On Tue, Jun 27, 2017 at 05:09:49PM +1000, Bruce Kellett wrote: > On 27/06/2017 10:21 am, Russell Standish wrote: > >No, you are just dealing with a function from whatever set the ψ and ψ_α > >are drawn from to that same set. There's never been an assumption that > >ψ are numbers or functions, and initialy not even vectors, as that > >later follows by derivation. > > psi(t) is an ensemble, psi_a is an outcome.
ψ_α is still an ensemble, just a sub-ensemble of the original. Assume α is the result (A has the value 2.1 ± 0.15). Then all universes where A is greater than 1.95 and 2.25 will be in the ensemble ψ_α. > The projection produces > the outcome from the observer moment psi. It maps the ensemble to > one member of that ensemble. No. See above. It always maps to an ensemble with an infinite number of members. > Certainly, psi is not a number or a > function, it is a set of possible outcomes: the psi_a are single > outcomes, be they numbers, functions or vectors, but the are not > just further ensembles. > > >>Thus, for the sum to make sense you must assume linearity. > >If you are objecting that the use of the symbol '+' implies linearity > >where no such thing is assumed, then feel free to replace it with the > >symbol of your choice. Then once linearity is established, feel free > >to replace it back again to + so that the formulae following D.8 have > >a more usual notation. Fine - that is a presentational quibble. My > >taste is that it is unnecessarily cumbersome, but if you find it helps > >prevent confusion in your mind, please do so. > > > >> Now > >>linearity is at the bottom of most distinctive quantum behaviour > >>such as superposition, interference, and entanglement. It is not > >>surprising, therefore, that if you assume linearity at the start, > >>you can get QM with minimal further effort. > >> > >Except that I don't assume linearity from the outset. > > There seems to be some confusion between outcomes of observations > and sets of possible outcomes. The \P_A*psi is actually defined as a > superposition in (D.2), ad you then seek to determine the > probability of this superposition? You define the probability of a > set of outcomes by P_psi(\P_A*psi), which is P_psi(\Sigma psi_a). I > find it hard to interpret what this might mean -- the probability of > a superposition of measurement outcomes (with equal weights, what is > more)? > The weights aren't equal. They're denoted P(ψ_α). > You then talk about this as though you were still partitioning sets, > but the probability is not defined on a set, only on a > superposition. If it is a set, then (D.2) makes no sense. > > You then introduce, quite arbitrarily, multiple observers for each > observer moment. This then gives you a measure, which is then made > to be complex!! The number of observers for each observer moment, > even if there can be more than one, which is not proved, cannot be > complex. Why? Give me one good reason - other than it doesn't match your intuition, which is generally not a good reason. > So your introduction of a measure, or weight for each > superposition really does not make sense. You then conclude that V, > the set of all observer moment, is a vector space over the complex > numbers. > That's right. If a linear combination of observer moments is also and observer moment, then the set of all observer moments is a vector space. This is linear algebra 101. > I remain baffled. You start with an observer moment as a set of > consistent possibilities. But there is no specification of what > 'consistent' might mean. There doesn't need to be a specification. All we need to know is that some worlds correspond to the one we see, and some don't. We don't need a constructive procedure for determining which worlds are to be included, and in all likelihood, no such constructive procedure will be found anyway. > There is also no particular structure > imposed on this observer moment It satisfies set axioms, otherwise you cannot apply Kolmogorov's probability axioms. I have been criticised for this particular assumption before, however the Nothing (the book, after all is called "Theory of Nothing") is a set above all. > , and you conclude, after a number of > obscure manipulations, that the set of all observer moments is a > vector space over the complex numbers. I look in vain for the magic > that converts an unstructured ensemble into a linear vector space. The "magic" IMHO is to consider that observers are also drawn from a distribution according to some measure, rather than just being a single observer. This is forced onto us by the Multiverse nature of assumed reality. An observer cannot see a∧b, where a and b are disjoint, but two different observers in different branches of the Multiverse can. > This is surely a non-trivial restriction on the nature of observer > moments, but you do not restrict the possible generality, you only > project particular (unstructured) results from this observer moment. > What, exactly, has gone on to extract linearity? > > That is why I think that you have actually built linearity in from > the start -- there is no mechanism, engine, or procedure that > extracts this linearity. And what happens to things in the ensemble > that are not linear? > > This does not, to me, pass the sniff test (or, in the Australian > vernacular, the pub test). Can you wonder that I am sceptical that > you have actually proved anything? > Yes, but then you've also exhibited some fairly deep misunderstandings of the argument, possibly because you haven't read the whole book, or alternatively, because my presentation is not sufficiently clear. But at least we can iterate on that, modulo your patience, until we arrive at a genuine disagreement :). -- ---------------------------------------------------------------------------- Dr Russell Standish Phone 0425 253119 (mobile) Principal, High Performance Coders Visiting Senior Research Fellow hpco...@hpcoders.com.au Economics, Kingston University http://www.hpcoders.com.au ---------------------------------------------------------------------------- -- 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. 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