Categories: Physics, Quantum mechanics.

Lindhard function

The Lindhard function describes the response of jellium (i.e. a free electron gas) to an external perturbation, and is a quantum-mechanical alternative to the Drude model.

We start from the Kubo formula for the electron density operator n^\hat{n}, which describes the change in n^\Expval{\hat{n}} due to a time-dependent perturbation H^1\hat{H}_1:

δ ⁣n^ ⁣(r,t)=iΘ(tt)[n^I(r,t),H^1,I(t)]0dt\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 II refers to the interaction picture, and the expectation 0\Expval{}_0 is for a thermal equilibrium before the perturbation was applied. Now consider a harmonic H^1\hat{H}_1:

H^1,S(t)=ei(ω+iη)tU(r)n^S(r)dr\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 SS is the Schrödinger picture, η\eta is a positive infinitesimal to ensure convergence later, and U(r)U(\vb{r}) is an arbitrary potential function. The Kubo formula becomes:

δ ⁣n^ ⁣(r,t)=χ(r,r;t,t)U(r)ei(ω+iη)tdtdr\begin{aligned} \delta\!\Expval{ {\hat{n}}}\!(\vb{r}, t) = \iint_{-\infty}^\infty \chi(\vb{r}, \vb{r}'; t, t') \: U(\vb{r}') \: e^{i (\omega + i \eta) t'} \dd{t'} \dd{\vb{r}'} \end{aligned}

Here, χ\chi is the density-density correlation function, i.e. a two-particle Green’s function:

χ(r,r;t,t)iΘ(tt)[n^I(r,t),n^I(r,t)]0\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 UU) is spatially uniform, so that χ\chi only depends on the difference rr\vb{r} - \vb{r}'. We then take its Fourier transform r ⁣ ⁣rq\vb{r}\!-\!\vb{r}' \to \vb{q}:

χ(q;t,t)=χ(rr;t,t)eiq(rr)dr=iΘ(t ⁣ ⁣t)(2π)2D[n^I(q1,t),n^I(q2,t)]0eiq1reiq2reiq(rr)dq1dq2dr\begin{aligned} \chi(\vb{q}; t, t') &= \int_{-\infty}^\infty \chi(\vb{r} - \vb{r}'; t, t') \: e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{r}} \\ &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: e^{i \vb{q}_1 \cdot \vb{r}} e^{i \vb{q}_2 \cdot \vb{r}'} e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}} \end{aligned}

Where both n^I\hat{n}_I have been written as inverse Fourier transforms, giving a factor (2π)2D(2 \pi)^{-2 D}, with DD being the number of spatial dimensions. We rearrange to get a Dirac delta function δ\delta:

χ(q;t,t)=iΘ(t ⁣ ⁣t)(2π)2D[n^I(q1,t),n^I(q2,t)]0ei(q1q)rei(q2+q)rdq1dq2dr=iΘ(t ⁣ ⁣t)(2π)D[n^I(q1,t),n^I(q2,t)]0δ(q1 ⁣ ⁣q)ei(q2+q)rdq1dq2=iΘ(t ⁣ ⁣t)(2π)D[n^I(q,t),n^I(q2,t)]0ei(q2+q)rdq2\begin{aligned} \chi(\vb{q}; t, t') &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: e^{i (\vb{q}_1 - \vb{q}) \cdot \vb{r}} e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}} \\ &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \iint \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: \delta(\vb{q}_1 \!-\! \vb{q}) \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2} \\ &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \int \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \end{aligned}

On the left, r\vb{r}' does not appear, so it must also disappear on the right. If we choose an arbitrary (hyper)cube of volume VV in real space, then clearly Vdr=V\int_V \dd{\vb{r}'} = V. Therefore:

χ(q;t,t)=iΘ(t ⁣ ⁣t)(2π)D1VV[n^I(q,t),n^I(q2,t)]0ei(q2+q)rdq2dr\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}

For VV \to \infty we get a Dirac delta function, but in fact the conclusion holds for finite VV too:

χ(q;t,t)=iΘ(t ⁣ ⁣t)1V[n^I(q,t),n^I(q2,t)]0δ(q2 ⁣+ ⁣q)dq2=iΘ(t ⁣ ⁣t)1V[n^I(q,t),n^I(q,t)]0\begin{aligned} \chi(\vb{q}; t, t') &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \int_{-\infty}^\infty \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: \delta(\vb{q}_2 \!+\! \vb{q}) \dd{\vb{q}_2} \\ &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0 \end{aligned}

Similarly, if the unperturbed Hamiltonian H^0\hat{H}_0 is time-independent, χ\chi only depends on the time difference ttt - t'. Note that δn^\delta{\Expval{\hat{n}}} already has the form of a Fourier transform, which gives us an opportunity to rewrite χ\chi in the Lehmann representation:

χ(q,ω)=1ZVνννn^S(q)ννn^S(q)ν(ω+iη)+EνEν(eβEνeβEν)\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 ν\Ket{\nu} and ν\Ket{\nu'} are many-electron eigenstates of H^0\hat{H}_0, and ZZ is the grand partition function. According to the convolution theorem δn^(q,ω)=χ(q,ω)U(q)\delta{\Expval{\hat{n}}}(\vb{q}, \omega) = \chi(\vb{q}, \omega) \: U(\vb{q}). In anticipation, we swap ν\nu and ν\nu'' in the second term, so the general response function is written as:

χ(q,ω)=1ZVνν(νn^(q)ννn^(q)ν(ω+iη)+EνEννn^(q)ννn^(q)ν(ω+iη)+EνEν)eβEν\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}

All operators are in the Schrödinger picture from now on, hence we dropped the subscript SS.

To proceed, we need to rewrite n^(q)\hat{n}(\vb{q}) somehow. If we neglect electron-electron interactions, the single-particle states are simply plane waves, in which case:

n^(q)=σkc^σ,kc^σ,k+qn^(q)=n^(q)\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}

Starting from the general definition of n^\hat{n}, we write out the field operators Ψ^(r)\hat{\Psi}(\vb{r}), and insert the known non-interacting single-electron orbitals ψk(r)=eikr/V\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}:

n^(r)Ψ^(r)Ψ^(r)=kkψk(r)ψk(r)c^kc^k=1Vkkei(kk)rc^kc^k\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 δ\delta:

n^(q)=1Vkkc^kc^kei(kkq)rdr=(2π)DVkkc^kc^kδ(k ⁣ ⁣k ⁣ ⁣q)\begin{aligned} \hat{n}(\vb{q}) = \frac{1}{V} \int_{-\infty}^\infty \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: e^{i (\vb{k}' - \vb{k} - \vb{q})\cdot \vb{r}} \dd{\vb{r}} = \frac{(2 \pi)^D}{V} \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q}) \end{aligned}

If we impose periodic boundary conditions on our DD-dimensional hypercube of volume VV, then k\vb{k} becomes discrete, with per-value spacing 2π/V1/D2 \pi / V^{1/D} along each axis.

Consequently, each orbital ψk\psi_\vb{k} uniquely occupies a volume (2π)D/V(2 \pi)^D / V in k\vb{k}-space, so we make the approximation kV/(2π)Ddk\sum_{\vb{k}} \approx V / (2 \pi)^D \int_{-\infty}^\infty \dd{\vb{k}}. This becomes exact for VV \to \infty, in which case k\vb{k} also becomes continuous again, which is what we want for jellium.

We apply this standard trick from condensed matter physics to n^\hat{n}, and VV cancels out:

n^(q)=(2π)DVV(2π)Dkc^kc^kδ(k ⁣ ⁣k ⁣ ⁣q)dk=kc^kc^k+q\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 kkq\vb{k}' \equiv \vb{k} - \vb{q} to show that n^(q)=n^(q)\hat{n}(-\vb{q}) = \hat{n}{}^\dagger(\vb{q}), which can also be understood as a consequence of n^(r)\hat{n}(\vb{r}) being real:

n^(q)=kc^kc^kq=kc^k+qc^k=n^(q)\begin{aligned} \hat{n}(-\vb{q}) = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} - \vb{q}} = \sum_{\vb{k}'} \hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'} = \hat{n}^\dagger(\vb{q}) \end{aligned}

The summation variable k\vb{k} has an associated spin σ\sigma, and n^\hat{n} does not carry any spin.

When neglecting interactions, it is tradition to rename χ\chi to χ0\chi_0. We insert n^\hat{n}, suppressing spin:

χ0=1ZVkkνν(νc^kc^k+qννc^k+qc^kν(ω+iη)+EνEννc^k+qc^kννc^kc^k+qν(ω+iη)+EνEν)eβEν\begin{aligned} \chi_0 &= \frac{1}{Z V} \sum_{\vb{k} \vb{k}'} \sum_{\nu \nu'} \bigg( \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'} \matrixel{\nu'}{\hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'}}{\nu}} {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}} - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'} \matrixel{\nu'}{\hat{c}_{\vb{k}'}^\dagger \hat{c}_{\vb{k}' + \vb{q}}}{\nu}} {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu} \end{aligned}

Here, νc^kc^k+qν\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'} is only nonzero if ν\Ket{\nu'} is contructed from ν\Ket{\nu} by moving an electron from k\vb{k} to k ⁣+ ⁣q\vb{k} \!+\! \vb{q}, and analogously for the other inner products. As a result, k=k\vb{k} = \vb{k}' (and σ=σ\sigma = \sigma').

For the same reason, the energy difference Eν ⁣ ⁣EνE_\nu \!-\! E_{\nu'} can simply be replaced by the cost of the single-particle excitation ξk ⁣ ⁣ξk+q\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}}, where ξk\xi_{\vb{k}} is the energy of a k\vb{k}-orbital. Therefore:

χ0=1ZVkνν(νc^kc^k+qννc^k+qc^kν(ω+iη)+ξkξk+qνc^k+qc^kννc^kc^k+qν(ω+iη)+ξkξk+q)eβEν\begin{aligned} \chi_0 &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu \nu'} \bigg( \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'} \matrixel{\nu'}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'} \matrixel{\nu'}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu} \end{aligned}

Notice that we have eliminated all dependence on ν\Ket{\nu'}, so we remove it by ννν=1\sum_{\nu} \Ket{\nu} \Bra{\nu} = 1:

χ0=1ZVkν(νc^kc^k+qc^k+qc^kν(ω+iη)+ξkξk+qνc^k+qc^kc^kc^k+qν(ω+iη)+ξkξk+q)eβEν=1ZVkνν[c^kc^k+q,c^k+qc^k]eβH^0ν(ω+iη)+ξkξk+q\begin{aligned} \chi_0 &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu} \bigg( \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu} \\ &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu} \frac{\matrixel{\nu}{\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}} {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0}}{\nu}} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \end{aligned}

Where we recognized the commutator, and eliminated EνE_\nu using H^0n=Eνν\hat{H}_0 \Ket{n} = E_\nu \Ket{\nu}. The resulting expression has the form of a matrix trace Tr\Tr and a thermal expectation 0\Expval{}_0:

χ0=1ZVkTr ⁣([c^kc^k+q,c^k+qc^k]eβH^0)(ω+iη)+ξkξk+q=1Vk[c^kc^k+q,c^k+qc^k]0(ω+iη)+ξkξk+q\begin{aligned} \chi_0 &= \frac{1}{Z V} \sum_{\vb{k}} \frac{\Tr\!\big(\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}} {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0} \big)} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} = \frac{1}{V} \sum_{\vb{k}} \frac{\expval{\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}}_0} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \end{aligned}

This commutator can be evaluated, and in this particular case it turns out to be:

[c^kc^k+q,c^k+qc^k]=c^kc^kc^k+qc^k+q\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}

In general, for any single-particle states labeled by mm, nn, oo and pp, we have:

[c^mc^n,c^oc^p]=c^mc^nc^oc^pc^oc^pc^mc^n=c^m({c^n,c^o}c^oc^n)c^pc^o({c^p,c^m}c^mc^p)c^n\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:

[c^mc^n,c^oc^p]=c^m(δnoc^oc^n)c^pc^o(δpmc^mc^p)c^n=c^mc^pδnoc^mc^oc^nc^pc^oc^nδpm+c^oc^mc^pc^n=c^mc^pδnoc^oc^nδpm\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=km = p = \vb{k} and n=o=k ⁣+ ⁣qn = o = \vb{k} \!+\! \vb{q}, so the Kronecker deltas are unnecessary.

We substitute this result into χ0\chi_0, and reintroduce the spin index σ\sigma associated with k\vb{k}:

χ0(q,ω)=1Vσkc^σ,kc^σ,kc^σ,k+qc^σ,k+q0(ω+iη)+ξkξk+q\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 c^σ.kc^σ.k\hat{c}_{\sigma.\vb{k}}^\dagger \hat{c}_{\sigma.\vb{k}} simply counts the number of electrons in state (σ,k)(\sigma, \vb{k}), which is given by the Fermi-Dirac distribution nFn_F. This gives us the Lindhard response function:

χ0(q,ω)=1VσknF(ξk)nF(ξk+q)(ω+iη)+ξkξk+q\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}

From this, we would like to get the dielectric function εr\varepsilon_r. Recall its definition, where UtotU_\mathrm{tot}, UextU_\mathrm{ext}, and UindU_\mathrm{ind} are the total, external and induced potentials, respectively:

Utot=Uext+Uind=Uextεr\begin{aligned} U_\mathrm{tot} = U_\mathrm{ext} + U_\mathrm{ind} = \frac{U_\mathrm{ext}}{\varepsilon_r} \end{aligned}

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 Φtot=qeUtot\Phi_\mathrm{tot} = q_e U_\mathrm{tot}, where qe<0q_e < 0 is the charge of an electron.

From the Lindhard response function χ0\chi_0, we get the induced particle density offset δn^\delta{\Expval{\hat{n}}} caused by a potential UU. The density δn^\delta{\Expval{\hat{n}}} should be self-consistent, implying U=UtotU = U_\mathrm{tot}. In other words, we have a linear relation δn^=χ0Utot\delta{\Expval{\hat{n}}} = \chi_0 U_\mathrm{tot}, so the standard formula for εr\varepsilon_r gives:

εr(q,ω)=1Uee(q)VσknF(ξk)nF(ξk+q)(ω+iη)+ξkξk+q\begin{aligned} \boxed{ \varepsilon_r(\vb{q}, \omega) = 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 Uee(q)=qe2/(ε0q2)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.


  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.