Categories: Mathematics, Optics, Physics.

# Kramers-Kronig relations

Let $$\chi(t)$$ be a complex function describing the response of a system to an impulse $$f(t)$$ starting at $$t = 0$$. The Kramers-Kronig relations connect the real and imaginary parts of $$\chi(t)$$, such that one can be reconstructed from the other. Suppose we can only measure $$\chi_r(t)$$ or $$\chi_i(t)$$:

\begin{aligned} \chi(t) = \chi_r(t) + i \chi_i(t) \end{aligned}

Assuming that the system was at rest until $$t = 0$$, the response $$\chi(t)$$ cannot depend on anything from $$t < 0$$, since the known impulse $$f(t)$$ had not started yet, This principle is called causality, and to enforce it, we use the Heaviside step function $$\Theta(t)$$ to create a causality test for $$\chi(t)$$:

\begin{aligned} \chi(t) = \chi(t) \: \Theta(t) \end{aligned}

If we Fourier transform this equation, then it will become a convolution in the frequency domain thanks to the convolution theorem, where $$A$$, $$B$$ and $$s$$ are constants from the FT definition:

\begin{aligned} \tilde{\chi}(\omega) %= \hat{\mathcal{F}}\{\chi_c(t) \: \Theta(t)\} = (\tilde{\chi} * \tilde{\Theta})(\omega) = B \int_{-\infty}^\infty \tilde{\chi}(\omega') \: \tilde{\Theta}(\omega - \omega') \dd{\omega'} \end{aligned}

We look up the FT of the step function $$\tilde{\Theta}(\omega)$$, which involves the signum function $$\mathrm{sgn}(t)$$, the Dirac delta function $$\delta$$, and the Cauchy principal value $$\pv{}$$. We arrive at:

\begin{aligned} \tilde{\chi}(\omega) &= \frac{A B}{|s|} \: \pv{\int_{-\infty}^\infty \tilde{\chi}(\omega') \Big( \pi \delta(\omega - \omega') + i \:\mathrm{sgn} \frac{1}{\omega - \omega'} \Big) \dd{\omega'}} \\ &= \Big( \frac{1}{2} \frac{2 \pi A B}{|s|} \Big) \tilde{\chi}(\omega) + i \Big( \frac{\mathrm{sgn}(s)}{2 \pi} \frac{2 \pi A B}{|s|} \Big) \: \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} \end{aligned}

From the definition of the Fourier transform we know that $$2 \pi A B / |s| = 1$$:

\begin{aligned} \tilde{\chi}(\omega) &= \frac{1}{2} \tilde{\chi}(\omega) + \mathrm{sgn}(s) \frac{i}{2 \pi} \: \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} \end{aligned}

We isolate this equation for $$\tilde{\chi}(\omega)$$ to get the final version of the causality test:

\begin{aligned} \boxed{ \tilde{\chi}(\omega) = - \mathrm{sgn}(s) \frac{i}{\pi} \: \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} } \end{aligned}

By inserting $$\tilde{\chi}(\omega) = \tilde{\chi}_r(\omega) + i \tilde{\chi}_i(\omega)$$ and splitting the equation into real and imaginary parts, we get the Kramers-Kronig relations:

\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}

If the time-domain response function $$\chi(t)$$ is real (so far we have assumed it to be complex), then we can take advantage of the fact that the FT of a real function satisfies $$\tilde{\chi}(-\omega) = \tilde{\chi}^*(\omega)$$, i.e. $$\tilde{\chi}_r(\omega)$$ is even and $$\tilde{\chi}_i(\omega)$$ is odd. We multiply the fractions by $$(\omega' + \omega)$$ above and below:

\begin{aligned} \tilde{\chi}_r(\omega) &= \mathrm{sgn}(s) \bigg( \frac{1}{\pi} \: \pv{\int_{-\infty}^\infty \frac{\omega' \tilde{\chi}_i(\omega')}{{\omega'}^2 - \omega^2} \dd{\omega'}} + \frac{\omega}{\pi} \: \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_i(\omega')}{{\omega'}^2 - \omega^2} \dd{\omega'}} \bigg) \\ \tilde{\chi}_i(\omega) &= - \mathrm{sgn}(s) \bigg( \frac{1}{\pi} \: \pv{\int_{-\infty}^\infty \frac{\omega' \tilde{\chi}_r(\omega')}{{\omega'}^2 - \omega^2} \dd{\omega'}} + \frac{\omega}{\pi} \: \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_r(\omega')}{{\omega'}^2 - \omega^2} \dd{\omega'}} \bigg) \end{aligned}

For $$\tilde{\chi}_r(\omega)$$, the second integrand is odd, so we can drop it. Similarly, for $$\tilde{\chi}_i(\omega)$$, the first integrand is odd. We therefore find the following variant of the Kramers-Kronig relations:

\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 \omega}{\pi} \: \pv{\int_0^\infty \frac{\tilde{\chi}_r(\omega')}{{\omega'}^2 - \omega^2} \dd{\omega'}} \end{aligned} } \end{aligned}

To reiterate: this version is only valid if $$\chi(t)$$ is real in the time domain.