Categories: Complex analysis, Mathematics, Optics, Physics.

Kramers-Kronig relations

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

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

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

χ(t)=χ(t)Θ(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 AA, BB and ss are constants from the FT definition:

χ~(ω)=(χ~Θ~)(ω)=Bχ~(ω)Θ~(ωω)dω\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 FT of the step function Θ~(ω)\tilde{\Theta}(\omega), which involves the signum function sgn(t)\mathrm{sgn}(t), the Dirac delta function δ\delta, and the Cauchy principal value P\pv{}. We arrive at:

χ~(ω)=ABsPχ~(ω)(πδ(ωω)+isgn1ωω)dω=(122πABs)χ~(ω)+i(sgn(s)2π2πABs)Pχ~(ω)ωωdω\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 2πAB/s=12 \pi A B / |s| = 1:

χ~(ω)=12χ~(ω)+sgn(s)i2πPχ~(ω)ωωdω\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 χ~(ω)\tilde{\chi}(\omega) to get the final version of the causality test:

χ~(ω)=sgn(s)iπPχ~(ω)ωωdω\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 χ~(ω)=χ~r(ω)+iχ~i(ω)\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:

χ~r(ω)=sgn(s)1πPχ~i(ω)ωωdωχ~i(ω)=sgn(s)1πPχ~r(ω)ωωdω\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 χ(t)\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 χ~(ω)=χ~(ω)\tilde{\chi}(-\omega) = \tilde{\chi}^*(\omega), i.e. χ~r(ω)\tilde{\chi}_r(\omega) is even and χ~i(ω)\tilde{\chi}_i(\omega) is odd. We multiply the fractions by (ω+ω)(\omega' + \omega) above and below:

χ~r(ω)=sgn(s)(1πPωχ~i(ω)ω2ω2dω+ωπPχ~i(ω)ω2ω2dω)χ~i(ω)=sgn(s)(1πPωχ~r(ω)ω2ω2dω+ωπPχ~r(ω)ω2ω2dω)\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 χ~r(ω)\tilde{\chi}_r(\omega), the second integrand is odd, so we can drop it. Similarly, for χ~i(ω)\tilde{\chi}_i(\omega), the first integrand is odd. We therefore find the following variant of the Kramers-Kronig relations:

χ~r(ω)=sgn(s)2πP0ωχ~i(ω)ω2ω2dωχ~i(ω)=sgn(s)2ωπP0χ~r(ω)ω2ω2dω\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 χ(t)\chi(t) is real in the time domain.


  1. M. Wubs, Optical properties of solids: Kramers-Kronig relations, 2013, unpublished.