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3 files changed, 554 insertions, 1 deletions

diff --git a/content/know/concept/kubo-formula/index.pdc b/content/know/concept/kubo-formula/index.pdc
index 9e52835..f0208da 100644
--- a/content/know/concept/kubo-formula/index.pdc
+++ b/content/know/concept/kubo-formula/index.pdc
@@ -223,5 +223,5 @@ $$\begin{aligned}
     *Many-body quantum theory in condensed matter physics*,
     2016, Oxford.
 2.  K.S. Thygesen,
-    *Linear response theory*,
+    *Advanced solid state physics: linear response theory*,
     2013, unpublished.
diff --git a/content/know/concept/lawson-criterion/index.pdc b/content/know/concept/lawson-criterion/index.pdc
new file mode 100644
index 0000000..59801ee
--- /dev/null
+++ b/content/know/concept/lawson-criterion/index.pdc
@@ -0,0 +1,133 @@
+---
+title: "Lawson criterion"
+firstLetter: "L"
+publishDate: 2021-10-06
+categories:
+- Physics
+- Plasma physics
+
+date: 2021-10-04T14:49:24+02:00
+draft: false
+markup: pandoc
+---
+
+# Lawson criterion
+
+For sustained nuclear fusion to be possible,
+the **Lawson criterion** must be met,
+from which some required properties
+of the plasma and the reactor chamber can be deduced.
+
+Suppose that a reactor generates a given power $P_\mathrm{fus}$ by nuclear fusion,
+but that it leaks energy at a rate $P_\mathrm{loss}$ in an unusable way.
+If an auxiliary input power $P_\mathrm{aux}$ sustains the fusion reaction,
+then the following inequality must be satisfied
+in order to have harvestable energy:
+
+$$\begin{aligned}
+    P_\mathrm{loss}
+    \le P_\mathrm{fus} + P_\mathrm{aux}
+\end{aligned}$$
+
+We can rewrite $P_\mathrm{aux}$ using the definition
+of the **energy gain factor** $Q$,
+which is the ratio of the output and input powers of the fusion reaction:
+
+$$\begin{aligned}
+    Q
+    \equiv \frac{P_\mathrm{fus}}{P_\mathrm{aux}}
+    \quad \implies \quad
+    P_\mathrm{aux}
+    = \frac{P_\mathrm{fus}}{Q}
+\end{aligned}$$
+
+Returning to the inequality, we can thus rearrange its right-hand side as follows:
+
+$$\begin{aligned}
+    P_\mathrm{loss}
+    \le P_\mathrm{fus} + \frac{P_\mathrm{fus}}{Q}
+    = P_\mathrm{fus} \Big( 1 + \frac{1}{Q} \Big)
+    = P_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big)
+\end{aligned}$$
+
+We assume that the plasma has equal species densities $n_i = n_e$,
+so its total density $n = 2 n_i$.
+Then $P_\mathrm{fus}$ is as follows,
+where $f_{ii}$ is the frequency
+with which a given ion collides with other ions,
+and $E_\mathrm{fus}$ is the energy released by a single fusion reaction:
+
+$$\begin{aligned}
+    P_\mathrm{fus}
+    = f_{ii} n_i E_\mathrm{fus}
+    = \big( n_i \expval{\sigma v} \big) n_i E_\mathrm{fus}
+    = \frac{n^2}{4} \expval{\sigma v} E_\mathrm{fus}
+\end{aligned}$$
+
+Where $\expval{\sigma v}$ is the mean product
+of the velocity $v$ and the collision cross-section $\sigma$.
+
+Furthermore, assuming that both species have the same temperature $T_i = T_e = T$,
+the total energy density $W$ of the plasma is given by:
+
+$$\begin{aligned}
+    W
+    = \frac{3}{2} k_B T_i n_i + \frac{3}{2} k_B T_e n_e
+    = 3 k_B T n
+\end{aligned}$$
+
+Where $k_B$ is Boltzmann's constant.
+From this, we can define the **confinement time** $\tau_E$
+as the characteristic lifetime of energy in the reactor, before leakage.
+Therefore:
+
+$$\begin{aligned}
+    \tau_E
+    \equiv \frac{W}{P_\mathrm{loss}}
+    \quad \implies \quad
+    P_\mathrm{loss}
+    = \frac{3 n k_B T}{\tau_E}
+\end{aligned}$$
+
+Inserting these new expressions for $P_\mathrm{fus}$ and $P_\mathrm{loss}$
+into the inequality, we arrive at:
+
+$$\begin{aligned}
+    \frac{3 n k_B T}{\tau_E}
+    \le \frac{n^2}{4} \expval{\sigma v} E_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big)
+\end{aligned}$$
+
+This can be rearranged to the form below,
+which is the original Lawson criterion:
+
+$$\begin{aligned}
+    n \tau_E
+    \ge \frac{Q}{Q + 1} \frac{12 k_B T}{\expval{\sigma v} E_\mathrm{fus}}
+\end{aligned}$$
+
+However, it turns out that the highest fusion power density
+is reached when $T$ is at the minimum of $T^2 / \expval{\sigma v}$.
+Therefore, we multiply by $T$ to get the Lawson triple product:
+
+$$\begin{aligned}
+    \boxed{
+        n T \tau_E
+        \ge \frac{Q}{Q + 1} \frac{12 k_B T^2}{\expval{\sigma v} E_\mathrm{fus}}
+    }
+\end{aligned}$$
+
+For some reason,
+it is often assumed that the fusion is infinitely profitable $Q \to \infty$,
+in which case the criterion reduces to:
+
+$$\begin{aligned}
+    n T \tau_E
+    \ge \frac{12 k_B T^2}{\expval{\sigma v} E_\mathrm{fus}}
+\end{aligned}$$
+
+
+
+## References
+1.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.
diff --git a/content/know/concept/lindhard-function/index.pdc b/content/know/concept/lindhard-function/index.pdc
new file mode 100644
index 0000000..d38dc2e
--- /dev/null
+++ b/content/know/concept/lindhard-function/index.pdc
@@ -0,0 +1,420 @@
+---
+title: "Lindhard function"
+firstLetter: "L"
+publishDate: 2021-10-12
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-09-23T16:21:57+02:00
+draft: false
+markup: pandoc
+---
+
+# Lindhard function
+
+We start from the [Kubo formula](/know/concept/kubo-formula/)
+for the electron density operator $\hat{n}$,
+which describes the change in $\expval{\hat{n}}$
+due to a time-dependent perturbation $\hat{H}_1$:
+
+$$\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}$$
+
+Where $\Theta$ is the [Heaviside step function](/know/concept/heaviside-step-function/),
+and the subscript $I$ refers to the [interaction picture](/know/concept/interaction-picture/).
+Notice from the limits that the perturbation is switched on at $t = -\infty$.
+Now, let us consider the following harmonic $\hat{H}_1$ in the Schrödinger picture:
+
+$$\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}}
+\end{aligned}$$
+
+Where $\eta$ is a tiny positive number,
+which represents a gradual switching-on of $\hat{H}_1$,
+eliminating transient effects
+and helping the convergence of an integral later.
+
+We assume that $V(\vb{r})$ varies slowly compared to the electrons' wavefunctions,
+so we argue that $\hat{V}_S$ is practically time-independent,
+because the total number of electrons is conserved,
+and $\hat{n}$ is only weakly perturbed by $\hat{H}_1$.
+
+Because $\hat{H}_1$ starts at $t = -\infty$,
+we can always shift the time axis such that the point of interest is at $t = 0$.
+We thus have, without loss of generality:
+
+$$\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'}
+\end{aligned}$$
+
+The expectation values $\expval{}_0$ are calculated for $\ket{0}$,
+which was the state at $t = -\infty$.
+Note that if $\ket{0}$ is an eigenstate of $\hat{H}_{0,S}$,
+there is no difference which picture (Schrödinger or interaction) $\ket{0}$ is in,
+because any operator $\hat{A}$ then satisfies:
+
+$$\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}
+\end{aligned}$$
+
+Therefore, we will assume that $\ket{0}$ is an eigenstate of $\hat{H}_{0,S}$.
+Next, we insert the identity operator $\hat{I} = \sum_{j} \ket{j} \bra{j}$,
+where $\ket{j}$ are all the eigenstates of $\hat{H}_{0,S}$:
+
+$$\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'}
+\end{aligned}$$
+
+Using the fact that $\ket{0}$ and $\ket{j}$
+are eigenstates of $\hat{H}_{0,S}$,
+and that we chose $t = 0$, we find:
+
+$$\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) 
+    \\
+    \matrixel{j}{\hat{n}_I(0)}{0}
+    &= \matrixel{j}{\hat{n}_S(0)}{0} \:\exp\!(0 - 0)
+    = \matrixel{j}{\hat{n}_S(0)}{0}
+\end{aligned}$$
+
+We define $\omega_{j0} \equiv (E_j \!-\! E_0)/\hbar$
+and insert the above expressions into $\delta\expval{\hat{n}}$,
+yielding:
+
+$$\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)
+\end{aligned}$$
+
+These integrals are [Fourier transforms](/know/concept/fourier-transform/),
+and are straightforward to evaluate. The first is:
+
+$$\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
+    \\
+    &= \frac{1}{- i (\omega - \omega_{j0}) + \eta}
+    = \frac{i}{\omega - \omega_{j0} + i \eta}
+\end{aligned}$$
+
+The other integral simply has the opposite sign in front of $\omega_{j0}$.
+We thus arrive at:
+
+$$\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)
+\end{aligned}$$
+
+Inserting the definition $\hat{V}_S = \int V(\vb{r}') \:\hat{n}(\vb{r}') \dd{\vb{r}'}$
+leads us to the following formula for $\delta\expval{\hat{n}}$,
+which has the typical form of a linear response,
+with response function $\chi$:
+
+$$\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}
+    }
+\end{aligned}$$
+
+By definition, $\ket{j}$ are eigenstates
+of the many-electron Hamiltonian $\hat{H}_{0,S}$,
+which is only solvable if we crudely neglect
+any and all electron-electron interactions.
+Therefore, to continue, we neglect those interactions.
+According to tradition, we then rename $\chi$ to $\chi_0$.
+
+The well-known ground state of a non-interacting electron gas
+is the Fermi sea $\ket{\mathrm{FS}}$, given below,
+together with $\hat{n}_S$ in the language of the
+[second quantization](/know/concept/second-quantization/):
+
+$$\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
+\end{aligned}$$
+
+For now, we ignore thermal excitations,
+i.e. we set the temperature $T = 0$.
+In $\chi_0 = \chi$, we thus insert the above $\hat{n}_S$,
+and replace $\ket{0}$ with $\ket{\mathrm{FS}}$, yielding:
+
+$$\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}
+\end{aligned}$$
+
+This inner product is only nonzero if
+$\ket{j} = \hat{c}_a \hat{c}_b^\dagger \ket{\mathrm{FS}}$
+with $a = \alpha$ and $b = \beta$,
+or in other words,
+only if $\ket{j}$ is a single-electron excitation of $\ket{\mathrm{FS}}$.
+Furthermore, in $\ket{\mathrm{FS}}$,
+$\alpha$ must be filled, and $\beta$ must be empty.
+Let $f_\alpha \in \{0,1\}$ be the occupation number of orbital $\alpha$, then:
+
+$$\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}
+\end{aligned}$$
+
+In $\chi_0$, the sum over $j$ becomes a sum over $a$ and $b$
+(this implicitly eliminates all $\ket{j}$ that are not single-electron excitations),
+and $E_j\!-\!E_0$ becomes the cost of the excitation $\epsilon_b \!-\! \epsilon_a$,
+where $\epsilon_a$ is the energy of orbital $a$.
+Therefore, we find:
+
+$$\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)
+\end{aligned}$$
+
+Because $f_a, f_b \in \{0, 1\}$ when $T = 0$, we know that $f_a^2 (1 \!-\! f_b)^2 = f_a (1 \!-\! f_b)$.
+We then swap the indices $a$ and $b$ in the second term, leading to:
+
+$$\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}
+\end{aligned}$$
+
+To proceed, we make the radical assumption that $\vb{H}_{0,S}$
+has continuous translational symmetry,
+or in other words, that the external potential is uniform in space.
+Clearly, this is not realistic,
+so our conclusions will be more qualitative than quantitative.
+
+In that case, the wavefunction of a non-interacting particle is simply a plane wave,
+so we insert $\psi_a(\vb{r}) = \exp\!(i \vb{k}_a \cdot \vb{r})$
+and $\psi_b(\vb{r}) = \exp\!(i \vb{k}_b \cdot \vb{r})$, yielding:
+
+$$\begin{aligned}
+    \chi_0
+    &= \sum_{a b} \Big( f_a - f_b \Big)
+    \frac{\exp\!\big( \!-\! i \vb{k}_a \cdot \vb{r} + i \vb{k}_b \cdot \vb{r} + i \vb{k}_a \cdot \vb{r}' - i \vb{k}_b \cdot \vb{r}' \big)}
+    {(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a}
+    \\
+    &= \sum_{a b} \Big( f_a - f_b \Big)
+    \frac{\exp\!\big( \!-\! i (\vb{k}_a \!-\! \vb{k}_b) \cdot (\vb{r} \!-\! \vb{r}')\big)}
+    {(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a}
+\end{aligned}$$
+
+Here, we see that $\chi_0$ only depends on the differences
+$\vb{r}\!-\!\vb{r}'$ and $\vb{k}_a\!-\!\vb{k}_b$.
+Therefore, we define $\vb{q}' \equiv \vb{k}_b\!-\!\vb{k}_a$
+and rename $\vb{k}_a \to \vb{k}$.
+We thus have:
+
+$$\begin{aligned}
+    \chi_0(\vb{r}\!-\!\vb{r}')
+    &= \sum_{\vb{k} \vb{q}'} \Big( f_k - f_{k+q'} \Big)
+    \frac{\exp\!\big( i \vb{q'} \cdot (\vb{r} \!-\! \vb{r}')\big)}
+    {(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k}
+\end{aligned}$$
+
+The summation goes over all $\vb{k}$ and $\vb{q}'$
+where $\vb{k}$ is inside the Fermi sphere, and $\vb{k}\!+\!\vb{q}'$ is outside.
+Let $k_F$ be the Fermi radius,
+then we convert this sum into an integral,
+which means introducing a factor of $1/(2 \pi)^{3}$
+as usual in solid state physics:
+
+$$\begin{aligned}
+    \chi_0(\vb{r})
+    &= \sum_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}}
+    \frac{(f_k - f_{k+q'}) \exp\!\big( i \vb{q}' \cdot \vb{r} \big)}
+    {(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k}
+    \\
+    &= \frac{1}{(2 \pi)^3} \iint_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}}
+    \frac{(f_k - f_{k+q'}) \exp\!\big( i \vb{q}' \cdot \vb{r} \big)}
+    {(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} \dd{\vb{k}} \dd{\vb{q}'}
+\end{aligned}$$
+
+Fourier transforming the position $\vb{r}$ into the wavevector $\vb{q}$,
+we recognize an integral that can be evaluated
+to a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta(\vb{q})$:
+
+$$\begin{aligned}
+    \chi_0(\vb{q})
+    &= \frac{1}{(2 \pi)^3} \int_{-\infty}^\infty \iint_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}}
+    \frac{(f_k - f_{k+q'}) \exp\!\big( i (\vb{q}' \!-\! \vb{q}) \cdot \vb{r} \big)}{(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k}
+    \dd{\vb{k}} \dd{\vb{q}'} \dd{\vb{r}}
+    \\
+    &= \iint_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}}
+    \frac{(f_k - f_{k+q'}) \:\delta(\vb{q}'\!-\!\vb{q})}{(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} \dd{\vb{k}} \dd{\vb{q}'}
+\end{aligned}$$
+
+This delta functions eliminates the integral over $\vb{q}'$,
+giving the following linear response $\chi_0$
+of a non-interacting electron gas in a uniform potential:
+
+$$\begin{aligned}
+    \boxed{
+        \chi_0(\vb{q}, \omega)
+        = \int_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}| > k_F}}
+        \frac{f_k - f_{k+q}}{(\omega + i \eta) \hbar - \epsilon_{k+q} + \epsilon_k} \dd{\vb{k}}
+    }
+\end{aligned}$$
+
+The resulting electron density change $\delta{\expval{\hat{n}}}$ is as follows,
+where we use the [convolution theorem](/know/concept/convolution-theorem/)
+to convert the convolution in $\vb{r}$-space into a product in $\vb{q}$-space:
+
+$$\begin{gathered}
+    \delta\expval{\hat{n}}(\vb{r}, \omega)
+    = \int_{-\infty}^\infty \chi_0(\vb{r}\!-\!\vb{r}', \omega) \: V(\vb{r}') \dd{r'}
+    \\
+    \implies \qquad
+    \boxed{
+        \delta\expval{\hat{n}}(\vb{q}, \omega)
+        = \chi_0(\vb{q}, \omega) \: V(\vb{q})
+    }
+\end{gathered}$$
+
+So far, we have neglected electron-electron interactions,
+but now we approximately correct this.
+We split the effective potential $\vb{V}_\mathrm{eff}$ felt by the electrons
+into the external potential $V_\mathrm{ext}$
+and the internal interactions $V_\mathrm{int}$,
+such that:
+
+$$\begin{aligned}
+    V_\mathrm{eff}(\vb{r})
+    = V_\mathrm{ext} + V_\mathrm{int}
+\end{aligned}$$
+
+We approximate $V_\mathrm{int}$ as follows,
+where $V_{ee}$ represents electron-electron interactions:
+
+$$\begin{aligned}
+    V_\mathrm{int}(\vb{r})
+    \approx \int_{-\infty}^\infty V_{ee}(\vb{r} \!-\! \vb{r}') \: \delta{n}(\vb{r}') \dd{\vb{r}'}
+    \qquad\quad
+    V_{ee}(\vb{r} \!-\! \vb{r}')
+    = \frac{e^2}{4 \pi \varepsilon_0} \frac{1}{|\vb{r} - \vb{r}'|}
+\end{aligned}$$
+
+Consequently, $V_\mathrm{int}$ satisfies Poisson's equation,
+which has a well-known Fourier transform:
+
+$$\begin{aligned}
+    \nabla^2 V_\mathrm{int}(\vb{r})
+    = - \frac{\delta{n}(\vb{r})}{\varepsilon_0}
+    \quad \implies \quad
+    V_\mathrm{int}(\vb{q})
+    = \frac{e^2}{\varepsilon_0 |\vb{q}|^2} \delta{n}(\vb{q})
+\end{aligned}$$
+
+Meanwhile, from all of the above calculations,
+we can write $\delta{n}$ as follows,
+where $\chi$ and $\chi_0$ are the
+(unknown) interacting and (known) non-interacting response functions:
+
+$$\begin{aligned}
+    \delta{n}(\vb{q})
+    = \chi V_\mathrm{ext}
+    \approx \chi_0 V_\mathrm{eff}
+\end{aligned}$$
+
+Keep in mind that we are treating interactions as a perturbation to $V_\mathrm{ext}$,
+therefore $V_\mathrm{ext} \approx V_\mathrm{eff}$.
+With this, $V_\mathrm{eff}$ becomes as follows in $\vb{q}$-space,
+where we have used the convolution theorem
+to get the product $\delta{n} (\vb{q}) V_{ee}(\vb{q})$:
+
+$$\begin{aligned}
+    V_\mathrm{eff}(\vb{q})
+    = V_\mathrm{ext} + \chi_0 V_\mathrm{eff} V_{ee}
+    \qquad \quad
+    V_{ee}(\vb{q})
+    = \frac{e^2}{\varepsilon_0 |\vb{q}|^2}
+\end{aligned}$$
+
+Isolating this equation for $V_\mathrm{ext}$
+yields the definition of the relative permittivity $\varepsilon_r$:
+
+$$\begin{aligned}
+    V_\mathrm{ext}
+    = (1 - \chi_0 V_{ee}) V_\mathrm{eff}
+    \equiv \varepsilon_r V_\mathrm{eff}
+\end{aligned}$$
+
+Therefore, by inserting all the above expressions,
+we arrive at the following dielectric function $\varepsilon_r$
+for a non-interacting electron gas in a uniform potential:
+
+$$\begin{aligned}
+    \boxed{
+        \varepsilon_r(\vb{q}, \omega)
+        = 1 - \frac{e^2}{\varepsilon_0 |\vb{q}|^2} \sum_{k} \frac{f_{k-q} - f_k}{\hbar (\omega + i \eta) + E_{k-q} - E_k}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  K.S. Thygesen,
+    *Advanced solid state physics: linear response theory*,
+    2013, unpublished.
+2.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
+3.  G. Grosso, G.P. Parravicini,
+    *Solid state physics*,
+    2nd edition, Elsevier.