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Linear Component of Polarization - Mathematical transformation

  1. Oct 5, 2014 #1

    I'm currently going through Agrawal's book 'Nonlinear Fiber Optics' and got stuck with some mathematical cosmetics (pp. 40). It is the substition of:
    [tex] \vec{P_L}(\vec{r},t) = \frac{1}{2} \hat{x} \left(P_L \exp{(-i \omega_0 t)} + c.c.\right) [/tex]
    [tex] \vec{P_L}(\vec{r},t) = \epsilon_0 \int_{-\infty}^{\infty} \chi^{(1)} (t-t') \cdot E(\vec{r},t') dt' [/tex]
    According to the book this should result in:
    [tex] P_L(\vec{r},t) = \epsilon_0 \int_{-\infty}^{\infty}\chi^{(1)}_{xx} (t-t') \cdot E(\vec{r},t') \exp{(i \omega_0 (t-t'))} dt' [/tex]
    under the assumption that [itex] \chi [/itex] was diagonal, lumping [itex] \hat{x} [/itex] and [itex] \chi^{(1)} (t-t') [/itex] together makes sense. But what I don't get is how to integrate the exponentials into the integral. It looks like the shift theorem, but the sum of the two exponentials leaves me puzzled. Can anyone give me a hint?

    Thank you very much in advance,
    Last edited: Oct 5, 2014
  2. jcsd
  3. Oct 10, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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