path: root/source/know/concept/lindhard-function/index.md

---
title: "Lindhard function"
sort_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:

$$\begin{aligned}
    \chi(\vb{q}; t, t')
    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \frac{1}{V} \int_V \int_{-\infty}^\infty
    \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0
    \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \dd{\vb{r}'}
\end{aligned}$$

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):

$$\begin{aligned}
    \chi(\vb{q}; t, t')
    &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \int_{-\infty}^\infty
    \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: \delta(\vb{q}_2 \!+\! \vb{q}) \dd{\vb{q}_2}
    \\
    &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0
\end{aligned}$$

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
$$t\!-\!t' \to \omega\!+\!i \eta$$,
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-particle eigenstates of $$\hat{H}_0$$,
and $$Z$$ is the [grand partition function](/know/concept/grand-canonical-ensemble/).
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:

$$\begin{aligned}
    \chi(\vb{q}, \omega)
    = \frac{1}{Z V} \sum_{\nu \nu'} \bigg(
    \frac{\matrixel{\nu}{\hat{n}(\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(-\vb{q})}{\nu}}
    {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}}
    - \frac{\matrixel{\nu}{\hat{n}(-\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(\vb{q})}{\nu}}
    {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu}
\end{aligned}$$


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:

$$\begin{aligned}
    \hat{n}(\vb{q})
    = \sum_{\sigma \vb{k}} \hat{c}_{\sigma,\vb{k}}^\dagger \hat{c}_{\sigma,\vb{k} + \vb{q}}
    \qquad \qquad
    \hat{n}(-\vb{q})
    = \hat{n}^\dagger(\vb{q})
\end{aligned}$$


{% include proof/start.html id="proof-density" -%}
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 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:

$$\begin{aligned}
    \hat{n}(\vb{q})
    &= \frac{(2 \pi)^D}{V} \frac{V}{(2 \pi)^D} \sum_{\vb{k}} \int_{-\infty}^\infty
    \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q}) \dd{\vb{k}'}
    = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}
\end{aligned}$$

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.
{% include proof/end.html id="proof-density" %}


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:

$$\begin{aligned}
    \comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}
    = \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}} - \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k} + \vb{q}}
\end{aligned}$$


{% include proof/start.html id="proof-commutator" -%}
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.
{% include proof/end.html id="proof-commutator" %}


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**:

$$\begin{aligned}
    \boxed{
        \chi_0(\vb{q}, \omega)
        = \frac{1}{V} \sum_{\sigma \vb{k}}
        \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}
        {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
    }
\end{aligned}$$

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:

$$\begin{aligned}
    \boxed{
        \chi_0(\vb{q}, \omega)
        = \frac{2}{(2 \pi)^{D}} \int_{-\infty}^{\infty}
        \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}
        {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \dd{\vb{k}}
    }
\end{aligned}$$

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!

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:

$$\begin{aligned}
    \boxed{
        \varepsilon_r(\vb{q}, \omega)
        = 1 - U_{ee}(\vb{q}) \frac{2}{(2 \pi)^{D}}
        \int_{-\infty}^{\infty} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \dd{\vb{k}}
    }
\end{aligned}$$

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,
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.