--- title: "Lindhard function" sort_title: "Lindhard function" date: 2026-09-24 # Originally 2022-01-24, major rewrite categories: - Physics - Quantum mechanics layout: "concept" --- The **Lindhard function** describes how an external perturbation affects a gas of free electrons, and can be regarded as the backbone of quantum mechanical models of metals. ## Derivation Consider the following general harmonic perturbation $$\hat{H}_1$$, where $$U(\vb{r})$$ is an arbitrary potential, $$\hat{n}(\vb{r})$$ is the electron density operator, and $$\eta \to 0^{+}$$ is needed to ensure convergence: $$\begin{aligned} \hat{H}_{1,S}(t) = e^{i (\omega + i \eta) t} \int_{-\infty}^\infty U(\vb{r}) \: \hat{n}_S(\vb{r}) \dd{\vb{r}} \end{aligned}$$ Here, the subscript $$S$$ refers to the Schrödinger picture. The first-order effect of this perturbation on the expected density $$\Expval{\hat{n}}$$ is then given by the [Kubo formula](/know/concept/kubo-formula/) below, where the subscript $$I$$ refers to the [interaction picture](/know/concept/interaction-picture/), and the expectation $$\Expval{}_0$$ is taken with respect to the thermal equilibrium that existed before the perturbation was applied: $$\begin{aligned} \delta\!\Expval{\hat{n}(\vb{r}, t)} &= -\frac{i}{\hbar} \int_{-\infty}^\infty \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{r}, t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'} \\ &= \iint_{-\infty}^\infty \chi(\vb{r}, \vb{r}'; t, t') \: U(\vb{r}') \: e^{i (\omega + i \eta) t'} \dd{t'} \dd{\vb{r}'} \end{aligned}$$ Where $$\Theta$$ is a [Heaviside step function](/know/concept/heaviside-step-function/), and the response $$\chi$$ 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 $$U$$) was spatially uniform, so that $$\chi$$ only depends on the difference $$\vb{r} - \vb{r}'$$. We then take its [Fourier transform](/know/concept/fourier-transform/) $$\vb{r} - \vb{r}' \to \vb{q}$$: $$\begin{aligned} \chi(\vb{q}; t, t') &= \int_{-\infty}^\infty \chi(\vb{r} - \vb{r}'; t, t') \: e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{r}} \\ &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2 D}} \iiint_{-\infty}^\infty \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: e^{i \vb{q}_1 \cdot \vb{r}} e^{i \vb{q}_2 \cdot \vb{r}'} e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}} \end{aligned}$$ Where both $$\hat{n}_I$$ have been written as inverse Fourier transforms, giving a factor $$(2 \pi)^{-D}$$ each, where $$D$$ is the number of spatial dimenions. We rearrange to get a [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta$$: $$\begin{aligned} \chi(\vb{q}; t, t') &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2 D}} \iiint \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: e^{i (\vb{q}_1 - \vb{q}) \cdot \vb{r}} e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}} \\ &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \iint \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: \delta(\vb{q}_1 \!-\! \vb{q}) \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2} \\ &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \int \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \end{aligned}$$ On the left, $$\vb{r}'$$ does not appear, so it must also disappear on the right. To handle this properly, we consider an arbitrary finite piece of volume $$V$$ of this infinite system. Clearly $$\int_V \dd{\vb{r}'} = V$$, so: $$\begin{aligned} \chi(\vb{q}; t, t') &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \frac{1}{V} \int_V \int_{-\infty}^\infty \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \dd{\vb{r}'} \end{aligned}$$ This gives a Dirac delta function when we recover the full infinite system by letting $$V \to \infty$$ (we will take this limit properly later, but beware that some authors set $$V = 1$$ until then): $$\begin{aligned} \chi(\vb{q}; t, t') &= -\frac{i}{\hbar V} \Theta(t - t') \int_{-\infty}^\infty \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: \delta(\vb{q}_2 \!+\! \vb{q}) \dd{\vb{q}_2} \\ &= -\frac{i}{\hbar V} \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0 \end{aligned}$$ Similarly, if the unperturbed Hamiltonian $$\hat{H}_0$$ is time-independent, then $$\chi$$ 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: $$\begin{aligned} \chi(\vb{q}, \omega) = \frac{1}{Z V} \sum_{\nu \nu'} \bigg( \frac{\matrixel{\nu}{\hat{n}(\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(-\vb{q})}{\nu}} {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}} - \frac{\matrixel{\nu}{\hat{n}(-\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(\vb{q})}{\nu}} {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu} \end{aligned}$$ Next, we need to rewrite $$\hat{n}(\vb{q})$$ somehow. To do this, we must neglect electron-electron interactions, leaving us with a simple [Fermi gas](/know/concept/fermi-gas/). In that model, $$V$$ is a cube with periodic boundary conditions, meaning that $$\vb{q}$$ has a discrete spectrum of allowed values, and the single-electron states are just plane waves. This lets us write, in the [second quantization](/know/concept/second-quantization/): $$\begin{aligned} \hat{n}(\vb{q}) = \sum_{\sigma \vb{k}} \hat{c}_{\sigma,\vb{k}}^\dagger \hat{c}_{\sigma,\vb{k} + \vb{q}} \qquad \qquad \hat{n}(-\vb{q}) = \hat{n}^\dagger(\vb{q}) \end{aligned}$$ {% include proof/start.html id="proof-density" -%} Starting from the definition of $$\hat{n}(\vb{r})$$, we write out the field operators $$\hat{\Psi}(\vb{r})$$, and insert the known non-interacting single-electron orbitals $$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$$, omitting spin: $$\begin{aligned} \hat{n}(\vb{r}) \equiv \hat{\Psi}{}^\dagger(\vb{r}) \hat{\Psi}(\vb{r}) = \sum_{\vb{k} \vb{k}'} \psi_{\vb{k}}^*(\vb{r}) \: \psi_{\vb{k}'}(\vb{r})\: \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} = \frac{1}{V} \sum_{\vb{k} \vb{k}'} e^{i (\vb{k}' - \vb{k}) \cdot \vb{r}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \end{aligned}$$ Taking the Fourier transfom yields a Dirac delta function $$\delta$$: $$\begin{aligned} \hat{n}(\vb{q}) = \frac{1}{V} \int_{-\infty}^\infty \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: e^{i (\vb{k}' - \vb{k} - \vb{q})\cdot \vb{r}} \dd{\vb{r}} = \frac{(2 \pi)^D}{V} \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q}) \end{aligned}$$ Thanks to our periodic boundary conditions, [this trick](/know/concept/discrete-spectrum-summation/) lets us turn a sum into an integral: $$\begin{aligned} \hat{n}(\vb{q}) &= \frac{(2 \pi)^D}{V} \frac{V}{(2 \pi)^D} \sum_{\vb{k}} \int_{-\infty}^\infty \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q}) \dd{\vb{k}'} = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \end{aligned}$$ 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: $$\begin{aligned} \hat{n}(-\vb{q}) = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} - \vb{q}} = \sum_{\vb{k}'} \hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'} = \hat{n}^\dagger(\vb{q}) \end{aligned}$$ {% 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: $$\begin{aligned} \chi_0(\vb{q}, \omega) &= \frac{1}{Z V} \sum_{\vb{k} \vb{k}'} \sum_{\nu \nu'} \bigg( \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'} \matrixel{\nu'}{\hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'}}{\nu}} {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}} - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'} \matrixel{\nu'}{\hat{c}_{\vb{k}'}^\dagger \hat{c}_{\vb{k}' + \vb{q}}}{\nu}} {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu} \end{aligned}$$ 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: $$\begin{aligned} \chi_0(\vb{q}, \omega) &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu \nu'} \bigg( \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'} \matrixel{\nu'}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'} \matrixel{\nu'}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu} \end{aligned}$$ Notice that we have eliminated all dependence on $$\Ket{\nu'}$$, so we remove it by $$\sum_{\nu'} \Ket{\nu'} \Bra{\nu'} = 1$$: $$\begin{aligned} \chi_0(\vb{q}, \omega) &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu} \bigg( \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu} \\ &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu} \frac{\matrixel{\nu}{\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}} {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0}}{\nu}} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \end{aligned}$$ 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 therefore a thermal expectation $$\Expval{}_0$$: $$\begin{aligned} \chi_0(\vb{q}, \omega) &= \frac{1}{Z V} \sum_{\vb{k}} \frac{\Tr\!\big(\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}} {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0} \big)} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \\ &= \frac{1}{V} \sum_{\vb{k}} \frac{\expval{\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}}_0} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \end{aligned}$$ The 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 we can go further by exploiting the Fermi gas' periodic boundary conditions, which [enable us](/know/concept/discrete-spectrum-summation/) to turn the sum into an integral. Furthermore, electrons are spin-1/2 particles, so $$\sum_{\sigma}$$ simply gives a constant factor $$2$$: $$\begin{aligned} \boxed{ \chi_0(\vb{q}, \omega) = \frac{2}{(2 \pi)^D} \int_{-\infty}^{\infty} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \dd{\vb{k}} } \end{aligned}$$ Beware that some authors define $$\chi_0$$ as the response of the *charge* density to an *electric* potential, leading to an extra factor $$e_0^2$$, where $$e_0$$ is the electron charge. In contrast, our result describes the *particle* density response due to an *energy* potential. ## Evaluation Let us consider $$D = 3$$ (calculations in 1D or 2D are also possible), and rewrite the integrand by substituting $$\vb{k}' \equiv - \vb{k} - \vb{q}$$ in the second term and using $$\xi_{-\vb{k}} = \xi_{\vb{k}}$$: $$\begin{aligned} \chi_0(\vb{q}, \omega) &= \frac{2}{(2 \pi)^3} \int_{-\infty}^{\infty} \bigg( \frac{n_F(\xi_{\vb{k}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} - \frac{n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) \dd{\vb{k}} \\ &= \frac{1}{4 \pi^3} \bigg( \int_{-\infty}^{\infty} \frac{n_F(\xi_{\vb{k}})}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} + \hbar (\omega + i \eta)} \dd{\vb{k}} - \int_{-\infty}^{\infty} \frac{n_F(\xi_{- \vb{k}'})}{\xi_{- \vb{k}' - \vb{q}} - \xi_{- \vb{k}'} + \hbar (\omega + i \eta)} \dd{\vb{k}'} \bigg) \\ &= \frac{1}{4 \pi^3} \int_{-\infty}^{\infty} n_F(\xi_{\vb{k}}) \bigg( \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} + \hbar (\omega + i \eta)} + \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} - \hbar (\omega + i \eta)} \bigg) \dd{\vb{k}} \end{aligned}$$ At absolute zero $$T = 0$$, the Fermi gas has a well-known ground state, in which the occupied single-particle plane wave states form a sphere in $$\vb{k}$$-space with radius $$k_F$$. In that case, $$n_F(\xi_\vb{k})$$ reduces to a Heaviside step function $$\Theta(k_F - |\vb{k}|)$$. We switch to [spherical coordinates](/know/concept/spherical-coordinates/) $$(|\vb{k}|, \theta, \varphi)$$ and adjust $$|\vb{k}|$$'s integration limit based on the step function's demand that $$|\vb{k}| < k_F$$: $$\begin{aligned} \chi_0(\vb{q}, \omega) &= \frac{2 \pi}{4 \pi^3} \int_0^{k_F} \! \int_0^\pi \bigg( \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} + \hbar (\omega \!+\! i \eta)} + \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} - \hbar (\omega \!+\! i \eta)} \bigg) |\vb{k}|^2 \sin{\theta} \dd{\theta} \dd{|\vb{k}|} \end{aligned}$$ Where we have already used that $$\varphi$$ does not appear in the integrand to get a factor $$2 \pi$$. Next, let us take a closer look at the excitation energy in the denominator. Thanks to symmetry, we can always choose our coordinate system such that $$\vb{q}$$ points in the $$\theta = 0$$ direction, yielding: $$\begin{aligned} \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} &= \frac{\hbar^2}{2 m} \Big( (\vb{k} \cdot \vb{k}) - (\vb{k} \!+\! \vb{q}) \cdot (\vb{k} \!+\! \vb{q}) \Big) \\ &= - \frac{\hbar^2}{2 m} \Big( |\vb{q}|^2 + 2 |\vb{k}| |\vb{q}| \cos{\theta} \Big) \end{aligned}$$ By defining the dimensionless variables $$x \equiv |\vb{q}| / (2 k_F)$$ and $$\kappa \equiv |\vb{k}| / k_F$$, and recognizing the Fermi energy $$\xi_F = \hbar^2 k_F^2 / (2 m)$$, this turns into: $$\begin{aligned} \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} &= - 4 \xi_F (x^2 + x \kappa \cos{\theta}) \end{aligned}$$ We insert this back into $$\chi_0$$, and abbreviate $$y \equiv \hbar (\omega + i \eta) / (4 \xi_F)$$ and $$c \equiv \cos{\theta}$$, giving: $$\begin{aligned} \chi_0(\vb{q}, \omega) &= \frac{k_F^3}{8 \pi^2 \xi_F} \int_0^1 \! \int_{-1}^1 \bigg( \frac{1}{-x^2 - x \kappa c + y} + \frac{1}{-x^2 - x \kappa c - y} \bigg) \kappa^2 \dd{c} \dd{\kappa} \end{aligned}$$ Evaluating the integral over $$c$$ gives a result that involves the complex natural logarithm $$\ln$$: $$\begin{aligned} \chi_0 &= \frac{-k_F^3}{8 \pi^2 \xi_F} \int_0^1 \! \int_{-1}^1 \bigg( \frac{1}{x \kappa c + x^2 - y} + \frac{1}{x \kappa c + x^2 + y} \bigg) \kappa^2 \dd{c} \dd{\kappa} \\ &= \frac{-m k_F}{4 \pi^2 \hbar^2} \int_0^1 \bigg[ \frac{1}{x \kappa} \ln(x \kappa c + x^2 - y) + \frac{1}{x \kappa} \ln(x \kappa c + x^2 + y) \bigg]_{-1}^1 \kappa^2 \dd{\kappa} \\ &= \frac{-m k_F}{4 \pi^2 \hbar^2} \frac{1}{x} \int_0^1 \bigg( \ln\!\bigg( \frac{\kappa + x - y / x}{-\kappa + x - y / x} \bigg) + \ln\!\bigg( \frac{\kappa + x + y / x}{-\kappa + x + y / x} \bigg) \bigg) \kappa \dd{\kappa} \end{aligned}$$ We abbreviate this further by defining $$z_{+} \equiv x + y / x$$ and $$z_{-} \equiv x - y / x$$, yielding: $$\begin{aligned} \chi_0 &= \frac{-m k_F}{4 \pi^2 \hbar^2} \frac{1}{x} \int_0^1 \ln\!\bigg( \frac{(\kappa + z_{-})(\kappa + z_{+})}{(\kappa - z_{-})(\kappa - z_{+})} \bigg) \kappa \dd{\kappa} \end{aligned}$$ Looking up this integral leads to the expression below. Note that all terms disappear for $$\kappa = 0$$: $$\begin{aligned} \chi_0 &= \frac{-m k_F}{4 \pi^2 \hbar^2} \frac{1}{x} \bigg[ \kappa (z_{+} \!+\! z_{-}) - z_{+}^2 \tanh^{-1}\!\bigg(\!\frac{\kappa}{z_{+}}\!\bigg) - z_{-}^2 \tanh^{-1}\!\bigg(\!\frac{\kappa}{z_{-}}\!\bigg) + \frac{1}{2} \kappa^2 \ln\!\bigg( \frac{(\kappa + z_{+})(\kappa + z_{-})}{(\kappa - z_{-})(\kappa - z_{+})} \bigg) \bigg]_0^1 \\ &= \frac{-m k_F}{\pi^2 \hbar^2} \frac{1}{4 x} \bigg( 2 x - z_{+}^2 \tanh^{-1}\!\bigg(\!\frac{1}{z_{+}}\!\bigg) - z_{-}^2 \tanh^{-1}\!\bigg(\!\frac{1}{z_{-}}\!\bigg) + \frac{1}{2} \ln\!\bigg( \frac{(1 + z_{+})(1 + z_{-})}{(1 - z_{-})(1 - z_{+})} \bigg) \bigg) \\ &= \frac{-m k_F}{\pi^2 \hbar^2} \frac{1}{8 x} \bigg( 4 x - z_{+}^2 \ln\!\bigg( \frac{z_{+} + 1}{z_{+} - 1} \bigg) - z_{-}^2 \ln\!\bigg( \frac{z_{-} + 1}{z_{-} - 1} \bigg) + \ln\!\bigg( \frac{(z_{+} + 1)(z_{-} + 1)}{(z_{+} - 1)(z_{-} - 1)} \bigg) \bigg) \\ &= \frac{-m k_F}{\pi^2 \hbar^2} \bigg( \frac{1}{2} + \frac{1}{8 x} (1 - z_{+}^2) \ln\!\bigg( \frac{z_{+} + 1}{z_{+} - 1} \bigg) + \frac{1}{8 x} (1 - z_{-}^2) \ln\!\bigg( \frac{z_{-} + 1}{z_{-} - 1} \bigg) \bigg) \end{aligned}$$ We thus arrive at this expression for the response function of non-interacting electrons: $$\begin{aligned} \boxed{ \chi_0(\vb{q}, \omega) = - \frac{m k_F}{\pi^2 \hbar^2} \: \mathcal{F}\bigg(\frac{|\vb{q}|}{2 k_F}, \: \frac{\hbar (\omega \!+\! i \eta)}{4 \xi_F}\bigg) } \end{aligned}$$ Where $$\mathcal{F}$$ (which is sometimes also called the Lindhard function), is defined like so: $$\begin{aligned} \boxed{ \mathcal{F}(x, y) \equiv \frac{1}{2} + \frac{1}{8 x} \bigg( 1 - \Big(x \!+\! \frac{y}{x}\Big)^{\!2} \bigg) \ln\!\bigg( \frac{x + y/x + 1}{x + y/x - 1} \bigg) + \frac{1}{8 x} \bigg( 1 - \Big(x \!-\! \frac{y}{x}\Big)^{\!2} \bigg) \ln\!\bigg( \frac{x - y/x + 1}{x - y/x - 1} \bigg) } \end{aligned}$$ Keep in mind that these are complex logarithms, whose branch cut should lie along the negative real axis, such that the results' imaginary parts are always between $$-\pi$$ and $$\pi$$. That way, it is consistent with alternative derivations that use the [Sokhotski-Plemelj theorem](/know/concept/sokhotski-plemelj-theorem/) (e.g. Mihaila's paper). ## Interpretation 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 $$\Phi_\mathrm{tot} = U / e_0$$ (where $$e_0 < 0$$). 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 $$\rho_\mathrm{ind} = e_0^2 \chi_0 \Phi_\mathrm{tot}$$, so the standard formula for $$\varepsilon_r$$ yields: $$\begin{aligned} \boxed{ \varepsilon_r(\vb{q}, \omega) = 1 - U_{ee}(\vb{q}) \: \chi_0(\vb{q}, \omega) } \end{aligned}$$ Where $$U_{ee}(\vb{q}) = e_0^2 / (\varepsilon_0 |\vb{q}|^2)$$ is Coulomb repulsion. This is the **Lindhard dielectric function** of a free non-interacting electron gas, at any temperature and for any dimensionality. Lindhard theory can reproduce several notable discoveries in various limits. For long wavelengths $$\vb{q} \to 0$$, i.e. $$x \to 0$$, it can be shown that $$\mathcal{F}(x, y) \approx - x^2 / (3 y^2)$$, and therefore: $$\begin{aligned} \chi_0(\vb{q}, \omega) &\approx \frac{m k_F}{\pi^2 \hbar^2} \frac{|\vb{q}|^2}{4 k_F^2} \frac{16 \xi_F^2}{3 \hbar^2 \omega^2} = \frac{k_F^3}{3 \pi^2 m} \frac{|\vb{q}|^2}{\omega^2} \end{aligned}$$ For a Fermi gas, we know that $$k_F^3 = 3 \pi^2 n$$, where $$n$$ is the unperturbed density. Upon insertion into $$\varepsilon_r$$, this gives the same result as the [Drude model](/know/concept/drude-model/), predicting *plasma oscillations*, also known as *plasmons* (when treated as quantized particles) or [*cold Langmuir waves*](/know/concept/langmuir-waves/): $$\begin{aligned} \boxed{ \varepsilon_r(0, \omega) = 1 - \frac{\omega_p^2}{\omega^2} } \qquad \qquad \boxed{ \omega_p \equiv \sqrt{\frac{e_0^2 n}{\varepsilon_0 m}} } \end{aligned}$$ Meanwhile, in the static limit $$\omega \to 0$$, i.e. $$y \to 0$$, we arrive at a function that is also encountered when using the *Hartree-Fock method* to approximate the ground state of [jellium](/know/concept/jellium/): $$\begin{aligned} \boxed{ \mathcal{F}(x, 0) = \frac{1}{2} + \frac{1}{4 x} (1 - x^2) \ln\!\bigg( \frac{x + 1}{x - 1} \bigg) } \end{aligned}$$ If we assume small $$|\vb{q}|$$ and $$x$$ again, it can be shown that $$\mathcal{F}(x, 0) \approx 1$$, in which case: $$\begin{aligned} \chi_0(\vb{q}, 0) = - \frac{m k_F^3}{\pi^2 \hbar^2 k_F^2} = - \frac{3 \pi^2 m n}{\pi^2 \hbar^2 k_F^2} = - \frac{3 m n}{\hbar^2 k_F^2} = - \frac{3 n}{2 \xi_F} \end{aligned}$$ The resulting dielectric function also appears in the derivation of *Thomas-Fermi screening*, which can be regarded as the zero-temperature limit of [Debye screening](/know/concept/debye-length/): $$\begin{aligned} \boxed{ \varepsilon_r(\vb{q}, 0) = 1 + \frac{k_\mathrm{TF}^2}{|\vb{q}|^2} } \qquad \qquad \boxed{ k_\mathrm{TF} \equiv \sqrt{\frac{3 e_0^2 n}{2 \varepsilon_0 \xi_F}} } \end{aligned}$$ But probably the greatest achievement of Lindhard theory is its appearance in the [random phase approximation](/know/concept/random-phase-approximation/) (where $$\chi_0$$ is sometimes known as the **retarded pair-bubble**) for treating electron-electron Coulomb interactions in the Fermi gas model. So although we neglected interactions in our derivation, the result is an essential building block of interacting models. ## 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. 4. G.F. Giuliani, G. Vignale, *Quantum theory of the electron liquid*, 2005, Cambridge. 5. B. Mihaila, [*Lindhard function of a $$d$$-dimensional Fermi gas*](https://doi.org/10.48550/arXiv.1111.5337), 2011, arXiv.