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Dipole moment operator.

  1. Oct 5, 2008 #1

    Is dipole operator symmetric or antisimmetric? Or, in other words, do the transition dipole moments from state 1 to the state 2, and from the state 2 to the state 1 have different sign? I.e.

    mu_12 = - mu_21.

    As far as I understand, the diagonal elements for the dipole operator should be zero (since the transition dipole moments from state "n" to the same state "n" should ne zero). However, in this case the dipole moments of all excited states should be zeros! This does not look to be physical. How one can resolve this contradiction?

    Thank you in advance.
  2. jcsd
  3. Oct 5, 2008 #2


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    I assume you mean the electric dipole operator, which is [tex]\vec r[/tex].
    This has odd parity, so that it is zero for any non-degenerate ground or excited state.
    It is not zero for transitions from an excited state to a lower state of different parity.
    In any event, mu_12=mu_21.
  4. Oct 5, 2008 #3
    Thank you for the answer. That what you say is an agreement with what I have. But I still does not understand the logic. You say that the dipole operator has odd party. I am not sure what exactly it means. Does it mean that the dipole operator is an odd function with respect to all Cartesian coordinates of the system? I also do not understand why in the case of the "odd parity" of the dipole operator the diagonal matrix elements should be zeros (for non-degenerate states)? Are wave-functions of non-degenerate states always even? If yes, why?

    Thank you.
  5. Oct 6, 2008 #4


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    The "parity operation" is taking each Cartesian coordinate to its negative.
    This changes [tex]\vec r[/tex] to [tex]-{\vec r}[/tex].
    A non-degenerate eigenstate has either even or odd parity, but then [tex|\psi|^2[/tex] will always have even parity since (-1)*(-1)=+1.
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