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I'm having some problems with the definition of the real and imaginary parts of the Lorentz model, we start from the definition of the permitivity $$\chi=\chi'+i\chi''=\frac{\omega_p^2}{\omega^2-\omega^2-i\omega \gamma}=\omega_p^2\frac{\omega_0^2-\omega^2}{(\omega_0^2-\omega^2)^2+(\omega \gamma)^2}+i\omega_p^2\frac{\omega \gamma}{(\omega_0^2-\omega^2)^2+(\omega \gamma)^2}$$

So we have,

\begin{align} \chi'=\omega_p^2\frac{\omega_0^2-\omega^2}{(\omega_0^2-\omega^2)^2+(\omega \gamma)^2} \quad \xrightarrow{} \quad \text{Refractive index}\\ \chi''=\omega_p^2\frac{\omega \gamma}{(\omega_0^2-\omega^2)^2+(\omega \gamma)^2}\quad \xrightarrow{} \quad \text{Absorption} \qquad \, \end{align}

This results are correct, you can see them in different sites for example MIT-Lorentz-Oscillator, and as we expect, for the absorption we get a Lorentzian, and for the refractive index we get a dispersive distribution, as is illustrated at the image,enter image description here
In the image are represented the permeabilities but are proportional. My doubt comes when if you plot the before expressions $\chi(\omega)$ you don't get the Lorentzian or the dispersive plots, in fact $\chi'$ is a even function, and $\chi''$ is a odd function, that doesnt match with the linked image... And I don't know what I'm doing wrong

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    $\begingroup$ What does it look like when you try to plot them? The parity of the function is irrelevant because we only consider $\omega > 0$. $\endgroup$
    – user318039
    Commented Nov 10, 2021 at 18:15
  • $\begingroup$ That was my error, I was looking the function around zero, but we are interested in the part around $\omega_0$, my confusion comes from that the imaginary plot with $\omega_0 \rightarrow 0$ is almost exactly a dispersive plot around zero. Thanks! $\endgroup$
    – Euler
    Commented Nov 11, 2021 at 6:51
  • $\begingroup$ The functions may seem as if their parity was wrong, however the submitted picture does not show the functions around $\omega=0$. We're looking at high values of $\omega$. If you were to look at negative values, you'd see that the parity is actually correct. $\endgroup$
    – Dio
    Commented Nov 19, 2022 at 2:00

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Yeah, the imaginary part (loss) is supposed to be an odd function. For negative frequency, you have gain. Try plotting it!

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