Improve knowledge base

2 files changed, 336 insertions, 83 deletions

diff --git a/source/know/concept/kubo-formula/index.md b/source/know/concept/kubo-formula/index.md
index 41fda3c..774ddf7 100644
--- a/source/know/concept/kubo-formula/index.md
+++ b/source/know/concept/kubo-formula/index.md
@@ -105,7 +105,7 @@ Where we have defined the **retarded correlation function** $$C^R_{A H_1}(t, t')
 $$\begin{aligned}
     \boxed{
         C^R_{A H_1}(t, t')
-        \equiv - \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0
+        \equiv - \frac{i}{\hbar} \Theta(t - t') \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0
     }
 \end{aligned}$$
 
diff --git a/source/know/concept/lindhard-function/index.md b/source/know/concept/lindhard-function/index.md
index 5f11d36..1e3c675 100644
--- a/source/know/concept/lindhard-function/index.md
+++ b/source/know/concept/lindhard-function/index.md
@@ -1,77 +1,77 @@
 ---
 title: "Lindhard function"
 sort_title: "Lindhard function"
-date: 2022-01-24 # Originally 2021-10-12, major rewrite
+date: 2026-09-24 # Originally 2022-01-24, major rewrite
 categories:
 - Physics
 - Quantum mechanics
 layout: "concept"
 ---
 
-The **Lindhard function** describes the response of
-[jellium](/know/concept/jellium) (i.e. a free electron gas)
-to an external perturbation, and is a quantum-mechanical
-alternative to the [Drude model](/know/concept/drude-model/).
+The **Lindhard function** describes how an external perturbation affects
+a gas of free electrons, and can be regarded as
+the backbone of quantum mechanical models of metals.
 
-We start from the [Kubo formula](/know/concept/kubo-formula/)
-for the electron density operator $$\hat{n}$$,
-which describes the change in $$\Expval{\hat{n}}$$
-due to a time-dependent perturbation $$\hat{H}_1$$:
 
-$$\begin{aligned}
-    \delta\!\Expval{\hat{n}(\vb{r}, t)}
-    = -\frac{i}{\hbar} \int_{-\infty}^\infty \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{r}, t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
-\end{aligned}$$
 
-Where the subscript $$I$$ refers to the [interaction picture](/know/concept/interaction-picture/),
-and the expectation $$\Expval{}_0$$ is for
-a thermal equilibrium before the perturbation was applied.
-Now consider a harmonic $$\hat{H}_1$$:
+## Derivation
+
+Consider the following general harmonic perturbation $$\hat{H}_1$$,
+where $$U(\vb{r})$$ is an arbitrary potential,
+$$\hat{n}(\vb{r})$$ is the electron density operator,
+and $$\eta \to 0^{+}$$ is needed to ensure convergence:
 
 $$\begin{aligned}
     \hat{H}_{1,S}(t)
     = e^{i (\omega + i \eta) t} \int_{-\infty}^\infty U(\vb{r}) \: \hat{n}_S(\vb{r}) \dd{\vb{r}}
 \end{aligned}$$
 
-Where $$S$$ is the Schrödinger picture,
-$$\eta$$ is a positive infinitesimal to ensure convergence later,
-and $$U(\vb{r})$$ is an arbitrary potential function.
-The Kubo formula becomes:
+Here, the subscript $$S$$ refers to the Schrödinger picture.
+The first-order effect of this perturbation
+on the expected density $$\Expval{\hat{n}}$$
+is then given by the [Kubo formula](/know/concept/kubo-formula/) below,
+where the subscript $$I$$ refers to
+the [interaction picture](/know/concept/interaction-picture/),
+and the expectation $$\Expval{}_0$$ is taken with respect to
+the thermal equilibrium that existed before the perturbation was applied:
 
 $$\begin{aligned}
     \delta\!\Expval{\hat{n}(\vb{r}, t)}
-    = \iint_{-\infty}^\infty \chi(\vb{r}, \vb{r}'; t, t') \: U(\vb{r}') \: e^{i (\omega + i \eta) t'} \dd{t'} \dd{\vb{r}'}
+    &= -\frac{i}{\hbar} \int_{-\infty}^\infty \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{r}, t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
+    \\
+    &= \iint_{-\infty}^\infty \chi(\vb{r}, \vb{r}'; t, t') \: U(\vb{r}') \: e^{i (\omega + i \eta) t'} \dd{t'} \dd{\vb{r}'}
 \end{aligned}$$
 
-Here, $$\chi$$ is the density-density correlation function,
-i.e. a two-particle [Green's function](/know/concept/greens-functions/):
+Where $$\Theta$$ is a [Heaviside step function](/know/concept/heaviside-step-function/),
+and the response $$\chi$$ is the density-density correlation function
+(i.e. a two-particle [Green's function](/know/concept/greens-functions/)):
 
 $$\begin{aligned}
     \chi(\vb{r}, \vb{r}'; t, t')
     \equiv - \frac{i}{\hbar} \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{r}, t)}{\hat{n}_I(\vb{r}', t')}}_0
 \end{aligned}$$
 
-Let us assume that the unperturbed system (i.e. without $$U$$) is spatially uniform,
+Let us assume that the unperturbed system (i.e. without $$U$$) was spatially uniform,
 so that $$\chi$$ only depends on the difference $$\vb{r} - \vb{r}'$$.
 We then take its [Fourier transform](/know/concept/fourier-transform/)
-$$\vb{r}\!-\!\vb{r}' \to \vb{q}$$:
+$$\vb{r} - \vb{r}' \to \vb{q}$$:
 
 $$\begin{aligned}
     \chi(\vb{q}; t, t')
     &= \int_{-\infty}^\infty \chi(\vb{r} - \vb{r}'; t, t') \: e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{r}}
     \\
-    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint
+    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2 D}} \iiint_{-\infty}^\infty
     \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0
     \: e^{i \vb{q}_1 \cdot \vb{r}} e^{i \vb{q}_2 \cdot \vb{r}'} e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}}
 \end{aligned}$$
 
 Where both $$\hat{n}_I$$ have been written as inverse Fourier transforms,
-giving a factor $$(2 \pi)^{-2 D}$$, with $$D$$ being the number of spatial dimensions.
+giving a factor $$(2 \pi)^{-D}$$ each, where $$D$$ is the number of spatial dimenions.
 We rearrange to get a [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta$$:
 
 $$\begin{aligned}
     \chi(\vb{q}; t, t')
-    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint
+    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2 D}} \iiint
     \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0
     \: e^{i (\vb{q}_1 - \vb{q}) \cdot \vb{r}} e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}}
     \\
@@ -85,8 +85,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 On the left, $$\vb{r}'$$ does not appear, so it must also disappear on the right.
-If we choose an arbitrary (hyper)cube of volume $$V$$ in real space,
-then clearly $$\int_V \dd{\vb{r}'} = V$$. Therefore:
+To handle this properly, we consider an arbitrary finite piece of volume $$V$$
+of this infinite system. Clearly $$\int_V \dd{\vb{r}'} = V$$, so:
 
 $$\begin{aligned}
     \chi(\vb{q}; t, t')
@@ -95,22 +95,23 @@ $$\begin{aligned}
     \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \dd{\vb{r}'}
 \end{aligned}$$
 
-This gives a Dirac delta function for $$V \to \infty$$
-(a limit that we will take properly later,
+This gives a Dirac delta function when we recover
+the full infinite system by letting $$V \to \infty$$
+(we will take this limit properly later,
 but beware that some authors set $$V = 1$$ until then):
 
 $$\begin{aligned}
     \chi(\vb{q}; t, t')
-    &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \int_{-\infty}^\infty
+    &= -\frac{i}{\hbar V} \Theta(t - t') \int_{-\infty}^\infty
     \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: \delta(\vb{q}_2 \!+\! \vb{q}) \dd{\vb{q}_2}
     \\
-    &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0
+    &= -\frac{i}{\hbar V} \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0
 \end{aligned}$$
 
 Similarly, if the unperturbed Hamiltonian $$\hat{H}_0$$ is time-independent,
-$$\chi$$ only depends on the time difference $$t\!-\!t'$$.
+then $$\chi$$ only depends on the time difference $$t - t'$$.
 Note that $$\delta{\Expval{\hat{n}}}$$ already has the form of a Fourier transform
-$$t\!-\!t' \to \omega\!+\!i \eta$$,
+$$t - t' \to \omega + i \eta$$,
 which gives us an opportunity to rewrite $$\chi$$
 in the [Lehmann representation](/know/concept/lehmann-representation/):
 
@@ -137,10 +138,13 @@ $$\begin{aligned}
     {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu}
 \end{aligned}$$
 
-
-To proceed, we need to rewrite $$\hat{n}(\vb{q})$$ somehow.
-If we neglect electron-electron interactions,
-the single-particle states are simply plane waves, in which case:
+Next, we need to rewrite $$\hat{n}(\vb{q})$$ somehow.
+To do this, we must neglect electron-electron interactions,
+leaving us with a simple [Fermi gas](/know/concept/fermi-gas/).
+In that model, $$V$$ is a cube with periodic boundary conditions,
+meaning that $$\vb{q}$$ has a discrete spectrum of allowed values,
+and the single-electron states are just plane waves.
+This lets us write, in the [second quantization](/know/concept/second-quantization/):
 
 $$\begin{aligned}
     \hat{n}(\vb{q})
@@ -152,10 +156,11 @@ $$\begin{aligned}
 
 
 {% include proof/start.html id="proof-density" -%}
-Starting from the general definition of $$\hat{n}$$,
+Starting from the definition of $$\hat{n}(\vb{r})$$,
 we write out the field operators $$\hat{\Psi}(\vb{r})$$,
 and insert the known non-interacting single-electron orbitals
-$$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$$:
+$$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$$,
+omitting spin:
 
 $$\begin{aligned}
     \hat{n}(\vb{r})
@@ -173,19 +178,9 @@ $$\begin{aligned}
     = \frac{(2 \pi)^D}{V} \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q})
 \end{aligned}$$
 
-If we impose periodic boundary conditions
-on our $$D$$-dimensional hypercube of volume $$V$$,
-then $$\vb{k}$$ becomes discrete,
-with per-value spacing $$2 \pi / V^{1/D}$$ along each axis.
-
-Consequently, each orbital $$\psi_\vb{k}$$ uniquely occupies
-a volume $$(2 \pi)^D / V$$ in $$\vb{k}$$-space, so we make the approximation
-$$\sum_{\vb{k}} \approx V / (2 \pi)^D \int_{-\infty}^\infty \dd{\vb{k}}$$.
-This is exact in the limit $$V \to \infty$$,
-in which case $$\vb{k}$$ also becomes a continuous variable again.
-
-We apply this standard trick from condensed matter physics to $$\hat{n}$$,
-and $$V$$ cancels out:
+Thanks to our periodic boundary conditions,
+[this trick](/know/concept/discrete-spectrum-summation/)
+lets us turn a sum into an integral:
 
 $$\begin{aligned}
     \hat{n}(\vb{q})
@@ -204,9 +199,6 @@ $$\begin{aligned}
     = \sum_{\vb{k}'} \hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'}
     = \hat{n}^\dagger(\vb{q})
 \end{aligned}$$
-
-The summation variable $$\vb{k}$$ has an associated spin $$\sigma$$,
-and $$\hat{n}$$ does not carry any spin.
 {% include proof/end.html id="proof-density" %}
 
 
@@ -214,7 +206,7 @@ When neglecting interactions, it is tradition to rename $$\chi$$ to $$\chi_0$$.
 We insert $$\hat{n}$$, suppressing spin:
 
 $$\begin{aligned}
-    \chi_0
+    \chi_0(\vb{q}, \omega)
     &= \frac{1}{Z V} \sum_{\vb{k} \vb{k}'} \sum_{\nu \nu'} \bigg(
     \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'}
     \matrixel{\nu'}{\hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'}}{\nu}}
@@ -230,14 +222,14 @@ by moving an electron from $$\vb{k}$$ to $$\vb{k} \!+\! \vb{q}$$,
 and analogously for the other inner products.
 As a result, $$\vb{k} = \vb{k}'$$ (and $$\sigma = \sigma'$$).
 
-For the same reason, the energy difference $$E_\nu \!-\! E_{\nu'}$$
+For the same reason, the energy difference $$E_\nu - E_{\nu'}$$
 can simply be replaced by the cost of the single-particle excitation
-$$\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}}$$,
+$$\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}$$,
 where $$\xi_{\vb{k}}$$ is the energy of a $$\vb{k}$$-orbital.
 Therefore:
 
 $$\begin{aligned}
-    \chi_0
+    \chi_0(\vb{q}, \omega)
     &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu \nu'} \bigg(
     \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'}
     \matrixel{\nu'}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}}
@@ -248,10 +240,10 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Notice that we have eliminated all dependence on $$\Ket{\nu'}$$,
-so we remove it by $$\sum_{\nu} \Ket{\nu} \Bra{\nu} = 1$$:
+so we remove it by $$\sum_{\nu'} \Ket{\nu'} \Bra{\nu'} = 1$$:
 
 $$\begin{aligned}
-    \chi_0
+    \chi_0(\vb{q}, \omega)
     &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu} \bigg(
     \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}}
     {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
@@ -266,20 +258,21 @@ $$\begin{aligned}
 
 Where we recognized the commutator,
 and eliminated $$E_\nu$$ using $$\hat{H}_0 \Ket{n} = E_\nu \Ket{\nu}$$.
-The resulting expression has the form of a matrix trace $$\Tr$$
-and a thermal expectation $$\Expval{}_0$$:
+The resulting expression has the form of a matrix trace $$\Tr$$,
+and therefore a thermal expectation $$\Expval{}_0$$:
 
 $$\begin{aligned}
-    \chi_0
+    \chi_0(\vb{q}, \omega)
     &= \frac{1}{Z V} \sum_{\vb{k}} \frac{\Tr\!\big(\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}
     {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0} \big)}
     {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
-    = \frac{1}{V} \sum_{\vb{k}}
+    \\
+    &= \frac{1}{V} \sum_{\vb{k}}
     \frac{\expval{\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}}_0}
     {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
 \end{aligned}$$
 
-This commutator can be evaluated,
+The commutator can be evaluated,
 and in this particular case it turns out to be:
 
 $$\begin{aligned}
@@ -341,51 +334,305 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-This is its most general form, but for practical calculations
-we need to formally take the limit $$V \to \infty$$
-and then use $$\sum_{\vb{k}} = V / (2 \pi)^{D} \int_{-\infty}^{\infty} \dd{\vb{k}}$$.
+This is its most general form, but we can go further
+by exploiting the Fermi gas' periodic boundary conditions,
+which [enable us](/know/concept/discrete-spectrum-summation/)
+to turn the sum into an integral.
 Furthermore, electrons are spin-1/2 particles,
-so each orbital contains two, meaning
-$$\sum_{\sigma}$$ simply gives a constant factor:
+so $$\sum_{\sigma}$$ simply gives a constant factor $$2$$:
 
 $$\begin{aligned}
     \boxed{
         \chi_0(\vb{q}, \omega)
-        = \frac{2}{(2 \pi)^{D}} \int_{-\infty}^{\infty}
+        = \frac{2}{(2 \pi)^D} \int_{-\infty}^{\infty}
         \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}
         {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \dd{\vb{k}}
     }
 \end{aligned}$$
 
+Beware that some authors define $$\chi_0$$ as the response
+of the *charge* density to an *electric* potential,
+leading to an extra factor $$e_0^2$$, where $$e_0$$ is the electron charge.
+In contrast, our result describes the *particle* density response
+due to an *energy* potential.
+
+
+
+## Evaluation
+
+Let us consider $$D = 3$$ (calculations in 1D or 2D are also possible),
+and rewrite the integrand by substituting $$\vb{k}' \equiv - \vb{k} - \vb{q}$$
+in the second term and using $$\xi_{-\vb{k}} = \xi_{\vb{k}}$$:
+
+$$\begin{aligned}
+    \chi_0(\vb{q}, \omega)
+    &= \frac{2}{(2 \pi)^3} \int_{-\infty}^{\infty} \bigg(
+    \frac{n_F(\xi_{\vb{k}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+    - \frac{n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+    \bigg) \dd{\vb{k}}
+    \\
+    &= \frac{1}{4 \pi^3} \bigg(
+    \int_{-\infty}^{\infty} \frac{n_F(\xi_{\vb{k}})}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} + \hbar (\omega + i \eta)} \dd{\vb{k}}
+    - \int_{-\infty}^{\infty} \frac{n_F(\xi_{- \vb{k}'})}{\xi_{- \vb{k}' - \vb{q}} - \xi_{- \vb{k}'} + \hbar (\omega + i \eta)} \dd{\vb{k}'}
+    \bigg)
+    \\
+    &= \frac{1}{4 \pi^3} \int_{-\infty}^{\infty} n_F(\xi_{\vb{k}}) \bigg(
+    \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} + \hbar (\omega + i \eta)}
+    + \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} - \hbar (\omega + i \eta)}
+    \bigg) \dd{\vb{k}}
+\end{aligned}$$
+
+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 $$|\vb{k}| < k_F$$:
+
+$$\begin{aligned}
+    \chi_0(\vb{q}, \omega)
+    &= \frac{2 \pi}{4 \pi^3} \int_0^{k_F} \! \int_0^\pi
+    \bigg( \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} + \hbar (\omega \!+\! i \eta)}
+    + \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} - \hbar (\omega \!+\! i \eta)} \bigg) |\vb{k}|^2 \sin{\theta} \dd{\theta} \dd{|\vb{k}|}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}
+    &= \frac{\hbar^2}{2 m} \Big( (\vb{k} \cdot \vb{k}) - (\vb{k} \!+\! \vb{q}) \cdot (\vb{k} \!+\! \vb{q}) \Big)
+    \\
+    &= - \frac{\hbar^2}{2 m} \Big( |\vb{q}|^2 + 2 |\vb{k}| |\vb{q}| \cos{\theta} \Big)
+\end{aligned}$$
+
+By defining the dimensionless variables
+$$x \equiv |\vb{q}| / (2 k_F)$$ and $$\kappa \equiv |\vb{k}| / k_F$$,
+and recognizing the Fermi energy $$\xi_F = \hbar^2 k_F^2 / (2 m)$$,
+this turns into:
+
+$$\begin{aligned}
+    \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}
+    &= - 4 \xi_F (x^2 + x \kappa \cos{\theta})
+\end{aligned}$$
+
+We insert this back into $$\chi_0$$, and abbreviate
+$$y \equiv \hbar (\omega + i \eta) / (4 \xi_F)$$
+and $$c \equiv \cos{\theta}$$, giving:
+
+$$\begin{aligned}
+    \chi_0(\vb{q}, \omega)
+    &= \frac{k_F^3}{8 \pi^2 \xi_F} \int_0^1 \! \int_{-1}^1 \bigg(
+    \frac{1}{-x^2 - x \kappa c + y} + \frac{1}{-x^2 - x \kappa c - y}
+    \bigg) \kappa^2 \dd{c} \dd{\kappa}
+\end{aligned}$$
+
+Evaluating the integral over $$c$$ gives
+a result that involves the complex natural logarithm $$\ln$$:
+
+$$\begin{aligned}
+    \chi_0
+    &= \frac{-k_F^3}{8 \pi^2 \xi_F} \int_0^1 \! \int_{-1}^1 \bigg(
+    \frac{1}{x \kappa c + x^2 - y} + \frac{1}{x \kappa c + x^2 + y}
+    \bigg) \kappa^2 \dd{c} \dd{\kappa}
+    \\
+    &= \frac{-m k_F}{4 \pi^2 \hbar^2} \int_0^1 \bigg[
+    \frac{1}{x \kappa} \ln(x \kappa c + x^2 - y) + \frac{1}{x \kappa} \ln(x \kappa c + x^2 + y)
+    \bigg]_{-1}^1 \kappa^2 \dd{\kappa}
+    \\
+    &= \frac{-m k_F}{4 \pi^2 \hbar^2} \frac{1}{x} \int_0^1 \bigg(
+    \ln\!\bigg( \frac{\kappa + x - y / x}{-\kappa + x - y / x} \bigg)
+    + \ln\!\bigg( \frac{\kappa + x + y / x}{-\kappa + x + y / x} \bigg)
+    \bigg) \kappa \dd{\kappa}
+\end{aligned}$$
+
+We abbreviate this further by defining $$z_{+} \equiv x + y / x$$
+and $$z_{-} \equiv x - y / x$$, yielding:
+
+$$\begin{aligned}
+    \chi_0
+    &= \frac{-m k_F}{4 \pi^2 \hbar^2} \frac{1}{x} \int_0^1
+    \ln\!\bigg( \frac{(\kappa + z_{-})(\kappa + z_{+})}{(\kappa - z_{-})(\kappa - z_{+})} \bigg)
+    \kappa \dd{\kappa}
+\end{aligned}$$
+
+Looking up this integral leads to the expression below.
+Note that all terms disappear for $$\kappa = 0$$:
+
+$$\begin{aligned}
+    \chi_0
+    &= \frac{-m k_F}{4 \pi^2 \hbar^2} \frac{1}{x} \bigg[
+    \kappa (z_{+} \!+\! z_{-})
+    - z_{+}^2 \tanh^{-1}\!\bigg(\!\frac{\kappa}{z_{+}}\!\bigg)
+    - z_{-}^2 \tanh^{-1}\!\bigg(\!\frac{\kappa}{z_{-}}\!\bigg)
+    + \frac{1}{2} \kappa^2 \ln\!\bigg( \frac{(\kappa + z_{+})(\kappa + z_{-})}{(\kappa - z_{-})(\kappa - z_{+})} \bigg)
+    \bigg]_0^1
+    \\
+    &= \frac{-m k_F}{\pi^2 \hbar^2} \frac{1}{4 x} \bigg(
+    2 x
+    - z_{+}^2 \tanh^{-1}\!\bigg(\!\frac{1}{z_{+}}\!\bigg)
+    - z_{-}^2 \tanh^{-1}\!\bigg(\!\frac{1}{z_{-}}\!\bigg)
+    + \frac{1}{2} \ln\!\bigg( \frac{(1 + z_{+})(1 + z_{-})}{(1 - z_{-})(1 - z_{+})} \bigg)
+    \bigg)
+    \\
+    &= \frac{-m k_F}{\pi^2 \hbar^2} \frac{1}{8 x} \bigg(
+    4 x
+    - z_{+}^2 \ln\!\bigg( \frac{z_{+} + 1}{z_{+} - 1} \bigg)
+    - z_{-}^2 \ln\!\bigg( \frac{z_{-} + 1}{z_{-} - 1} \bigg)
+    + \ln\!\bigg( \frac{(z_{+} + 1)(z_{-} + 1)}{(z_{+} - 1)(z_{-} - 1)} \bigg)
+    \bigg)
+    \\
+    &= \frac{-m k_F}{\pi^2 \hbar^2} \bigg(
+    \frac{1}{2}
+    + \frac{1}{8 x} (1 - z_{+}^2) \ln\!\bigg( \frac{z_{+} + 1}{z_{+} - 1} \bigg)
+    + \frac{1}{8 x} (1 - z_{-}^2) \ln\!\bigg( \frac{z_{-} + 1}{z_{-} - 1} \bigg)
+    \bigg)
+\end{aligned}$$
+
+We thus arrive at this expression
+for the response function of non-interacting electrons:
+
+$$\begin{aligned}
+    \boxed{
+        \chi_0(\vb{q}, \omega)
+        = - \frac{m k_F}{\pi^2 \hbar^2} \: \mathcal{F}\bigg(\frac{|\vb{q}|}{2 k_F}, \: \frac{\hbar (\omega \!+\! i \eta)}{4 \xi_F}\bigg)
+    }
+\end{aligned}$$
+
+Where $$\mathcal{F}$$ (which is sometimes also called the Lindhard function),
+is defined like so:
+
+$$\begin{aligned}
+    \boxed{
+        \mathcal{F}(x, y)
+        \equiv \frac{1}{2}
+        + \frac{1}{8 x} \bigg( 1 - \Big(x \!+\! \frac{y}{x}\Big)^{\!2} \bigg) \ln\!\bigg( \frac{x + y/x + 1}{x + y/x - 1} \bigg)
+        + \frac{1}{8 x} \bigg( 1 - \Big(x \!-\! \frac{y}{x}\Big)^{\!2} \bigg) \ln\!\bigg( \frac{x - y/x + 1}{x - y/x - 1} \bigg)
+    }
+\end{aligned}$$
+
+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).
+
+
+
+## Interpretation
+
 From this, we would like to get the
 [dielectric function](/know/concept/dielectric-function/) $$\varepsilon_r$$.
 When an external [electric field](/know/concept/electric-field/) is applied,
 the electrons respond and thereby modify the net field inside the material.
 We include this effect in our *energy* potential $$U$$,
 such that the net *electric* potential
-$$\Phi_\mathrm{tot} = U / q_e$$,
-where $$q_e < 0$$ is the charge of an electron.
+$$\Phi_\mathrm{tot} = U / e_0$$ (where $$e_0 < 0$$).
 This is not the same as including direct electron-electron interactions!
 
 We thus have a linear relation for the induced *particle* density
 $$\delta\!\Expval{\hat{n}(\vb{q}, \omega)} = \chi_0(\vb{q}, \omega) \: U(\vb{q})$$
 thanks to the [convolution theorem](/know/concept/convolution-theorem/).
 The corresponding induced *charge* density is given by
-$$\rho_\mathrm{ind} = q_e^2 \chi_0 \Phi_\mathrm{tot}$$,
+$$\rho_\mathrm{ind} = e_0^2 \chi_0 \Phi_\mathrm{tot}$$,
 so the standard formula for $$\varepsilon_r$$ yields:
 
 $$\begin{aligned}
     \boxed{
         \varepsilon_r(\vb{q}, \omega)
-        = 1 - U_{ee}(\vb{q}) \frac{2}{(2 \pi)^{D}}
-        \int_{-\infty}^{\infty} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \dd{\vb{k}}
+        = 1 - U_{ee}(\vb{q}) \: \chi_0(\vb{q}, \omega)
     }
 \end{aligned}$$
 
-Where $$U_{ee}(\vb{q}) = q_e^2 / (\varepsilon_0 |\vb{q}|^2)$$ is Coulomb repulsion.
+Where $$U_{ee}(\vb{q}) = e_0^2 / (\varepsilon_0 |\vb{q}|^2)$$ is Coulomb repulsion.
 This is the **Lindhard dielectric function** of a free non-interacting electron gas,
 at any temperature and for any dimensionality.
 
+Lindhard theory can reproduce several notable discoveries in various limits.
+For long wavelengths $$\vb{q} \to 0$$, i.e. $$x \to 0$$,
+it can be shown that $$\mathcal{F}(x, y) \approx - x^2 / (3 y^2)$$,
+and therefore:
+
+$$\begin{aligned}
+    \chi_0(\vb{q}, \omega)
+    &\approx \frac{m k_F}{\pi^2 \hbar^2} \frac{|\vb{q}|^2}{4 k_F^2} \frac{16 \xi_F^2}{3 \hbar^2 \omega^2}
+    = \frac{k_F^3}{3 \pi^2 m} \frac{|\vb{q}|^2}{\omega^2}
+\end{aligned}$$
+
+For a Fermi gas, we know that $$k_F^3 = 3 \pi^2 n$$,
+where $$n$$ is the unperturbed density.
+Upon insertion into $$\varepsilon_r$$,
+this gives the same result as the [Drude model](/know/concept/drude-model/),
+predicting *plasma oscillations*, also known as
+*plasmons* (when treated as quantized particles)
+or [*cold Langmuir waves*](/know/concept/langmuir-waves/):
+
+$$\begin{aligned}
+    \boxed{
+        \varepsilon_r(0, \omega)
+        = 1 - \frac{\omega_p^2}{\omega^2}
+    }
+    \qquad \qquad
+    \boxed{
+        \omega_p
+        \equiv \sqrt{\frac{e_0^2 n}{\varepsilon_0 m}}
+    }
+\end{aligned}$$
+
+Meanwhile, in the static limit $$\omega \to 0$$, i.e. $$y \to 0$$,
+we arrive at a function that is also encountered
+when using the *Hartree-Fock method* to approximate
+the ground state of [jellium](/know/concept/jellium/):
+
+$$\begin{aligned}
+    \boxed{
+        \mathcal{F}(x, 0)
+        = \frac{1}{2}
+        + \frac{1}{4 x} (1 - x^2) \ln\!\bigg( \frac{x + 1}{x - 1} \bigg)
+    }
+\end{aligned}$$
+
+If we assume small $$|\vb{q}|$$ and $$x$$ again,
+it can be shown that $$\mathcal{F}(x, 0) \approx 1$$, in which case:
+
+$$\begin{aligned}
+    \chi_0(\vb{q}, 0)
+    = - \frac{m k_F^3}{\pi^2 \hbar^2 k_F^2}
+    = - \frac{3 \pi^2 m n}{\pi^2 \hbar^2 k_F^2}
+    = - \frac{3 m n}{\hbar^2 k_F^2}
+    = - \frac{3 n}{2 \xi_F}
+\end{aligned}$$
+
+The resulting dielectric function
+also appears in the derivation of *Thomas-Fermi screening*,
+which can be regarded as the zero-temperature limit
+of [Debye screening](/know/concept/debye-length/):
+
+$$\begin{aligned}
+    \boxed{
+        \varepsilon_r(\vb{q}, 0)
+        = 1 + \frac{k_\mathrm{TF}^2}{|\vb{q}|^2}
+    }
+    \qquad \qquad
+    \boxed{
+        k_\mathrm{TF}
+        \equiv \sqrt{\frac{3 e_0^2 n}{2 \varepsilon_0 \xi_F}}
+    }
+\end{aligned}$$
+
+But probably the greatest achievement of Lindhard theory is its appearance
+in the [random phase approximation](/know/concept/random-phase-approximation/)
+(where $$\chi_0$$ is sometimes known as the **retarded pair-bubble**)
+for treating electron-electron Coulomb interactions in the Fermi gas model.
+So although we neglected interactions in our derivation,
+the result is an essential building block of interacting models.
+
 
 
 ## References
@@ -398,3 +645,9 @@ at any temperature and for any dimensionality.
 3.  G. Grosso, G.P. Parravicini,
     *Solid state physics*,
     2nd edition, Elsevier.
+4.  G.F. Giuliani, G. Vignale,
+    *Quantum theory of the electron liquid*,
+    2005, Cambridge.
+5.  B. Mihaila,
+    [*Lindhard function of a $$d$$-dimensional Fermi gas*](https://doi.org/10.48550/arXiv.1111.5337),
+    2011, arXiv.