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Atomic Physics: Optical molasses

  1. Aug 31, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi

    I have a question regarding the force that it used in Optical Molasses. The force is generally given by
    [tex]
    F = \frac{\hbar k}{2} \frac{s_0}{1 + s_0+ (2\delta/\Gamma)^2}
    [/tex]
    where s0 is the saturation parameter. The force saturates to
    [tex]
    F = \frac{\hbar k}{2}
    [/tex]
    It is normally said that this result is a consequence of the fact that the atom can only absorb and subsequently emit a photon twice during its excited state lifetime. However this latter statement does not make sense to me intuitively. I would say that once the atom has absorbed a photon, then -- on average -- it takes a time [itex]1/\Gamma[/itex] to emit it. So the remaining factor 1/2, where does this come in when looking at the problem like this?

    Best,
    Niles.
     
    Last edited: Aug 31, 2012
  2. jcsd
  3. Sep 1, 2012 #2
    Just to clarify, my question is the following: The intuitive picture often presented is that when an atom absorbs a photon, on average a time [itex]\tau=1/\Gamma[/itex] passes by until the atom emits the photon again. So the force is (under the condition that the atom is in resonance with the light so that the probability of absorbing a photon is maximal)
    [tex]
    F = \frac{\Delta p}{\Delta t} = \frac{\hbar k }{\tau} = \hbar k \Gamma
    [/tex]
    But we know that the maximum force is given by (see my OP)
    [tex]
    F = \hbar k \frac{\Gamma }{2}
    [/tex]
    So my question is, where does the factor of 1/2 come from, when this framework is used to interpret the force?

    Best,
    Niles.
     
    Last edited: Sep 1, 2012
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