Improve knowledge baseHEADmaster

1 files changed, 187 insertions, 28 deletions

diff --git a/source/know/concept/lindhard-function/index.md b/source/know/concept/lindhard-function/index.md
index 1e3c675..67af408 100644
--- a/source/know/concept/lindhard-function/index.md
+++ b/source/know/concept/lindhard-function/index.md
@@ -376,30 +376,46 @@ $$\begin{aligned}
     - \int_{-\infty}^{\infty} \frac{n_F(\xi_{- \vb{k}'})}{\xi_{- \vb{k}' - \vb{q}} - \xi_{- \vb{k}'} + \hbar (\omega + i \eta)} \dd{\vb{k}'}
     \bigg)
     \\
-    &= \frac{1}{4 \pi^3} \int_{-\infty}^{\infty} n_F(\xi_{\vb{k}}) \bigg(
+    &= \frac{1}{4 \pi^3} \int_{-\infty}^{\infty} \bigg(
     \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} + \hbar (\omega + i \eta)}
     + \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} - \hbar (\omega + i \eta)}
-    \bigg) \dd{\vb{k}}
+    \bigg) n_F(\xi_{\vb{k}}) \dd{\vb{k}}
+\end{aligned}$$
+
+Considering the system's symmetry, it is natural
+to switch to [spherical coordinates](/know/concept/spherical-coordinates/)
+$$(|\vb{k}|, \theta, \varphi)$$:
+
+$$\begin{aligned}
+    \chi_0
+    &= \frac{1}{4 \pi^3} \int_0^\infty \!\! \int_0^\pi \! \int_{-\pi}^\pi \! \bigg(
+    \frac{1}{\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}} + \hbar (\omega \!+\! i \eta)}
+    + \frac{1}{\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}} - \hbar (\omega \!+\! i \eta)}
+    \bigg) n_F(\xi_{\vb{k}}) \, |\vb{k}|^2 \sin{\theta} \dd{\varphi} \dd{\theta} \dd{|\vb{k}|}
+    \\
+    &= \frac{2 \pi}{4 \pi^3} \int_0^\infty \!\! \int_0^\pi \bigg(
+    \frac{1}{\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}} + \hbar (\omega \!+\! i \eta)}
+    + \frac{1}{\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}} - \hbar (\omega \!+\! i \eta)}
+    \bigg) n_F(\xi_{\vb{k}}) \, |\vb{k}|^2 \sin{\theta} \dd{\theta} \dd{|\vb{k}|}
 \end{aligned}$$
 
 At absolute zero $$T = 0$$, the Fermi gas has a well-known ground state,
 in which the occupied single-particle plane wave states
 form a sphere in $$\vb{k}$$-space with radius $$k_F$$.
-In that case, $$n_F(\xi_\vb{k})$$ reduces to
-a Heaviside step function $$\Theta(k_F - |\vb{k}|)$$.
-We switch to [spherical coordinates](/know/concept/spherical-coordinates/)
-$$(|\vb{k}|, \theta, \varphi)$$ and adjust $$|\vb{k}|$$'s
-integration limit based on the step function's demand that $$|\vb{k}| < k_F$$:
+This is a good approximation for nonzero $$T$$ too, because in practice,
+the electrons' Fermi level $$\xi_F \equiv \hbar^2 k_F^2 / (2 m)$$
+is huge compared to thermal excitations.
+Formally, $$n_F(\xi_\vb{k}) \approx \Theta(k_F - |\vb{k}|)$$,
+so we adjust the integration limit:
 
 $$\begin{aligned}
-    \chi_0(\vb{q}, \omega)
-    &= \frac{2 \pi}{4 \pi^3} \int_0^{k_F} \! \int_0^\pi
-    \bigg( \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} + \hbar (\omega \!+\! i \eta)}
-    + \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} - \hbar (\omega \!+\! i \eta)} \bigg) |\vb{k}|^2 \sin{\theta} \dd{\theta} \dd{|\vb{k}|}
+    \chi_0
+    &\approx \frac{1}{2 \pi^2} \int_0^{k_F} \! \int_0^\pi \bigg(
+    \frac{1}{\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}} + \hbar (\omega \!+\! i \eta)}
+    + \frac{1}{\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}} - \hbar (\omega \!+\! i \eta)}
+    \bigg) |\vb{k}|^2 \sin{\theta} \dd{\theta} \dd{|\vb{k}|}
 \end{aligned}$$
 
-Where we have already used that $$\varphi$$ does not appear
-in the integrand to get a factor $$2 \pi$$.
 Next, let us take a closer look at the excitation energy in the denominator.
 Thanks to symmetry, we can always choose our coordinate system
 such that $$\vb{q}$$ points in the $$\theta = 0$$ direction, yielding:
@@ -426,14 +442,20 @@ $$y \equiv \hbar (\omega + i \eta) / (4 \xi_F)$$
 and $$c \equiv \cos{\theta}$$, giving:
 
 $$\begin{aligned}
-    \chi_0(\vb{q}, \omega)
+    \chi_0
     &= \frac{k_F^3}{8 \pi^2 \xi_F} \int_0^1 \! \int_{-1}^1 \bigg(
     \frac{1}{-x^2 - x \kappa c + y} + \frac{1}{-x^2 - x \kappa c - y}
     \bigg) \kappa^2 \dd{c} \dd{\kappa}
 \end{aligned}$$
 
-Evaluating the integral over $$c$$ gives
-a result that involves the complex natural logarithm $$\ln$$:
+From now on, we need to treat $$\chi_0$$'s real and imaginary parts separately
+using the [Sokhotski-Plemelj theorem](/know/concept/sokhotski-plemelj-theorem/).
+The following is valid for the real part,
+but is (for now) still written as a complex expression.
+The point is that, to the best of my knowledge,
+it is not practical to extract the imaginary part
+from the "real part's complex result" so to speak.
+So, hiding the [principal value](/know/concept/cauchy-principal-value/):
 
 $$\begin{aligned}
     \chi_0
@@ -494,13 +516,23 @@ $$\begin{aligned}
     \bigg)
 \end{aligned}$$
 
-We thus arrive at this expression
+Finally, we also know that the [density of states](/know/concept/density-of-states/)
+$$g$$ per unit volume of a Fermi gas is given by:
+
+$$\begin{aligned}
+    g(\xi_F)
+    = \frac{2}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \xi_F^{1/2}
+    = \frac{m k_F}{\pi^2 \hbar^2}
+\end{aligned}$$
+
+Now we officially take the real part,
+and arrive at the following expression
 for the response function of non-interacting electrons:
 
 $$\begin{aligned}
     \boxed{
-        \chi_0(\vb{q}, \omega)
-        = - \frac{m k_F}{\pi^2 \hbar^2} \: \mathcal{F}\bigg(\frac{|\vb{q}|}{2 k_F}, \: \frac{\hbar (\omega \!+\! i \eta)}{4 \xi_F}\bigg)
+        \Real{\chi_0(\vb{q}, \omega)}
+        = - g(\xi_F) \: \mathcal{F}\bigg(\frac{|\vb{q}|}{2 k_F}, \: \frac{\hbar \omega}{4 \xi_F}\bigg)
     }
 \end{aligned}$$
 
@@ -511,17 +543,144 @@ $$\begin{aligned}
     \boxed{
         \mathcal{F}(x, y)
         \equiv \frac{1}{2}
-        + \frac{1}{8 x} \bigg( 1 - \Big(x \!+\! \frac{y}{x}\Big)^{\!2} \bigg) \ln\!\bigg( \frac{x + y/x + 1}{x + y/x - 1} \bigg)
-        + \frac{1}{8 x} \bigg( 1 - \Big(x \!-\! \frac{y}{x}\Big)^{\!2} \bigg) \ln\!\bigg( \frac{x - y/x + 1}{x - y/x - 1} \bigg)
+        + \frac{1}{8 x} \bigg( 1 - \Big(x + \frac{y}{x}\Big)^{\!2} \bigg) \ln\!\bigg| \frac{x + y/x + 1}{x + y/x - 1} \bigg|
+        + \frac{1}{8 x} \bigg( 1 - \Big(x - \frac{y}{x}\Big)^{\!2} \bigg) \ln\!\bigg| \frac{x - y/x + 1}{x - y/x - 1} \bigg|
     }
 \end{aligned}$$
 
-Keep in mind that these are complex logarithms,
-whose branch cut should lie along the negative real axis,
-such that the results' imaginary parts are always between $$-\pi$$ and $$\pi$$.
-That way, it is consistent with alternative derivations
-that use the [Sokhotski-Plemelj theorem](/know/concept/sokhotski-plemelj-theorem/)
-(e.g. Mihaila's paper).
+Now we move on to $$\chi_0$$'s imaginary part.
+Recall the following expression from earlier:
+
+$$\begin{aligned}
+    \chi_0
+    &= \frac{- k_F^3}{8 \pi^2 \xi_F} \frac{1}{x} \int_0^1 \! \int_{-1}^1 \bigg(
+    \frac{1}{x + y/x + \kappa c} + \frac{1}{x - y/x + \kappa c}
+    \bigg) \kappa^2 \dd{c} \dd{\kappa}
+\end{aligned}$$
+
+The Sokhotski-Plemelj theorem then states that the imaginary part is as shown below.
+There is a nasty detail here regarding the positive infinitesimal $$\eta \to 0^{+}$$:
+the first input term's denominator contains $$+ i \eta$$,
+so the theorem can be applied normally,
+but the second term has $$-i \eta$$, meaning the theorem's output must be negated!
+
+$$\begin{aligned}
+    \Imag{\chi_0}
+    &= \frac{-2 m k_F^3}{8 \pi^2 \hbar^2 k_F^2} \frac{1}{x} \int_0^1 \! \int_{-1}^1 \bigg(
+    \!-\! \pi \delta(z_{+} + \kappa c) + \pi \delta(z_{-} + \kappa c)
+    \bigg) \kappa^2 \dd{c} \dd{\kappa}
+    \\
+    &= \frac{\pi m k_F}{4 \pi^2 \hbar^2} \frac{1}{x} \int_0^1 \bigg[
+    \frac{1}{\kappa} \Theta(z_{+} + \kappa c) - \frac{1}{\kappa} \Theta(z_{-} + \kappa c)
+    \bigg]_{-1}^1 \kappa^2 \dd{\kappa}
+    \\
+    &= \frac{\pi}{4 x} \: g(\xi_F) \int_0^1 \bigg(
+    \Theta(z_{+} + \kappa) - \Theta(z_{+} - \kappa) - \Theta(z_{-} + \kappa) + \Theta(z_{-} - \kappa)
+    \bigg) \kappa \dd{\kappa}
+\end{aligned}$$
+
+These integrals are easily evaluated
+according to the following straightforward relations:
+
+$$\begin{aligned}
+    \int_0^1 \Theta(\alpha - \kappa) \: \kappa \dd{\kappa}
+    &= \frac{1}{2} \big( \alpha^2 - (\alpha^2 - 1) \: \Theta(\alpha - 1) \big) \: \Theta(\alpha)
+    \\
+    \int_0^1 \Theta(\alpha + \kappa) \: \kappa \dd{\kappa}
+    &= \frac{1}{2} \big( \alpha^2 \: \Theta(\alpha) - \alpha^2 + 1 \big) \: \Theta(\alpha + 1)
+\end{aligned}$$
+
+Let us assume that $$y > 0$$ for now,
+such that $$z_{+} > 0$$ and $$z_{+} > z_{-}$$.
+We can then naively distinguish several options to be considered:
+
+1.  $$z_{+} > z_{-} > 1 > 0 > -1$$.
+2.  $$z_{+} > 1 > z_{-} > 0 > -1$$.
+3.  $$z_{+} > 1 > 0 > z_{-} > -1$$.
+4.  $$z_{+} > 1 > 0 > -1 > z_{-}$$.
+5.  $$1 > z_{+} > z_{-} > 0 > -1$$.
+6.  $$1 > z_{+} > 0 > z_{-} > -1$$.
+7.  $$1 > z_{+} > 0 > -1 > z_{-}$$.
+
+It can be shown that #7 is impossible
+due to the definitions of $$z_{+}$$ and $$z_{-}$$.
+A tedious investigation of the six remaining cases then yields:
+
+1.  In cases #1 or #4,
+    or, equivalently, if both $$|z_{+}| > 1$$ and $$|z_{-}| > 1$$,
+    the result vanishes:
+
+    $$\begin{aligned}
+        \Imag{\chi_0}
+        = 0
+    \end{aligned}$$
+
+    An alternative statement of those conditions can be found, namely:
+
+    $$\begin{aligned}
+        \begin{cases}
+            y > x^2 + x
+            &\mathrm{if} \: x < 1
+            \\
+            x^2 - x > y > 0
+            &\mathrm{if} \: x > 1
+        \end{cases}
+    \end{aligned}$$
+
+2.  In cases #2 or #3,
+    or, equivalently, if both $$|z_{+}| > 1$$ and $$|z_{-}| < 1$$,
+    the result is parabolic in $$y$$.
+    These conditions can be restated as $$x^2 + x > y > |x^2 - x|$$.
+
+    $$\begin{aligned}
+        \Imag{\chi_0}
+        &= -\frac{\pi}{8 x} \: g(\xi_F) \: (1 - z_{-}^2)
+    \end{aligned}$$
+
+3.  In cases #5 or #6,
+    or, equivalently, if both $$|z_{+}| < 1$$ and $$|z_{-}| < 1$$
+    the result is linear in $$y$$.
+    These conditions can be restated as $$-(x^2 - x) > y > 0$$.
+
+    $$\begin{aligned}
+        \Imag{\chi_0}
+        &= \frac{\pi}{8 x} \: g(\xi_F) \: (z_{-}^2 - z_{+}^2)
+        = - \frac{\pi}{2 x} \: g(\xi_F) \: y
+    \end{aligned}$$
+
+Repeating the analysis for $$y < 0$$ and putting everything together,
+we arrive at the following formula for the imaginary part
+of the non-interacting response function:
+
+$$\begin{gathered}
+    \boxed{
+        \Imag{\chi_0(\vb{q}, \omega)}
+        = - g(\xi_F) \: \mathcal{G}\bigg( \frac{|\vb{q}|}{2 k_F}, \: \frac{\hbar \omega}{4 \xi_F} \bigg)
+    }
+    \\
+    \boxed{
+        \mathcal{G}(x, y)
+        \equiv
+        \begin{cases}
+            0
+            &\mathrm{if} \: |y| > x^2 + x > 0
+            \\
+            \sgn(y) \displaystyle\frac{\pi}{8 x} \bigg( 1 - \Big( x - \displaystyle\frac{|y|}{x} \Big)^2 \bigg)
+            &\mathrm{if} \: x^2 + x > |y| > |x^2 - x|
+            \\
+            \displaystyle\frac{\pi}{2 x} y
+            &\mathrm{if} \: \!-\!(x^2 - x) > |y| > 0
+            \\
+            0
+            &\mathrm{if} \: x^2 - x > |y| > 0
+        \end{cases}
+    }
+\end{gathered}$$
+
+We are being a bit sloppy here by ignoring the boundaries,
+e.g. $$|y| = x^2 - x$$.
+Fortunately, they are all naturally continuous,
+so either of the two neighboring formulas can be used.
 
 
 
@@ -594,7 +753,7 @@ $$\begin{aligned}
     \boxed{
         \mathcal{F}(x, 0)
         = \frac{1}{2}
-        + \frac{1}{4 x} (1 - x^2) \ln\!\bigg( \frac{x + 1}{x - 1} \bigg)
+        + \frac{1}{4 x} (1 - x^2) \ln\!\bigg| \frac{x + 1}{x - 1} \bigg|
     }
 \end{aligned}$$