--- title: "Kramers-Kronig relations" sort_title: "Kramers-Kronig relations" date: 2021-02-25 categories: - Mathematics - Complex analysis - Physics - Optics layout: "concept" --- Let $$\chi(t)$$ be the response function of a system to an external impulse $$f(t)$$, which starts at $$t = 0$$. Assuming initial equilibrium, the principle of causality states that there is no response before the impulse, so $$\chi(t) = 0$$ for $$t < 0$$. To enforce this, we demand that $$\chi(t)$$ satisfies a **causality test**, where $$\Theta(t)$$ is the [Heaviside step function](/know/concept/heaviside-step-function/): $$\begin{aligned} \chi(t) = \chi(t) \: \Theta(t) \end{aligned}$$ If we take the [Fourier transform](/know/concept/fourier-transform/) (FT) $$\chi(t) \!\to\! \tilde{\chi}(\omega)$$ of this equation, the right-hand side becomes a convolution in the frequency domain thanks to the [convolution theorem](/know/concept/convolution-theorem/), where $$A$$, $$B$$ and $$s$$ are constants determined by how we choose to define our FT: $$\begin{aligned} \tilde{\chi}(\omega) &= (\tilde{\chi} * \tilde{\Theta})(\omega) \\ &= B \int_{-\infty}^\infty \tilde{\chi}(\omega') \: \tilde{\Theta}(\omega - \omega') \dd{\omega'} \end{aligned}$$ We look up the full expression for $$\tilde{\Theta}(\omega)$$, which involves the signum function $$\mathrm{sgn}(t)$$, the [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta$$, and the [Cauchy principal value](/know/concept/cauchy-principal-value/) $$\pv{}$$. Inserting that, we arrive at: $$\begin{aligned} \tilde{\chi}(\omega) &= \frac{A B}{|s|} \pv{\int_{-\infty}^\infty \tilde{\chi}(\omega') \bigg( \pi \delta(\omega - \omega') + i \frac{\mathrm{sgn}(s)}{\omega - \omega'} \bigg) \dd{\omega'}} \\ &= \bigg( \frac{2}{2} \frac{\pi A B}{|s|} \bigg) \tilde{\chi}(\omega) + i \: \mathrm{sgn}(s) \bigg( \frac{2 \pi}{2 \pi} \frac{A B}{|s|} \bigg) \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} \end{aligned}$$ From the definition of the FT we know that $$2 \pi A B / |s| = 1$$, so this reduces to: $$\begin{aligned} \tilde{\chi}(\omega) &= \frac{1}{2} \tilde{\chi}(\omega) + i \: \mathrm{sgn}(s) \frac{1}{2 \pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} \end{aligned}$$ We rearrange this equation a bit to get the final version of the causality test: $$\begin{aligned} \boxed{ \tilde{\chi}(\omega) = i \: \mathrm{sgn}(s) \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} } \end{aligned}$$ Next, we split $$\tilde{\chi}(\omega)$$ into its real and imaginary parts, i.e. $$\tilde{\chi}(\omega) = \tilde{\chi}_r(\omega) + i \tilde{\chi}_i(\omega)$$: $$\begin{aligned} \tilde{\chi}_r(\omega) + i \tilde{\chi}_i(\omega) = i \: \mathrm{sgn}(s) \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_r(\omega')}{\omega - \omega'} \dd{\omega'}} - \mathrm{sgn}(s) \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_i(\omega')}{\omega - \omega'} \dd{\omega'}} \end{aligned}$$ This equation can likewise be split into real and imaginary parts, leading to the **Kramers-Kronig relations**, which enable us to reconstruct $$\tilde{\chi}_r(\omega)$$ from $$\tilde{\chi}_i(\omega)$$ and vice versa: $$\begin{aligned} \boxed{ \begin{aligned} \tilde{\chi}_r(\omega) &= - \mathrm{sgn}(s) \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_i(\omega')}{\omega - \omega'} \dd{\omega'}} \\ \tilde{\chi}_i(\omega) &= \mathrm{sgn}(s) \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_r(\omega')}{\omega - \omega'} \dd{\omega'}} \end{aligned} } \end{aligned}$$ The sign of these expressions deserves special attention: it depends on an author's choice of FT definition via $$\mathrm{sgn}(s)$$, and, to make matters even more confusing, many also choose to use the opposite sign in the denominator, i.e. they write $$\omega' - \omega$$ instead of $$\omega - \omega'$$. In the special case where $$\chi(t)$$ is real, we can take advantage of the property that the FT of a real function always satisfies $$\tilde{\chi}(-\omega) = \tilde{\chi}^*(\omega)$$. Here, this means that $$\tilde{\chi}_r(\omega)$$ is even and $$\tilde{\chi}_i(\omega)$$ is odd. To use this fact, we simultaneously multiply and divide the integrands by $$\omega + \omega'$$: $$\begin{aligned} \tilde{\chi}_r(\omega) &= - \mathrm{sgn}(s) \frac{1}{\pi} \bigg( \!\pv{\int_{-\infty}^\infty \frac{\omega \tilde{\chi}_i(\omega')}{\omega^2 - {\omega'}^2} \dd{\omega'}} + \pv{\int_{-\infty}^\infty \frac{\omega' \tilde{\chi}_i(\omega')}{\omega^2 - {\omega'}^2} \dd{\omega'}} \bigg) \\ \tilde{\chi}_i(\omega) &= \mathrm{sgn}(s) \frac{1}{\pi} \bigg( \!\pv{\int_{-\infty}^\infty \frac{\omega \tilde{\chi}_r(\omega')}{\omega^2 - {\omega'}^2} \dd{\omega'}} + \pv{\int_{-\infty}^\infty \frac{\omega' \tilde{\chi}_r(\omega')}{\omega^2 - {\omega'}^2} \dd{\omega'}} \bigg) \end{aligned}$$ In $$\tilde{\chi}_r(\omega)$$'s equation, the first integrand is odd, so the integral's value is zero. Similarly, for $$\tilde{\chi}_i(\omega)$$, the second integrand is odd, so we drop it too. We thus arrive at the following common variant of the Kramers-Kronig relations, only valid for real $$\chi(t)$$: $$\begin{aligned} \boxed{ \begin{aligned} \tilde{\chi}_r(\omega) &= - \mathrm{sgn}(s) \frac{2}{\pi} \pv{\int_0^\infty \frac{\omega' \tilde{\chi}_i(\omega')}{\omega^2 - {\omega'}^2} \dd{\omega'}} \\ \tilde{\chi}_i(\omega) &= \mathrm{sgn}(s) \frac{2}{\pi} \pv{\int_0^\infty \frac{\omega \tilde{\chi}_r(\omega')}{\omega^2 - {\omega'}^2} \dd{\omega'}} \end{aligned} } \end{aligned}$$ Note that we have modified the integration limits using the fact that the integrands are even, leading to an extra factor of $$2$$. ## References 1. M. Wubs, *Optical properties of solids: Kramers-Kronig relations*, 2013, unpublished.