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# 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}