On Sun, Sep 6, 2020 at 10:25 AM 'Brent Meeker' via Everything List < everything-list@googlegroups.com> wrote:
> On 9/5/2020 4:59 PM, Bruce Kellett wrote: > > > So why do you defend Carroll and Everett? Even self-locating uncertainty > is an essentially probabilistic idea. > > > I don't defend them. I criticize your argument against them because I > think it is unconvincing for the reasons I have given; essentially because > you cut off the MWI interpretation before the step in which it extracts > probabilistic statements by using self-locating uncertainty in the ensemble > of worlds. > I think that the only way this comment makes sense is if the number of worlds multiplies in proportion to the Born probabilities on each interaction. That is an even bigger departure from Everett than anything you might have accused me of doing. Let us revisit this problem using David Albert's example of Captain Kirk's transporter malfunction, so that when Kirk is beamed down to the surface of a planet, two "Kirks" arrive, one dressed in blue and the other in green. (One could make the same argument in terms of Bruno's WM duplication experiment.) If, after transportation, the Kirks re-ascend to the Enterprise and each copy again transports down: being duplicated in the same way. After N iterations, there are 2^N Kirks on the surface of the planet. If each carries a notebook in which he has recorded the sequence of colours of his outfits, all possible binary sequences of B and G will be recorded in some book or the other. A simple application of the binomial distribution shows that the notebook records peak around sequences showing approximately equal numbers of blue and green outfits. This is experimental verification of the probability of p(blue) = 0.5 = p(green). Now let us try to vary the probabilities, say to p(blue) = 0.9 and p(green) = 0.1. How do we do this? OK, we transport Kirk and, with probability p = 0.9, we colour one of the uniforms blue. The other must, therefore, be coloured green. But then we simply have two Kirks on the surface of the planet, one in a blue uniform and one in a green uniform -- exactly as we had before. It is easy to see that, no matter how we imagine that we have changed the relative probabilities of uniform colours, we must always end up with just one blue uniform and one green uniform. Our attempt to change the probabilities has failed. There is a way out, however. If, instead of simply duplicating the Kirks on transportation, the transporter manufactures 10 copies on the surface of the planet. Then we can suppose that 9 of these have blue uniforms, and the remaining Kirk is dressed in green. Iterating this procedure, we end up with 10^N Kirks on the surface of the planet, the vast majority of whom are dressed in blue. We have, thereby, changed the probability of a blue uniform for Kirk to 0.9 -- in the majority of cases. The trouble with this is that such a scenario cannot be reproduced with the Schrodinger equation. If the universal wave function is represented by a vector in Hilbert space, for a two-outcome experiment the Hilbert space is two-dimensional, and we cannot fit 10 independent basis vectors into such a two-dimensional space. So the multiple branches for each outcome solution is not available in quantum mechanics. We might be able to dream up a theory in which this multiplication of branches would work, but that is not Everett, and it is not quantum mechanics as we know it. (Carroll and Zurek attempt to get around this by expanding the dimensionality of the operative Hilbert space by "borrowing" degrees of freedom from environmental decoherence. I doubt that this is actually convincing, or even possible. Whatever, it is a hopelessly ad hoc violation of the underlying dynamics.) I can, therefore, see no way in which the Born rule can be made compatible with strictly deterministic Everettian Schrodinger evolution. Note that (pace Bruno) this conclusion does not depend on any 1p/3p confusion. It depends only on the details of the assumed dynamics. 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/CAFxXSLQZbZpG755LrRTV52OyOEO-rEDX9-izt%3DDY3N8saHyThQ%40mail.gmail.com.
