Categories: Physics, Quantum mechanics.

# Lindhard function

The Lindhard function describes the response of an electron gas to an external perturbation, and can be regarded as a quantum-mechanical alternative to the Drude model.

We start from the 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 $$\Theta$$ is the Heaviside step function, and the subscript $$I$$ refers to the 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:

\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}} \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:

\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'} \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:

\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} \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 $$\hat{H}_{0,S}$$:

\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'} \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:

\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) \\ \matrixel{j}{\hat{n}_I(0)}{0} &= \matrixel{j}{\hat{n}_S(0)}{0} \:\exp\!(0 - 0) = \matrixel{j}{\hat{n}_S(0)}{0} \end{aligned}

We define $$\omega_{j0} \equiv (E_j \!-\! E_0)/\hbar$$ and insert the above expressions into $$\delta\expval{\hat{n}}$$, yielding:

\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) \end{aligned}

These integrals are Fourier transforms, and are straightforward to evaluate. The first is:

\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 \\ &= \frac{1}{- i (\omega - \omega_{j0}) + \eta} = \frac{i}{\omega - \omega_{j0} + i \eta} \end{aligned}

The other integral simply has the opposite sign in front of $$\omega_{j0}$$. We thus arrive at:

\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) \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 $$\chi$$:

\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} } \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 $$\chi_0$$.

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:

\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 \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:

\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} \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:

\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} \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:

\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) \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:

\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} \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.

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:

\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} \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}

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:

\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}'} \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 $$\delta(\vb{q})$$:

\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}} \\ &= \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}'} \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:

\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}} } \end{aligned}

The resulting electron density change $$\delta{\expval{\hat{n}}}$$ is as follows, where we use the 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:

\begin{aligned} V_\mathrm{eff}(\vb{r}) = V_\mathrm{ext} + V_\mathrm{int} \end{aligned}

We approximate $$V_\mathrm{int}$$ as follows, where $$V_{ee}$$ represents electron-electron interactions:

\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}'|} \end{aligned}

Consequently, $$V_\mathrm{int}$$ satisfies Poisson’s equation, which has a well-known Fourier transform:

\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}) \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:

\begin{aligned} \delta{n}(\vb{q}) = \chi V_\mathrm{ext} \approx \chi_0 V_\mathrm{eff} \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 $$\delta{n} (\vb{q}) V_{ee}(\vb{q})$$:

\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} \end{aligned}

Isolating this equation for $$V_\mathrm{ext}$$ yields the definition of the relative permittivity $$\varepsilon_r$$:

\begin{aligned} V_\mathrm{ext} = (1 - \chi_0 V_{ee}) V_\mathrm{eff} \equiv \varepsilon_r V_\mathrm{eff} \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:

\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} } \end{aligned}

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.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.