> On 11 Jan 2019, at 10:54, [email protected] wrote: > > > > On Friday, January 11, 2019 at 9:07:50 AM UTC, Bruno Marchal wrote: > >> On 10 Jan 2019, at 22:08, [email protected] <javascript:> wrote: >> >> >> >> On Thursday, January 10, 2019 at 11:07:51 AM UTC, Bruno Marchal wrote: >> >>> On 9 Jan 2019, at 07:58, [email protected] <> wrote: >>> >>> >>> >>> On Monday, January 7, 2019 at 11:37:13 PM UTC, [email protected] >>> <http://gmail.com/> wrote: >>> >>> >>> On Monday, January 7, 2019 at 2:52:27 PM UTC, [email protected] <> wrote: >>> >>> >>> On Sunday, January 6, 2019 at 10:45:01 PM UTC, Bruce wrote: >>> On Mon, Jan 7, 2019 at 9:42 AM <[email protected] <>> wrote: >>> On Saturday, December 8, 2018 at 2:46:41 PM UTC, [email protected] <> >>> wrote: >>> On Thursday, December 6, 2018 at 5:46:13 PM UTC, [email protected] <> >>> wrote: >>> On Wednesday, December 5, 2018 at 10:13:57 PM UTC, [email protected] <> >>> wrote: >>> On Wednesday, December 5, 2018 at 9:42:51 PM UTC, Bruce wrote: >>> On Wed, Dec 5, 2018 at 10:52 PM <[email protected] <>> wrote: >>> On Wednesday, December 5, 2018 at 11:42:06 AM UTC, [email protected] <> >>> wrote: >>> On Tuesday, December 4, 2018 at 9:57:41 PM UTC, Bruce wrote: >>> On Wed, Dec 5, 2018 at 2:36 AM <[email protected] <>> 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 [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.
