From 8a9fb5fef2a97af3274290e512816e1a4cac0c02 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Mon, 24 Jan 2022 19:29:00 +0100
Subject: Rewrite "Lindhard function", split off "dielectric function"
---
content/know/concept/lindhard-function/index.pdc | 535 +++++++++++------------
1 file changed, 258 insertions(+), 277 deletions(-)
(limited to 'content/know/concept/lindhard-function/index.pdc')
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:
+
+
+
+
+
+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:
$$\begin{aligned}
- \chi_0
- &= \sum_{a b} \Big( f_a (1 \!-\! f_b) - f_b (1 \!-\! f_a) \Big)
- \frac{\psi_a^*(\vb{r}) \psi_b(\vb{r}) \psi_a(\vb{r}') \psi_b^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a}
- \\
- &= \sum_{a b} \Big( f_a - f_b \Big)
- \frac{\psi_a^*(\vb{r}) \psi_b(\vb{r}) \psi_a(\vb{r}') \psi_b^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a}
+ \hat{n}(-\vb{q})
+ = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} - \vb{q}}
+ = \sum_{\vb{k}'} \hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'}
+ = \hat{n}^\dagger(\vb{q})
\end{aligned}$$
-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.
+The summation variable $\vb{k}$ has an associated spin $\sigma$,
+and $\hat{n}$ does not carry any spin.
+
+
-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:
+When neglecting interactions, it is tradition to rename $\chi$ to $\chi_0$.
+We insert $\hat{n}$, suppressing spin:
$$\begin{aligned}
\chi_0
- &= \sum_{a b} \Big( f_a - f_b \Big)
- \frac{\exp\!\big( \!-\! i \vb{k}_a \cdot \vb{r} + i \vb{k}_b \cdot \vb{r} + i \vb{k}_a \cdot \vb{r}' - i \vb{k}_b \cdot \vb{r}' \big)}
- {(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a}
- \\
- &= \sum_{a b} \Big( f_a - f_b \Big)
- \frac{\exp\!\big( \!-\! i (\vb{k}_a \!-\! \vb{k}_b) \cdot (\vb{r} \!-\! \vb{r}')\big)}
- {(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a}
+ &= \frac{1}{Z V} \sum_{\vb{k} \vb{k}'} \sum_{\nu \nu'} \bigg(
+ \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'}
+ \matrixel{\nu'}{\hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'}}{\nu}}
+ {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}}
+ - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'}
+ \matrixel{\nu'}{\hat{c}_{\vb{k}'}^\dagger \hat{c}_{\vb{k}' + \vb{q}}}{\nu}}
+ {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu}
\end{aligned}$$
-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:
-
-$$\begin{aligned}
- \chi_0(\vb{r}\!-\!\vb{r}')
- &= \sum_{\vb{k} \vb{q}'} \Big( f_k - f_{k+q'} \Big)
- \frac{\exp\!\big( i \vb{q'} \cdot (\vb{r} \!-\! \vb{r}')\big)}
- {(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k}
-\end{aligned}$$
+Here, $\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'}$
+is only nonzero if $\ket{\nu'}$ is contructed from $\ket{\nu}$
+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 $\sigma = \sigma'$).
-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:
+For the same reason, the energy difference $E_\nu \!-\! E_{\nu'}$
+can simply be replaced by the cost of the single-particle excitation
+$\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}}$,
+where $\xi_{\vb{k}}$ is the energy of a $\vb{k}$-orbital.
+Therefore:
$$\begin{aligned}
- \chi_0(\vb{r})
- &= \sum_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}}
- \frac{(f_k - f_{k+q'}) \exp\!\big( i \vb{q}' \cdot \vb{r} \big)}
- {(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k}
- \\
- &= \frac{1}{(2 \pi)^3} \iint_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}}
- \frac{(f_k - f_{k+q'}) \exp\!\big( i \vb{q}' \cdot \vb{r} \big)}
- {(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} \dd{\vb{k}} \dd{\vb{q}'}
+ \chi_0
+ &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu \nu'} \bigg(
+ \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'}
+ \matrixel{\nu'}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}}
+ {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+ - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'}
+ \matrixel{\nu'}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}}
+ {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu}
\end{aligned}$$
-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/) $\delta(\vb{q})$:
+Notice that we have eliminated all dependence on $\ket{\nu'}$,
+so we remove it by $\sum_{\nu} \ket{\nu} \bra{\nu} = 1$:
$$\begin{aligned}
- \chi_0(\vb{q})
- &= \frac{1}{(2 \pi)^3} \int_{-\infty}^\infty \iint_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}}
- \frac{(f_k - f_{k+q'}) \exp\!\big( i (\vb{q}' \!-\! \vb{q}) \cdot \vb{r} \big)}{(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k}
- \dd{\vb{k}} \dd{\vb{q}'} \dd{\vb{r}}
+ \chi_0
+ &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu} \bigg(
+ \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}}
+ {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+ - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}}
+ {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu}
\\
- &= \iint_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}}
- \frac{(f_k - f_{k+q'}) \:\delta(\vb{q}'\!-\!\vb{q})}{(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} \dd{\vb{k}} \dd{\vb{q}'}
+ &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu}
+ \frac{\matrixel{\nu}{\comm*{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}
+ {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0}}{\nu}}
+ {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
\end{aligned}$$
-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:
+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 $\expval{}_0$:
$$\begin{aligned}
- \boxed{
- \chi_0(\vb{q}, \omega)
- = \int_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}| > k_F}}
- \frac{f_k - f_{k+q}}{(\omega + i \eta) \hbar - \epsilon_{k+q} + \epsilon_k} \dd{\vb{k}}
- }
+ \chi_0
+ &= \frac{1}{Z V} \sum_{\vb{k}} \frac{\Tr\!\big(\comm*{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}
+ {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0} \big)}
+ {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+ = \frac{1}{V} \sum_{\vb{k}}
+ \frac{\expval*{\comm*{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}}_0}
+ {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
\end{aligned}$$
-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:
+This commutator can be evaluated,
+and in this particular case it turns out to be:
$$\begin{aligned}
- V_\mathrm{eff}(\vb{r})
- = V_\mathrm{ext} + V_\mathrm{int}
+ \comm*{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}
+ = \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}} - \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k} + \vb{q}}
\end{aligned}$$
-We approximate $V_\mathrm{int}$ as follows,
-where $V_{ee}$ represents electron-electron interactions:
-
+
+
+
+
+
+In general, for any single-particle states labeled by $m$, $n$, $o$ and $p$, we have:
$$\begin{aligned}
- V_\mathrm{int}(\vb{r})
- \approx \int_{-\infty}^\infty V_{ee}(\vb{r} \!-\! \vb{r}') \: \delta{n}(\vb{r}') \dd{\vb{r}'}
- \qquad\quad
- V_{ee}(\vb{r} \!-\! \vb{r}')
- = \frac{e^2}{4 \pi \varepsilon_0} \frac{1}{|\vb{r} - \vb{r}'|}
+ \comm*{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p}
+ &= \hat{c}_m^\dagger \hat{c}_n \hat{c}_o^\dagger \hat{c}_p - \hat{c}_o^\dagger \hat{c}_p \hat{c}_m^\dagger \hat{c}_n
+ \\
+ &= \hat{c}_m^\dagger \big( \acomm*{\hat{c}_n}{\hat{c}_o^\dagger} - \hat{c}_o^\dagger \hat{c}_n \big) \hat{c}_p
+ - \hat{c}_o^\dagger \big( \acomm*{\hat{c}_p}{\hat{c}_m^\dagger} - \hat{c}_m^\dagger \hat{c}_p \big) \hat{c}_n
\end{aligned}$$
-Consequently, $V_\mathrm{int}$ satisfies Poisson's equation,
-which has a well-known Fourier transform:
+Using the standard fermion anticommutation relations, this becomes:
$$\begin{aligned}
- \nabla^2 V_\mathrm{int}(\vb{r})
- = - \frac{\delta{n}(\vb{r})}{\varepsilon_0}
- \quad \implies \quad
- V_\mathrm{int}(\vb{q})
- = \frac{e^2}{\varepsilon_0 |\vb{q}|^2} \delta{n}(\vb{q})
+ \comm*{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p}
+ &= \hat{c}_m^\dagger \big( \delta_{no} - \hat{c}_o^\dagger \hat{c}_n \big) \hat{c}_p
+ - \hat{c}_o^\dagger \big( \delta_{pm} - \hat{c}_m^\dagger \hat{c}_p \big) \hat{c}_n
+ \\
+ &= \hat{c}_m^\dagger \hat{c}_p \: \delta_{no} - \hat{c}_m^\dagger \hat{c}_o^\dagger \hat{c}_n \hat{c}_p
+ - \hat{c}_o^\dagger \hat{c}_n \: \delta_{pm} + \hat{c}_o^\dagger \hat{c}_m^\dagger \hat{c}_p \hat{c}_n
+ \\
+ &= \hat{c}_m^\dagger \hat{c}_p \: \delta_{no} - \hat{c}_o^\dagger \hat{c}_n \: \delta_{pm}
\end{aligned}$$
-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:
+In this case, $m = p = \vb{k}$ and $n = o = \vb{k} \!+\! \vb{q}$,
+so the Kronecker deltas are unnecessary.
+
+
+
+We substitute this result into $\chi_0$,
+and reintroduce the spin index $\sigma$ associated with $\vb{k}$:
$$\begin{aligned}
- \delta{n}(\vb{q})
- = \chi V_\mathrm{ext}
- \approx \chi_0 V_\mathrm{eff}
+ \chi_0(\vb{q}, \omega)
+ = \frac{1}{V} \sum_{\sigma \vb{k}}
+ \frac{\expval*{\hat{c}_{\sigma,\vb{k}}^\dagger \hat{c}_{\sigma,\vb{k}} - \hat{c}_{\sigma,\vb{k}+\vb{q}}^\dagger \hat{c}_{\sigma,\vb{k}+\vb{q}}}_0}
+ {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
\end{aligned}$$
-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 $\delta{n} (\vb{q}) V_{ee}(\vb{q})$:
+The operator $\hat{c}_{\sigma.\vb{k}}^\dagger \hat{c}_{\sigma.\vb{k}}$
+simply counts the number of electrons in state $(\sigma, \vb{k})$,
+which is given by the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/) $n_F$.
+This gives us the **Lindhard response function**:
$$\begin{aligned}
- V_\mathrm{eff}(\vb{q})
- = V_\mathrm{ext} + \chi_0 V_\mathrm{eff} V_{ee}
- \qquad \quad
- V_{ee}(\vb{q})
- = \frac{e^2}{\varepsilon_0 |\vb{q}|^2}
+ \boxed{
+ \chi_0(\vb{q}, \omega)
+ = \frac{1}{V} \sum_{\sigma \vb{k}}
+ \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}
+ {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+ }
\end{aligned}$$
-Isolating this equation for $V_\mathrm{ext}$
-yields the definition of the relative permittivity $\varepsilon_r$:
+From this, we would like to get the
+[dielectric function](/know/concept/dielectric-function/) $\varepsilon_r$.
+Recall its definition, where $U_\mathrm{tot}$, $U_\mathrm{ext}$, and $U_\mathrm{ind}$
+are the total, external and induced potentials, respectively:
$$\begin{aligned}
- V_\mathrm{ext}
- = (1 - \chi_0 V_{ee}) V_\mathrm{eff}
- \equiv \varepsilon_r V_\mathrm{eff}
+ U_\mathrm{tot}
+ = U_\mathrm{ext} + U_\mathrm{ind}
+ = \frac{U_\mathrm{ext}}{\varepsilon_r}
\end{aligned}$$
-Therefore, by inserting all the above expressions,
-we arrive at the Lindhard dielectric function $\varepsilon_r$
-for a non-interacting electron gas in a uniform potential:
+Note that these are all *energy* potentials:
+this choice is justified because all energy potentials
+are caused by electric fields in this case.
+The *electric* potential is recoverable as
+$\Phi_\mathrm{tot} = q_e U_\mathrm{tot}$,
+where $q_e < 0$ is the charge of an electron.
+
+From the Lindhard response function $\chi_0$,
+we get the induced particle density offset $\delta{\expval{\hat{n}}}$
+caused by a potential $U$.
+The density $\delta{\expval{\hat{n}}}$ should be self-consistent,
+implying $U = U_\mathrm{tot}$.
+In other words, we have a linear relation
+$\delta{\expval{\hat{n}}} = \chi_0 U_\mathrm{tot}$,
+so the standard formula for $\varepsilon_r$ gives:
$$\begin{aligned}
\boxed{
\varepsilon_r(\vb{q}, \omega)
- = 1 - \frac{e^2}{\varepsilon_0 |\vb{q}|^2} \sum_{k} \frac{f_{k-q} - f_k}{\hbar (\omega + i \eta) + E_{k-q} - E_k}
+ = 1 - \frac{U_{ee}(\vb{q})}{V}
+ \sum_{\sigma \vb{k}} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
}
\end{aligned}$$
+Where $U_{ee}(\vb{q}) = q_e^2 / (\varepsilon_0 |\vb{q}|^2)$
+is Coulomb repulsion.
+This is the **Lindhard dielectric function** of a free
+non-interacting electron gas,
+at any temperature and for any dimensionality.
+
## References
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