Rewrite "Lindhard function", split off "dielectric function"

8 files changed, 490 insertions, 364 deletions

diff --git a/content/know/concept/convolution-theorem/index.pdc b/content/know/concept/convolution-theorem/index.pdc
index 2712c21..a14e960 100644
--- a/content/know/concept/convolution-theorem/index.pdc
+++ b/content/know/concept/convolution-theorem/index.pdc
@@ -26,7 +26,7 @@ and $A$ and $B$ are constants from its definition:
 $$\begin{aligned}
     \boxed{
         \begin{aligned}
-            A \cdot (f * g)(x) &= \hat{\mathcal{F}}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\
+            A \cdot (f * g)(x) &= \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\
             B \cdot (\tilde{f} * \tilde{g})(k) &= \hat{\mathcal{F}}\{f(x) \: g(x)\}
         \end{aligned}
     }
@@ -41,12 +41,12 @@ We expand the right-hand side of the theorem and
 rearrange the integrals:
 
 $$\begin{aligned}
-    \hat{\mathcal{F}}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\}
-    &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \exp(i s k x') \dd{x'} \Big) \exp(-i s k x) \dd{k}
+    \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\}
+    &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \exp\!(i s k x') \dd{x'} \Big) \exp\!(-i s k x) \dd{k}
     \\
-    &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k (x - x')) \dd{k} \Big) \dd{x'}
+    &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \exp\!(- i s k (x - x')) \dd{k} \Big) \dd{x'}
     \\
-    &= A \int_{-\infty}^\infty g(x') f(x - x') \dd{x'}
+    &= A \int_{-\infty}^\infty g(x') \: f(x - x') \dd{x'}
     = A \cdot (f * g)(x)
 \end{aligned}$$
 
@@ -55,11 +55,11 @@ this time starting from a product in the $x$-domain:
 
 $$\begin{aligned}
     \hat{\mathcal{F}}\{f(x) \: g(x)\}
-    &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp(- i s x k') \dd{k'} \Big) \exp(i s k x) \dd{x}
+    &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp\!(- i s x k') \dd{k'} \Big) \exp\!(i s k x) \dd{x}
     \\
-    &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp(i s x (k - k')) \dd{x} \Big) \dd{k'}
+    &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp\!(i s x (k - k')) \dd{x} \Big) \dd{k'}
     \\
-    &= B \int_{-\infty}^\infty \tilde{g}(k') \tilde{f}(k - k') \dd{k'}
+    &= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'}
     = B \cdot (\tilde{f} * \tilde{g})(k)
 \end{aligned}$$
 </div>
@@ -73,10 +73,10 @@ the convolution theorem can also be stated using
 the [Laplace transform](/know/concept/laplace-transform/):
 
 $$\begin{aligned}
-    \boxed{(f * g)(t) = \hat{\mathcal{L}}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}}
+    \boxed{(f * g)(t) = \hat{\mathcal{L}}{}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}}
 \end{aligned}$$
 
-Because the inverse Laplace transform $\hat{\mathcal{L}}^{-1}$ is
+Because the inverse Laplace transform $\hat{\mathcal{L}}{}^{-1}$ is
 unpleasant, the theorem is often stated using the forward transform
 instead:
 
@@ -95,20 +95,20 @@ because we set both $f(t)$ and $g(t)$ to zero for $t < 0$:
 
 $$\begin{aligned}
     \hat{\mathcal{L}}\{(f * g)(t)\}
-    &= \int_0^\infty \Big( \int_0^\infty g(t') f(t - t') \dd{t'} \Big) \exp(- s t) \dd{t}
+    &= \int_0^\infty \Big( \int_0^\infty g(t') f(t - t') \dd{t'} \Big) \exp\!(- s t) \dd{t}
     \\
-    &= \int_0^\infty \Big( \int_0^\infty f(t - t')  \exp(- s t) \dd{t} \Big) g(t') \dd{t'}
+    &= \int_0^\infty \Big( \int_0^\infty f(t - t')  \exp\!(- s t) \dd{t} \Big) g(t') \dd{t'}
 \end{aligned}$$
 
 Then we define a new integration variable $\tau = t - t'$, yielding:
 
 $$\begin{aligned}
     \hat{\mathcal{L}}\{(f * g)(t)\}
-    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'}
+    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp\!(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'}
     \\
-    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s \tau) \dd{\tau} \Big) g(t') \exp(- s t') \dd{t'}
+    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp\!(- s \tau) \dd{\tau} \Big) g(t') \exp\!(- s t') \dd{t'}
     \\
-    &= \int_0^\infty \tilde{f}(s) g(t') \exp(- s t') \dd{t'}
+    &= \int_0^\infty \tilde{f}(s) \: g(t') \exp\!(- s t') \dd{t'}
     = \tilde{f}(s) \: \tilde{g}(s)
 \end{aligned}$$
 </div>
diff --git a/content/know/concept/coulomb-logarithm/index.pdc b/content/know/concept/coulomb-logarithm/index.pdc
index 71b13a8..422f73b 100644
--- a/content/know/concept/coulomb-logarithm/index.pdc
+++ b/content/know/concept/coulomb-logarithm/index.pdc
@@ -84,8 +84,9 @@ and is thus barely affected by this inconsistency.
 
 You can easily convince yourself
 that the average time $\tau$ between "collisions"
-is related as follows to the cross-section $\sigma$,
-the density $n$, and relative velocity $|\vb{v}|$:
+is related like so to the cross-section $\sigma$,
+the total density $n$ of charged particles,
+and the relative velocity $|\vb{v}|$:
 
 $$\begin{aligned}
     \frac{1}{\tau}
diff --git a/content/know/concept/dielectric-function/index.pdc b/content/know/concept/dielectric-function/index.pdc
new file mode 100644
index 0000000..cd2a0b7
--- /dev/null
+++ b/content/know/concept/dielectric-function/index.pdc
@@ -0,0 +1,144 @@
+---
+title: "Dielectric function"
+firstLetter: "D"
+publishDate: 2022-01-24
+categories:
+- Physics
+- Electromagnetism
+- Quantum mechanics
+
+date: 2022-01-20T22:04:13+01:00
+draft: false
+markup: pandoc
+---
+
+# Dielectric function
+
+The **dielectric function** or **relative permittivity** $\varepsilon_r$
+is a measure of how strongly a given medium counteracts
+[electric fields](/know/concept/electric-field/) compared to a vacuum.
+Let $\vb{D}$ be the applied external field,
+and $\vb{E}$ the effective field inside the material:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{D} = \varepsilon_0 \varepsilon_r \vb{E}
+    }
+\end{aligned}$$
+
+If $\varepsilon_r$ is large, then $\vb{D}$ is strongly suppressed,
+because the material's electrons and nuclei move to create an opposing field.
+In order for $\varepsilon_r$ to be well defined, we only consider linear media,
+where the induced polarization $\vb{P}$ is proportional to $\vb{E}$.
+
+We would like to find an alternative definition of $\varepsilon_r$.
+Consider that the usual electric fields $\vb{E}$, $\vb{D}$, and $\vb{P}$
+can each be written as the gradient of an electrostatic potential like so,
+where $\Phi_\mathrm{tot}$, $\Phi_\mathrm{ext}$ and $\Phi_\mathrm{ind}$
+are the total, external and induced potentials, respectively:
+
+$$\begin{aligned}
+    \vb{E}
+    = -\nabla \Phi_\mathrm{tot}
+    \qquad \qquad
+    \vb{D}
+    = - \varepsilon_0 \nabla \Phi_\mathrm{ext}
+    \qquad \qquad
+    \vb{P}
+    = \varepsilon_0 \nabla \Phi_\mathrm{ind}
+\end{aligned}$$
+
+Such that $\Phi_\mathrm{tot} = \Phi_\mathrm{ext} + \Phi_\mathrm{ind}$.
+Inserting this into $\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}$
+then suggests defining:
+
+$$\begin{aligned}
+    \boxed{
+        \varepsilon_r
+        \equiv \frac{\Phi_\mathrm{ext}}{\Phi_\mathrm{tot}}
+    }
+\end{aligned}$$
+
+
+## From induced charge density
+
+A common way to calculate $\varepsilon_r$ is from
+the induced charge density $\rho_\mathrm{ind}$,
+i.e. the offset caused by the material's particles responding to the field.
+We start from [Gauss' law](/know/concept/maxwells-equations/) for $\vb{P}$:
+
+$$\begin{aligned}
+    \nabla \cdot \vb{P}
+    = \varepsilon_0 \nabla^2 \Phi_\mathrm{ind}(\vb{r})
+    = - \rho_\mathrm{ind}(\vb{r})
+\end{aligned}$$
+
+This is Poisson's equation, which has the following well-known
+[Fourier transform](/know/concept/fourier-transform/):
+
+$$\begin{aligned}
+    \Phi_\mathrm{ind}(\vb{q})
+    = \frac{\rho_\mathrm{ind}(\vb{q})}{\varepsilon_0 |\vb{q}|^2}
+    = V(\vb{q}) \: \rho_\mathrm{ind}(\vb{q})
+\end{aligned}$$
+
+Where $V(\vb{q})$ represents Coulomb interactions,
+and $V(0) = 0$ to ensure overall neutrality:
+
+$$\begin{aligned}
+    V(\vb{q})
+    = \frac{1}{\varepsilon_0 |\vb{q}|^2}
+    \qquad \implies \qquad
+    V(\vb{r} - \vb{r}')
+    = \frac{1}{4 \pi \varepsilon_0 |\vb{r} - \vb{r}'|}
+\end{aligned}$$
+
+The [convolution theorem](/know/concept/convolution-theorem/)
+then gives us the solution $\Phi_\mathrm{ind}$ in the $\vb{r}$-domain:
+
+$$\begin{aligned}
+    \Phi_\mathrm{ind}(\vb{r})
+    = (V * \rho_\mathrm{ind})(\vb{r})
+    = \int_{-\infty}^\infty V(\vb{r} - \vb{r}') \: \rho_\mathrm{ind}(\vb{r}') \dd{\vb{r}'}
+\end{aligned}$$
+
+To proceed, we need to find an expression for $\rho_\mathrm{ind}$
+that is proportional to $\Phi_\mathrm{tot}$ or $\Phi_\mathrm{ext}$,
+or some linear combination thereof.
+Such an expression must exist for a linear material.
+
+Suppose we can show that $\rho_\mathrm{ind} = C_\mathrm{ext} \Phi_\mathrm{ext}$,
+for some $C_\mathrm{ext}$, which may depend on $\vb{q}$. Then:
+
+$$\begin{aligned}
+    \Phi_\mathrm{tot}
+    = (1 + C_\mathrm{ext} V) \Phi_\mathrm{ext}
+    \quad \implies \quad
+    \boxed{
+        \varepsilon_r(\vb{q})
+        = \frac{1}{1 + C_\mathrm{ext}(\vb{q}) V(\vb{q})}
+    }
+\end{aligned}$$
+
+Similarly, suppose we can show that $\rho_\mathrm{ind} = C_\mathrm{tot} \Phi_\mathrm{tot}$,
+for some quantity $C_\mathrm{tot}$, then:
+
+$$\begin{aligned}
+    \Phi_\mathrm{ext}
+    = (1 - C_\mathrm{tot} V) \Phi_\mathrm{tot}
+    \quad \implies \quad
+    \boxed{
+        \varepsilon_r(\vb{q})
+        = 1 - C_\mathrm{tot}(\vb{q}) V(\vb{q})
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
+2.  M. Fox,
+    *Optical properties of solids*, 2nd edition,
+    Oxford.
diff --git a/content/know/concept/electric-field/index.pdc b/content/know/concept/electric-field/index.pdc
index 6162e0b..62ce1f5 100644
--- a/content/know/concept/electric-field/index.pdc
+++ b/content/know/concept/electric-field/index.pdc
@@ -31,7 +31,7 @@ since opposite charges attracts and like charges repel.
 If two opposite point charges with magnitude $q$
 are observed from far away,
 they can be treated as a single object called a **dipole**,
-which has an **electric dipole moment** $\vb{p}$ defined as follows,
+which has an **electric dipole moment** $\vb{p}$ defined like so,
 where $\vb{d}$ is the vector going from
 the negative to the positive charge (opposite direction of $\vb{E}$):
 
@@ -88,7 +88,7 @@ and that $\vb{M}$ has the opposite sign of $\vb{P}$.
 The polarization $\vb{P}$ is a function of $\vb{E}$.
 In addition to the inherent polarity
 of the material $\vb{P}_0$ (zero in most cases),
-there is a possibly nonlinear response
+there is a (possibly nonlinear) response
 to the applied $\vb{E}$-field:
 
 $$\begin{aligned}
@@ -101,10 +101,7 @@ $$\begin{aligned}
 Where the $\chi_e^{(n)}$ are the **electric susceptibilities** of the medium.
 For simplicity, we often assume that only the $n\!=\!1$ term is nonzero,
 which is the linear response to $\vb{E}$.
-In that case, we define
-the **relative permittivity** $\varepsilon_r \equiv 1 + \chi_e^{(1)}$
-and the **absolute permittivity** $\varepsilon \equiv \varepsilon_r \varepsilon_0$,
-so that:
+In that case, we define the **absolute permittivity** $\varepsilon$ so that:
 
 $$\begin{aligned}
     \vb{D}
@@ -114,14 +111,20 @@ $$\begin{aligned}
     = \varepsilon \vb{E}
 \end{aligned}$$
 
+I.e. $\varepsilon \equiv \varepsilon_r \varepsilon_0$,
+where $\varepsilon_r \equiv 1 + \chi_e^{(1)}$ is
+the [**dielectric function**](/know/concept/dielectric-function/)
+or **relative permittivity**,
+whose calculation is of great interest in physics.
+
 In reality, a material cannot respond instantly to $\vb{E}$,
 meaning that $\chi_e^{(1)}$ is a function of time,
 and that $\vb{P}$ is the convolution of $\chi_e^{(1)}(t)$ and $\vb{E}(t)$:
 
 $$\begin{aligned}
     \vb{P}(t)
-    = (\chi_e^{(1)} * \vb{E})(t)
-    = \int_{-\infty}^\infty \chi_e^{(1)}(t - \tau) \: \vb{E}(\tau) \:d\tau
+    = \varepsilon_0 \big(\chi_e^{(1)} * \vb{E}\big)(t)
+    = \varepsilon_0 \int_{-\infty}^\infty \chi_e^{(1)}(t - \tau) \: \vb{E}(\tau) \:d\tau
 \end{aligned}$$
 
 Note that this definition requires $\chi_e^{(1)}(t) = 0$ for $t < 0$
diff --git a/content/know/concept/greens-functions/index.pdc b/content/know/concept/greens-functions/index.pdc
index b3c9ede..92f0fcf 100644
--- a/content/know/concept/greens-functions/index.pdc
+++ b/content/know/concept/greens-functions/index.pdc
@@ -32,12 +32,9 @@ If the two operators are single-particle creation/annihilation operators,
 then we get the **single-particle Green's functions**,
 for which the symbol $G$ is used.
 
-The **time-ordered** or **causal Green's function** $G_{\nu \nu'}$
-is defined as follows,
+The **time-ordered** or **causal Green's function** $G_{\nu \nu'}$ is as follows,
 where $\mathcal{T}$ is the [time-ordered product](/know/concept/time-ordered-product/),
-the expectation value $\expval{}$ is
-with respect to thermodynamic equilibrium,
-$\nu$ and $\nu'$ are labels of single-particle states,
+$\nu$ and $\nu'$ are single-particle states,
 and $\hat{c}_\nu$ annihilates a particle from $\nu$, etc.:
 
 $$\begin{aligned}
@@ -47,6 +44,24 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
+The expectation value $\expval{}$ is
+with respect to thermodynamic equilibrium.
+This is sometimes in the [canonical ensemble](/know/concept/canonical-ensemble/)
+(for some two-particle Green's functions, see below),
+but usually in the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
+since we are adding/removing particles.
+In the latter case, we assume that the chemical potential $\mu$
+is already included in the Hamiltonian $\hat{H}$.
+Explicitly, for a complete set of many-particle states $\ket{\Psi_n}$, we have:
+
+$$\begin{aligned}
+    G_{\nu \nu'}(t, t')
+    &= -\frac{i}{\hbar Z} \Tr\!\Big( \mathcal{T} \Big\{ \hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t')\Big\} \: e^{- \beta \hat{H}} \Big)
+    \\
+    &= -\frac{i}{\hbar Z} \sum_{n}
+    \matrixel**{\Psi_n}{\mathcal{T} \Big\{ \hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t')\Big\} \: e^{- \beta \hat{H}}}{\Psi_n}
+\end{aligned}$$
+
 Arguably more prevalent are
 the **retarded Green's function** $G_{\nu \nu'}^R$
 and the **advanced Green's function** $G_{\nu \nu'}^A$
@@ -67,10 +82,10 @@ $$\begin{aligned}
 Where $\Theta$ is a [Heaviside function](/know/concept/heaviside-step-function/),
 and $[,]_{\mp}$ is a commutator for bosons,
 and an anticommutator for fermions.
-We are in the [Heisenberg picture](/know/concept/heisenberg-picture/),
-hence $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$ are time-dependent,
-but keep in mind that time-dependent Hamiltonians are allowed,
-so it might not be trivial.
+Depending on the context,
+we could either be in the [Heisenberg picture](/know/concept/heisenberg-picture/)
+or in the [interaction picture](/know/concept/interaction-picture/),
+hence $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$ are time-dependent.
 
 Furthermore, the **greater Green's function** $G_{\nu \nu'}^>$
 and **lesser Green's function** $G_{\nu \nu'}^<$ are:
@@ -146,16 +161,7 @@ $\expval*{\hat{A}(t) \hat{B}(t')}$ only depends on $t - t'$
 for arbitrary $\hat{A}$ and $\hat{B}$,
 and it trivially follows that the Green's functions do too.
 
-Suppose that the system started in thermodynamic equilibrium.
-This could sometimes be in the [canonical ensemble](/know/concept/canonical-ensemble/)
-(for two-particle Green's functions, see below),
-but usually it will be in the
-[grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
-since we are adding/removing particles.
-In the latter case, we assume that the chemical potential $\mu$
-is already included in the Hamiltonian.
-
-In any case, at equilibrium, we know that the
+In (grand) canonical equilibrium, we know that the
 [density operator](/know/concept/density-operator/)
 $\hat{\rho}$ is as follows:
 
@@ -163,9 +169,8 @@ $$\begin{aligned}
     \hat{\rho} = \frac{1}{Z} \exp\!(- \beta \hat{H})
 \end{aligned}$$
 
-Where $Z \equiv \Tr\!(\exp\!(- \beta \hat{H}))$ is the partition function.
-In that case, the expected value of the product
-of the time-independent operators $\hat{A}$ and $\hat{B}$ is calculated like so:
+The expected value of the product
+of the time-independent operators $\hat{A}$ and $\hat{B}$ is then:
 
 $$\begin{aligned}
     \expval*{\hat{A}(t) \hat{B}(t')}
diff --git a/content/know/concept/laplace-transform/index.pdc b/content/know/concept/laplace-transform/index.pdc
index bd7673b..5e91a04 100644
--- a/content/know/concept/laplace-transform/index.pdc
+++ b/content/know/concept/laplace-transform/index.pdc
@@ -33,6 +33,12 @@ This is solved by restricting the domain of $\tilde{f}(s)$
 to $s$ where $\mathrm{Re}\{s\} > s_0$,
 for an $s_0$ large enough to compensate for the growth of $f(t)$.
 
+The **inverse Laplace transform** $\hat{\mathcal{L}}{}^{-1}$ involves complex integration,
+and is therefore a lot more difficult to calculate.
+Fortunately, it is usually avoidable by rewriting a given $s$-space expression
+using [partial fraction decomposition](/know/concept/partial-fraction-decomposition/),
+and then looking up the individual terms.
+
 
 ## Derivatives
 
@@ -42,7 +48,7 @@ This is especially useful for transforming ODEs with variable coefficients:
 
 $$\begin{aligned}
     \boxed{
-        \tilde{f}'(s) = - \hat{\mathcal{L}}\{t f(t)\}
+        \tilde{f}{}'(s) = - \hat{\mathcal{L}}\{t f(t)\}
     }
 \end{aligned}$$
 
@@ -107,9 +113,9 @@ $$\begin{aligned}
     \hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\}
     &= \int_0^\infty f^{(n)}(t) \exp\!(- s t) \dd{t}
     \\
-    &= \big[ f^{(n - 1)}(t) \exp\!(- s t) \big]_0^\infty + s \int_0^\infty f^{(n-1)}(t) \exp\!(- s t) \dd{t}
+    &= \Big[ f^{(n - 1)}(t) \exp\!(- s t) \Big]_0^\infty + s \int_0^\infty f^{(n-1)}(t) \exp\!(- s t) \dd{t}
     \\
-    &= - f^{(n - 1)}(0) + s \big[ f^{(n - 2)}(t) \exp\!(- s t) \big]_0^\infty + s^2 \int_0^\infty f^{(n-2)}(t) \exp\!(- s t) \dd{t}
+    &= - f^{(n - 1)}(0) + s \Big[ f^{(n - 2)}(t) \exp\!(- s t) \Big]_0^\infty + s^2 \int_0^\infty f^{(n-2)}(t) \exp\!(- s t) \dd{t}
 \end{aligned}$$
 
 And so on.
diff --git a/content/know/concept/lindhard-function/index.pdc b/content/know/concept/lindhard-function/index.pdc
index 96244c9..aedf0f5 100644
--- a/content/know/concept/lindhard-function/index.pdc
+++ b/content/know/concept/lindhard-function/index.pdc
@@ -1,7 +1,7 @@
 ---
 title: "Lindhard function"
 firstLetter: "L"
-publishDate: 2021-10-12
+publishDate: 2022-01-24 # Originally 2021-10-12, major rewrite
 categories:
 - Physics
 - Quantum mechanics
@@ -13,9 +13,9 @@ markup: pandoc
 
 # Lindhard function
 
-The **Lindhard function** describes the response of an electron gas
-to an external perturbation,
-and can be regarded as a quantum-mechanical
+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/).
 
 We start from the [Kubo formula](/know/concept/kubo-formula/)
@@ -28,389 +28,370 @@ $$\begin{aligned}
     = -\frac{i}{\hbar} \int_{-\infty}^\infty \Theta(t - t') \expval{\comm{\hat{n}_I(\vb{r}, t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
 \end{aligned}$$
 
-Where $\Theta$ is the [Heaviside step function](/know/concept/heaviside-step-function/),
-and the subscript $I$ refers to the [interaction picture](/know/concept/interaction-picture/).
-Notice from the limits that the perturbation is switched on at $t = -\infty$.
-Now, let us consider the following harmonic $\hat{H}_1$ in the Schrödinger picture:
+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 $\hat{H}_1$:
 
 $$\begin{aligned}
     \hat{H}_{1,S}(t)
-    = g(t) \: \hat{V}_S
-    \qquad
-    g(t)
-    \equiv \exp\!(- i \omega t) \exp\!(\eta t)
-    \qquad
-    \hat{V}_S
-    \equiv \int_{-\infty}^\infty V(\vb{r}) \: \hat{n}(\vb{r}) \dd{\vb{r}}
+    = e^{i (\omega + i \eta) t} \int_{-\infty}^\infty U(\vb{r}) \: \hat{n}(\vb{r}) \dd{\vb{r}}
 \end{aligned}$$
 
-Where $\eta$ is a tiny positive number,
-which represents a gradual switching-on of $\hat{H}_1$,
-eliminating transient effects
-and helping the convergence of an integral later.
-
-We assume that $V(\vb{r})$ varies slowly compared to the electrons' wavefunctions,
-so we argue that $\hat{V}_S$ is practically time-independent,
-because the total number of electrons is conserved,
-and $\hat{n}$ is only weakly perturbed by $\hat{H}_1$.
-
-Because $\hat{H}_1$ starts at $t = -\infty$,
-we can always shift the time axis such that the point of interest is at $t = 0$.
-We thus have, without loss of generality:
+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:
 
 $$\begin{aligned}
-    \delta\expval{{\hat{n}}}(\vb{r})
-    = \delta\expval{{\hat{n}}}(\vb{r}, 0)
-    &= -\frac{i}{\hbar} \int_{-\infty}^\infty \Theta(- t')
-    \Big( \expval{\hat{n}_I \hat{V}_I}_0 - \expval{\hat{V}_I \hat{n}_I}_0 \Big) g(t') \dd{t'}
+    \delta\expval{{\hat{n}}}(\vb{r}, t)
+    = \iint_{-\infty}^\infty \chi(\vb{r}, \vb{r}'; t, t') \: U(\vb{r}') \: e^{i (\omega + i \eta) t'} \dd{t'} \dd{\vb{r}'}
 \end{aligned}$$
 
-The expectation values $\expval{}_0$ are calculated for $\ket{0}$,
-which was the state at $t = -\infty$.
-Note that if $\ket{0}$ is an eigenstate of $\hat{H}_{0,S}$,
-there is no difference which picture (Schrödinger or interaction) $\ket{0}$ is in,
-because any operator $\hat{A}$ then satisfies:
+Here, $\chi$ is the density-density correlation function,
+i.e. a two-particle [Green's function](/know/concept/greens-functions/):
 
 $$\begin{aligned}
-    \matrixel{0_I}{\hat{A}_I}{0_I}
-    &= \matrixel**{0_S}{\exp\!(-i \hat{H}_{0,S} t / \hbar) \:\: \hat{A}_I \: \exp\!(i \hat{H}_{0,S} t / \hbar)}{0_S}
-    \\
-    &= \matrixel{0_S}{\hat{A}_I}{0_S} \:\exp\!\big(i (E_0\!-\!E_0) t / \hbar\big) 
-    = \matrixel{0_S}{\hat{A}_I}{0_S}
+    \chi(\vb{r}, \vb{r}'; t, t')
+    \equiv - \frac{i}{\hbar} \Theta(t - t') \expval{\comm{\hat{n}_I(\vb{r}, t)}{\hat{n}_I(\vb{r}', t')}}_0
 \end{aligned}$$
 
-Therefore, we will assume that $\ket{0}$ is an eigenstate of $\hat{H}_{0,S}$.
-Next, we insert the identity operator $\hat{I} = \sum_{j} \ket{j} \bra{j}$,
-where $\ket{j}$ are all the eigenstates of $\hat{H}_{0,S}$:
+Let us assume that the unperturbed system (i.e. without $U$) is spatially uniform,
+so that $\chi$ only depends on the difference $\vb{r} - \vb{r}'$.
+We then take its [Fourier transform](/know/concept/fourier-transform/)
+$\vb{r}\!-\!\vb{r}' \to \vb{q}$:
 
 $$\begin{aligned}
-    \delta\expval{\hat{n}}(\vb{r})
-    &= -\frac{i}{\hbar} \sum_{j} \int \Theta(-t')
-    \Big( \matrixel{0}{\hat{n}_I}{j} \matrixel{j}{\hat{V}_I}{0} - \matrixel{0}{\hat{V}_I}{j} \matrixel{j}{\hat{n}_I}{0} \Big) g(t') \dd{t'}
+    \chi(\vb{q}; t, t')
+    &= \int_{-\infty}^\infty \chi(\vb{r} - \vb{r}'; t, t') \: e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{r}}
+    \\
+    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint
+    \expval{\comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0
+    \: e^{i \vb{q}_1 \cdot \vb{r}} e^{i \vb{q}_2 \cdot \vb{r}'} e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}}
 \end{aligned}$$
 
-Using the fact that $\ket{0}$ and $\ket{j}$
-are eigenstates of $\hat{H}_{0,S}$,
-and that we chose $t = 0$, we find:
+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.
+We rearrange to get a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$:
 
 $$\begin{aligned}
-    \matrixel{j}{\hat{V}_I(t')}{0}
-    &= \matrixel{j}{\hat{V}_S}{0} \:\exp\!\big(i (E_j \!-\! E_0) t' / \hbar\big) 
+    \chi(\vb{q}; t, t')
+    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint
+    \expval{\comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0
+    \: e^{i (\vb{q}_1 - \vb{q}) \cdot \vb{r}} e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}}
+    \\
+    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \iint
+    \expval{\comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0
+    \: \delta(\vb{q}_1 \!-\! \vb{q}) \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2}
     \\
-    \matrixel{j}{\hat{n}_I(0)}{0}
-    &= \matrixel{j}{\hat{n}_S(0)}{0} \:\exp\!(0 - 0)
-    = \matrixel{j}{\hat{n}_S(0)}{0}
+    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \int
+    \expval{\comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0
+    \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2}
 \end{aligned}$$
 
-We define $\omega_{j0} \equiv (E_j \!-\! E_0)/\hbar$
-and insert the above expressions into $\delta\expval{\hat{n}}$,
-yielding:
+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:
 
 $$\begin{aligned}
-    \delta\expval{\hat{n}}(\vb{r})
-    &= -\frac{i}{\hbar} \sum_{j} \bigg(
-    \matrixel{0}{\hat{n}_S}{j} \matrixel{j}{\hat{V}_S}{0} \int \Theta(-t') \exp\!(i \omega_{j0} t') \: g(t') \dd{t'}
-    \\
-    &\qquad\qquad\:\,
-    - \matrixel{0}{\hat{V}_S}{j} \matrixel{j}{\hat{n}_S}{0} \int \Theta(-t') \exp\!(-i \omega_{j0} t') \: g(t') \dd{t'} \bigg)
+    \chi(\vb{q}; t, t')
+    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \frac{1}{V} \int_V \int_{-\infty}^\infty
+    \expval{\comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0
+    \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \dd{\vb{r}'}
 \end{aligned}$$
 
-These integrals are [Fourier transforms](/know/concept/fourier-transform/),
-and are straightforward to evaluate. The first is:
+For $V \to \infty$ we get a Dirac delta function,
+but in fact the conclusion holds for finite $V$ too:
 
 $$\begin{aligned}
-    \int_{-\infty}^\infty \Theta(-t') \exp\!(i \omega_{j0} t') \: g(t') \dd{t'}
-    &= \int_{-\infty}^0 \exp\!\big(\!-\! i (\omega - \omega_{j0}) t' + \eta t' \big) \dd{t'}
-    \\
-    &= \bigg[ \frac{\exp\!\big(\!-\! i (\omega - \omega_{j0}) t' + \eta t' \big)}{- i (\omega - \omega_{j0}) + \eta} \bigg]_{-\infty}^0
+    \chi(\vb{q}; t, t')
+    &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \int_{-\infty}^\infty
+    \expval{\comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: \delta(\vb{q}_2 \!+\! \vb{q}) \dd{\vb{q}_2}
     \\
-    &= \frac{1}{- i (\omega - \omega_{j0}) + \eta}
-    = \frac{i}{\omega - \omega_{j0} + i \eta}
+    &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \expval{\comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0
 \end{aligned}$$
 
-The other integral simply has the opposite sign in front of $\omega_{j0}$.
-We thus arrive at:
+Similarly, if the unperturbed Hamiltonian $\hat{H}_0$ is time-independent,
+$\chi$ only depends on the time difference $t - t'$.
+Note that $\delta{\expval{\hat{n}}}$ already has the form of a Fourier transform,
+which gives us an opportunity to rewrite $\chi$
+in the [Lehmann representation](/know/concept/lehmann-representation/):
 
 $$\begin{aligned}
-    \delta\expval{\hat{n}}(\vb{r}, \omega)
-    &= \frac{1}{\hbar} \sum_{j} \bigg(
-    \frac{\matrixel{0}{\hat{n}_S}{j} \matrixel{j}{\hat{V}_S}{0}}{\omega - \omega_{j0} + i \eta}
-    - \frac{\matrixel{0}{\hat{V}_S}{j} \matrixel{j}{\hat{n}_S}{0}}{\omega + \omega_{j0} + i \eta} \bigg)
+    \chi(\vb{q}, \omega)
+    = \frac{1}{Z V} \sum_{\nu \nu'}
+    \frac{\matrixel{\nu}{\hat{n}_S(\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}_S(-\vb{q})}{\nu}}{\hbar (\omega + i \eta) + E_\nu - E_{\nu'}}
+    \Big( e^{-\beta E_\nu} - e^{- \beta E_{\nu'}} \Big)
 \end{aligned}$$
 
-Inserting the definition $\hat{V}_S = \int V(\vb{r}') \:\hat{n}(\vb{r}') \dd{\vb{r}'}$
-leads us to the following formula for $\delta\expval{\hat{n}}$,
-which has the typical form of a linear response,
-with response function $\chi$:
+Where $\ket{\nu}$ and $\ket{\nu'}$ are many-electron eigenstates of $\hat{H}_0$,
+and $Z$ is the [grand partition function](/know/concept/grand-canonical-ensemble/).
+According to the [convolution theorem](/know/concept/convolution-theorem/)
+$\delta{\expval{\hat{n}}}(\vb{q}, \omega) = \chi(\vb{q}, \omega) \: U(\vb{q})$.
+In anticipation, we swap $\nu$ and $\nu''$ in the second term,
+so the general response function is written as:
 
 $$\begin{aligned}
-    \boxed{
-        \begin{gathered}
-            \delta{n}(\vb{r}, \omega)
-            = \int_{-\infty}^\infty \chi(\vb{r}, \vb{r'}, \omega) \: V(\vb{r}') \dd{\vb{r}'}
-            \qquad\quad \mathrm{where}
-            \\
-            \chi(\vb{r}, \vb{r'}, \omega)
-            = \sum_{j} \bigg( \frac{\matrixel{0}{\hat{n}_S(\vb{r})}{j} \matrixel{j}{\hat{n}_S(\vb{r}')}{0}}{(\omega + i \eta) \hbar - E_j + E_0}
-            - \frac{\matrixel{0}{\hat{n}_S(\vb{r}')}{j} \matrixel{j}{\hat{n}_S(\vb{r})}{0}}{(\omega + i \eta) \hbar + E_j - E_0} \bigg)
-        \end{gathered}
-    }
+    \chi(\vb{q}, \omega)
+    = \frac{1}{Z V} \sum_{\nu \nu'} \bigg(
+    \frac{\matrixel{\nu}{\hat{n}(\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(-\vb{q})}{\nu}}
+    {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}}
+    - \frac{\matrixel{\nu}{\hat{n}(-\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(\vb{q})}{\nu}}
+    {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu}
 \end{aligned}$$
 
-By definition, $\ket{j}$ are eigenstates
-of the many-electron Hamiltonian $\hat{H}_{0,S}$,
-which is only solvable if we crudely neglect
-any and all electron-electron interactions.
-Therefore, to continue, we neglect those interactions.
-According to tradition, we then rename $\chi$ to $\chi_0$.
+All operators are in the Schrödinger picture from now on, hence we dropped the subscript $S$.
 
-The well-known ground state of a non-interacting electron gas
-is the Fermi sea $\ket{\mathrm{FS}}$, given below,
-together with $\hat{n}_S$ in the language of the
-[second quantization](/know/concept/second-quantization/):
+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:
 
 $$\begin{aligned}
-    \ket{\mathrm{FS}}
-    = \prod_\alpha \hat{c}_\alpha^\dagger \ket{0}
-    \qquad \quad
-    \hat{n}_S(\vb{r})
-    = \hat{\Psi}{}^\dagger(\vb{r}) \hat{\Psi}(\vb{r})
-    = \sum_{\alpha \beta} \psi_\alpha^*(\vb{r}) \psi_\beta(\vb{r})\: \hat{c}_\alpha^\dagger \hat{c}_\beta
+    \hat{n}(\vb{q})
+    = \sum_{\sigma \vb{k}} \hat{c}_{\sigma,\vb{k}}^\dagger \hat{c}_{\sigma,\vb{k} + \vb{q}}
+    \qquad \qquad
+    \hat{n}(-\vb{q})
+    = \hat{n}^\dagger(\vb{q})
 \end{aligned}$$
 
-For now, we ignore thermal excitations,
-i.e. we set the temperature $T = 0$.
-In $\chi_0 = \chi$, we thus insert the above $\hat{n}_S$,
-and replace $\ket{0}$ with $\ket{\mathrm{FS}}$, yielding:
+<div class="accordion">
+<input type="checkbox" id="proof-density"/>
+<label for="proof-density">Proof</label>
+<div class="hidden">
+<label for="proof-density">Proof.</label>
+Starting from the general definition of $\hat{n}$,
+we write out the field operators $\hat{\Psi}(\vb{r})$,
+and insert the known non-interacting single-electron orbitals
+$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$:
 
 $$\begin{aligned}
-    \matrixel{0}{\hat{n}_S}{j}
-    \quad\longrightarrow\quad \matrixel**{\mathrm{FS}}{\hat{\Psi}{}^\dagger \hat{\Psi}}{j}
-    = \sum_{\alpha \beta} \psi_\alpha^* \psi_\beta \matrixel**{\mathrm{FS}}{\hat{c}_\alpha^\dagger \hat{c}_\beta}{j}
+    \hat{n}(\vb{r})
+    \equiv \hat{\Psi}{}^\dagger(\vb{r}) \hat{\Psi}(\vb{r})
+    = \sum_{\vb{k} \vb{k}'} \psi_{\vb{k}}^*(\vb{r}) \: \psi_{\vb{k}'}(\vb{r})\: \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'}
+    = \frac{1}{V} \sum_{\vb{k} \vb{k}'} e^{i (\vb{k}' - \vb{k}) \cdot \vb{r}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'}
 \end{aligned}$$
 
-This inner product is only nonzero if
-$\ket{j} = \hat{c}_a \hat{c}_b^\dagger \ket{\mathrm{FS}}$
-with $a = \alpha$ and $b = \beta$,
-or in other words,
-only if $\ket{j}$ is a single-electron excitation of $\ket{\mathrm{FS}}$.
-Furthermore, in $\ket{\mathrm{FS}}$,
-$\alpha$ must be filled, and $\beta$ must be empty.
-Let $f_\alpha \in \{0,1\}$ be the occupation number of orbital $\alpha$, then:
+Taking the Fourier transfom yields a Dirac delta function $\delta$:
 
 $$\begin{aligned}
-    \matrixel{0}{\hat{n}_S}{j}
-    \longrightarrow \sum_{\alpha \beta} \psi_\alpha^* \psi_\beta \: f_\alpha (1 - f_\beta) \: \delta_{a \alpha} \delta_{b \beta}
+    \hat{n}(\vb{q})
+    = \frac{1}{V} \int_{-\infty}^\infty
+    \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: e^{i (\vb{k}' - \vb{k} - \vb{q})\cdot \vb{r}} \dd{\vb{r}}
+    = \frac{(2 \pi)^D}{V} \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q})
 \end{aligned}$$
 
-In $\chi_0$, the sum over $j$ becomes a sum over $a$ and $b$
-(this implicitly eliminates all $\ket{j}$ that are not single-electron excitations),
-and $E_j\!-\!E_0$ becomes the cost of the excitation $\epsilon_b \!-\! \epsilon_a$,
-where $\epsilon_a$ is the energy of orbital $a$.
-Therefore, we find:
+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 becomes exact for $V \to \infty$,
+in which case $\vb{k}$ also becomes continuous again,
+which is what we want for jellium.
+
+We apply this standard trick from condensed matter physics to $\hat{n}$,
+and $V$ cancels out:
 
 $$\begin{aligned}
-    \chi_0
-    &= \sum_{a b} \bigg( \sum_{\alpha \beta \kappa \mu} f_\alpha (1 \!-\! f_\beta) f_\kappa (1 \!-\! f_\mu)
-    \frac{\psi_\alpha^*(\vb{r}) \psi_\beta(\vb{r}) \psi_\kappa(\vb{r}') \psi_\mu^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a}
-    \delta_{a \alpha} \delta_{b \beta} \delta_{a \kappa} \delta_{b \mu}
-    \\
-    &\qquad\:\:\: - \sum_{\alpha \beta \kappa \mu} f_\alpha (1 \!-\! f_\beta) f_\kappa (1 \!-\! f_\mu)
-    \frac{\psi_\alpha^*(\vb{r'}) \psi_\beta(\vb{r'}) \psi_\kappa(\vb{r}) \psi_\mu^*(\vb{r})}{(\omega + i \eta) \hbar + \epsilon_b - \epsilon_a}
-    \delta_{a \alpha} \delta_{b \beta} \delta_{a \kappa} \delta_{b \mu} \bigg)
-    \\
-    &= \sum_{a b} \bigg( f_a^2 (1 \!-\! f_b)^2
-    \frac{\psi_a^*(\vb{r}) \psi_b(\vb{r}) \psi_a(\vb{r}') \psi_b^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a}
-    \\
-    &\qquad\:\: - f_a^2 (1 \!-\! f_b)^2
-    \frac{\psi_a^*(\vb{r'}) \psi_b(\vb{r'}) \psi_a(\vb{r}) \psi_b^*(\vb{r})}{(\omega + i \eta) \hbar + \epsilon_b - \epsilon_a}
-    \bigg)
+    \hat{n}(\vb{q})
+    &= \frac{(2 \pi)^D}{V} \frac{V}{(2 \pi)^D} \sum_{\vb{k}} \int_{-\infty}^\infty
+    \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q}) \dd{\vb{k}'}
+    = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}
 \end{aligned}$$
 
-Because $f_a, f_b \in \{0, 1\}$ when $T = 0$, we know that $f_a^2 (1 \!-\! f_b)^2 = f_a (1 \!-\! f_b)$.
-We then swap the indices $a$ and $b$ in the second term, leading to:
+For negated arguments, we simply define $\vb{k}' \equiv \vb{k} - \vb{q}$
+to show that $\hat{n}(-\vb{q}) = \hat{n}{}^\dagger(\vb{q})$,
+which can also be understood as a consequence of $\hat{n}(\vb{r})$ being real:
 
 $$\begin{aligned}
-    \chi_0
-    &= \sum_{a b} \Big( f_a (1 \!-\! f_b) - f_b (1 \!-\! f_a) \Big)
-    \frac{\psi_a^*(\vb{r}) \psi_b(\vb{r}) \psi_a(\vb{r}') \psi_b^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a}
-    \\
-    &= \sum_{a b} \Big( f_a - f_b \Big)
-    \frac{\psi_a^*(\vb{r}) \psi_b(\vb{r}) \psi_a(\vb{r}') \psi_b^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a}
+    \hat{n}(-\vb{q})
+    = \sum_{\vb{k}} \hat{c}