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diff --git a/content/know/concept/lindhard-function/index.pdc b/content/know/concept/lindhard-function/index.pdc index 96244c9..aedf0f5 100644 --- a/content/know/concept/lindhard-function/index.pdc +++ b/content/know/concept/lindhard-function/index.pdc @@ -1,7 +1,7 @@ --- title: "Lindhard function" firstLetter: "L" -publishDate: 2021-10-12 +publishDate: 2022-01-24 # Originally 2021-10-12, major rewrite categories: - Physics - Quantum mechanics @@ -13,9 +13,9 @@ markup: pandoc # Lindhard function -The **Lindhard function** describes the response of an electron gas -to an external perturbation, -and can be regarded as a quantum-mechanical +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/) @@ -28,389 +28,370 @@ $$\begin{aligned} = -\frac{i}{\hbar} \int_{-\infty}^\infty \Theta(t - t') \expval{\comm{\hat{n}_I(\vb{r}, t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'} \end{aligned}$$ -Where $\Theta$ is the [Heaviside step function](/know/concept/heaviside-step-function/), -and the subscript $I$ refers to the [interaction picture](/know/concept/interaction-picture/). -Notice from the limits that the perturbation is switched on at $t = -\infty$. -Now, let us consider the following harmonic $\hat{H}_1$ in the Schrödinger picture: +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 $\hat{H}_1$: $$\begin{aligned} \hat{H}_{1,S}(t) - = g(t) \: \hat{V}_S - \qquad - g(t) - \equiv \exp\!(- i \omega t) \exp\!(\eta t) - \qquad - \hat{V}_S - \equiv \int_{-\infty}^\infty V(\vb{r}) \: \hat{n}(\vb{r}) \dd{\vb{r}} + = e^{i (\omega + i \eta) t} \int_{-\infty}^\infty U(\vb{r}) \: \hat{n}(\vb{r}) \dd{\vb{r}} \end{aligned}$$ -Where $\eta$ is a tiny positive number, -which represents a gradual switching-on of $\hat{H}_1$, -eliminating transient effects -and helping the convergence of an integral later. - -We assume that $V(\vb{r})$ varies slowly compared to the electrons' wavefunctions, -so we argue that $\hat{V}_S$ is practically time-independent, -because the total number of electrons is conserved, -and $\hat{n}$ is only weakly perturbed by $\hat{H}_1$. - -Because $\hat{H}_1$ starts at $t = -\infty$, -we can always shift the time axis such that the point of interest is at $t = 0$. -We thus have, without loss of generality: +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: $$\begin{aligned} - \delta\expval{{\hat{n}}}(\vb{r}) - = \delta\expval{{\hat{n}}}(\vb{r}, 0) - &= -\frac{i}{\hbar} \int_{-\infty}^\infty \Theta(- t') - \Big( \expval{\hat{n}_I \hat{V}_I}_0 - \expval{\hat{V}_I \hat{n}_I}_0 \Big) g(t') \dd{t'} + \delta\expval{{\hat{n}}}(\vb{r}, t) + = \iint_{-\infty}^\infty \chi(\vb{r}, \vb{r}'; t, t') \: U(\vb{r}') \: e^{i (\omega + i \eta) t'} \dd{t'} \dd{\vb{r}'} \end{aligned}$$ -The expectation values $\expval{}_0$ are calculated for $\ket{0}$, -which was the state at $t = -\infty$. -Note that if $\ket{0}$ is an eigenstate of $\hat{H}_{0,S}$, -there is no difference which picture (Schrödinger or interaction) $\ket{0}$ is in, -because any operator $\hat{A}$ then satisfies: +Here, $\chi$ is the density-density correlation function, +i.e. a two-particle [Green's function](/know/concept/greens-functions/): $$\begin{aligned} - \matrixel{0_I}{\hat{A}_I}{0_I} - &= \matrixel**{0_S}{\exp\!(-i \hat{H}_{0,S} t / \hbar) \:\: \hat{A}_I \: \exp\!(i \hat{H}_{0,S} t / \hbar)}{0_S} - \\ - &= \matrixel{0_S}{\hat{A}_I}{0_S} \:\exp\!\big(i (E_0\!-\!E_0) t / \hbar\big) - = \matrixel{0_S}{\hat{A}_I}{0_S} + \chi(\vb{r}, \vb{r}'; t, t') + \equiv - \frac{i}{\hbar} \Theta(t - t') \expval{\comm{\hat{n}_I(\vb{r}, t)}{\hat{n}_I(\vb{r}', t')}}_0 \end{aligned}$$ -Therefore, we will assume that $\ket{0}$ is an eigenstate of $\hat{H}_{0,S}$. -Next, we insert the identity operator $\hat{I} = \sum_{j} \ket{j} \bra{j}$, -where $\ket{j}$ are all the eigenstates of $\hat{H}_{0,S}$: +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/) +$\vb{r}\!-\!\vb{r}' \to \vb{q}$: $$\begin{aligned} - \delta\expval{\hat{n}}(\vb{r}) - &= -\frac{i}{\hbar} \sum_{j} \int \Theta(-t') - \Big( \matrixel{0}{\hat{n}_I}{j} \matrixel{j}{\hat{V}_I}{0} - \matrixel{0}{\hat{V}_I}{j} \matrixel{j}{\hat{n}_I}{0} \Big) g(t') \dd{t'} + \chi(\vb{q}; t, t') + &= \int_{-\infty}^\infty \chi(\vb{r} - \vb{r}'; t, t') \: e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{r}} + \\ + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint + \expval{\comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: e^{i \vb{q}_1 \cdot \vb{r}} e^{i \vb{q}_2 \cdot \vb{r}'} e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}} \end{aligned}$$ -Using the fact that $\ket{0}$ and $\ket{j}$ -are eigenstates of $\hat{H}_{0,S}$, -and that we chose $t = 0$, we find: +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/) $\delta$: $$\begin{aligned} - \matrixel{j}{\hat{V}_I(t')}{0} - &= \matrixel{j}{\hat{V}_S}{0} \:\exp\!\big(i (E_j \!-\! E_0) t' / \hbar\big) + \chi(\vb{q}; t, t') + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint + \expval{\comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: e^{i (\vb{q}_1 - \vb{q}) \cdot \vb{r}} e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}} + \\ + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \iint + \expval{\comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: \delta(\vb{q}_1 \!-\! \vb{q}) \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2} \\ - \matrixel{j}{\hat{n}_I(0)}{0} - &= \matrixel{j}{\hat{n}_S(0)}{0} \:\exp\!(0 - 0) - = \matrixel{j}{\hat{n}_S(0)}{0} + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \int + \expval{\comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \end{aligned}$$ -We define $\omega_{j0} \equiv (E_j \!-\! E_0)/\hbar$ -and insert the above expressions into $\delta\expval{\hat{n}}$, -yielding: +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: $$\begin{aligned} - \delta\expval{\hat{n}}(\vb{r}) - &= -\frac{i}{\hbar} \sum_{j} \bigg( - \matrixel{0}{\hat{n}_S}{j} \matrixel{j}{\hat{V}_S}{0} \int \Theta(-t') \exp\!(i \omega_{j0} t') \: g(t') \dd{t'} - \\ - &\qquad\qquad\:\, - - \matrixel{0}{\hat{V}_S}{j} \matrixel{j}{\hat{n}_S}{0} \int \Theta(-t') \exp\!(-i \omega_{j0} t') \: g(t') \dd{t'} \bigg) + \chi(\vb{q}; t, t') + &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \frac{1}{V} \int_V \int_{-\infty}^\infty + \expval{\comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 + \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \dd{\vb{r}'} \end{aligned}$$ -These integrals are [Fourier transforms](/know/concept/fourier-transform/), -and are straightforward to evaluate. The first is: +For $V \to \infty$ we get a Dirac delta function, +but in fact the conclusion holds for finite $V$ too: $$\begin{aligned} - \int_{-\infty}^\infty \Theta(-t') \exp\!(i \omega_{j0} t') \: g(t') \dd{t'} - &= \int_{-\infty}^0 \exp\!\big(\!-\! i (\omega - \omega_{j0}) t' + \eta t' \big) \dd{t'} - \\ - &= \bigg[ \frac{\exp\!\big(\!-\! i (\omega - \omega_{j0}) t' + \eta t' \big)}{- i (\omega - \omega_{j0}) + \eta} \bigg]_{-\infty}^0 + \chi(\vb{q}; t, t') + &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \int_{-\infty}^\infty + \expval{\comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: \delta(\vb{q}_2 \!+\! \vb{q}) \dd{\vb{q}_2} \\ - &= \frac{1}{- i (\omega - \omega_{j0}) + \eta} - = \frac{i}{\omega - \omega_{j0} + i \eta} + &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \expval{\comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0 \end{aligned}$$ -The other integral simply has the opposite sign in front of $\omega_{j0}$. -We thus arrive at: +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/): $$\begin{aligned} - \delta\expval{\hat{n}}(\vb{r}, \omega) - &= \frac{1}{\hbar} \sum_{j} \bigg( - \frac{\matrixel{0}{\hat{n}_S}{j} \matrixel{j}{\hat{V}_S}{0}}{\omega - \omega_{j0} + i \eta} - - \frac{\matrixel{0}{\hat{V}_S}{j} \matrixel{j}{\hat{n}_S}{0}}{\omega + \omega_{j0} + i \eta} \bigg) + \chi(\vb{q}, \omega) + = \frac{1}{Z V} \sum_{\nu \nu'} + \frac{\matrixel{\nu}{\hat{n}_S(\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}_S(-\vb{q})}{\nu}}{\hbar (\omega + i \eta) + E_\nu - E_{\nu'}} + \Big( e^{-\beta E_\nu} - e^{- \beta E_{\nu'}} \Big) \end{aligned}$$ -Inserting the definition $\hat{V}_S = \int V(\vb{r}') \:\hat{n}(\vb{r}') \dd{\vb{r}'}$ -leads us to the following formula for $\delta\expval{\hat{n}}$, -which has the typical form of a linear response, -with response function $\chi$: +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: $$\begin{aligned} - \boxed{ - \begin{gathered} - \delta{n}(\vb{r}, \omega) - = \int_{-\infty}^\infty \chi(\vb{r}, \vb{r'}, \omega) \: V(\vb{r}') \dd{\vb{r}'} - \qquad\quad \mathrm{where} - \\ - \chi(\vb{r}, \vb{r'}, \omega) - = \sum_{j} \bigg( \frac{\matrixel{0}{\hat{n}_S(\vb{r})}{j} \matrixel{j}{\hat{n}_S(\vb{r}')}{0}}{(\omega + i \eta) \hbar - E_j + E_0} - - \frac{\matrixel{0}{\hat{n}_S(\vb{r}')}{j} \matrixel{j}{\hat{n}_S(\vb{r})}{0}}{(\omega + i \eta) \hbar + E_j - E_0} \bigg) - \end{gathered} - } + \chi(\vb{q}, \omega) + = \frac{1}{Z V} \sum_{\nu \nu'} \bigg( + \frac{\matrixel{\nu}{\hat{n}(\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(-\vb{q})}{\nu}} + {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}} + - \frac{\matrixel{\nu}{\hat{n}(-\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(\vb{q})}{\nu}} + {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu} \end{aligned}$$ -By definition, $\ket{j}$ are eigenstates -of the many-electron Hamiltonian $\hat{H}_{0,S}$, -which is only solvable if we crudely neglect -any and all electron-electron interactions. -Therefore, to continue, we neglect those interactions. -According to tradition, we then rename $\chi$ to $\chi_0$. +All operators are in the Schrödinger picture from now on, hence we dropped the subscript $S$. -The well-known ground state of a non-interacting electron gas -is the Fermi sea $\ket{\mathrm{FS}}$, given below, -together with $\hat{n}_S$ in the language of the -[second quantization](/know/concept/second-quantization/): +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: $$\begin{aligned} - \ket{\mathrm{FS}} - = \prod_\alpha \hat{c}_\alpha^\dagger \ket{0} - \qquad \quad - \hat{n}_S(\vb{r}) - = \hat{\Psi}{}^\dagger(\vb{r}) \hat{\Psi}(\vb{r}) - = \sum_{\alpha \beta} \psi_\alpha^*(\vb{r}) \psi_\beta(\vb{r})\: \hat{c}_\alpha^\dagger \hat{c}_\beta + \hat{n}(\vb{q}) + = \sum_{\sigma \vb{k}} \hat{c}_{\sigma,\vb{k}}^\dagger \hat{c}_{\sigma,\vb{k} + \vb{q}} + \qquad \qquad + \hat{n}(-\vb{q}) + = \hat{n}^\dagger(\vb{q}) \end{aligned}$$ -For now, we ignore thermal excitations, -i.e. we set the temperature $T = 0$. -In $\chi_0 = \chi$, we thus insert the above $\hat{n}_S$, -and replace $\ket{0}$ with $\ket{\mathrm{FS}}$, yielding: +<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 +$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$: $$\begin{aligned} - \matrixel{0}{\hat{n}_S}{j} - \quad\longrightarrow\quad \matrixel**{\mathrm{FS}}{\hat{\Psi}{}^\dagger \hat{\Psi}}{j} - = \sum_{\alpha \beta} \psi_\alpha^* \psi_\beta \matrixel**{\mathrm{FS}}{\hat{c}_\alpha^\dagger \hat{c}_\beta}{j} + \hat{n}(\vb{r}) + \equiv \hat{\Psi}{}^\dagger(\vb{r}) \hat{\Psi}(\vb{r}) + = \sum_{\vb{k} \vb{k}'} \psi_{\vb{k}}^*(\vb{r}) \: \psi_{\vb{k}'}(\vb{r})\: \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} + = \frac{1}{V} \sum_{\vb{k} \vb{k}'} e^{i (\vb{k}' - \vb{k}) \cdot \vb{r}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \end{aligned}$$ -This inner product is only nonzero if -$\ket{j} = \hat{c}_a \hat{c}_b^\dagger \ket{\mathrm{FS}}$ -with $a = \alpha$ and $b = \beta$, -or in other words, -only if $\ket{j}$ is a single-electron excitation of $\ket{\mathrm{FS}}$. -Furthermore, in $\ket{\mathrm{FS}}$, -$\alpha$ must be filled, and $\beta$ must be empty. -Let $f_\alpha \in \{0,1\}$ be the occupation number of orbital $\alpha$, then: +Taking the Fourier transfom yields a Dirac delta function $\delta$: $$\begin{aligned} - \matrixel{0}{\hat{n}_S}{j} - \longrightarrow \sum_{\alpha \beta} \psi_\alpha^* \psi_\beta \: f_\alpha (1 - f_\beta) \: \delta_{a \alpha} \delta_{b \beta} + \hat{n}(\vb{q}) + = \frac{1}{V} \int_{-\infty}^\infty + \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: e^{i (\vb{k}' - \vb{k} - \vb{q})\cdot \vb{r}} \dd{\vb{r}} + = \frac{(2 \pi)^D}{V} \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q}) \end{aligned}$$ -In $\chi_0$, the sum over $j$ becomes a sum over $a$ and $b$ -(this implicitly eliminates all $\ket{j}$ that are not single-electron excitations), -and $E_j\!-\!E_0$ becomes the cost of the excitation $\epsilon_b \!-\! \epsilon_a$, -where $\epsilon_a$ is the energy of orbital $a$. -Therefore, we find: +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: $$\begin{aligned} - \chi_0 - &= \sum_{a b} \bigg( \sum_{\alpha \beta \kappa \mu} f_\alpha (1 \!-\! f_\beta) f_\kappa (1 \!-\! f_\mu) - \frac{\psi_\alpha^*(\vb{r}) \psi_\beta(\vb{r}) \psi_\kappa(\vb{r}') \psi_\mu^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a} - \delta_{a \alpha} \delta_{b \beta} \delta_{a \kappa} \delta_{b \mu} - \\ - &\qquad\:\:\: - \sum_{\alpha \beta \kappa \mu} f_\alpha (1 \!-\! f_\beta) f_\kappa (1 \!-\! f_\mu) - \frac{\psi_\alpha^*(\vb{r'}) \psi_\beta(\vb{r'}) \psi_\kappa(\vb{r}) \psi_\mu^*(\vb{r})}{(\omega + i \eta) \hbar + \epsilon_b - \epsilon_a} - \delta_{a \alpha} \delta_{b \beta} \delta_{a \kappa} \delta_{b \mu} \bigg) - \\ - &= \sum_{a b} \bigg( f_a^2 (1 \!-\! f_b)^2 - \frac{\psi_a^*(\vb{r}) \psi_b(\vb{r}) \psi_a(\vb{r}') \psi_b^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a} - \\ - &\qquad\:\: - f_a^2 (1 \!-\! f_b)^2 - \frac{\psi_a^*(\vb{r'}) \psi_b(\vb{r'}) \psi_a(\vb{r}) \psi_b^*(\vb{r})}{(\omega + i \eta) \hbar + \epsilon_b - \epsilon_a} - \bigg) + \hat{n}(\vb{q}) + &= \frac{(2 \pi)^D}{V} \frac{V}{(2 \pi)^D} \sum_{\vb{k}} \int_{-\infty}^\infty + \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q}) \dd{\vb{k}'} + = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \end{aligned}$$ -Because $f_a, f_b \in \{0, 1\}$ when $T = 0$, we know that $f_a^2 (1 \!-\! f_b)^2 = f_a (1 \!-\! f_b)$. -We then swap the indices $a$ and $b$ in the second term, leading to: +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: $$\begin{aligned} - \chi_0 - &= \sum_{a b} \Big( f_a (1 \!-\! f_b) - f_b (1 \!-\! f_a) \Big) - \frac{\psi_a^*(\vb{r}) \psi_b(\vb{r}) \psi_a(\vb{r}') \psi_b^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a} - \\ - &= \sum_{a b} \Big( f_a - f_b \Big) - \frac{\psi_a^*(\vb{r}) \psi_b(\vb{r}) \psi_a(\vb{r}') \psi_b^*(\vb{r'})}{(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a} + \hat{n}(-\vb{q}) + = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} - \vb{q}} + = \sum_{\vb{k}'} \hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'} + = \hat{n}^\dagger(\vb{q}) \end{aligned}$$ -To proceed, we make the radical assumption that $\vb{H}_{0,S}$ -has continuous translational symmetry, -or in other words, that the external potential is uniform in space. -Clearly, this is not realistic, -so our conclusions will be more qualitative than quantitative. +The summation variable $\vb{k}$ has an associated spin $\sigma$, +and $\hat{n}$ does not carry any spin. +</div> +</div> -In that case, the wavefunction of a non-interacting particle is simply a plane wave, -so we insert $\psi_a(\vb{r}) = \exp\!(i \vb{k}_a \cdot \vb{r})$ -and $\psi_b(\vb{r}) = \exp\!(i \vb{k}_b \cdot \vb{r})$, yielding: +When neglecting interactions, it is tradition to rename $\chi$ to $\chi_0$. +We insert $\hat{n}$, suppressing spin: $$\begin{aligned} \chi_0 - &= \sum_{a b} \Big( f_a - f_b \Big) - \frac{\exp\!\big( \!-\! i \vb{k}_a \cdot \vb{r} + i \vb{k}_b \cdot \vb{r} + i \vb{k}_a \cdot \vb{r}' - i \vb{k}_b \cdot \vb{r}' \big)} - {(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a} - \\ - &= \sum_{a b} \Big( f_a - f_b \Big) - \frac{\exp\!\big( \!-\! i (\vb{k}_a \!-\! \vb{k}_b) \cdot (\vb{r} \!-\! \vb{r}')\big)} - {(\omega + i \eta) \hbar - \epsilon_b + \epsilon_a} + &= \frac{1}{Z V} \sum_{\vb{k} \vb{k}'} \sum_{\nu \nu'} \bigg( + \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'} + \matrixel{\nu'}{\hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'}}{\nu}} + {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}} + - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'} + \matrixel{\nu'}{\hat{c}_{\vb{k}'}^\dagger \hat{c}_{\vb{k}' + \vb{q}}}{\nu}} + {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu} \end{aligned}$$ -Here, we see that $\chi_0$ only depends on the differences -$\vb{r}\!-\!\vb{r}'$ and $\vb{k}_a\!-\!\vb{k}_b$. -Therefore, we define $\vb{q}' \equiv \vb{k}_b\!-\!\vb{k}_a$ -and rename $\vb{k}_a \to \vb{k}$. -We thus have: - -$$\begin{aligned} - \chi_0(\vb{r}\!-\!\vb{r}') - &= \sum_{\vb{k} \vb{q}'} \Big( f_k - f_{k+q'} \Big) - \frac{\exp\!\big( i \vb{q'} \cdot (\vb{r} \!-\! \vb{r}')\big)} - {(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} -\end{aligned}$$ +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 $\sigma = \sigma'$). -The summation goes over all $\vb{k}$ and $\vb{q}'$ -where $\vb{k}$ is inside the Fermi sphere, and $\vb{k}\!+\!\vb{q}'$ is outside. -Let $k_F$ be the Fermi radius, -then we convert this sum into an integral, -which means introducing a factor of $1/(2 \pi)^{3}$ -as usual in solid state physics: +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: $$\begin{aligned} - \chi_0(\vb{r}) - &= \sum_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}} - \frac{(f_k - f_{k+q'}) \exp\!\big( i \vb{q}' \cdot \vb{r} \big)} - {(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} - \\ - &= \frac{1}{(2 \pi)^3} \iint_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}} - \frac{(f_k - f_{k+q'}) \exp\!\big( i \vb{q}' \cdot \vb{r} \big)} - {(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} \dd{\vb{k}} \dd{\vb{q}'} + \chi_0 + &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu \nu'} \bigg( + \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'} + \matrixel{\nu'}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} + - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'} + \matrixel{\nu'}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu} \end{aligned}$$ -Fourier transforming the position $\vb{r}$ into the wavevector $\vb{q}$, -we recognize an integral that can be evaluated -to a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta(\vb{q})$: +Notice that we have eliminated all dependence on $\ket{\nu'}$, +so we remove it by $\sum_{\nu} \ket{\nu} \bra{\nu} = 1$: $$\begin{aligned} - \chi_0(\vb{q}) - &= \frac{1}{(2 \pi)^3} \int_{-\infty}^\infty \iint_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}} - \frac{(f_k - f_{k+q'}) \exp\!\big( i (\vb{q}' \!-\! \vb{q}) \cdot \vb{r} \big)}{(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} - \dd{\vb{k}} \dd{\vb{q}'} \dd{\vb{r}} + \chi_0 + &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu} \bigg( + \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} + - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu} \\ - &= \iint_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}'| > k_F}} - \frac{(f_k - f_{k+q'}) \:\delta(\vb{q}'\!-\!\vb{q})}{(\omega + i \eta) \hbar - \epsilon_{k+q'} + \epsilon_k} \dd{\vb{k}} \dd{\vb{q}'} + &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu} + \frac{\matrixel{\nu}{\comm*{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}} + {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0}}{\nu}} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \end{aligned}$$ -This delta functions eliminates the integral over $\vb{q}'$, -giving the following linear response $\chi_0$ -of a non-interacting electron gas in a uniform potential: +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 $\expval{}_0$: $$\begin{aligned} - \boxed{ - \chi_0(\vb{q}, \omega) - = \int_{\substack{|\vb{k}| < k_F \\ |\vb{k}+\vb{q}| > k_F}} - \frac{f_k - f_{k+q}}{(\omega + i \eta) \hbar - \epsilon_{k+q} + \epsilon_k} \dd{\vb{k}} - } + \chi_0 + &= \frac{1}{Z V} \sum_{\vb{k}} \frac{\Tr\!\big(\comm*{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}} + {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0} \big)} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} + = \frac{1}{V} \sum_{\vb{k}} + \frac{\expval*{\comm*{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}}_0} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \end{aligned}$$ -The resulting electron density change $\delta{\expval{\hat{n}}}$ is as follows, -where we use the [convolution theorem](/know/concept/convolution-theorem/) -to convert the convolution in $\vb{r}$-space into a product in $\vb{q}$-space: - -$$\begin{gathered} - \delta\expval{\hat{n}}(\vb{r}, \omega) - = \int_{-\infty}^\infty \chi_0(\vb{r}\!-\!\vb{r}', \omega) \: V(\vb{r}') \dd{r'} - \\ - \implies \qquad - \boxed{ - \delta\expval{\hat{n}}(\vb{q}, \omega) - = \chi_0(\vb{q}, \omega) \: V(\vb{q}) - } -\end{gathered}$$ - -So far, we have neglected electron-electron interactions, -but now we approximately correct this. -We split the effective potential $\vb{V}_\mathrm{eff}$ felt by the electrons -into the external potential $V_\mathrm{ext}$ -and the internal interactions $V_\mathrm{int}$, -such that: +This commutator can be evaluated, +and in this particular case it turns out to be: $$\begin{aligned} - V_\mathrm{eff}(\vb{r}) - = V_\mathrm{ext} + V_\mathrm{int} + \comm*{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} + = \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}} - \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \end{aligned}$$ -We approximate $V_\mathrm{int}$ as follows, -where $V_{ee}$ represents electron-electron interactions: - +<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: $$\begin{aligned} - V_\mathrm{int}(\vb{r}) - \approx \int_{-\infty}^\infty V_{ee}(\vb{r} \!-\! \vb{r}') \: \delta{n}(\vb{r}') \dd{\vb{r}'} - \qquad\quad - V_{ee}(\vb{r} \!-\! \vb{r}') - = \frac{e^2}{4 \pi \varepsilon_0} \frac{1}{|\vb{r} - \vb{r}'|} + \comm*{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p} + &= \hat{c}_m^\dagger \hat{c}_n \hat{c}_o^\dagger \hat{c}_p - \hat{c}_o^\dagger \hat{c}_p \hat{c}_m^\dagger \hat{c}_n + \\ + &= \hat{c}_m^\dagger \big( \acomm*{\hat{c}_n}{\hat{c}_o^\dagger} - \hat{c}_o^\dagger \hat{c}_n \big) \hat{c}_p + - \hat{c}_o^\dagger \big( \acomm*{\hat{c}_p}{\hat{c}_m^\dagger} - \hat{c}_m^\dagger \hat{c}_p \big) \hat{c}_n \end{aligned}$$ -Consequently, $V_\mathrm{int}$ satisfies Poisson's equation, -which has a well-known Fourier transform: +Using the standard fermion anticommutation relations, this becomes: $$\begin{aligned} - \nabla^2 V_\mathrm{int}(\vb{r}) - = - \frac{\delta{n}(\vb{r})}{\varepsilon_0} - \quad \implies \quad - V_\mathrm{int}(\vb{q}) - = \frac{e^2}{\varepsilon_0 |\vb{q}|^2} \delta{n}(\vb{q}) + \comm*{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p} + &= \hat{c}_m^\dagger \big( \delta_{no} - \hat{c}_o^\dagger \hat{c}_n \big) \hat{c}_p + - \hat{c}_o^\dagger \big( \delta_{pm} - \hat{c}_m^\dagger \hat{c}_p \big) \hat{c}_n + \\ + &= \hat{c}_m^\dagger \hat{c}_p \: \delta_{no} - \hat{c}_m^\dagger \hat{c}_o^\dagger \hat{c}_n \hat{c}_p + - \hat{c}_o^\dagger \hat{c}_n \: \delta_{pm} + \hat{c}_o^\dagger \hat{c}_m^\dagger \hat{c}_p \hat{c}_n + \\ + &= \hat{c}_m^\dagger \hat{c}_p \: \delta_{no} - \hat{c}_o^\dagger \hat{c}_n \: \delta_{pm} \end{aligned}$$ -Meanwhile, from all of the above calculations, -we can write $\delta{n}$ as follows, -where $\chi$ and $\chi_0$ are the -(unknown) interacting and (known) non-interacting response functions: +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 $\vb{k}$: $$\begin{aligned} - \delta{n}(\vb{q}) - = \chi V_\mathrm{ext} - \approx \chi_0 V_\mathrm{eff} + \chi_0(\vb{q}, \omega) + = \frac{1}{V} \sum_{\sigma \vb{k}} + \frac{\expval*{\hat{c}_{\sigma,\vb{k}}^\dagger \hat{c}_{\sigma,\vb{k}} - \hat{c}_{\sigma,\vb{k}+\vb{q}}^\dagger \hat{c}_{\sigma,\vb{k}+\vb{q}}}_0} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \end{aligned}$$ -Keep in mind that we are treating interactions as a perturbation to $V_\mathrm{ext}$, -therefore $V_\mathrm{ext} \approx V_\mathrm{eff}$. -With this, $V_\mathrm{eff}$ becomes as follows in $\vb{q}$-space, -where we have used the convolution theorem -to get the product $\delta{n} (\vb{q}) V_{ee}(\vb{q})$: +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**: $$\begin{aligned} - V_\mathrm{eff}(\vb{q}) - = V_\mathrm{ext} + \chi_0 V_\mathrm{eff} V_{ee} - \qquad \quad - V_{ee}(\vb{q}) - = \frac{e^2}{\varepsilon_0 |\vb{q}|^2} + \boxed{ + \chi_0(\vb{q}, \omega) + = \frac{1}{V} \sum_{\sigma \vb{k}} + \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})} + {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} + } \end{aligned}$$ -Isolating this equation for $V_\mathrm{ext}$ -yields the definition of the relative permittivity $\varepsilon_r$: +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: $$\begin{aligned} - V_\mathrm{ext} - = (1 - \chi_0 V_{ee}) V_\mathrm{eff} - \equiv \varepsilon_r V_\mathrm{eff} + U_\mathrm{tot} + = U_\mathrm{ext} + U_\mathrm{ind} + = \frac{U_\mathrm{ext}}{\varepsilon_r} \end{aligned}$$ -Therefore, by inserting all the above expressions, -we arrive at the Lindhard dielectric function $\varepsilon_r$ -for a non-interacting electron gas in a uniform potential: +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: $$\begin{aligned} \boxed{ \varepsilon_r(\vb{q}, \omega) - = 1 - \frac{e^2}{\varepsilon_0 |\vb{q}|^2} \sum_{k} \frac{f_{k-q} - f_k}{\hbar (\omega + i \eta) + E_{k-q} - E_k} + = 1 - \frac{U_{ee}(\vb{q})}{V} + \sum_{\sigma \vb{k}} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} } \end{aligned}$$ +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 |