From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'
---
source/know/concept/lindhard-function/index.md | 170 ++++++++++++-------------
1 file changed, 85 insertions(+), 85 deletions(-)
(limited to 'source/know/concept/lindhard-function')
diff --git a/source/know/concept/lindhard-function/index.md b/source/know/concept/lindhard-function/index.md
index 4252bd4..4033148 100644
--- a/source/know/concept/lindhard-function/index.md
+++ b/source/know/concept/lindhard-function/index.md
@@ -14,28 +14,28 @@ 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/)
-for the electron density operator $\hat{n}$,
-which describes the change in $\Expval{\hat{n}}$
-due to a time-dependent perturbation $\hat{H}_1$:
+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
+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 $\hat{H}_1$:
+Now consider a harmonic $$\hat{H}_1$$:
$$\begin{aligned}
\hat{H}_{1,S}(t)
= e^{i (\omega + i \eta) t} \int_{-\infty}^\infty U(\vb{r}) \: \hat{n}_S(\vb{r}) \dd{\vb{r}}
\end{aligned}$$
-Where $S$ is the Schrödinger picture,
-$\eta$ is a positive infinitesimal to ensure convergence later,
-and $U(\vb{r})$ is an arbitrary potential function.
+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:
$$\begin{aligned}
@@ -43,7 +43,7 @@ $$\begin{aligned}
= \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}$$
-Here, $\chi$ is the density-density correlation function,
+Here, $$\chi$$ is the density-density correlation function,
i.e. a two-particle [Green's function](/know/concept/greens-functions/):
$$\begin{aligned}
@@ -51,10 +51,10 @@ $$\begin{aligned}
\equiv - \frac{i}{\hbar} \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{r}, t)}{\hat{n}_I(\vb{r}', t')}}_0
\end{aligned}$$
-Let us assume that the unperturbed system (i.e. without $U$) is spatially uniform,
-so that $\chi$ only depends on the difference $\vb{r} - \vb{r}'$.
+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/)
-$\vb{r}\!-\!\vb{r}' \to \vb{q}$:
+$$\vb{r}\!-\!\vb{r}' \to \vb{q}$$:
$$\begin{aligned}
\chi(\vb{q}; t, t')
@@ -65,9 +65,9 @@ $$\begin{aligned}
\: 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}$$
-Where both $\hat{n}_I$ have been written as inverse Fourier transforms,
-giving a factor $(2 \pi)^{-2 D}$, with $D$ being the number of spatial dimensions.
-We rearrange to get a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$:
+Where both $$\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}
\chi(\vb{q}; t, t')
@@ -84,9 +84,9 @@ $$\begin{aligned}
\: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2}
\end{aligned}$$
-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:
+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:
$$\begin{aligned}
\chi(\vb{q}; t, t')
@@ -95,8 +95,8 @@ $$\begin{aligned}
\: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \dd{\vb{r}'}
\end{aligned}$$
-For $V \to \infty$ we get a Dirac delta function,
-but in fact the conclusion holds for finite $V$ too:
+For $$V \to \infty$$ we get a Dirac delta function,
+but in fact the conclusion holds for finite $$V$$ too:
$$\begin{aligned}
\chi(\vb{q}; t, t')
@@ -106,10 +106,10 @@ $$\begin{aligned}
&= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0
\end{aligned}$$
-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 $\chi$
+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/):
$$\begin{aligned}
@@ -119,11 +119,11 @@ $$\begin{aligned}
\Big( e^{-\beta E_\nu} - e^{- \beta E_{\nu'}} \Big)
\end{aligned}$$
-Where $\Ket{\nu}$ and $\Ket{\nu'}$ are many-electron eigenstates of $\hat{H}_0$,
-and $Z$ is the [grand partition function](/know/concept/grand-canonical-ensemble/).
+Where $$\Ket{\nu}$$ and $$\Ket{\nu'}$$ are many-electron eigenstates of $$\hat{H}_0$$,
+and $$Z$$ is the [grand partition function](/know/concept/grand-canonical-ensemble/).
According to the [convolution theorem](/know/concept/convolution-theorem/)
-$\delta{\Expval{\hat{n}}}(\vb{q}, \omega) = \chi(\vb{q}, \omega) \: U(\vb{q})$.
-In anticipation, we swap $\nu$ and $\nu''$ in the second term,
+$$\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}
@@ -135,9 +135,9 @@ $$\begin{aligned}
{\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu}
\end{aligned}$$
-All operators are in the Schrödinger picture from now on, hence we dropped the subscript $S$.
+All operators are in the Schrödinger picture from now on, hence we dropped the subscript $$S$$.
-To proceed, we need to rewrite $\hat{n}(\vb{q})$ somehow.
+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:
@@ -154,10 +154,10 @@ $$\begin{aligned}
-Starting from the general definition of $\hat{n}$,
-we write out the field operators $\hat{\Psi}(\vb{r})$,
+Starting from the general definition of $$\hat{n}$$,
+we write out the field operators $$\hat{\Psi}(\vb{r})$$,
and insert the known non-interacting single-electron orbitals
-$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$:
+$$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$$:
$$\begin{aligned}
\hat{n}(\vb{r})
@@ -166,7 +166,7 @@ $$\begin{aligned}
= \frac{1}{V} \sum_{\vb{k} \vb{k}'} e^{i (\vb{k}' - \vb{k}) \cdot \vb{r}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'}
\end{aligned}$$
-Taking the Fourier transfom yields a Dirac delta function $\delta$:
+Taking the Fourier transfom yields a Dirac delta function $$\delta$$:
$$\begin{aligned}
\hat{n}(\vb{q})
@@ -176,19 +176,19 @@ $$\begin{aligned}
\end{aligned}$$
If we impose periodic boundary conditions
-on our $D$-dimensional hypercube of volume $V$,
-then $\vb{k}$ becomes discrete,
-with per-value spacing $2 \pi / V^{1/D}$ along each axis.
-
-Consequently, each orbital $\psi_\vb{k}$ uniquely occupies
-a volume $(2 \pi)^D / V$ in $\vb{k}$-space, so we make the approximation
-$\sum_{\vb{k}} \approx V / (2 \pi)^D \int_{-\infty}^\infty \dd{\vb{k}}$.
-This becomes exact for $V \to \infty$,
-in which case $\vb{k}$ also becomes continuous again,
+on our $$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}$$ also becomes continuous again,
which is what we want for jellium.
-We apply this standard trick from condensed matter physics to $\hat{n}$,
-and $V$ cancels out:
+We apply this standard trick from condensed matter physics to $$\hat{n}$$,
+and $$V$$ cancels out:
$$\begin{aligned}
\hat{n}(\vb{q})
@@ -197,9 +197,9 @@ $$\begin{aligned}
= \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}
\end{aligned}$$
-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:
+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:
$$\begin{aligned}
\hat{n}(-\vb{q})
@@ -208,13 +208,13 @@ $$\begin{aligned}
= \hat{n}^\dagger(\vb{q})
\end{aligned}$$
-The summation variable $\vb{k}$ has an associated spin $\sigma$,
-and $\hat{n}$ does not carry any spin.
+The summation variable $$\vb{k}$$ has an associated spin $$\sigma$$,
+and $$\hat{n}$$ does not carry any spin.
-When neglecting interactions, it is tradition to rename $\chi$ to $\chi_0$.
-We insert $\hat{n}$, suppressing spin:
+When neglecting interactions, it is tradition to rename $$\chi$$ to $$\chi_0$$.
+We insert $$\hat{n}$$, suppressing spin:
$$\begin{aligned}
\chi_0
@@ -227,16 +227,16 @@ $$\begin{aligned}
{\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu}
\end{aligned}$$
-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 $\vb{k} \!+\! \vb{q}$,
+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 $\sigma = \sigma'$).
+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
-$\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}}$,
-where $\xi_{\vb{k}}$ is the energy of a $\vb{k}$-orbital.
+$$\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}}$$,
+where $$\xi_{\vb{k}}$$ is the energy of a $$\vb{k}$$-orbital.
Therefore:
$$\begin{aligned}
@@ -250,8 +250,8 @@ $$\begin{aligned}
{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu}
\end{aligned}$$
-Notice that we have eliminated all dependence on $\Ket{\nu'}$,
-so we remove it by $\sum_{\nu} \Ket{\nu} \Bra{\nu} = 1$:
+Notice that we have eliminated all dependence on $$\Ket{\nu'}$$,
+so we remove it by $$\sum_{\nu} \Ket{\nu} \Bra{\nu} = 1$$:
$$\begin{aligned}
\chi_0
@@ -268,9 +268,9 @@ $$\begin{aligned}
\end{aligned}$$
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 $\Expval{}_0$:
+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 $$\Expval{}_0$$:
$$\begin{aligned}
\chi_0
@@ -295,7 +295,7 @@ $$\begin{aligned}
-In general, for any single-particle states labeled by $m$, $n$, $o$ and $p$, we have:
+In general, for any single-particle states labeled by $$m$$, $$n$$, $$o$$ and $$p$$, we have:
$$\begin{aligned}
\comm{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p}
&= \hat{c}_m^\dagger \hat{c}_n \hat{c}_o^\dagger \hat{c}_p - \hat{c}_o^\dagger \hat{c}_p \hat{c}_m^\dagger \hat{c}_n
@@ -317,13 +317,13 @@ $$\begin{aligned}
&= \hat{c}_m^\dagger \hat{c}_p \: \delta_{no} - \hat{c}_o^\dagger \hat{c}_n \: \delta_{pm}
\end{aligned}$$
-In this case, $m = p = \vb{k}$ and $n = o = \vb{k} \!+\! \vb{q}$,
+In this case, $$m = p = \vb{k}$$ and $$n = o = \vb{k} \!+\! \vb{q}$$,
so the Kronecker deltas are unnecessary.
-We substitute this result into $\chi_0$,
-and reintroduce the spin index $\sigma$ associated with $\vb{k}$:
+We substitute this result into $$\chi_0$$,
+and reintroduce the spin index $$\sigma$$ associated with $$\vb{k}$$:
$$\begin{aligned}
\chi_0(\vb{q}, \omega)
@@ -332,9 +332,9 @@ $$\begin{aligned}
{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
\end{aligned}$$
-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/) $n_F$.
+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**:
$$\begin{aligned}
@@ -347,8 +347,8 @@ $$\begin{aligned}
\end{aligned}$$
From this, we would like to get the
-[dielectric function](/know/concept/dielectric-function/) $\varepsilon_r$.
-Recall its definition, where $U_\mathrm{tot}$, $U_\mathrm{ext}$, and $U_\mathrm{ind}$
+[dielectric function](/know/concept/dielectric-function/) $$\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}
@@ -361,17 +361,17 @@ 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 $U = U_\mathrm{tot}$.
+$$\Phi_\mathrm{tot} = q_e U_\mathrm{tot}$$,
+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:
+$$\delta{\Expval{\hat{n}}} = \chi_0 U_\mathrm{tot}$$,
+so the standard formula for $$\varepsilon_r$$ gives:
$$\begin{aligned}
\boxed{
@@ -381,7 +381,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $U_{ee}(\vb{q}) = q_e^2 / (\varepsilon_0 |\vb{q}|^2)$
+Where $$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,
--
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