# The Relationship Between Causality and Kramers-Kronig Relations

The Kramers-Kronig relations tell us that the real and imaginary parts of causal response functions are related. The relations are of immense importance to optical spectroscopists, who use them to obtain, for example, the optical conductivity from the reflectivity. It is often said in textbooks that the Kramers-Kronig relations (KKRs) follow from causality. The proof of this statement usually uses contour integration and the role of causality is then a little obscured. Here, I hope to use intuitive arguments to show the meaning behind the KKRs.

If one imagines applying a sudden force to a simple harmonic oscillator (SHO) and then watches its response, one would expect that the response will look something like this:

We expect the SHO to oscillate for a little while and eventually stop due to friction of some kind. Let’s call the function in the plot above $\chi(t)$. Because $\chi(t)$ is zero before we “kick” the system, we can play a little trick and write $\chi(t) = \chi_s(t)+\chi_a(t)$ where the symmetrized and anti-symmetrized parts are plotted below:

Since for $t<0$, the symmetrized and anti-symmetrized parts will cancel out perfectly, we recover our original spectrum. Just to convince you (as if you needed convincing!) that this works, I have explicitly plotted this:

Now let’s see what happens when we take this over to the frequency domain, where the KKRs apply, by doing a Fourier transform. We can write the following:

$\tilde{\chi}(\omega) = \int_{-\infty}^\infty e^{i \omega t} \chi(t) \mathrm{d}t$ $= \int_{-\infty}^\infty (\mathrm{cos}(\omega t) + i \mathrm{sin}(\omega t)) (\chi_s (t)+\chi_a(t))\mathrm{d}t$

where in the last step I’ve used Euler’s identity for the exponential and I’ve decomposed $\chi(t)$ into its symmetrized and anti-symmetrized parts as before. Now, there is something immediately apparent in the last integral. Because the domain of integration is from $-\infty$ to $\infty$, the area under the curve of any odd (a.k.a anti-symmetric) function will necessarily be zero. Lastly, noting that anti-symmetric $\times$ symmetric = anti-symmetric and symmetric (anti-symmetric) $\times$ symmetric (anti-symmetric) = symmetric, we can write for the equation above:

$\tilde{\chi}(\omega) = \int_{-\infty}^\infty \mathrm{cos}(\omega t) \chi_s(t) \mathrm{d}t$ + $i \int_{-\infty}^\infty \mathrm{sin}(\omega t) \chi_a(t) \mathrm{d}t$ = $\tilde{\chi}_s(\omega) + i \tilde{\chi}_a(\omega)$

Before I continue, some remarks are in order:

1. Even though we now have two functions in the frequency domain (i.e. $\tilde{\chi}_s(\omega)$ and  $\tilde{\chi}_a(\omega)$), they actually derive from one function in the time-domain, $\chi(t)$. We just symmetrized and anti-symmetrized the function artificially.
2. We actually know the relationship between the symmetric and anti-symmetric functions in the time-domain because of causality.
3. The symmetrized part of $\chi(t)$ corresponds to the real part of $\tilde{\chi}(\omega)$. The anti-symmetrized part of $\chi(t)$ corresponds to the imaginary part of $\tilde{\chi}(\omega)$.

With this correspondence, the question then naturally arises:

How do we express the relationship between the real and imaginary parts of $\tilde{\chi}(\omega)$, knowing the relationship between the symmetrized and anti-symmetrized functions in the time-domain?

This actually turns out to not be too difficult and involves just a little more math. First, let us express the relationship between the symmetrized and anti-symmetrized parts of $\chi(t)$ mathematically.

$\chi_s(t) = \mathrm{sgn}(t) \times \chi_a(t)$

where $\mathrm{sgn} (t)$ just changes the sign of the $t<0$ part of the plot and is shown below.

Now let’s take this expression over to the frequency domain. Here, we must use the convolution theorem. This theorem says that if we have two functions multiplied by each other, e.g. $h(t) = f(t)g(t)$, the Fourier transform of this product is expressed as a convolution in the frequency domain as so:

$\tilde{h}(\omega)$=$\mathcal{F}(f(t)g(t)) = \int \tilde{f}(\omega-\omega')\tilde{g}(\omega') \mathrm{d}\omega'$

where $\mathcal{F}$ means Fourier transform. Therefore, all we have left to do is figure out the Fourier transform of $\mathrm{sgn}(t)$. The answer is given here (in terms of frequency not angular frequency!), but it is a fun exercise to work it out on your own. The answer is:

$\mathcal{F}(sgn(t)) = \frac{2}{i\omega}$

With this answer, and using the convolution theorem, we can write:

$\tilde{\chi_s}(\omega) = \int_{-\infty}^{\infty} \frac{2}{i(\omega-\omega')} \tilde{\chi}_a(\omega')\mathrm{d}\omega'$

Hence, up to some factors of $2\pi$ and $i$, we can now see better what is behind the KKRs without using contour integration. We can also see why it is always said that the KKRs are a result of causality. Thinking about the KKRs this way has definitely aided in my thinking about response functions more generally.

I hope to write a post in the future talking a little more about the connection between the imaginary part of the response function and dissipation. Let me know if you think this will be helpful.

A lot of this post was inspired by this set of notes, which I found to be very valuable.

### 3 responses to “The Relationship Between Causality and Kramers-Kronig Relations”

1. Dmytro Solonenko

Dear Anshul,

please, keep posting. I find your excerpts into math around physics extremely insightful and easy to grasp.
I would love to read the second part of the post about relation between the imaginary part of the response function and dissipation.

Kind regards,
Dmytro

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