Improve knowledge baseHEADmaster

1 files changed, 45 insertions, 41 deletions

diff --git a/source/know/concept/lindhard-function/index.md b/source/know/concept/lindhard-function/index.md
index fd620df..5f11d36 100644
--- a/source/know/concept/lindhard-function/index.md
+++ b/source/know/concept/lindhard-function/index.md
@@ -19,7 +19,7 @@ which describes the change in $$\Expval{\hat{n}}$$
 due to a time-dependent perturbation $$\hat{H}_1$$:
 
 $$\begin{aligned}
-    \delta\!\Expval{ {\hat{n}}}\!(\vb{r}, t)
+    \delta\!\Expval{\hat{n}(\vb{r}, t)}
     = -\frac{i}{\hbar} \int_{-\infty}^\infty \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{r}, t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
 \end{aligned}$$
 
@@ -39,7 +39,7 @@ and $$U(\vb{r})$$ is an arbitrary potential function.
 The Kubo formula becomes:
 
 $$\begin{aligned}
-    \delta\!\Expval{ {\hat{n}}}\!(\vb{r}, t)
+    \delta\!\Expval{\hat{n}(\vb{r}, t)}
     = \iint_{-\infty}^\infty \chi(\vb{r}, \vb{r}'; t, t') \: U(\vb{r}') \: e^{i (\omega + i \eta) t'} \dd{t'} \dd{\vb{r}'}
 \end{aligned}$$
 
@@ -95,8 +95,9 @@ $$\begin{aligned}
     \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \dd{\vb{r}'}
 \end{aligned}$$
 
-For $$V \to \infty$$ we get a Dirac delta function,
-but in fact the conclusion holds for finite $$V$$ too:
+This gives a Dirac delta function for $$V \to \infty$$
+(a limit that we will take properly later,
+but beware that some authors set $$V = 1$$ until then):
 
 $$\begin{aligned}
     \chi(\vb{q}; t, t')
@@ -107,8 +108,9 @@ $$\begin{aligned}
 \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,
+$$\chi$$ only depends on the time difference $$t\!-\!t'$$.
+Note that $$\delta{\Expval{\hat{n}}}$$ already has the form of a Fourier transform
+$$t\!-\!t' \to \omega\!+\!i \eta$$,
 which gives us an opportunity to rewrite $$\chi$$
 in the [Lehmann representation](/know/concept/lehmann-representation/):
 
@@ -119,12 +121,12 @@ $$\begin{aligned}
     \Big( e^{-\beta E_\nu} - e^{- \beta E_{\nu'}} \Big)
 \end{aligned}$$
 
-Where $$\Ket{\nu}$$ and $$\Ket{\nu'}$$ are many-electron eigenstates of $$\hat{H}_0$$,
+Where $$\Ket{\nu}$$ and $$\Ket{\nu'}$$ are many-particle 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:
+To get ready for the calculations ahead,
+we swap $$\nu$$ and $$\nu'$$ in the second term,
+so the response function is as shown below.
+All operators are in the Schrödinger picture from now on:
 
 $$\begin{aligned}
     \chi(\vb{q}, \omega)
@@ -135,7 +137,6 @@ $$\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 $$S$$.
 
 To proceed, we need to rewrite $$\hat{n}(\vb{q})$$ somehow.
 If we neglect electron-electron interactions,
@@ -180,9 +181,8 @@ 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.
+This is exact in the limit $$V \to \infty$$,
+in which case $$\vb{k}$$ also becomes a continuous variable again.
 
 We apply this standard trick from condensed matter physics to $$\hat{n}$$,
 and $$V$$ cancels out:
@@ -341,45 +341,49 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-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 $$U_\mathrm{ind}$$
-are the total, external and induced potentials, respectively:
+This is its most general form, but for practical calculations
+we need to formally take the limit $$V \to \infty$$
+and then use $$\sum_{\vb{k}} = V / (2 \pi)^{D} \int_{-\infty}^{\infty} \dd{\vb{k}}$$.
+Furthermore, electrons are spin-1/2 particles,
+so each orbital contains two, meaning
+$$\sum_{\sigma}$$ simply gives a constant factor:
 
 $$\begin{aligned}
-    U_\mathrm{tot}
-    = U_\mathrm{ext} + U_\mathrm{ind}
-    = \frac{U_\mathrm{ext}}{\varepsilon_r}
+    \boxed{
+        \chi_0(\vb{q}, \omega)
+        = \frac{2}{(2 \pi)^{D}} \int_{-\infty}^{\infty}
+        \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}
+        {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \dd{\vb{k}}
+    }
 \end{aligned}$$
 
-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
-$$\Phi_\mathrm{tot} = q_e U_\mathrm{tot}$$,
+From this, we would like to get the
+[dielectric function](/know/concept/dielectric-function/) $$\varepsilon_r$$.
+When an external [electric field](/know/concept/electric-field/) is applied,
+the electrons respond and thereby modify the net field inside the material.
+We include this effect in our *energy* potential $$U$$,
+such that the net *electric* potential
+$$\Phi_\mathrm{tot} = U / q_e$$,
 where $$q_e < 0$$ is the charge of an electron.
+This is not the same as including direct electron-electron interactions!
 
-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:
+We thus have a linear relation for the induced *particle* density
+$$\delta\!\Expval{\hat{n}(\vb{q}, \omega)} = \chi_0(\vb{q}, \omega) \: U(\vb{q})$$
+thanks to the [convolution theorem](/know/concept/convolution-theorem/).
+The corresponding induced *charge* density is given by
+$$\rho_\mathrm{ind} = q_e^2 \chi_0 \Phi_\mathrm{tot}$$,
+so the standard formula for $$\varepsilon_r$$ yields:
 
 $$\begin{aligned}
     \boxed{
         \varepsilon_r(\vb{q}, \omega)
-        = 1 - \frac{U_{ee}(\vb{q})}{V}
-        \sum_{\sigma \vb{k}} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+        = 1 - U_{ee}(\vb{q}) \frac{2}{(2 \pi)^{D}}
+        \int_{-\infty}^{\infty} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \dd{\vb{k}}
     }
 \end{aligned}$$
 
-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,
+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.