# Linear Component of Polarization - Mathematical transformation

1. Oct 5, 2014

### spookyfw

Hello,

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:
$$\vec{P_L}(\vec{r},t) = \frac{1}{2} \hat{x} \left(P_L \exp{(-i \omega_0 t)} + c.c.\right)$$
into
$$\vec{P_L}(\vec{r},t) = \epsilon_0 \int_{-\infty}^{\infty} \chi^{(1)} (t-t') \cdot E(\vec{r},t') dt'$$
According to the book this should result in:
$$P_L(\vec{r},t) = \epsilon_0 \int_{-\infty}^{\infty}\chi^{(1)}_{xx} (t-t') \cdot E(\vec{r},t') \exp{(i \omega_0 (t-t'))} dt'$$
under the assumption that $\chi$ was diagonal, lumping $\hat{x}$ and $\chi^{(1)} (t-t')$ 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,
spookyfw

Last edited: Oct 5, 2014
2. Oct 10, 2014