On Sunday, November 4, 2018 at 7:50:30 AM UTC+11, agrays...@gmail.com wrote: > > > > On Thursday, November 1, 2018 at 11:22:46 PM UTC, Pierz wrote: >> >> >> >> On Monday, October 15, 2018 at 9:40:39 PM UTC+11, agrays...@gmail.com >> wrote: >>> >>> >>> >>> On Sunday, October 14, 2018 at 5:08:42 PM UTC, smitra wrote: >>>> >>>> On 14-10-2018 15:24, agrays...@gmail.com wrote: >>>> > In a two state system, such as a qubit, what forces the >>>> interpretation >>>> > that the system is in both states simultaneously before measurement, >>>> > versus the interpretation that we just don't what state it's in >>>> before >>>> > measurement? Is the latter interpretation equivalent to Einstein >>>> > Realism? And if so, is this the interpretation allegedly falsified by >>>> > Bell experiments? AG >>>> >>>> It is indeed inconsistent with QM itself as Bell has shown. Experiments >>>> have later demonstrated that the Bell inequalities are violated in >>>> precisely the way predicted by QM. This then rules out local hidden >>>> variables, therefore the information about the outcome of a measurement >>>> is not already present locally in the environment. >>>> >>>> Saibal >>>> >>> >>> What puzzles me is this; why would the Founders assume that a system in >>> a superposition is in all component states simultaneously -- contradicting >>> the intuitive appeal of Einstein realism -- when that assumption is not >>> used in calculating probabilities (since the component states are >>> orthogonal)? AG >>> >> >> I think because of interference. Consider the paradigmatic double slit, >> with the single electron going through it. It sure looks like the electron >> was in two place at once, doesn't it? >> > > *Yes, that's my assessment how the erroneous interpretation took hold, but > only if you restrict yourself to the particle interpretation. If the > electron travels as a wave, it can go through both slits simultaneously and > interfere with itself. This is my preferred interpretation; the only one > that makes sense. AG* >
Although as stated I think "being in two states at once" is a manner of speaking quasi-classically about non-classical phenomena, it seems you still have a very classical imagination of what's going on here, but I have my doubts: 1 - You say it makes sense, but I'm not sure that an electron "travelling as a wave" but being measured as a particle makes an awful lot more sense! 2 - Schrödinger initially thought of his equation (the one that applies to double slits) as being the equation for a physical wave, as you seem to be doing. However he was forced eventually to accept that it was something a lot more abstract than that. The statistical interpretation formulated by Born superseded any such notion. Interference happens whenever a quantum system can reach the same state via more than one history. In the case of quantum computers, complex interfering superpositions are constructed in which it is impossible to conceive of the "wave function" as literally describing some kind of mechanical wave. > > I'm not sure what you mean by "that assumption is not used in calculating >> probabilities". >> > > *If the operator whose eigenvalues are being measured has a well defined > mathematical form -- e.g., not like |alive> -- it has specific eigenvectors > and eigenvalues, and the state function can be written as superposition of > these eigenvectors. It can be shown that eigenvectors with distinct > eigenvalues are orthogonal, meaning the Kronecker delta applies to their > mutual inner products. Therefore, to calculate the probability of observing > a particular eigenvalue, one must take the inner product of the wf with the > eigenvector which has that eigenvalue. Due to the orthogonality, all terms > drop out except for the term in the superposition which contains the > eigenvector whose eigenvalue you want to measure. As you should see, there > is nothing in this process of calculating probabilities that in any way > implies, assumes, or uses, the concept that the system is simultaneously in > ALL component states of the superposition (written as a sum of > eigenvectors). AG* > > Sure, but this relates to measurement *outcomes* not to the question fo what state the system is in while not being measured. Clearly the fact that the vector spans more than one dimension expresses a state that *in a mathematical sense* is a combination of more than one component state. If it weren't for interference and entanglement (per Bell), no doubt scientists would simply consider this combination of states a measure of our ignorance of the underlying reaility (the hidden variables). But those three elements of quantum weirdness make it impossible to sustain that view. If you take a sum-over-histories approach it's explicitly assumed the >> electron went via all possible paths. >> > > *I don't know that method, but offhand POSSIBLE PATHS might have nothing > to do with, and possibly independent of SUPERPOSITIONS OF STATE. AG* > > That's *very* offhand. Possible paths are mathematically equivalent to the superposition that emerges from the Schrödinger picture. It's silly to say it might be independent of superposition if you don't know anything about the approach. > I don't see what the orthogonality of the basis vectors (and hence >> component states) has to do with the question of interpretation of >> superposition. >> > > *Explained in detail above. AG* > > Clearly the system will be measured in only one state, and this is what >> the orthogonal vectors represent. However the quantum state itself >> typically spans more than one dimension of the vector space - that's what a >> superposition is. However I think when physicists say that the >> superposition is in all states simultaneously, it's only in a manner of >> speaking - a way of conveying the mathematical situation in natural >> language that is inherently classical. >> > > > *It's a totally misleading way to discuss the quantum superpositions. > Even classically, say for the vector space of "little pointy things" in a > plane, each vector can be expressed in uncountably many bases, both > orthogonal and non-orthogonal. So to claim that one basis is somehow > preferred, and the vector being expressed as a sum or superposition in that > basis, is simultaneously in all components of that particular basis, make > no sense whatsoever. AG* > To take up your classical example, the classical system is in a well-defined state regardless of which basis is ultimately chosen to specify that state. In order to tell you the location of Paris (or some little pointy thing), I need some coordinate basis, but Paris is still located *somewhere* before I specify that basis. In some (trivial) sense it's simultaneously at all the coordinates that could be specified were we to be specify its location in every one of the possible bases we could choose. The argument is that a quantum state is equally well-defined, with no *underlying* state that is not included in the component vectors. The difference with the Paris example of course is that it is not in a superposition within those different bases - it has one and just one coordinate in each of them. If there is no underlying reality beyond the state - the hidden variables ruled out by Bell - then IMO it is not "totally misleading" to say it is in some sense in all the states of the superposition simultaneously. It is *somewhat *misleading however if you don't understand QM because of the possible confusion between a coherent and a decohered state. They are obviously quite different things, the difference being expressed in the density matrices of pure and mixed states. A system evolving in isolation shouldn't be muddled up with a system post measurement. > Reading Born's exchange of letters with Einstein (I'm proud to say Born >> was my great grandfather), it's clear that Born had a conception of QM that >> was still very realistic in the Einstein sense. Though they disagreed >> significantly and somewhat heatedly, Born still seems to have regarded QM >> probabilities as classical probabilities in disguise. >> > > *Einstein realism seems to have been falsified due to Bell experiments. If > that's the case, it would mean that BEFORE measurement of a quantum system, > it is not only NOT in all states of a superposition simultaneously for the > reasons I have argued (nothing to do with Bell), but ALSO has no local > preexisting value. AG* > > I don't think he would ever have endorsed the notion that a particle is >> truly in all of the states of the superposition simultaneously. >> > > *Thanks for your input. AG * > >> >> > -- 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 email@example.com. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.