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---
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.
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