> On 16 Aug 2018, at 21:58, [email protected] wrote: > > > > On Thursday, August 16, 2018 at 10:05:31 AM UTC, Bruno Marchal wrote: > >> On 15 Aug 2018, at 13:33, Bruce Kellett <[email protected] >> <javascript:>> wrote: >> >> From: Bruno Marchal <[email protected] <javascript:>> >>>> On 15 Aug 2018, at 01:48, Bruce Kellett <[email protected] >>>> <javascript:>> wrote: >>>> >>>> From: Bruno Marchal <[email protected] <javascript:>> >>>>>> On 14 Aug 2018, at 04:30, Bruce Kellett <[email protected] >>>>>> <javascript:>> wrote: >>>>>> >>>>>>> If they are space separated, I am not sure I can make sense of being in >>>>>>> the same branch. >>>>>> >>>>>> You appear to be referring to the presence of quantum fluctuations >>>>>> continually splitting the classical Alice and Bob into multiple copies >>>>>> -- the point that Jason has made. >>>>> >>>>> That points is correct, but I was alluding to the infinity of Bob and >>>>> Alice couples associated with the singlet state. That is needed to tackle >>>>> the case where Alice and Bob makes non orthogonal measurements. >>>> >>>> I was trying to make sense of the suggestion of many Alices and Bobs >>>> before any measurement. That can easily be implemented by having Alice >>>> select her measurement angle according to the time of some radioactive >>>> decay. Since an infinity of decay times is possible, we get a >>>> superposition of an infinite number of copies of Alice. >>> >>> OK. But we have this in our context too. >>> >>>> But this makes not difference to the basic argument -- one just picks out >>>> a typical Alice. >>> >>> How? >> >> Do you really no know how to pick out a typical component from an ensemble? > > > I cannot when the elements cannot be distinguished. Alice cannot do that, but > each Bob and Alice picks their counterparts by doing their measurements, but > that take some times. > > > >> >>>> You are wrong when you claim that an infinity of couples are required to >>>> make sense of measurements made at arbitrary angles. >>> >>> Why? >> >> Because that is not how angular momentum operators in quantum mechanics work. >> >>>> The singlet state is rotationally symmetric, >>> >>> That’s why. >> >> That's why what? > > > That is why a singlet state describe a collection of situations withAlice’s > particles spin well defined in all directions (and the opposite for Bob). But > none know which one. > > >> >>>> and can be expressed in any base. But this does not mean that there >>>> actually exists a copy of the observer for each of the potential bases. >>>> That idea makes no sense at all; it is not part of quantum mechanics in >>>> any possible formulation. >>> >>> ? >>> >>> That would contradict the complementary principle. A well localised >>> particle is a particle having almost all possible momenta in many different >>> histories. >> >> For fuck's sake, Bruno. Do you understand nothing of elementary quantum >> mechanics? > > No comment. > > > > >> The angular momentum operators do not commute, sure, so that if one has a >> precise measurement in one direction, one has no knowledge of the projection >> in an orthogonal direction. But the possible values of any such operator on >> the spin-1/2 state are +1 or -1 (in units of hbar/2). So there is no >> infinity as there is in the case of the complementarity of position and >> momentum operators! > > No problem with this, but Alice can choose to measure that spin in any > direction. > > > >> >> Besides, it is possible to have exact values for both the total angular >> momentum operator (L^2) and any particular component, say L_z if we are >> measuring in that direction, and that is all we require here. See the >> Wikipedia article: >> >> https://en.wikipedia.org/wiki/Angular_momentum_operator#Uncertainty_principle >> >> <https://en.wikipedia.org/wiki/Angular_momentum_operator#Uncertainty_principle> >> >> >>>>> The singlet state does not single out one base, despite the notation. It >>>>> describes an infinite of Alice and Bob right at the start. >>>> >>>> Sure, the singlet state does not single out one base. But that does not >>>> mean that it describes an infinity of observers. Just because you can >>>> measure at any angle does not mean that there is actually an infinity of >>>> observers making all those possible measurements. That notion is just >>>> crazy. >>> >>> ? >>> >>> It is just what the wave described literally. >> >> No, it is not. Look up some reference on the application of the uncertainty >> principle to angular momentum operators. (Such as the Wikipedia article >> above.) > > I do not see any problem between what I said and that wiki pages, which is > rather neutral on the interpretations. They do not provide the “many-worlds” > view on this, and some links there suggests they use the Copenhagen > formulation. > > You seem to reintroduce implicitly some collapse in the picture. That’s my > feeling, as this is not clear. When measuring a spin: there are two possible > values *for all possible direction of the spin*. That makes infinitely many > worlds. Same for an electronic orbital. There are as many world that the > possible position of the electron in the orbitals. > > > In such scenario, you'd have to include all bound electrons in the universe, > bound to atoms and molecules, from which the number of possible worlds and > Alice's would hugely metastasize.
Yes. Well, all those atoms to which I am casually not related, that is basically those out of my light cone. Yes, with both QM and computationalism, we are somehow multiplied by our factual ignorance, when due to isolation. > I am fine with this scenario provided you add immaculate conception to the > multiple universes and Alice's. AG By definition, the universal wave is immaculate, it seems to me. Bruno > Are you OK with this? I try to figure out what is your interpretation of the > SWE. > > Bruno > > > > >> >> 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 [email protected] <javascript:>. >> To post to this group, send email to [email protected] >> <javascript:>. >> Visit this group at https://groups.google.com/group/everything-list >> <https://groups.google.com/group/everything-list>. >> For more options, visit https://groups.google.com/d/optout >> <https://groups.google.com/d/optout>. > > > -- > 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] > <mailto:[email protected]>. > To post to this group, send email to [email protected] > <mailto:[email protected]>. > Visit this group at https://groups.google.com/group/everything-list > <https://groups.google.com/group/everything-list>. > For more options, visit https://groups.google.com/d/optout > <https://groups.google.com/d/optout>. -- 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 post to this group, send email to [email protected]. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
