimate of the probability of
obtaining a zero. Say there are r zeros in someone's observed
sequence. Then a good estimate of the probability of getting a
zero will be p = r/N. If one compares this with the Born rule
prediction, which is |a|^2, then in the majority of cases the
experimentalist will find the Born rule to be disconfirmed: his
estimate p = r/N will not equal |a|^2. Since the same applies for
any sequence in the set of 2^N: the value of r for different
experimentalists, will range from zero to one. But all have the
same Born rule probability, |a|^2. This is what happens when the
data from such an experiment is used for theory confirmation. MWI
is not confirmed, whereas the single world model is confirmed on
every occasion.
The problem is made more acute if you consider that we can take
completely different values for the amplitudes a and b, but we
will get the same sequences, and the same set of values of p =
r/N, even though the Born probabilities |a|^2 are wildly
different. Of course, there is a value of r in the set of
sequences that agrees with the Born probability, simply because
the set of sequences covers all possibilities, including the set
of zeros and ones that would be obtained by a single
experimentalist in a single world scenario. In the single world,
the Born rule is confirmed in every case, Though in the many-world
case, the majority of experimentalists will find that the Born
rule is violated.
Bruce
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