---
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 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:

$$\begin{aligned}
    \tilde{\chi}(\omega)
    = (\tilde{\chi} * \tilde{\Theta})(\omega)
    = B \int_{-\infty}^\infty \tilde{\chi}(\omega') \: \tilde{\Theta}(\omega - \omega') \dd{\omega'}
\end{aligned}$$

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.