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diff --git a/source/know/concept/kramers-kronig-relations/index.md b/source/know/concept/kramers-kronig-relations/index.md index 711023e..68e27dc 100644 --- a/source/know/concept/kramers-kronig-relations/index.md +++ b/source/know/concept/kramers-kronig-relations/index.md @@ -10,124 +10,145 @@ categories: 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 $$\chi_i(t)$$: +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/): $$\begin{aligned} - \chi(t) = \chi_r(t) + i \chi_i(t) + \chi(t) + = \chi(t) \: \Theta(t) \end{aligned}$$ -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 +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 from the FT definition: +where $$A$$, $$B$$ and $$s$$ are constants determined by +how we choose to define our FT: $$\begin{aligned} \tilde{\chi}(\omega) - = (\tilde{\chi} * \tilde{\Theta})(\omega) - = B \int_{-\infty}^\infty \tilde{\chi}(\omega') \: \tilde{\Theta}(\omega - \omega') \dd{\omega'} + &= (\tilde{\chi} * \tilde{\Theta})(\omega) + \\ + &= B \int_{-\infty}^\infty \tilde{\chi}(\omega') \: \tilde{\Theta}(\omega - \omega') \dd{\omega'} \end{aligned}$$ -We look up the FT of the step function $$\tilde{\Theta}(\omega)$$, +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 $$\pv{}$$. -We arrive at: +and the [Cauchy principal value](/know/concept/cauchy-principal-value/) $$\pv{}$$. +Inserting that, we arrive at: $$\begin{aligned} \tilde{\chi}(\omega) &= \frac{A B}{|s|} \pv{\int_{-\infty}^\infty \tilde{\chi}(\omega') - \Big( \pi \delta(\omega - \omega') + i \:\mathrm{sgn} \frac{1}{\omega - \omega'} \Big) \dd{\omega'}} + \bigg( \pi \delta(\omega - \omega') + i \frac{\mathrm{sgn}(s)}{\omega - \omega'} \bigg) \dd{\omega'}} \\ - &= \Big( \frac{1}{2} \frac{2 \pi A B}{|s|} \Big) \tilde{\chi}(\omega) - + i \Big( \frac{\mathrm{sgn}(s)}{2 \pi} \frac{2 \pi A B}{|s|} \Big) + &= \bigg( \frac{2}{2} \frac{\pi A B}{|s|} \bigg) \tilde{\chi}(\omega) + + i \: \mathrm{sgn}(s) \bigg( \frac{2 \pi}{2 \pi} \frac{A B}{|s|} \bigg) \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} \end{aligned}$$ -From the definition of the Fourier transform we know that -$$2 \pi A B / |s| = 1$$: +From the definition of the FT we know that +$$2 \pi A B / |s| = 1$$, so this reduces to: $$\begin{aligned} \tilde{\chi}(\omega) &= \frac{1}{2} \tilde{\chi}(\omega) - + \mathrm{sgn}(s) \frac{i}{2 \pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} + + i \: \mathrm{sgn}(s) \frac{1}{2 \pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} \end{aligned}$$ -We isolate this equation for $$\tilde{\chi}(\omega)$$ -to get the final version of the causality test: +We rearrange this equation a bit to get the final version of the causality test: $$\begin{aligned} \boxed{ \tilde{\chi}(\omega) - = - \mathrm{sgn}(s) \frac{i}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} + = i \: \mathrm{sgn}(s) \frac{1}{\pi} + \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} } \end{aligned}$$ -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: +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: $$\begin{aligned} \boxed{ \begin{aligned} \tilde{\chi}_r(\omega) - &= \mathrm{sgn}(s) \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_i(\omega')}{\omega' - \omega} \dd{\omega'}} + &= - \mathrm{sgn}(s) \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_i(\omega')}{\omega - \omega'} \dd{\omega'}} \\ \tilde{\chi}_i(\omega) - &= - \mathrm{sgn}(s) \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_r(\omega')}{\omega' - \omega} \dd{\omega'}} + &= \mathrm{sgn}(s) \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_r(\omega')}{\omega - \omega'} \dd{\omega'}} \end{aligned} } \end{aligned}$$ -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: +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 $$\omega + \omega'$$: $$\begin{aligned} \tilde{\chi}_r(\omega) - &= \mathrm{sgn}(s) \bigg( \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\omega' \tilde{\chi}_i(\omega')}{ {\omega'}^2 - \omega^2} \dd{\omega'}} - + \frac{\omega}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_i(\omega')}{ {\omega'}^2 - \omega^2} \dd{\omega'}} \bigg) + &= - \mathrm{sgn}(s) \frac{1}{\pi} + \bigg( \!\pv{\int_{-\infty}^\infty \frac{\omega \tilde{\chi}_i(\omega')}{\omega^2 - {\omega'}^2} \dd{\omega'}} + + \pv{\int_{-\infty}^\infty \frac{\omega' \tilde{\chi}_i(\omega')}{\omega^2 - {\omega'}^2} \dd{\omega'}} \bigg) \\ \tilde{\chi}_i(\omega) - &= - \mathrm{sgn}(s) \bigg( \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\omega' \tilde{\chi}_r(\omega')}{ {\omega'}^2 - \omega^2} \dd{\omega'}} - + \frac{\omega}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_r(\omega')}{ {\omega'}^2 - \omega^2} \dd{\omega'}} \bigg) + &= \mathrm{sgn}(s) \frac{1}{\pi} + \bigg( \!\pv{\int_{-\infty}^\infty \frac{\omega \tilde{\chi}_r(\omega')}{\omega^2 - {\omega'}^2} \dd{\omega'}} + + \pv{\int_{-\infty}^\infty \frac{\omega' \tilde{\chi}_r(\omega')}{\omega^2 - {\omega'}^2} \dd{\omega'}} \bigg) \end{aligned}$$ -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: +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 $$\chi(t)$$: $$\begin{aligned} \boxed{ \begin{aligned} \tilde{\chi}_r(\omega) - &= \mathrm{sgn}(s) \frac{2}{\pi} \pv{\int_0^\infty \frac{\omega' \tilde{\chi}_i(\omega')}{ {\omega'}^2 - \omega^2} \dd{\omega'}} + &= - \mathrm{sgn}(s) \frac{2}{\pi} + \pv{\int_0^\infty \frac{\omega' \tilde{\chi}_i(\omega')}{\omega^2 - {\omega'}^2} \dd{\omega'}} \\ \tilde{\chi}_i(\omega) - &= - \mathrm{sgn}(s) \frac{2 \omega}{\pi} \pv{\int_0^\infty \frac{\tilde{\chi}_r(\omega')}{ {\omega'}^2 - \omega^2} \dd{\omega'}} + &= \mathrm{sgn}(s) \frac{2}{\pi} + \pv{\int_0^\infty \frac{\omega \tilde{\chi}_r(\omega')}{\omega^2 - {\omega'}^2} \dd{\omega'}} \end{aligned} } \end{aligned}$$ -To reiterate: this version is only valid if $$\chi(t)$$ is real in the time domain. +Note that we have modified the integration limits +using the fact that the integrands are even, +leading to an extra factor of $$2$$. |