On Tue, Feb 18, 2025 at 6:18 PM Quentin Anciaux <[email protected]> wrote:
> Le mar. 18 févr. 2025, 07:44, Quentin Anciaux <[email protected]> a > écrit : > >> Le mar. 18 févr. 2025, 03:42, Bruce Kellett <[email protected]> a >> écrit : >> >>> On Tue, Feb 18, 2025 at 12:00 PM Quentin Anciaux <[email protected]> >>> wrote: >>> >>>> 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. >>>> >>> >>> It has occurred to me what is wrong with this example. Instead of >>> considering a two-outcome experiment, where we get either a zero or a one, >>> you have considered a ten-outcome experiment, with one zero and nine ones. >>> This is not equivalent to the binary case under consideration. In the >>> binary case, we get 2^N possible sequences after N trials, (not the 10^N >>> sequences as in your example). Because there are only two possible outcomes >>> in my example, the majority of sequences will have approximately a 50/50 >>> split of zeros and ones. The majority of observers are then going to use >>> their data to estimate a probability of 0.5 for getting a zero. Now this >>> bears absolutely no relation to the actual Born probability, which is >>> a^2.The majority of observers estimate p = 0.5, whatever the value of a. >>> This is because there are only 2^N possible binary sequences of length N, >>> and we get the same 2^N sequences whatever the values of a and b. That is >>> why I say the amplitudes have no effect. >>> >>> Bruce >>> >> >> There are only two possible outcome in my example, seeing 1 or seeing 0. >> > > Bruce, > > Your response misses the core issue. In my example, there are still only > two possible outcomes per trial: seeing 0 or seeing 1. > It is amazing what a little sophistry and a dose of self deception can get you! Ten copie means ten different outcomes. The fact that some labels are duplicated does not alter this fact. The key difference is that the number of observers in each outcome is not > equal, which is precisely what happens in MWI due to the wavefunction > amplitudes. > > Your claim that "each observer sees only one set of data" is trivial—it’s > true in any probability framework, classical or quantum. The issue is why > most observers will see outcomes that align with the Born rule rather than > a uniform 50/50 split. > You are missing the point. Each observer sees only one data set and that dta set gives an estimate of the probability. And those probabilities do not confirm the Born rule. You say that in the standard 2-outcome case, the majority of sequences will > have an approximately 50/50 split of zeros and ones. That is only true if > each sequence contributes equally, but this is exactly what’s in question. > In MWI, the amplitude determines the number of copies of an observer > experiencing each outcome, just as in my thought experiment. > Unfortunately, that is not true. Inthe two-outcome case, there are 2^N binary sequences of length N after N trials. That exhausts the possibilities, so you must get the same set of sequences for any input amplitudes a and b. So the amplitudes cannot give multiple copies as you assume. There are only ever 2^N distinct sequences for any input amplitudes. > You also claim that my example "assumes" the Born rule by setting 0.9 and > 0.1 at the outset. That’s not an assumption—it’s a demonstration of what > happens when different numbers of observers exist in different branches. > > My example does not assume the Born rule. It is an illustration of > asymmetric duplication, which mirrors how observer instances are > distributed in MWI. There are still only two possible outcomes per > trial—seeing 0 or seeing 1—but the number of observers experiencing each > outcome is not equal. > > Your argument hinges on the assumption that every sequence is equally > likely because all 2ⁿ sequences exist. But this only holds if each branch > contains the same number of observer copies, which is precisely what is in > question. > It is not in question in standard quantum mechanics, or in Everettian multiple worlds. There is no way that you get anything other than all 2^N sequences if all outcomes are produced on each trial. In my example, just like in MWI, the duplication process creates more > observer instances in one outcome than in the other. > But your example has ten possible outcomes - - you assume ten copies are made in each trial. That is just wrong, because it doesn't happen. The linearity of unitary QM ensures that one and only one copy of each outcome is produced on each trial. 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]. 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