# [Physics] Proving susceptibility in Lorentz Model satisy Kramers-Kronig relations

optics

My instructor asked me to prove that the real and imaginary parts of the electric susceptibility derived from Lorentz Model satisfy the Kramers-Kronig relations using the residue theorem. The problem is that my complex calculus is pretty rusty and I do not know which poles contribute exactly. There are 5 poles in total 4 from the susceptibility and 1 from the denominator(see the expression please). I took the integral using the principal value option in Mathematica and it turned out not as expected. Is this analytically tractable easily?

$$\chi(\omega) = \frac{\omega_{p}^2}{(\omega_0^2-\omega^2)+i\gamma\omega}$$

The Kramers-Kronig relations are

$$\chi_r(\omega) = \frac{1}{\pi} P \int_{-\infty}^{\infty} d\bar{\omega} \frac{\chi_i(\bar{\omega})}{\bar{\omega}-\omega}$$
$$\chi_i(\omega) = -\frac{1}{\pi} P \int_{-\infty}^{\infty} d\bar{\omega} \frac{\chi_r(\bar{\omega})}{\bar{\omega}-\omega}$$

I assume all the variables involved are real. The roots of the denominator of $$\chi(\omega)$$ are $$\frac{1}{2}\Big(i\gamma\pm\sqrt{-\gamma^2+4\omega_0^2}\Big)$$ which lies in the lower (upper) half of the complex plane for $$\gamma<(>)0$$. You need to further specify the sign of $$\gamma$$. $$\gamma<0$$ leads to the relationship in your question, while $$\gamma>0$$ put negative sign to the left hand sides of the relationship, and $$\gamma=0$$ destroys the relationship.
Suppose $$\gamma<0$$. $$\chi$$ is analytic on the upper half complex plane and by Cauchy's Integral Theorem, $$0 = \frac{1}{2\pi i}\oint_C \frac{\chi(\bar\omega)}{\bar\omega-\omega}\mathrm d\bar\omega,$$ where the contour $$C$$ runs along the real axis from $$-R<0$$ to $$R>0$$ with an infinitesimally small semicircle running clockwise around and above $$\omega$$, then describes the large semicircle circle counter-clockwise in the upper half complex plane with radius $$R$$. The clockwise integral around the small circle above $$\omega$$ approaches $$-\frac{1}{2}\chi(\omega)$$ while the integral on the large semicircle with radius $$R$$ approaches $$0$$ as $$R\rightarrow\infty$$ as the magnitude of the integrand is $$O\big(\frac{1}{R^3}\big)$$. Therefore $$\frac{1}{2}\big(\chi_r(\omega)+i\chi_i(\omega)\big) = \frac{1}{2\pi i}P\int_{-\infty}^\infty \frac{\chi_r(\bar\omega)+i\chi_i(\bar\omega)}{\bar\omega-\omega}\mathrm d\bar\omega.$$ Equating the real and imaginary parts of the equation, leads to the desired result.
To explicitly verify the relationship, we use $$\chi_i=\frac{1}{2i}(\chi-\chi^*). \tag 1$$ The poles of $$\chi$$ lies in the lower half complex plane. Evaluate $$\frac{1}{\pi}P\int_{-\infty}^\infty\frac{\chi(\bar\omega)}{\bar\omega-\omega}\mathrm d\bar\omega$$ using the contour integral described in the general proof above and get $$i\chi(\omega)$$. The poles of $$\chi^*$$ lies in the upper half complex plane. So we evaluate $$\frac{1}{\pi}P\int_{-\infty}^\infty\frac{\chi(\bar\omega)^*}{\bar\omega-\omega}\mathrm d\bar\omega$$ using the previous contour reflected with respect to the real axis and get $$-i\chi(\omega)^*$$. Then evaluate $$\frac{1}{\pi}P\int_{-\infty}^\infty\frac{\chi_i(\bar\omega)}{\bar\omega-\omega}\mathrm d\bar\omega$$ using Equation $$(1)$$, we arrives at the desired first relationship.
The second relationship is derived explicitly similarly with $$\chi_r=\frac{1}{2}(\chi+\chi^*)$$.