---
title: "Lindhard function"
firstLetter: "L"
publishDate: 2022-01-24 # Originally 2021-10-12, major rewrite
categories:
- Physics
- Quantum mechanics

date: 2021-09-23T16:21:57+02:00
draft: false
markup: pandoc
---

# Lindhard function

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 $\hat{H}_1$:

$$\begin{aligned}
    \hat{H}_{1,S}(t)
    = e^{i (\omega + i \eta) t} \int_{-\infty}^\infty U(\vb{r}) \: \hat{n}(\vb{r}) \dd{\vb{r}}
\end{aligned}$$

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">
<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.