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Superposition of the electron

  1. Jun 29, 2009 #1
    If the electron is in a superposition in such a way that there is no angular momentum about the nucleus, then classicaly the charge of the electron would be zero from inside the orbit correct?

    If the charge was changing anywhere there would be angular momentum.
  2. jcsd
  3. Jun 29, 2009 #2
    Probably not. If we consider a hydrogen atom, there is no net charge outside, so Gauss's law would give ∫E*n dS = 0 on any surface outside the atom. But the electron does have charge, and ∫E*n dS = -q/e0 for the surface integral around a bare electron, where -q is the charge on an electron. So if there are no field lines outside that atom, and we know an electron is inside, where are its field lines? If we perform the integral outside the atom, and also include the surface just outside the nucleus. we will find ∫E*n dS = -q/e0, so there are field lines between the electron cloud and the nucleus. This is independent of the n and l of the electron state. It is true that in an n=1 (ground) state, there is no orbital angular momentum, so l = 0 and there is no orbital magnetic moment.

    ∂ ∫ ∏ ∑
  4. Jun 29, 2009 #3
    There is charge outside the atom at a specific period of time. There is no net charge because over a period of time the charges cancel themselves out.

    If there was never a charge outside the atom then chemical reactions would not happen.

    There must be angular momentum if something is moving around a point even it is just a charge difference.
  5. Jun 29, 2009 #4
    If you were say 5 Angstroms outside a neutral hydrogen atom, you will not see any charge unless you ionize it. That requires about 13.6 eV. Water ions are actually H3O+ and (OH)-. To ionize hydrogen or water requires work.

    Have you ever watched a pendulum, or a Foucault pendulum? How much angular momentum does it have in the horizontal plane?

    For a hydrogen atom without electron spin, look at the Klein Gordon solution for the hydrogen atom. There is no angular momentum in the 1s, 2s, 3s etc. states. With the Dirac solution, the only angular momentum comes from the electron spin, which is not orbital.
  6. Jun 30, 2009 #5


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    "And yet it does move", as Galileo said. You still have van der Waals forces, which are certainly electrical in origin. (specifically, an effect of long-range correlation of electronic motion)

    Which is all just yet another example of how applying classical electrostatics to explain atomic and molecular phenomena just doesn't work. It leads you wrong more often than not. Which is my immediate reaction to these kinds of questions - "Don't think about it that way!"

    Considering electrons as some kind of classical or semi-classical charge density cloud inevitably leads you wrong on the dynamical effects of electronic behavior. Considering them to be moving point-charges inevitably leads you wrong on the delocalized nature of bound electrons. Basically repeating the wave-or-particle dilemma.
  7. Jun 30, 2009 #6
    Unless you analyze it in the de Broglie-Bohm interpretation where you have both moving point charges correlating with each other and a separate distinct wave which affects the trajectories of the moving charges and accounts for the 'delocalized nature of bound electrons'. And then everything is crystal clear.

    I'm sorry to go on about it, but the de Broglie estate is paying me to propagandize on their late master's behalf.
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