On Tue, Feb 11, 2025 at 11:27 PM Quentin Anciaux <[email protected]> wrote:
> Bruce, > > I'll still give it a try to get a discussion (dumb me). > > If your response boils down to "this is nonsense" and "you’re not clever > enough," then you’re not engaging with the actual argument. The question is > not whether the Schrödinger equation explicitly encodes the Born rule—it > does not, just as it does not encode classical probability either. The > question is whether MWI can recover the Born rule without adding collapse, > and there are multiple serious approaches to doing so. > > Your claim that "MWI does not match experiments because it cannot get the > Born rule" is just an assertion. The Schrödinger equation does evolve > amplitudes, and those amplitudes do determine the structure of the > wavefunction. You dismiss measure as meaningless, yet every quantum > experiment confirms that the statistics follow . If naive branch counting > were correct, experiments would contradict the Born rule—but they do not. > That means something in MWI must account for it. > > Saying "all branches exist equally" ignores what "equally" even means in a > probabilistic context. Probability is not about "some things happen while > others don’t"—that’s a description, not an explanation. Classical > probability arises because there are more ways for some outcomes to occur > than others. In MWI, the weight of a branch is not a degree of > existence—it’s a statement about how many copies of an observer find > themselves in that outcome. > > If you have a counterargument, provide one—just dismissing the approach as > "fantasy" without addressing the core point doesn’t make your position > stronger. If you want to argue that MWI cannot recover the Born rule, then > you need to explain why all proposed derivations (Deutsch-Wallace, Zurek’s > envariance, self-locating uncertainty, etc.) are fundamentally flawed, not > just assert that they don’t count. > Many others have pointed out the deficiencies of the arguments by Deutsch-Wallace, Zurek, and many others. The problems usually boil down to the fact that these attempts implicitly assume the Born rule from the outset. For example, as soon as you involve separate non-interacting worlds, and rely on decoherence to give (approximate) orthogonality, then you have assumed that small amplitudes correspond to low probability -- which is just the Born rule. Similar considerations apply to other arguments. The paper by Kent that I referenced earlier looks at many of the arguments and points out the many problems. As far as your basic argument goes, there is no evidence that the Schrodinger equation itself "evolves the amplitude", or that it gives different numbers of observers on branches according to the amplitudes. The idea of "branch weight" is just a made-up surrogate for assuming a probabilistic interpretation; namely, the Born rule. The position I am taking tries to avoid all these spurious additional assumptions/interpretations. We take the Schrodinger equation with the Everettian proposal that all outcomes occur on every trial, and see where that takes us. In the binary case, with repeated trials on similarly prepared systems, we get the 2^N binary strings. We get the same 2^N strings whatever amplitudes the initial wave function started with. There is only one copy of the initial observer on every such binary sequence. That observer can count the number of zeros in his/her string to estimate the probability. Since the string is independent of the amplitudes, the same proportion of ones will be found for the same string in every case. Since the Born probability varies according to the original amplitude, we find that this simplest version of many worlds is in conflict with the Born rule. Other conflicts with the Born rule are evident in other ways -- I have mentioned some of them previously. To go beyond this you have to introduce complications that are not inherent in the original Schrodinger equation and are largely incompatible with simple unitary state evolution. 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/CAFxXSLR50MgHVjm91s%3Dnz5urL98RSLGAC6FANmj23OMQbjOBXg%40mail.gmail.com.
