On Thu, Mar 5, 2020 at 12:51 PM 'Brent Meeker' via Everything List < everything-list@googlegroups.com> wrote:
> On 3/4/2020 5:25 PM, Bruce Kellett wrote: > > On Thu, Mar 5, 2020 at 11:59 AM 'Brent Meeker' via Everything List < > everything-list@googlegroups.com> wrote: > >> On 3/4/2020 4:34 PM, Bruce Kellett wrote: >> >> >> The crux of the matter is that all branches are equivalent when both >> outcomes occur on every trial, so all observers will infer that their >> observed relative frequencies reflect the actual probabilities. Since there >> are observers for all possibilities for p in the range [0,1], and not all >> can be correct, no sensible probability value can be assigned to such >> duplication experiments. >> >> The problem is even worse in quantum mechanics, where you measure a state >> such as >> >> |psi> = a|0> + b|1>. >> >> When both outcomes occur on every trial, the result of a sequence of N >> trials is all possible binary strings of length N, (all 2^N of them). You >> then notice that this set of all possible strings is obtained whatever >> non-zero values of a and b you assume. The assignment of some propbability >> relation to the coefficients is thus seen to be meaningless -- all >> probabilities occur equal for any non-zero choices of a and b. >> >> >> But E(number|0>) = aN >> > > Where does this come from? The weight of each branch is a^x*b^y for a > branch with x zeros and y ones. > > But this weight is external to the branch, and the 1p probability > estimates from within the branch are necessarily independent of the overall > coefficient. The expectation for the number of zeros within any branch > depends on the branch, but is independent of both a and b. > > > Sorry, I see I didn't make it clear I was assuming the Born rule. I was > just pointing out that this makes an assignment of probabilities to the > multiple worlds which is the same as looking at a single world as a member > of an ensemble. > So you are taking the probability of each branch as (a^x*b^y)^2. (Note that a and b are amplitudes, not probabilities, so your expectation above should presumably be E(#0) = a^2*N. For a = b = 1/sqrt(2), this just means that the expected number of zeros equals the expected number of ones, namely, N/2. Which is rather trivial, given that there is exactly one zero and one one on each trial -- independent of the amplitudes! But one cannot just assume the Born rule in this case -- one has to use the data to verify the probabilistic predictions. And the observers on the majority of branches will get data that disconfirms the Born rule. (For any value of the probability, the proportion of observes who get data consistent with this value decreases as N becomes large.) 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 on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLSJPDHjtjawS5Wk7vrCncQTgAo8UR5y6q8DiURbN3d0pA%40mail.gmail.com.
