From: g.scholten@nospam.gmx.de   
      
   [I already tried to send this on Monday, September 11, this is the   
   second attempt]   
      
   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!"   
      
   This might be true for Quantum Field Theory ("Second quantization"), but   
   not for Quantum Mechanics ("First quantization"). Lagrangian densities   
   are a property of field theories, not of mechanics. In Quantum   
   Mechanics, there is a Lagrangian function, but no Lagrangian density.   
      
   Below, you described how you think your "paradox" can be solve in   
   Quantum Field Theory (or in QED, which is the application of Quantum   
   Field Theory to electromagnetic interaction), but this "paradox" can   
   already be solved in Quantum Mechanics, in the way I described it.   
      
      
   > There is nothing said about how we should imagine things.   
      
   You're wrong. Quantum Mechanics clearly tells us that imaging a   
   classical charge clould with charge density -e |psi(x)|^2 (1) would be   
   wrong.   
      
      
   > Besides,   
   > that wouldn't change the equations one bit, of course.   
      
   You're wrong again. The fact that there is no classical charge cloud   
   with charge density (1) makes clear that there is no shielding effect as   
   you assumed it.   
      
      
   >> 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   
      
   The fact that the hydrogen atom is neutral means that if we assume a   
   third charged particle besides the proton and the electron, e.g. a   
   second electron, this third particle is not attracted or repelled by the   
   hydrogen atom. This does not in any way contradict the fact that the   
   electron that is part of the hydrogen atom does not shield the   
   attractive force from the proton on that electron itself.   
      
      
   > so also when using the   
   > expectation value of an operator, you will have to get that   
   > result (and you get it, presumably).   
      
   No problem: when considering a large distance much larger than Bohr   
   radius, the probability is very high that the distance between proton   
   and electron is much lower, and so, the probability is very high that a   
   third charged particles does not feel any attraction or repulsion.   
      
      
   >> Therefore, there is no such shielding effect as you described it.   
   >   
   > Yes there is, Hydrogen is neutral.   
      
   You're wrong. The neutrality of a hydrogen atom concerns the effect on a   
   third particle, it does not imply a self-shielding of the electron.   
      
      
   >>    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?   
      
   Because of the reasons I explained. There is no classical charge clould   
   with charge density (1).   
      
      
   >> 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!   
      
   This just follows from the classical situation by applying the rules of   
   quantization.   
      
      
   >> 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.   
      
   This is how you think that the "paradox" can be solved in QFT/QED.   
   However, it can be already solved in Quantum Mechanics, in the way I   
   described it. And as we will see below, both solutions turn out to be   
   equivalent.   
      
      
   > 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.   
      
   BTW: this is equivalent to what I said. Let's consider some distance r0,   
   e,g, one Bohr radius. The average value  at that distance falls   
   off faster than 1/r0^2, because of the shielding effect of the   
   electron's charge:  is determined by the average value  of   
   the charge in the range r < r0, where  is given by the proton's   
   charge + the average value of the electron's charge in the range r < r0.   
      
   However, the attraction from the proton on the electron at r = r0 is not   
   given by : let   
      
   P(r0) = \int_r0^\infty |psi(r)|^2 r^2 dr   
      
   be the probability that the electron is outside the region r < r0, then   
   P(r0) is also the probability that there no electron charge in the   
   region r < r0, and therefore, the probabilily that the electric field   
   E(r0) at r = r0 is higher than , namely proportional to 1/r0^2.   
      
      
   [continued in next message]   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)