On Thu, Mar 5, 2020 at 11:59 AM 'Brent Meeker' via Everything List < [email protected]> 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. I suspect that you are mixing the 1p and 3p viewpoints. Or else you are using the expectation for a single outcome per trial (not that for which both outcomes occur on every trial.) Bruce > and Var(number|0>) = abN. The fraction x within one std-deviation of > the expected number is a constant > > F( a-sqrt[ab/N]<x<a+sqrt[ab/N])=1/e > > So that fraction become more an more sharply confined around a as N->oo. > > Brent > -- 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 on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLQRTO79k1M5P8LcQLJBqWhop_Hw9Ti6%2BrnTjR38yuT0CQ%40mail.gmail.com.
