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diff --git a/source/know/concept/kubo-formula/index.md b/source/know/concept/kubo-formula/index.md index 41fda3c..774ddf7 100644 --- a/source/know/concept/kubo-formula/index.md +++ b/source/know/concept/kubo-formula/index.md @@ -105,7 +105,7 @@ Where we have defined the **retarded correlation function** $$C^R_{A H_1}(t, t') $$\begin{aligned} \boxed{ C^R_{A H_1}(t, t') - \equiv - \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 + \equiv - \frac{i}{\hbar} \Theta(t - t') \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 } \end{aligned}$$ diff --git a/source/know/concept/lindhard-function/index.md b/source/know/concept/lindhard-function/index.md index 5f11d36..1e3c675 100644 --- a/source/know/concept/lindhard-function/index.md +++ b/source/know/concept/lindhard-function/index.md @@ -1,77 +1,77 @@ --- title: "Lindhard function" sort_title: "Lindhard function" -date: 2022-01-24 # Originally 2021-10-12, major rewrite +date: 2026-09-24 # Originally 2022-01-24, major rewrite categories: - Physics - Quantum mechanics layout: "concept" --- -The **Lindhard function** describes the response of -[jellium](/know/concept/jellium) (i.e. a free electron gas) -to an external perturbation, and is a quantum-mechanical -alternative to the [Drude model](/know/concept/drude-model/). +The **Lindhard function** describes how an external perturbation affects +a gas of free electrons, and can be regarded as +the backbone of quantum mechanical models of metals. -We start from the [Kubo formula](/know/concept/kubo-formula/) -for the electron density operator $$\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$$: +## Derivation + +Consider the following general harmonic perturbation $$\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}$$ -Where $$S$$ is the Schrödinger picture, -$$\eta$$ is a positive infinitesimal to ensure convergence later, -and $$U(\vb{r})$$ is an arbitrary potential function. -The Kubo formula becomes: +Here, the subscript $$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)} - = \iint_{-\infty}^\infty \chi(\vb{r}, \vb{r}'; t, t') \: U(\vb{r}') \: e^{i (\omega + i \eta) t'} \dd{t'} \dd{\vb{r}'} + &= -\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}$$ -Here, $$\chi$$ is the density-density correlation function, -i.e. a two-particle [Green's function](/know/concept/greens-functions/): +Where $$\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$$) is spatially uniform, +Let us assume that the unperturbed system (i.e. without $$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}$$: +$$\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 + &= -\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)^{-2 D}$$, with $$D$$ being the number of spatial dimensions. +giving a factor $$(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)^{2D}} \iiint + &= -\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}} \\ @@ -85,8 +85,8 @@ $$\begin{aligned} \end{aligned}$$ 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: +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: $$\begin{aligned} \chi(\vb{q}; t, t') @@ -95,22 +95,23 @@ $$\begin{aligned} \: 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$$ -(a limit that we will take properly later, +This gives a Dirac delta function when we recover +the full infinite system by letting $$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} \Theta(t \!-\! t') \frac{1}{V} \int_{-\infty}^\infty + &= -\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} \Theta(t \!-\! t') \frac{1}{V} \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0 + &= -\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$$ is time-independent, -$$\chi$$ only depends on the time difference $$t\!-\!t'$$. +then $$\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$$, +$$t - t' \to \omega + i \eta$$, which gives us an opportunity to rewrite $$\chi$$ in the [Lehmann representation](/know/concept/lehmann-representation/): @@ -137,10 +138,13 @@ $$\begin{aligned} {\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})$$ somehow. -If we neglect electron-electron interactions, -the single-particle states are simply plane waves, in which case: +Next, we need to rewrite $$\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}) @@ -152,10 +156,11 @@ $$\begin{aligned} {% include proof/start.html id="proof-density" -%} -Starting from the general definition of $$\hat{n}$$, +Starting from the definition of $$\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}$$: +$$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$$, +omitting spin: $$\begin{aligned} \hat{n}(\vb{r}) @@ -173,19 +178,9 @@ $$\begin{aligned} = \frac{(2 \pi)^D}{V} \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q}) \end{aligned}$$ -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: +Thanks to our periodic boundary conditions, +[this trick](/know/concept/discrete-spectrum-summation/) +lets us turn a sum into an integral: $$\begin{aligned} \hat{n}(\vb{q}) @@ -204,9 +199,6 @@ $$\begin{aligned} = \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}$$ does not carry any spin. {% include proof/end.html id="proof-density" %} @@ -214,7 +206,7 @@ When neglecting interactions, it is tradition to rename $$\chi$$ to $$\chi_0$$. We insert $$\hat{n}$$, suppressing spin: $$\begin{aligned} - \chi_0 + \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}} @@ -230,14 +222,14 @@ 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 $$E_\nu \!-\! E_{\nu'}$$ +For the same reason, the energy difference $$E_\nu - E_{\nu'}$$ can simply be replaced by the cost of the single-particle excitation -$$\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}}$$, +$$\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}$$, where $$\xi_{\vb{k}}$$ is the energy of a $$\vb{k}$$-orbital. Therefore: $$\begin{aligned} - \chi_0 + \chi_0(\vb{q}, \omega) &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu \nu'} \bigg( \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'} \matrixel{\nu'}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}} @@ -248,10 +240,10 @@ $$\begin{aligned} \end{aligned}$$ Notice that we have eliminated all dependence on $$\Ket{\nu'}$$, -so we remove it by $$\sum_{\nu} \Ket{\nu} \Bra{\nu} = 1$$: +so we remove it by $$\sum_{\nu'} \Ket{\nu'} \Bra{\nu'} = 1$$: $$\begin{aligned} - \chi_0 + \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}}} @@ -266,20 +258,21 @@ $$\begin{aligned} Where we recognized the commutator, and eliminated $$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$$: +The resulting expression has the form of a matrix trace $$\Tr$$, +and therefore a thermal expectation $$\Expval{}_0$$: $$\begin{aligned} - \chi_0 + \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{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}$$ -This commutator can be evaluated, +The commutator can be evaluated, and in this particular case it turns out to be: $$\begin{aligned} @@ -341,51 +334,305 @@ $$\begin{aligned} } \end{aligned}$$ -This is its most general form, but for practical calculations -we need to formally take the limit $$V \to \infty$$ -and then use $$\sum_{\vb{k}} = V / (2 \pi)^{D} \int_{-\infty}^{\infty} \dd{\vb{k}}$$. +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 each orbital contains two, meaning -$$\sum_{\sigma}$$ simply gives a constant factor: +so $$\sum_{\sigma}$$ simply gives a constant factor $$2$$: $$\begin{aligned} \boxed{ \chi_0(\vb{q}, \omega) - = \frac{2}{(2 \pi)^{D}} \int_{-\infty}^{\infty} + = \frac{2}{(2 \pi)^D} \int_{-\infty}^{\infty} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \dd{\vb{k}} } \end{aligned}$$ +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 -$$\Phi_\mathrm{tot} = U / q_e$$, -where $$q_e < 0$$ is the charge of an electron. +$$\Phi_\mathrm{tot} = U / e_0$$ (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 -$$\rho_\mathrm{ind} = q_e^2 \chi_0 \Phi_\mathrm{tot}$$, +$$\rho_\mathrm{ind} = e_0^2 \chi_0 \Phi_\mathrm{tot}$$, so the standard formula for $$\varepsilon_r$$ yields: $$\begin{aligned} \boxed{ \varepsilon_r(\vb{q}, \omega) - = 1 - U_{ee}(\vb{q}) \frac{2}{(2 \pi)^{D}} - \int_{-\infty}^{\infty} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \dd{\vb{k}} + = 1 - U_{ee}(\vb{q}) \: \chi_0(\vb{q}, \omega) } \end{aligned}$$ -Where $$U_{ee}(\vb{q}) = q_e^2 / (\varepsilon_0 |\vb{q}|^2)$$ is Coulomb repulsion. +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 @@ -398,3 +645,9 @@ at any temperature and for any dimensionality. 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. |