8.14: Appendix V- Kramers-Krönig Relations

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Suppose ${\hat\chi}(\omega)\equiv {\hat G}(\omega)$ is analytic in the UHP¹⁹. Then for all $\nu$, we must have $$\impi{d\nu\over 2\pi}\,{{\hat\chi}(\nu)\over\nu-\omega+i\eps}=0\ ,\label{kka}$$ where $\eps$ is a positive infinitesimal. The reason is simple: just close the contour in the UHP, assuming ${\hat\chi}(\omega)$ vanishes sufficiently rapidly that Jordan’s lemma can be applied. Clearly this is an extremely weak restriction on ${\hat\chi}(\omega)$, given the fact that the denominator already causes the integrand to vanish as $|\omega|^{-1}$.

Let us examine the function $${1\over \nu-\omega+i\eps}={\nu-\omega\over (\nu-\omega)^2+\eps^2}-\, {i\eps\over (\nu-\omega)^2+\eps^2}\ .$$ which we have separated into real and imaginary parts. Under an integral sign, the first term, in the limit $\eps\to 0$, is equivalent to taking a principal part of the integral. That is, for any function $F(\nu)$ which is regular at $\nu=\omega$, $$\lim_{\eps\to 0}\impi {d\nu\over 2\pi}\,{\nu-\omega\over (\nu-\omega)^2+\eps^2} \,F(\nu)\equiv\wp\!\! \impi {d\nu\over 2\pi}\,{F(\nu)\over\nu-\omega}\ .$$ The principal part symbol $\wp$ means that the singularity at $\nu=\omega$ is elided, either by smoothing out the function $1/(\nu-\eps)$ as above, or by simply cutting out a region of integration of width $\eps$ on either side of $\nu=\omega$.

The imaginary part is more interesting. Let us write $$h(u)\equiv {\eps\over u^2+\eps^2}\ .$$ For $|u|\gg\eps$, $h(u)\simeq \eps/u^2$, which vanishes as $\eps\to 0$. For $u=0$, $h(0)=1/\eps$ which diverges as $\eps\to 0$. Thus, $h(u)$ has a huge peak at $u=0$ and rapidly decays to $0$ as one moves off the peak in either direction a distance greater that $\eps$. Finally, note that $$\impi du\,h(u)=\pi\ ,$$ a result which itself is easy to show using contour integration. Putting it all together, this tells us that $$\lim_{\eps\to 0} {\eps\over u^2+\eps^2}=\pi\delta(u)\ .$$ Thus, for positive infinitesimal $\eps$, $${1\over u\pm i\eps}={\wp\over u} \mp i\pi\delta(u)\ ,$$ a most useful result.

We now return to our initial result [kka], and we separate ${\hat\chi}(\omega)$ into real and imaginary parts: $${\hat\chi}(\omega)={\hat\chi}'(\omega)+i{\hat\chi}''(\omega) \ .$$ (In this equation, the primes do not indicate differentiation with respect to argument.) We therefore have, for every real value of $\omega$, $$0=\impi{d\nu\over 2\pi}\,\Big[\chi'(\nu)+i\chi''(\nu)\Big]\, \Big[{\wp\over \nu-\omega}-i\pi\delta(\nu-\omega)\Big]\ .$$ Taking the real and imaginary parts of this equation, we derive the Kramers-Krönig relations: $$\begin{aligned} \chi'(\omega)&=&+\wp\!\!\impi{d\nu\over\pi}\,{{\hat\chi}''(\nu)\over\nu-\omega}\\ \chi''(\omega)&=&-\wp\!\!\impi{d\nu\over\pi}\,{{\hat\chi}'(\nu)\over\nu-\omega}\ .\end{aligned}$$