> On 19 Jun 2018, at 19:07, Brent Meeker <[email protected]> wrote: > > > > On 6/18/2018 10:21 PM, Bruce Kellett wrote: >> From: Brent Meeker <[email protected] <mailto:[email protected]> >>> On 6/17/2018 10:41 PM, Bruce Kellett wrote: >>>>> But the lens doesn't send one color to one photoreceptor and another >>>>> color to a different photorecptor. It focuses a spot of light on several >>>>> photorecptors and the one with the right pigment fires its neuron. So it >>>>> is energy detection. >>>> >>>> But if you use a different position basis the lens will no longer focus >>>> point objects to points on the retina. >>>> >>>>>> I don't know enough about the physics of calorimeters as used in HEP to >>>>>> comment here. But if temperature changes are measured by bimetals or >>>>>> strain gauges, position comes into it in an essential way. >>>>> >>>>> Most work by measuring a voltage. But you miss the point. Those >>>>> position measurements are not essential in the QM sense. They are just >>>>> changing one classical value into another. Temperature is the first >>>>> classical level. >>>> >>>> Fair enough. I suppose I am just very conscious of the fact that in a >>>> different position basis all of this physics will be very different. The >>>> classical universe will not look the same at all. >>> >>> I guess I don't understand your idea of "position basis". My understanding >>> of linear algebra is that any basis that spans the space can be used to >>> represent any relation between structures. Why should choosing a different >>> basis make any difference to the physics aside from the simplicity of its >>> representation. It's just a coordinate basis in Hilbert space. Or are you >>> thinking of bases different from position, e.g. momentum, energy, >>> live/dead,... >> >> Yes, there does seem to be a degree of miscommunication. I am not think of >> different variables such as energy, momentum, or the like. These are not >> different bases, they are different variables and they inhabit different >> Hilbert spaces. So a change of base in one Hilbert space does not take you >> to another space.
? >> >> No, what I am considering is the possibility of different bases in a single >> space, such as position space. If you assume an eignevector interpretation >> of a set of basis vectors, then a different basis will correspond to the >> eigenvalues of some different operator. It still acts in the same, position, >> space, so it must be regarded as a position operator, but it will have quite >> different physical properties from the usual position operator that we use >> from classical mechanics, where the eigenvectors are delta functions along >> the real line. >> >> Because this will be a different operator, it will correspond to different >> physics. For instance, if the position eigenvalues are superpositions of >> delta functions, corresponding to superpositions of different points, the >> point interactions of particles that we assume in constructing the >> interaction Hamiltonian will be replaced by some set of interactions between >> superpositions of points. This why I suggest that the physics will be >> different. If the physics is the same under this basis change, why is there >> any question about the preferred basis? The point is that a change of basis >> does not mean that we simply go to measure some other variable. I think >> Schlosshauer makes this mistake, if I remember correctly; he seems to >> suggest that the basis choice is between position or energy in most cases. >> That is just wrong. > > I think you're wrong about position operators. I agree. The Hilbert space is always the same. > Sure, we usually think of dividing space into little bins and a position > operator has eigenvectors that are 1 in some bin and zero in the other. But > we could do the same analysis in the Fourier transform of that space and the > delta function locations would be integrals over the wave numbers. It would > be the same physics. Yes. > There would still be localized interactions. The Hamiltonian would be > written as an interaction of a superposition of points, except they would all > destructive interfere except at one location. So the physics would be the > same. > > Consider the paradigmatic two slit experiment. The pattern you get on the > screen, which is predicted by the Schroedinger equation, is described in your > idea of position space by a lot of little bins that have different degrees of > probability, so if you put a detector there you get a certain count rate. > But that pattern on the screen is a certain wavelet and if you transformed > the Schroedinger equation to wavelet space that whole pattern would be just > one point in the space and it would be the eigenvector of the two slit > experiment. > > The problem of the preferred basis arises in trying to explain why we measure > position of needles but not momentum or energy and why we don't see > superpositions of different needle positions. That's a question of which > measurement operator has eigenvectors stable against decoherence. Not which > basis we express the operator in. A basis doesn't have to consist of the > eigenvectors of the mesaurement operator, that's just mathematically > convenient and independent of the physics. No matter what basis we use to > write the operator in, it has the same eigenvectors. > I agree with Brent here. Bruno > Brent > > >> >> 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] >> <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] > <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.
