Bruce, Consider the following thought experiment, which directly parallels MWI and illustrates why your argument assumes what it tries to prove.
Imagine we have a machine that can perfectly duplicate an observer, just as MWI implies happens during quantum measurements. The experiment works as follows: 1. The observer enters a sealed box. 2. Inside the box, they are duplicated into 10 copies. 3. Each duplicate is placed in an identical room with only one visible difference: One of them sees a 0 written on the wall. Nine of them see a 1 written on the wall. 4. The observer, upon exiting the box, can only report what they personally experienced. At each step, both events (seeing 0 and seeing 1) occur—so in a naive branch-counting view, after trials, we should get total sequences. However, the key insight is that these sequences are not equiprobable because the number of observers seeing each outcome is different at each step. After many trials, the vast majority of observer sequences will contain approximately 90% ones and 10% zeros, despite every individual trial having both outcomes. The reason is simple: there are exponentially more observers who experienced the outcome "1" than those who experienced "0". Now, compare this directly to MWI and the Born rule: Your argument assumes that all sequences in MWI should contribute equally to probability estimates, independent of amplitudes. But this ignores the role of measure—just like in our duplication experiment, some sequences have far more observers experiencing them than others. The amplitudes of the wavefunction dictate how many copies of an observer exist in each branch, meaning low-amplitude branches contain exponentially fewer copies of an observer. Consequently, an observer drawn at random from all existing copies will overwhelmingly experience frequencies aligned with the Born rule, not a uniform distribution. By assuming all sequences contribute equally, you implicitly assume that the amplitude does not matter—which is exactly what needs to be proven, not assumed. The duplication analogy makes it clear: even with deterministic, mechanical duplication, sequences are not equiprobable when observer count is taken into account. If your logic were correct, then our duplication experiment should also result in uniform distributions of 0s and 1s across sequences, which is clearly false. The fact that the amplitudes affect observer frequencies is exactly what leads to Born-rule probabilities in MWI, rather than a naive uniform distribution. Quentin Le mar. 18 févr. 2025, 01:09, Bruce Kellett <[email protected]> a écrit : > On Tue, Feb 18, 2025 at 11:01 AM Quentin Anciaux <[email protected]> > wrote: > >> >> You didn’t prove that MWI is inconsistent with the Born rule, you assumed >> it by asserting that all 2^N sequences contribute equally, which is not how >> MWI works. The amplitude coefficients do matter, they determine the measure >> of each sequence, which affects the relative frequency of observed outcomes. >> >> Your argument rests on the assumption that sequences exist independently >> of their amplitudes, but you haven’t justified why the observer should >> expect a uniform distribution rather than one weighted by the >> wavefunction’s structure. This is precisely the question that needs to be >> answered, not assumed away. >> > > The basic premise of MWI is that every possible outcome of an experiment > actually occurs, albeit on a separate branch with a separate copy of the > experimenter. This means that N trials on the binary state, give 2^N binary > sequences, covering all possible binary sequences of length N. This same > set of binary sequences is obtained for any values of the original > amplitudes. > > If you disagree with this simple mathematics, then I challenge you to > point out where it is wrong. And that does not mean just assuming that it > has something to do with the emplitues. The mathematics of the Schrodinger > equation says that they play no role in the formation of these 2^N > sequences. > > Bruce > > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > To view this discussion visit > https://groups.google.com/d/msgid/everything-list/CAFxXSLQ6fVGCZRQsFRcVyjbmktK6ozOnKpCcN1%3Did-U2YmZqxA%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLQ6fVGCZRQsFRcVyjbmktK6ozOnKpCcN1%3Did-U2YmZqxA%40mail.gmail.com?utm_medium=email&utm_source=footer> > . > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/CAMW2kAoF13tzJsrfuaTVca9V5q2Jh3oF%3D%3D4w_9L9jTVDophVtA%40mail.gmail.com.
