On Sun, Feb 23, 2020 at 4:30 PM 'Brent Meeker' via Everything List < everything-list@googlegroups.com> wrote:
> On 2/22/2020 7:11 PM, Bruce Kellett wrote: > > > The trouble is that such a procedure is entirely arbitrary. The only > probability that one could objectively assign to say, W, on each Bernoulli > trial is one, since W certainly occurs for each trial. In other words, > there is no natural probability associated with this duplication process, > so imposing one is ad hoc and arbitrary. > > In MWI, it seems that Carroll gives the conventional answer -- weights are > arbitrarily assigned to branches according to the branch amplitude (modulus > of the coefficient squared). This is arbitrary too, designed merely to give > the same answer that is naturally obtained in the single world case. What I > object to about this is not only its arbitrariness, but also the fact that > it is advertised as a "derivation" from the SWE, when it is no such thing. > It is arbitrarily imposed. > > > It's imposed. But it's hardly arbitrary. First, it agrees with > experiment. Second, Gleason's theorem shows it's the only mathematically > consistent measure that could be used as a probability. And Zurek tries to > prove it's implied by symmetry considerations. > It is imposed in such a way as to agree with experiment, yes. But that is how the Born rule was arrived at in the first instance. Consequently, MWI is no better than Copenhagen in this respect. Gleason's theorem is only of use if you first assume that there is a probability distribution -- that the theory is probabilistic. But Everett is deterministic, not probabilistic. Zurek tries to explain it by symmetry considerations. But the basis of his argument is the assumption that equal amplitudes imply equal probabilities. Since the amplitudes do not affect the Everett branching, this is ad hoc -- not a derivation. 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/CAFxXSLT-eCT5NePLEZVSw%3Dc%2BYafeNoW8nCYvLJQtHeDiEe4Gwg%40mail.gmail.com.
