From 075683cdf4588fe16f41d9f7b46b9720b42b2553 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Wed, 17 Jul 2024 10:01:43 +0200
Subject: Improve knowledge base

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 .../know/concept/kramers-kronig-relations/index.md | 133 ++++++++++++---------
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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 $$2$$.
 
 
 
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