> On 11 Jan 2019, at 10:54, agrayson2...@gmail.com wrote: > > > > On Friday, January 11, 2019 at 9:07:50 AM UTC, Bruno Marchal wrote: > >> On 10 Jan 2019, at 22:08, agrays...@gmail.com <javascript:> wrote: >> >> >> >> On Thursday, January 10, 2019 at 11:07:51 AM UTC, Bruno Marchal wrote: >> >>> On 9 Jan 2019, at 07:58, agrays...@gmail.com <> wrote: >>> >>> >>> >>> On Monday, January 7, 2019 at 11:37:13 PM UTC, agrays...@gmail.com >>> <http://gmail.com/> wrote: >>> >>> >>> On Monday, January 7, 2019 at 2:52:27 PM UTC, agrays...@gmail.com <> wrote: >>> >>> >>> On Sunday, January 6, 2019 at 10:45:01 PM UTC, Bruce wrote: >>> On Mon, Jan 7, 2019 at 9:42 AM <agrays...@gmail.com <>> wrote: >>> On Saturday, December 8, 2018 at 2:46:41 PM UTC, agrays...@gmail.com <> >>> wrote: >>> On Thursday, December 6, 2018 at 5:46:13 PM UTC, agrays...@gmail.com <> >>> wrote: >>> On Wednesday, December 5, 2018 at 10:13:57 PM UTC, agrays...@gmail.com <> >>> wrote: >>> On Wednesday, December 5, 2018 at 9:42:51 PM UTC, Bruce wrote: >>> On Wed, Dec 5, 2018 at 10:52 PM <agrays...@gmail.com <>> wrote: >>> On Wednesday, December 5, 2018 at 11:42:06 AM UTC, agrays...@gmail.com <> >>> wrote: >>> On Tuesday, December 4, 2018 at 9:57:41 PM UTC, Bruce wrote: >>> On Wed, Dec 5, 2018 at 2:36 AM <agrays...@gmail.com <>> wrote: >>> >>> Thanks, but I'm looking for a solution within the context of interference >>> and coherence, without introducing your theory of consciousness. Mainstream >>> thinking today is that decoherence does occur, but this seems to imply >>> preexisting coherence, and therefore interference among the component >>> states of a superposition. If the superposition is expressed using >>> eigenfunctions, which are mutually orthogonal -- implying no mutual >>> interference -- how is decoherence possible, insofar as coherence, IIUC, >>> doesn't exist using this basis? AG >>> >>> I think you misunderstand the meaning of "coherence" when it is used off an >>> expansion in terms of a set of mutually orthogonal eigenvectors. The >>> expansion in some eigenvector basis is written as >>> >>> |psi> = Sum_i (a_i |v_i>) >>> >>> where |v_i> are the eigenvectors, and i ranges over the dimension of the >>> Hilbert space. The expansion coefficients are the complex numbers a_i. >>> Since these are complex coefficients, they contain inherent phases. It is >>> the preservation of these phases of the expansion coefficients that is >>> meant by "maintaining coherence". So it is the coherence of the particular >>> expansion that is implied, and this has noting to do with the mutual >>> orthogonality or otherwise of the basis vectors themselves. In decoherence, >>> the phase relationships between the terms in the original expansion are >>> lost. >>> >>> Bruce >>> >>> I appreciate your reply. I was sure you could ascertain my error -- >>> confusing orthogonality with interference and coherence. Let me have your >>> indulgence on a related issue. AG >>> >>> Suppose the original wf is expressed in terms of p, and its superposition >>> expansion is also expressed in eigenfunctions with variable p. Does the >>> phase of the original wf carry over into the eigenfunctions as identical >>> for each, or can each component in the superposition have different phases? >>> I ask this because the probability determined by any complex amplitude is >>> independent of its phase. TIA, AG >>> >>> The phases of the coefficients are independent of each other. >>> >>> When I formally studied QM, no mention was made of calculating the phases >>> since, presumably, they don't effect probability calculations. Do you have >>> a link which explains how they're calculated? TIA, AG >>> >>> I found some links on physics.stackexchange.com >>> <http://physics.stackexchange.com/> which show that relative phases can >>> effect probabilities, but none so far about how to calculate any phase >>> angle. AG >>> >>> Here's the answer if anyone's interested. But what's the question? How are >>> wf phase angles calculated? Clearly, if you solve for the eigenfunctions of >>> some QM operator such as the p operator, any phase angle is possible; its >>> value is completely arbitrary and doesn't effect a probability calculation. >>> In fact, IIUC, there is not sufficient information to solve for a unique >>> phase. So, I conclude,that the additional information required to uniquely >>> determine a phase angle for a wf, lies in boundary conditions. If the >>> problem of specifying a wf is defined as a boundary value problem, then, I >>> believe, a unique phase angle can be calculated. CMIIAW. AG >>> >>> Bruce >>> >>> I could use a handshake on this one. Roughly speaking, if one wants to >>> express the state of a system as a superposition of eigenstates, how does >>> one calculate the phase angles of the amplitudes for each eigenstate? AG >>> >>> One doesn't. The phases are arbitrary unless one interferes the system with >>> some other system. >>> >>> Bruce >>> >>> If the phases are arbitrary and the system interacts with some other >>> system, the new phases presumably are also arbitrary. So there doesn't seem >>> to be any physical significance, yet this is the heart of decoherence >>> theory as I understand it. What am I missing? TIA, AG >>> >>> Also, as we discussed, the phase angles determine interference. If they >>> can be chosen arbitrarily, it seems as if interference has no physical >>> significance. AG >>> >>> Puzzling, isn't it? We have waves in Wave Mechanics. Waves interfere with >>> each other, even if they're probability waves, and this is one of the core >>> features of Wave Mechanics. So phase angles must relate to degrees of >>> interference. But if the phase angles are arbitrary; ERGO, so is the >>> interference; arbitrary and thus NOT well defined. What am I missing? TIA, >>> AG >> >> >> The *global* phase angle is arbitrary: Psi = e^phi Psi. >> >> The relative phase angle is not arbitrary: you can distinguish all states up >> + e^phi down, when phi varies. >> >> All this follows from the Born rule. >> >> Bruno >> >> What about the case where the superposition is a sum of many eigenstates? > > That is always the case. > > ??? >> How do you calculate the phase angle of each eigenstate? I don't see how >> Born's rule helps. AG > By looking at the interference obtained when preparing many particles in that > superposition states. > > How can you prepare a system in any superposition state if you don't know the > phase angles beforehand?

? That is the point of the preparation. It is enough to rotate the polariser (say) in some special direction. I can prepare particles in some “up” state, and then I make them passing a polariser with a relative angle alpha, so that I can get the state sin(alpha) up + cos(alpha) down. I can verify this by measuring the density corresponding to the probabilities sin^2(alpha) of being up, and cos^2(alpha) of being down, + some other direction to make the difference with a mixture (as already explained once). > You fail to distinguish measuring or assuming the phase angles from > calculating them. One doesn't need Born's rule to calculate them. Maybe what > Bruce meant is that you can never calculate them, but you can prepare a > system with any relative phase angles. AG I have no clue what you mean by calculating. I postulate QM, and talk about experience done with state which have been prepared, as we can only do that in QM. I think that all what Bruce said about this is correct. We cannot distinguish Psi from e^phi Psi, but there is no problem distinguishing up + e^phi down from up + down. Bruno > > You will find more explanation on all this in David Albert’s book, which > minimises well the use of mathematics. > > Bruno > > > > > >> >> -- >> 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-li...@googlegroups.com <javascript:>. >> To post to this group, send email to everyth...@googlegroups.com >> <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 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. 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. For more options, visit https://groups.google.com/d/optout.