On Mon, Feb 10, 2025 at 9:41 PM Quentin Anciaux <allco...@gmail.com> wrote:
> Bruce, > > Yes, every possible experience is lived by some version of me in MWI, but > that does not mean all experiences are equally likely or subjectively > equivalent. The measure of a branch determines how many copies of me > experience a given outcome. In practice, my conscious experience will > overwhelmingly be shaped by the branches with higher measure, not by the > rare and improbable ones. > You cannot prove this. It is pure speculation. For example, if a quantum event has a 1% probability, then there will be > branches where I observe it, but they will be exponentially fewer than > those where I do not. The measure is not just an abstract number—it > reflects the relative weight of different outcomes in the wavefunction. > This is why, as an observer, I will almost always see frequencies matching > the Born rule, because the majority of my copies exist in branches where > this distribution holds. > No they don't. Your argument assumes that since all branches exist, they must be > equiprobable, but this ignores the fact that measure determines how many > copies of an observer exist in each branch. In a lottery, every ticket > exists, but some are printed in larger quantities. Saying "all branches > exist, so they must be equal" is as flawed as saying "all lottery tickets > exist, so all should win equally." > > Ultimately, my conscious experience is not determined by the mere > existence of branches, but by the relative number of copies of me in each. > Low-measure branches do exist, but they are not representative of my > experience. This is why MWI naturally leads to the Born probabilities, > without assuming collapse or introducing an arbitrary rule. > > Your reasoning collapses probability into mere branch-counting, but > probability is about where observers actually find themselves, not about an > abstract collection of sequences. > Like Russell, you have not even begun to understand the argument I am making. It has nothing to do with weights or the number of observers on each branch. Let me recast the argument. We have a binary wave function: |psi> = a|0> + b|1>. For convenience I have taken a spin-half system, or photon polarizations. Then we can use a = cos(theta) and b=sin(theta) so that a^ +b^2 = 1 is easily maintained and it is simple to rotate things to alter the magnitudes of the coefficients. Now we run N trials of measuring this system at some angle. Since the basic MWI principle is that every possibility is realized on every trial, we get 2^N sequences of results, covering all possible binary sequences of length N. Note particularly that we get exactly the same set of sequences for any angle theta. (We must, because there are only 2^N possible sequences.) The procedure is now to estimate the probability coefficient of the original wave function from our measured sequence (which is simply one of the 2^N). We do this by counting the number of zeros and/or ones in the sequence. Then p = n_zero/N The weight of the sequence, whatever it is, does not enter into this calculation of the probability, which is why I can reasonably take all sequences to have the same weight (although I do not do this, and it is not necessary). The point of this exercise is that the probability estimate that I get (p), is unlikely to be the Born probability which is a^2. As N becomes large, the law of large numbers implies that a large majority of the sequences will have approximately equal numbers of zeros and ones (independently of the coefficients a and b.). Consequently, the estimated probability will be 0.5 in nearly every case. This is only the Born probability for a set of angles of measure zero, so the majority of experimenters are going to find results that do not conform to the Born rule, and thus find that QM is disconfirmed. This follows directly from the requirement that every result be found on every trial ,which is an essential feature of MWI, so MWI is disconfirmed -- it is not a viable interpretation of QM. 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 everything-list+unsubscr...@googlegroups.com. To view this discussion visit https://groups.google.com/d/msgid/everything-list/CAFxXSLQshEssmB9vHes%2B7foaotfY08EDBdGTcfw7Ot78SUgBHQ%40mail.gmail.com.
