From: jos.bergervoet@xs4all.nl   
      
   On 9/2/2017 7:15 PM, Gregor Scholten wrote:   
   > Jos Bergervoet  wrote:   
   >   
   >> In solving the hydrogen atom we assume a 1/r electric potential.   
   >> But since the electron wave function squared is a source for the   
   >> electric field, this should be a screened potential. But if we   
   >> do that, then it will be quite different from 1/r. At large   
   >> distances it will vanish, and at distances around the Bohr radius   
   >> the potential will already be significantly reduced.   
   >   
   > You mean, you interpret the wave function of the electron that way that   
   > there would be a classical charge cloud with charge density   
   >   
   > rho(x) = -e |psi(x)|^2  (1)   
   >   
   > This interpretation is wrong, though. According to quantum mechanics,   
   > one should rather imagine the electron as a point charge with an   
   > uncertain position.   
      
   This seems to be a bit far-fetched. What quantum mechanics   
   says is merely this:   
      "This is the Lagrangian density, so now you can compute   
       the action. Have fun with it!"   
   There is nothing said about how we should imagine things. Besides,   
   that wouldn't change the equations one bit, of course.   
      
   > Instead of a certain classical charge density, there   
   > is a charge density operator   
   >   
   > \op{rho}(x) = -e delta(x - \op{x0})   
   >   
   > with \op{x0} being the position operator of the electron. Calculating   
   > the average value  =  yields   
   >   
   >  = -e |psi(x)|^2   
   >   
   > what is similar to (1), though, but this is only an average value, not a   
   > classical certain value.   
      
   To resolve the paradox as presented this does not help. You   
   cannot refute the fact that hydrogen in fact *is* neutral and   
   has vanishing field at large distance, so also when using the   
   expectation value of an operator, you will have to get that   
   result (and you get it, presumably). So it is the *other*   
   branch of the paradox that we are stuck with!   
      
   > Therefore, there is no such shielding effect as you described it.   
      
   Yes there is, Hydrogen is neutral.   
      
   >    There   
   > is no portion of the electron charge inside the Bohr radius that would   
   > shield the charge of the proton and reduce the potential outside the   
   > Bohr radius. The electron always feels the full charge of the proton,   
   > independently from the distance.   
      
   That is the other branch of the paradox! It must always feel   
   the full 1/r potential in the Schrodinger equation, in order   
   to get our well-known correct results. So why *does* it have   
   to feel the full potential?   
      
   > You also can see this by considering the two-particles wave function   
   > Psi(xe, xp) of electron and proton. In the corresponding Schroedinger   
   > equation, a two-particles potential operator   
   >   
   > \op{W} = -e^2 / (4pi eps0 |\op{xe} - \op{xp}|)    (2)   
      
   Yes, I was wondering about that, but the difficulty is how   
   to choose the equation! Because the QED Lagrangian density   
   does not directly tell us how the more simplistic descriptions   
   should be derived from it! (Well, of course the QCD Lagrangian   
   doesn't either, nor does string theory, so I don't mean to   
   blame QED in any way..)   
      
   > occurs, with \op{xe} and \op{xp} being the position operators of   
   > electron and positron. This is due to the rule for the transition from   
   > classical theory to quantized theory: if in the classical theory a   
   > classical two-particles potential W(x1,x2) occurs in the Hamilton   
   > function, in the quantized theory a corresponding two-particles   
   > potential operator like (2) occurs in the Hamilton operator.   
      
   I think the whole point is how to find the potential in the   
   first place, even for the one-particle Schrodinger equation.   
      
   In Weinberg (the QT of F's, part I,) we read in Sect. 13.6 ("The   
   External Field Approximation") that this potential is in fact   
   already summing a lot of multi-photon exchange diagrams. So the   
   potential in the Schrodinger equation is not meant as an expectation   
   value of a field, it is just a computational tool to represent a lot   
   of multi-particle interactions (obtained through the Bethe-Salpeter   
   equation). So this happens to give rise to an 1/r function in some   
   equation (the Schrodinger equation), whereas the expectation value   
   of the EM field in full QFT is falling off much faster than 1/r^n.   
   There is no contradiction, the two are not the same thing!   
      
   This, I believe, resolves the paradox.   
      
   --   
   Jos   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)