On Thu, Jun 29, 2017 at 03:19:40PM +1000, Bruce Kellett wrote: > On 28/06/2017 2:26 pm, Russell Standish wrote: > >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. > > That might be what you meant, but that is not what the average > reader (such as I am) is going to take from the text. You say the > projections divides the observer moment (set) into a discrete set of > outcomes (indexed by a). You then wish to calculate the probability > that outcome a is observed. Now observers observe one outcome -- > even if there is some associated measurement error, there is still > one outcome -- one observer does not see 2.0, 2.1, and 2.2. At > least, that would be a very strange way of talking about an > observation. Other observer might see that, and one observer might > see a range of results on repeated measurements, but that is not > what you appeared to be talking about. >
Any measurement on a continuum will have an uncertainty. A measurement of 2.0 ± 0.1 will be compatible with an infinite variety of observer moments, observing the value in the range 1.9 to 2.1. The average reader should remember that, at least if they're scientist. > > >>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. > > Not necessarily. A photon polarization measurement is dichotomous -- > you see a photon downstream of the polarizer, or you do not. No > uncertainty involved. > A discrete partitioning of the "Nothing" will still involve infinite sized ensembles. Just in this case, there is not way of disciminating them - with this observable at least. > >> 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). > > Again, that is not what the text says; (D.2) is a sum over outcomes > with equal weights. Yes - at D.2, there is no possibility of weighting - we're just aggregating disjoint outcomes. > > I find the notation confusing again. You have A contained in S, with > probability P_psi(\P_A*psi). A is original defined as an observable, > which divides the observer moment into a set of discrete outcomes. > But S is the set of possible outcomes: a is a member of S, so A > contained in S seems to be a different A -- the operator is not a > subset of the outcomes. To make something out of this, I took the > latter use of A to be the set of possible outcomes a -- not every > operator has the same set of possible outcomes. > Yes - you are right, it is confusing. A is used for two unrelated concepts. Initially it is the observable in paragraph 1, then it is related to the notion of an event coming in from the Kolmogorov axioms. After the first paragraph, A never refers to an observable again - so a different letter would be appropriate, if indeed a letter is needed at all. > >>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. > > People/observers can be complicated beings, but that does not mean > that you can have a complex number of them. Since all you are really > wanting, ISTM, is to give each observer moment a weight. The n umber > of observers observing tis moment seems a rather arbitrary source > for this weight, because it is not ever determinable. Why not just > give a weight, which can be complex if you wish, but it has nothing > to do with multiple observers. > Yes - that is exactly what I do. The talk around numbers of observers is to motivate the move to drawing them from some measure. > >>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. > > Again, that is not what your text says. You say that linearity comes > from the additive property of measure, and that is really what (D.7) > appears to be about. Except that it is not the additivity of measure > that is doing the work there, it is the additivity of probabilities > for disjoint observations (when the probability measure is > normalized to unity). > > I think you have to do more that just asserting that a linear > combination of observer moments is also an observer moment. The > notion of an observer moment has become opaque. An observer moment > is the set of possibilities consistent with what is known at that > point in time. So it is complete in itself -- how can you add two > observer moments? You clearly cannot add them for a single observer, > because adding two moments in time is not a defined operation -- it > would not be an observer moment since no observer observes two > moments in time simultaneously. Other observers at that time? Again, > if you add observer moments for different observers, you have no > guarantee that there is another observer who has just this > combination of possibilities consistent with what they know at that > point in time. That would have to be proved, rather than just > asserted. In fact, ISTM that such a result would require that every > observer knows everything at every time, and that everything that is > ever possible is part of the set of things consistent with what is > known by that observer at that point in time -- and the notion of > observer moments become otiose. > It is the passage 1p -> 1p plural -> 3p, where 3p is all observer moments, not any single one. That could be expressed more clearly. > Even if you widen definitions to this extent, it does not follow > that the projections for observable A are linear -- if you say that > there will always be an observable C, such that the set of outcomes > for A plus the set of outcomes for B, will be contained in the set > of outcomes for C. And even if you do want to claim this, such a > result would be of no use for quantum mechanics, since the important > thing there is that the set of outcomes for operator A in themselves > form a possible basis for a linear vector space. In QM, linear > combinations of the eigenvectors of one operator, if independent and > complete, can form a set of eigenfunctions for a different operator > in the same linear space. But that condition is not met by just any > two arbitrary operators. > The set of outcomes of any observable X will be the "certain event". The projections over these sets of outcomes will just be the identity. Adding the projection over the certain event for two observable A & B will just be twice the projection over the certain event for observable C. However, I suspect this is not what you're alluding to. "But that condition is not met by just any two arbitrary operators." My knowledge of functional analysis is some 3 decades old, but what I remember is that this is only possible for unbounded linear operators. All bounded operators have a set of eigenfunctions that span the whole linear space, ie is complete (and independent, obviously). Now it is true that classical quantum theory uses unbounded operators with gay abandon, much to the horror of mathematicians, who tend to be rather more careful about such things. x and ∂/∂x being classic examples. My own attitude is the x and ∂/∂x are not actually physical observables, as they correspond to observing position and momentum to infinite precision. When you modify the operators to take finite measurement resolution into account, you end up back with a bounded operator. But maybe I'm missing your point. > I understand that in your theory, an observer moment must be shown > to be a Hilbert space, ^ a vector in a Hilbert space > possibly a product Hilbert space -- a > separate component for different classes of operator: in QM the > Hilbert space for spin measurements on an electron is not the same > Hilbert space as that for position measurements. In the product > space, linearity is required in each term of the vector product. I > don't see that you have established this. > I'm guessing you mean tensor product, or cartesian product here. The vector product a × b is a 3D vector at right angles to a and b, with magnitude |a||b|sin(θ). I haven't established this, because it is not needed to make contact with the regular set of axioms assumed in quantum theory. But to the extent it can be shown within regular QM, it will be shown within my theory. > > >>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. > > That seems as though you ultimately rely on observation. But if you > cannot make predictions about what might or might not be observed, > then you don't have a scientific theory. Because the aim of science > is to constructively predict what we might see. So you seem to be > claiming that science is impossible, and if possible, useless (we > don't need such a constructive procedure). > Quite the opposite, actually. Science can be considered as the process of finding laws, or regularities in observed data. The various Occam's razor theorems I considered earlier in the book indicate that these regularities exist, and are even rather common. Furthermore, Bayesian inductive schemes will find them quite effectively (not that I discuss that in the book, but it is well known in computer science). However, that is rather different to having an effective procedure for determining which infinite length descriptions correspond to my current observer moment. > >> 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. > > A set consisting of what? > Of infinitely long strings, drawn from your alphabet of choice (binary is nice), that I call "descriptions" in my book (to the annoyance of some people, it must be admitted). > >>, 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. > > So what? if observations a and b are disjoint (in separate > universes), they cannot be part of a single observer moment. And it > is observer moments in one universe that must obey quantum > mechanics. Even in MWI, the separated universe do not actually play > any role in our observations. (Decoherence and all that.) Part of > the problem you face, it seems, is that the separate observers you > wish to consider must all be related -- their observer moments must > essentially be branches of the multiverse stemming from the same > quantum event -- possible outcomes of the observation at time t. If > not, then there is no necessary relation between the different > observer moments and linearity (in your terms) cannot be > established. > Yes the most interesting cases are where the different OMs are related to a single ancestor OM. -- ---------------------------------------------------------------------------- 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. To post to this group, send email to everything-list@googlegroups.com. Visit this group at https://groups.google.com/group/everything-list. 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