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Polarizaition and dipole moment?

  1. May 19, 2009 #1


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    I am learning a simple model for two-level system in which the dipole moment is written as

    [tex]d = d_{12}|1\rangle\langle 2| + d_{21} |2\rangle\langle 1|[/tex]

    where d is the dipole moment element, "1" stands for ground state and "2" stands for excited state. For a system has the particle density be N, the polarization is

    [tex]P = N \rangle d \langle = d_{12}\rho_{12} + d_{21} \rho_{21}[/tex]

    where [tex]\rho_{ij}[/tex] is the element of the density matrix. Now we apply an external field, with magnitue proportional to g, to the system, and suppose we'v already got the density matrix element is of the form (read it in a paper)

    [tex]\rho_{12} = \rho_{12}^{(0)} + \rho_{12}^{(1)}ge^{-i\delta t} + \rho_{21}^{(1)*}g^*e^{i\delta t}[/tex]

    where [tex]\rho_{12}^{(0)}[/tex] is the density matrix element without external field, [tex]\rho_{12}^{(1)}[/tex] and [tex]\rho_{21}^{(1)*}[/tex] is the first-order of the corresponding matrix element when there is a very weak field of magnitude g. * stands for complex conjugate.

    My question is:

    1) In E&M, the linear susceptibility,[tex]\chi[/tex], should be related to polarization in the form

    [tex]P = \chi \vec{E}[/tex]

    But in the text, it gives

    [tex]\chi = \frac{N|d_{12}|^2}{\hbar}\rho_{12}[/tex]

    how to get this?

    2) In the paper, when

    [tex]\rho_{12} = \rho_{12}^{(0)} + \rho_{12}^{(1)}ge^{-i\delta t} + \rho_{21}^{(1)*}g^*e^{i\delta t}[/tex]

    [tex]\chi = \frac{N|d_{12}|^2}{\hbar}\rho_{12}^{(1)}[/tex]

    why only keep [tex]\rho_{12}^{(1)}[/tex]? Where is [tex]\rho_{12}^{(0)}[/tex] and [tex]\rho_{21}^{(1)*}[/tex]?
  2. jcsd
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