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---
title: "Kramers-Kronig relations"
firstLetter: "K"
publishDate: 2021-02-25
categories:
- Mathematics
- Complex analysis
- Physics
- Optics
date: 2021-02-25T15:20:24+01:00
draft: false
markup: pandoc
---
# 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](/know/concept/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](/know/concept/fourier-transform/) this equation,
then it will become a convolution in the frequency domain
thanks to the [convolution theorem](/know/concept/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](/know/concept/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.
## References
1. M. Wubs,
*Optical properties of solids: Kramers-Kronig relations*, 2013,
unpublished.
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