[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

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

Best Answer

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 $\chi_r=\frac{1}{2}(\chi+\chi^*)$.

Related Question