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diff --git a/source/know/concept/lindhard-function/index.md b/source/know/concept/lindhard-function/index.md new file mode 100644 index 0000000..e3df901 --- /dev/null +++ b/source/know/concept/lindhard-function/index.md @@ -0,0 +1,400 @@ +--- +title: "Lindhard function" +date: 2022-01-24 # Originally 2021-10-12, major rewrite +categories: +- Physics +- Quantum mechanics +layout: "concept" +--- + +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/) +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 +a thermal equilibrium before the perturbation was applied. +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. +The Kubo formula becomes: + +$$\begin{aligned} + \delta\!\Expval{ {\hat{n}}}\!(\vb{r}, t) + = \iint_{-\infty}^\infty \chi(\vb{r}, \vb{r}'; t, t') \: U(\vb{r}') \: e^{i (\omega + i \eta) t'} \dd{t'} \dd{\vb{r}'} +\end{aligned}$$ + +Here, $\chi$ is the density-density correlation function, +i.e. a two-particle [Green's function](/know/concept/greens-functions/): + +$$\begin{aligned} + \chi(\vb{r}, \vb{r}'; t, t') + \equiv - \frac{i}{\hbar} \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{r}, t)}{\hat{n}_I(\vb{r}', t')}}_0 +\end{aligned}$$ + +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/) +$\vb{r}\!-\!\vb{r}' \to \vb{q}$: + +$$\begin{aligned} + \chi(\vb{q}; t, t') + &= \int_{-\infty}^\infty \chi(\vb{r} - \vb{r}'; t, t') \: e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{r}} + \\ + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint + \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: e^{i \vb{q}_1 \cdot \vb{r}} e^{i \vb{q}_2 \cdot \vb{r}'} e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}} +\end{aligned}$$ + +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/) $\delta$: + +$$\begin{aligned} + \chi(\vb{q}; t, t') + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint + \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: e^{i (\vb{q}_1 - \vb{q}) \cdot \vb{r}} e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}} + \\ + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \iint + \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: \delta(\vb{q}_1 \!-\! \vb{q}) \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2} + \\ + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \int + \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} +\end{aligned}$$ + +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: + +$$\begin{aligned} + \chi(\vb{q}; t, t') + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \frac{1}{V} \int_V \int_{-\infty}^\infty + \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \dd{\vb{r}'} +\end{aligned}$$ + +For $V \to \infty$ we get a Dirac delta function, +but in fact the conclusion holds for finite $V$ too: + +$$\begin{aligned} + \chi(\vb{q}; t, t') + &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \int_{-\infty}^\infty + \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: \delta(\vb{q}_2 \!+\! \vb{q}) \dd{\vb{q}_2} + \\ + &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0 +\end{aligned}$$ + +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 $\chi$ +in the [Lehmann representation](/know/concept/lehmann-representation/): + +$$\begin{aligned} + \chi(\vb{q}, \omega) + = \frac{1}{Z V} \sum_{\nu \nu'} + \frac{\matrixel{\nu}{\hat{n}_S(\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}_S(-\vb{q})}{\nu}}{\hbar (\omega + i \eta) + E_\nu - E_{\nu'}} + \Big( e^{-\beta E_\nu} - e^{- \beta E_{\nu'}} \Big) +\end{aligned}$$ + +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: + +$$\begin{aligned} + \chi(\vb{q}, \omega) + = \frac{1}{Z V} \sum_{\nu \nu'} \bigg( + \frac{\matrixel{\nu}{\hat{n}(\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(-\vb{q})}{\nu}} + {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}} + - \frac{\matrixel{\nu}{\hat{n}(-\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(\vb{q})}{\nu}} + {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu} +\end{aligned}$$ + +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, +the single-particle states are simply plane waves, in which case: + +$$\begin{aligned} + \hat{n}(\vb{q}) + = \sum_{\sigma \vb{k}} \hat{c}_{\sigma,\vb{k}}^\dagger \hat{c}_{\sigma,\vb{k} + \vb{q}} + \qquad \qquad + \hat{n}(-\vb{q}) + = \hat{n}^\dagger(\vb{q}) +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-density"/> +<label for="proof-density">Proof</label> +<div class="hidden" markdown="1"> +<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 +$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$: + +$$\begin{aligned} + \hat{n}(\vb{r}) + \equiv \hat{\Psi}{}^\dagger(\vb{r}) \hat{\Psi}(\vb{r}) + = \sum_{\vb{k} \vb{k}'} \psi_{\vb{k}}^*(\vb{r}) \: \psi_{\vb{k}'}(\vb{r})\: \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} + = \frac{1}{V} \sum_{\vb{k} \vb{k}'} e^{i (\vb{k}' - \vb{k}) \cdot \vb{r}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} +\end{aligned}$$ + +Taking the Fourier transfom yields a Dirac delta function $\delta$: + +$$\begin{aligned} + \hat{n}(\vb{q}) + = \frac{1}{V} \int_{-\infty}^\infty + \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: e^{i (\vb{k}' - \vb{k} - \vb{q})\cdot \vb{r}} \dd{\vb{r}} + = \frac{(2 \pi)^D}{V} \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q}) +\end{aligned}$$ + +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: + +$$\begin{aligned} + \hat{n}(\vb{q}) + &= \frac{(2 \pi)^D}{V} \frac{V}{(2 \pi)^D} \sum_{\vb{k}} \int_{-\infty}^\infty + \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q}) \dd{\vb{k}'} + = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}} +\end{aligned}$$ + +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: + +$$\begin{aligned} + \hat{n}(-\vb{q}) + = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} - \vb{q}} + = \sum_{\vb{k}'} \hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'} + = \hat{n}^\dagger(\vb{q}) +\end{aligned}$$ + +The summation variable $\vb{k}$ has an associated spin $\sigma$, +and $\hat{n}$ does not carry any spin. +</div> +</div> + +When neglecting interactions, it is tradition to rename $\chi$ to $\chi_0$. +We insert $\hat{n}$, suppressing spin: + +$$\begin{aligned} + \chi_0 + &= \frac{1}{Z V} \sum_{\vb{k} \vb{k}'} \sum_{\nu \nu'} \bigg( + \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'} + \matrixel{\nu'}{\hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'}}{\nu}} + {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}} + - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'} + \matrixel{\nu'}{\hat{c}_{\vb{k}'}^\dagger \hat{c}_{\vb{k}' + \vb{q}}}{\nu}} + {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu} +\end{aligned}$$ + +Here, $\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'}$ +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 $\sigma = \sigma'$). + +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: + +$$\begin{aligned} + \chi_0 + &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu \nu'} \bigg( + \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'} + \matrixel{\nu'}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} + - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'} + \matrixel{\nu'}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu} +\end{aligned}$$ + +Notice that we have eliminated all dependence on $\Ket{\nu'}$, +so we remove it by $\sum_{\nu} \Ket{\nu} \Bra{\nu} = 1$: + +$$\begin{aligned} + \chi_0 + &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu} \bigg( + \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} + - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu} + \\ + &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu} + \frac{\matrixel{\nu}{\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}} + {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} +\end{aligned}$$ + +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 $\Expval{}_0$: + +$$\begin{aligned} + \chi_0 + &= \frac{1}{Z V} \sum_{\vb{k}} \frac{\Tr\!\big(\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}} + {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0} \big)} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} + = \frac{1}{V} \sum_{\vb{k}} + \frac{\expval{\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}}_0} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} +\end{aligned}$$ + +This commutator can be evaluated, +and in this particular case it turns out to be: + +$$\begin{aligned} + \comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} + = \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}} - \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k} + \vb{q}} +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-commutator"/> +<label for="proof-commutator">Proof</label> +<div class="hidden"> +<label for="proof-commutator">Proof.</label> +In general, for any single-particle states labeled by $m$, $n$, $o$ and $p$, we have: +$$\begin{aligned} + \comm{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p} + &= \hat{c}_m^\dagger \hat{c}_n \hat{c}_o^\dagger \hat{c}_p - \hat{c}_o^\dagger \hat{c}_p \hat{c}_m^\dagger \hat{c}_n + \\ + &= \hat{c}_m^\dagger \big( \acomm{\hat{c}_n}{\hat{c}_o^\dagger} - \hat{c}_o^\dagger \hat{c}_n \big) \hat{c}_p + - \hat{c}_o^\dagger \big( \acomm{\hat{c}_p}{\hat{c}_m^\dagger} - \hat{c}_m^\dagger \hat{c}_p \big) \hat{c}_n +\end{aligned}$$ + +Using the standard fermion anticommutation relations, this becomes: + +$$\begin{aligned} + \comm{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p} + &= \hat{c}_m^\dagger \big( \delta_{no} - \hat{c}_o^\dagger \hat{c}_n \big) \hat{c}_p + - \hat{c}_o^\dagger \big( \delta_{pm} - \hat{c}_m^\dagger \hat{c}_p \big) \hat{c}_n + \\ + &= \hat{c}_m^\dagger \hat{c}_p \: \delta_{no} - \hat{c}_m^\dagger \hat{c}_o^\dagger \hat{c}_n \hat{c}_p + - \hat{c}_o^\dagger \hat{c}_n \: \delta_{pm} + \hat{c}_o^\dagger \hat{c}_m^\dagger \hat{c}_p \hat{c}_n + \\ + &= \hat{c}_m^\dagger \hat{c}_p \: \delta_{no} - \hat{c}_o^\dagger \hat{c}_n \: \delta_{pm} +\end{aligned}$$ + +In this case, $m = p = \vb{k}$ and $n = o = \vb{k} \!+\! \vb{q}$, +so the Kronecker deltas are unnecessary. +</div> +</div> + +We substitute this result into $\chi_0$, +and reintroduce the spin index $\sigma$ associated with $\vb{k}$: + +$$\begin{aligned} + \chi_0(\vb{q}, \omega) + = \frac{1}{V} \sum_{\sigma \vb{k}} + \frac{\expval{\hat{c}_{\sigma,\vb{k}}^\dagger \hat{c}_{\sigma,\vb{k}} - \hat{c}_{\sigma,\vb{k}+\vb{q}}^\dagger \hat{c}_{\sigma,\vb{k}+\vb{q}}}_0} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} +\end{aligned}$$ + +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**: + +$$\begin{aligned} + \boxed{ + \chi_0(\vb{q}, \omega) + = \frac{1}{V} \sum_{\sigma \vb{k}} + \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} + } +\end{aligned}$$ + +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: + +$$\begin{aligned} + U_\mathrm{tot} + = U_\mathrm{ext} + U_\mathrm{ind} + = \frac{U_\mathrm{ext}}{\varepsilon_r} +\end{aligned}$$ + +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 +$\Phi_\mathrm{tot} = q_e U_\mathrm{tot}$, +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: + +$$\begin{aligned} + \boxed{ + \varepsilon_r(\vb{q}, \omega) + = 1 - \frac{U_{ee}(\vb{q})}{V} + \sum_{\sigma \vb{k}} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} + } +\end{aligned}$$ + +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 +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. |