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L=0 state

  1. Aug 20, 2012 #1
    What is the classical analog for l=0 state?
    Angular momentum = 0 , what kind of orbits is that?
     
  2. jcsd
  3. Aug 20, 2012 #2
    Classically you cannot have an orbit with zero angular momentum. Consider a non-quantized (or classical) version of the Bohr model. Check this for exact calculations:

    http://en.wikipedia.org/wiki/Bohr_model

    The centrifugal force experienced by the electron will be balanced by the electrostatic attraction it experiences from the nucleus. This force balance condition will give you the velocity of the electron as a function of the radius. This functional dependence goes like [itex]1/\sqrt{r}[/itex]. The angular momentum of this electron is given by [itex]L = mvr[/itex]. Therefore, the angular momentum depends on radius as [itex]\sqrt{r}[/itex]. Consequently, the angular momentum is zero when radius is zero. Hence we do not really have an orbit.
     
  4. Aug 20, 2012 #3
    Thanks.
    I find the l=0 state to be one that the quantum and classical pictures differ most strikingly.
     
  5. Aug 21, 2012 #4

    DrDu

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    For angular momentum zero, in a centrosymmetric potential, the particle will move from -r to +r and back again along a line of constant angle phi. Hence it falls through the center. I do not see why this should be classically forbidden. If an obstacle (like a nucleus) happens to be in the center, the particle may or may not get reflected. In classical mechanics, the particle either gets completely reflected or not reflected at all, while in QM (like in the hydrogen atom) you usually observe a superposition of unreflected and reflected paths. Furthermore in QM, the angle phi is undetermined, which does not mean that it changes in time.
     
  6. Aug 21, 2012 #5
    Yes, that is a good example.
     
  7. Aug 21, 2012 #6
    Thanks.
    Nuclear size being 10-15m, the electrostatic attraction tends to negative infinity. Will the electron speed exceed speed of light?
     
  8. Aug 22, 2012 #7

    DrDu

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    No, you have to consider relativistic corrections (i.e. the Dirac equation) in the immediate vicinity of the nucleus.
     
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