> On 14 Aug 2018, at 04:12, Bruce Kellett <bhkell...@optusnet.com.au> wrote: > > From: Bruno Marchal <marc...@ulb.ac.be <mailto:marc...@ulb.ac.be>> >>> On 12 Aug 2018, at 14:59, Bruce Kellett <bhkell...@optusnet.com.au >>> <mailto:bhkell...@optusnet.com.au>> wrote: >>> No, Price is wrong. He collapses the wave function in a non-local manner, >>> even though he doesn't seem to realize it. Let me try again. The state is >>> >>> |psi>= (|u>|d> - |d>|u>). >>> >>> Let Alice interact with particle 1 at one end: >>> >>> |Alice>|psi> = |Alice>|u>|d> - |Alice>|d>|u> >>> >>> Alice interacts only with particle 1 (locally), so |Alice>|u>|d> --> |Alice >>> sees u>|u>|d>, and similarly for the other component. >>> Now Bob interacts with a different state. He does not see |psi> as above, >>> but rather >>> >>> |Alice sees u>|u>|d>|Bob> - |Alice sees d>|d>}u>|Bob> >>> >>> The, if Bob measures along the same axis, he gets down for Alice's up, or >>> up for Alice's down. If he measures at some different angle, he gets the >>> appropriate rotated results. But Bob NEVER sees the original unaltered >>> rotationally symmetric singlet state: >> >> I have never disagreed with this. But Bob’s view is only a part of the >> picture. Any particular Bob cannot see the symmetry; because it is part of >> the symmetry. > > But it is Bob's view here that is relevant to the formation of the > correlations.
It is each Bob’s view. Yes. > Bob measures that state he sees, not some phantom symmetric state. Which Bob? The whole question is there. > Alice's measurement destroys the symmetry of the singlet state In her view, yes. It makes the symmetry no more viable to her. It can be restored in principle if Alice get amnesic. Nothing is really destroyed. > -- it is not preserved magically in any many-worlds picture. There is no > 'outside view' in MWI There is the universal wave function, at least if you accept the idea that the cosmology obey QM. > as there is in your classical M/W duplication thought experiments. There is > no view where the symmetry is preserved. Not even Tegmark's imaginary bird > view preserves the symmetry. You are mistaken here. ? The side view is only a (big) rotation (unitary transformation) on a (big) Hilbert space or von Neuman algebra. > >>> Alice's measurement (assuming Alice measures first in some frame) collapses >>> the state non-locally to affect the state that Bob sees. >> >> That makes no sense to me. > > Then you don't understand quantum mechanics. According to Bohr, that means I understand it :) You are reitrocuing physical FTL influence (not sigma, but still influence). That makes no sense to me. > >>> Since the original state is non-separable, the fact that Alice has >>> interacted with it changes the whole state. >> >> No measurement makes any change in any state, except local memories. Even >> Bohr acknowledge this in his reply to EPR (but then get irrational). > > Bohr did not really understand what Einstein was on about. He thought it was > determinism, whereas Einstein was always concerned about locality: "No spooky > action at a distance”. Bohr’s reply to EPR is concerned with locality. In that response, Bohr admits that the collapse cannot be a physical phenomenon, but then get quite fuzzy and unclear on this. > But never mind Bohr or Einstein, the measurement changes the state by > destroying the original symmetry. A measurement does not make any change, except in the mind of the observer. You talk like if Bohr’s idea of perturbation induced by measurement was till accepted. > The singlet state is different from the conventional collapse picture used to > explain the formation of spots on the screen in the double slit experiment. > Everett removed the collapse in the double slit case, but he did not remove > it for the entangled state. Tell me, where in your many-worlds picture is the > measured singlet state still symmetrical? In the whole multiverse interpretation the universal wave. It cannot be seen by any observer as they belong to the wave, and their first person experiences have differentiated. > > > >>> This is the calculation as Price and Tipler give it, and this calculation >>> is clearly non-local. >> >> It is non local, but does not involve any physical FTL or instantaneous >> action at a distance. There would be some FTL in case the collapse are real. > > > You are still obsessing about FTL. Just because some people asserts that there are still FTL in the MW. > The collapse is instantaneous, There is no collapse in the MW. Everett makes this clear. There is no “measurement”. In the MW, all measurements are only entanglement. > but there is no physical FTL. Good. We agree then. But you might use FTL for no signalling. I use FTL for no physical influence at a distance (instantaneous). Unless there is only one “world”. > Maudlin finally got to understand this in his 2011 book. He adopts the > relativistically covariant "flash GRW" idea for the non-local collapse. I will order it. > >>> Going to the GHZ state will not change anything. What you have to do is >>> show how to re-interpret this calculation so that Bob sees the original >>> singlet AFTER Alice has measured her particle. I insist that the original >>> non-separability of the state makes any such demonstration impossible. >> >> I agree with this. >> >>> And even if it were possible, it would not reproduce the known quantum >>> correlations; the non-separability and the above non-local reduction of the >>> state is an essential part of quantum mechanics. >> >> Nor with this. > > Maybe you mean to say that you do not agree with the above? No, I agree. Once a measurement is done in some branch, the Bob and Alice that we can associate with that branch cannot recover the singlet state. But with the MW that does not mean that the singlet state has collapsed (and indeed with amnesia the singlet state is recoverable, but that is impossible in practice). 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 everything-list+unsubscr...@googlegroups.com > <mailto:everything-list+unsubscr...@googlegroups.com>. > To post to this group, send email to everything-list@googlegroups.com > <mailto:everything-list@googlegroups.com>. > 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. 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