From: jos.bergervoet@xs4all.nl   
      
   On 1/10/2017 10:31 PM, Roland Franzius wrote:   
   > Am 08.01.2017 um 17:07 schrieb Jos Bergervoet:   
   >> On 1/6/2017 5:04 PM, Roland Franzius wrote:   
   >>> Am 06.01.2017 um 05:07 schrieb Rich L.:   
   >>>> On Wednesday, January 4, 2017 at 2:14:30 PM UTC-6, SEKI wrote:   
   >>>>> ...   
   >>>>> My basic point is as follows:   
   >>>>> Without any cohesive force or some sort   
   >>      ...   
   >>>> There is no mechanism in QM that provides any "cohesive force" or   
   >>     ...   
   >>> In classical physics the distribution function for one particle is a   
   >>> function of six canonically chosen variables   
   >>     [ ... snip ]   
   >>> What is different in quantum mechanics?   
   >>>   
   >>> Not so much.   
   >>   
   >> I would say at least these two things:   
   >>     1. Those six variables cannot simultaneously be chosen   
   >>        independently at the start.   
   >>     2. The evolution for the members of the distribution slightly   
   >>        differs from Newton's laws (or else the hydrogen atom would   
   >>        still have a radiation dipole in its ground state!)   
   >   
   > It# always a bit difficult, to explain, why a wrong perception of the   
   > scientific correct and accepted quantum theory fails at any point, if   
   > you dig just one Angstroem deep.   
      
   Then you will have to dig deeper.   
      
   > 1. Momentum and position distributions can be chosen freely from the set   
   > of pure states that are in the common domain of p and x operators (x   
   > psi(x) and k F(psi)(k) both square integrable).   
      
   And that very much limits your choice, compared with the   
   classical case! You can no longer get any arbitrary combined   
   momentum-position distribution.   
      
   The clearest proof that your momentum-position distribution   
   *cannot* be chosen freely is to attempt to make it a product   
   of two Gaussian distributions, with the product of their   
   widths less than 1/2 (or 1/2 \hbar if you prefer). This will   
   not be possible by the uncertainty principle. So not every   
   function is allowed, QED.   
      
   One can also say that there is much more information in a   
   real function of six variables (the classical distribution)   
   than in a complex function of 3 variables, Psi(x,y,z).   
      
   Explicitly, one can discretize space, e.g. on a 10x10x10   
   lattice. Then the QM wave function Psi(x,y,z) consists of   
   1000 complex numbers, i.e. 2000 real numbers, whereas an   
   arbitrary position and momentum distribution (over the   
   spectrum of the discretized operators) contains 1000000   
   real numbers! That simply is more information.   
      
     ...   
   > The reason is that any Fourier composition can be placed anywhere in   
   > space just by a complex gauge transform fron the tranlsatin group   
   > repesentation.  Any position space distribution can be boosted to a   
   > moving distribution by a boost from the unitary representation of the   
   > Galilei group. The last transform is a bit tricky for Schrödinger theory   
   > since it involves a transform of d/dt -> d/dt + v.grad as known from   
   > diffusion theory. So it disappeares from sight if on focuses on time   
   > independent theory.   
      
   This will not help you. Your new situation will have   
   a distribution in p with unchanged shape, only centered   
   around another point in p-space. So you still can't   
   make arbitrary shapes based on that approach.   
      
     ...   
   >> To be honest, I think that full QFT needs a wave functional   
   >> describing distributions over field shapes, not distributions   
   >> over point particle in phase space, but your analogy at least   
   >> suffices for simple Schrodinger-like wave functions in QM.   
   >   
   > Yes, that is selfexplaining. Eigenstate series of the interaction   
   > Hamiltonian as a basis are not available under any circumstances except   
   > the small oscillation models and the three exactly solvable models   
   > presented in the first course.   
      
   With discretized space (e.g. on  a lattice) Hamiltonian   
   eigenstates will also be available. (But one has the problems   
   of the lattice, like broken Lorentz invariance..)   
      
   >>     ...   
   >>>          the pure states -   
   >>> classical the trajectories with fixed starting position, momentum and   
   >>> spin - are replaced with pure states that are eigenstates of some   
   >>> observables, that can be given sharp values in an experiment dictating   
   >>> the starting distribution at time zero.   
   >>   
   >> But they cannot all be given arbitrary sharp values! You   
   >> essentially only have to specify half of the starting values,   
   >> compared to the classical case..   
   >   
   > No wave theory has pointlike concentrated states in their domain of the   
   > energetic form.  The pointlike delta-distribution is not square   
   > integrable, neither in space nor in k-space.   
      
   So again that is *different* from the classical ideal of   
   point-particles! Of course point particles led to infinities   
   in several ways in classical physics as well, so we do not   
   have to mourn their absence in QM for too long..   
      
     ...   
   >>> The apparent difference to canonical mechanics is now, that the pure   
   >>> quantum states evolving in a given dynmical environment display - as   
   >>> functions in Hilbert space - a statistical non-delta distribution over   
   >>> position and momentum variables   
   >>   
   >> Of course Hilbert space is already described by *only* momentum   
   >> variables or position variables (the difference is just the kind   
   >> of basis vectors you construct.) So if you use both simultaneously,   
   >> then your possible distributions over them are highly constrained.   
   >> Of course that also applies to then classical case.   
   >   
   > Inappropriate conclusions from a poorly understood theory.   
      
   So what exactly don't you understand about the claim? In   
   QM the complex amplitude distribution over the spectrum   
   of only one of the two (either momentum or position)   
   suffices to completely describe the state. This differs   
   from having an arbitrary real distribution function over   
   the two spectra (which contains *much more information*,   
   as explained above).   
      
     ..   
   >>> Even if energy, angular momentum and spin components  are constantly   
   >>> delta-sharp in atomic states, people reason much about momentum and   
   >>> position distributions.   
   >>   
   >> So what's wrong? Another basis where you need to use a sum over   
   >> the basis vectors instead of using just one. No big deal, what is   
   >> your point?   
   >   
   > The point is that the observable algebra of movement on a 2-sphere with   
   > Hamiltonian L^2 of the angles theta, phi and the angular momentum   
   > d/dtheta, d/dphi allows sharp values of i d/dphi and L^2 at finite   
   > energy, but no delta in position space. Its nonsense to postulate any   
   > principles like "use any base" without specifying the context.   
      
   The position space in this case is (phi,theta) and that   
   is in fact the most widely used basis to set up the theory!   
   I don't see why you suggest there is anything wrong with   
   it: the Ylm's are in fact *defined* on that angle-basis,   
      
   [continued in next message]   
      
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