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.


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