---
title: "Kramers-Kronig relations"
firstLetter: "K"
publishDate: 2021-02-25
categories:
- Mathematics
- 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.