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The electric dipole approximation

  1. May 28, 2012 #1
    I am trying to understand the elctric dipole approximation when an atom interacts with an electromagnetic wave.

    I know that if the size of the atom is much much smaller than the wavelength of the radiation, then the dot product od the wavevector and the position vector becomes constant.

    I can't see why that has to be so.

    Any ideas?
  2. jcsd
  3. May 28, 2012 #2
    In a first approximation we can ignore the quantity [itex] \vec{k} \cdot \vec{x} = \frac{2 \pi}{\lambda} \hat{k} \cdot \vec{x} [/itex] because EM-induced atomic transitions involve radiation of lenght [itex] \approx 10^{3} [/itex] angstroms, and the integral is essentially in a domain with a characteristic lenght of 1 angstrom (the Bohr radius).
  4. May 28, 2012 #3
    In the case the atom is much smaller than the wave-length, the wave phase is almost the same anywhere at the atom, so exp(ik·x) is approximated by 1. That's the dipole approximation.
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