diff options
Diffstat (limited to 'source/know/concept/lindhard-function/index.md')
| -rw-r--r-- | source/know/concept/lindhard-function/index.md | 170 |
1 files changed, 85 insertions, 85 deletions
diff --git a/source/know/concept/lindhard-function/index.md b/source/know/concept/lindhard-function/index.md index 4252bd4..4033148 100644 --- a/source/know/concept/lindhard-function/index.md +++ b/source/know/concept/lindhard-function/index.md @@ -14,28 +14,28 @@ 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/) -for the electron density operator $\hat{n}$, -which describes the change in $\Expval{\hat{n}}$ -due to a time-dependent perturbation $\hat{H}_1$: +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 +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 $\hat{H}_1$: +Now consider a harmonic $$\hat{H}_1$$: $$\begin{aligned} \hat{H}_{1,S}(t) = e^{i (\omega + i \eta) t} \int_{-\infty}^\infty U(\vb{r}) \: \hat{n}_S(\vb{r}) \dd{\vb{r}} \end{aligned}$$ -Where $S$ is the Schrödinger picture, -$\eta$ is a positive infinitesimal to ensure convergence later, -and $U(\vb{r})$ is an arbitrary potential function. +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: $$\begin{aligned} @@ -43,7 +43,7 @@ $$\begin{aligned} = \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}$$ -Here, $\chi$ is the density-density correlation function, +Here, $$\chi$$ is the density-density correlation function, i.e. a two-particle [Green's function](/know/concept/greens-functions/): $$\begin{aligned} @@ -51,10 +51,10 @@ $$\begin{aligned} \equiv - \frac{i}{\hbar} \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{r}, t)}{\hat{n}_I(\vb{r}', t')}}_0 \end{aligned}$$ -Let us assume that the unperturbed system (i.e. without $U$) is spatially uniform, -so that $\chi$ only depends on the difference $\vb{r} - \vb{r}'$. +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/) -$\vb{r}\!-\!\vb{r}' \to \vb{q}$: +$$\vb{r}\!-\!\vb{r}' \to \vb{q}$$: $$\begin{aligned} \chi(\vb{q}; t, t') @@ -65,9 +65,9 @@ $$\begin{aligned} \: 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}$$ -Where both $\hat{n}_I$ have been written as inverse Fourier transforms, -giving a factor $(2 \pi)^{-2 D}$, with $D$ being the number of spatial dimensions. -We rearrange to get a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$: +Where both $$\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} \chi(\vb{q}; t, t') @@ -84,9 +84,9 @@ $$\begin{aligned} \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \end{aligned}$$ -On the left, $\vb{r}'$ does not appear, so it must also disappear on the right. -If we choose an arbitrary (hyper)cube of volume $V$ in real space, -then clearly $\int_V \dd{\vb{r}'} = V$. Therefore: +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: $$\begin{aligned} \chi(\vb{q}; t, t') @@ -95,8 +95,8 @@ $$\begin{aligned} \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \dd{\vb{r}'} \end{aligned}$$ -For $V \to \infty$ we get a Dirac delta function, -but in fact the conclusion holds for finite $V$ too: +For $$V \to \infty$$ we get a Dirac delta function, +but in fact the conclusion holds for finite $$V$$ too: $$\begin{aligned} \chi(\vb{q}; t, t') @@ -106,10 +106,10 @@ $$\begin{aligned} &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0 \end{aligned}$$ -Similarly, if the unperturbed Hamiltonian $\hat{H}_0$ 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 $\chi$ +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/): $$\begin{aligned} @@ -119,11 +119,11 @@ $$\begin{aligned} \Big( e^{-\beta E_\nu} - e^{- \beta E_{\nu'}} \Big) \end{aligned}$$ -Where $\Ket{\nu}$ and $\Ket{\nu'}$ are many-electron eigenstates of $\hat{H}_0$, -and $Z$ is the [grand partition function](/know/concept/grand-canonical-ensemble/). +Where $$\Ket{\nu}$$ and $$\Ket{\nu'}$$ are many-electron eigenstates of $$\hat{H}_0$$, +and $$Z$$ is the [grand partition function](/know/concept/grand-canonical-ensemble/). According to the [convolution theorem](/know/concept/convolution-theorem/) -$\delta{\Expval{\hat{n}}}(\vb{q}, \omega) = \chi(\vb{q}, \omega) \: U(\vb{q})$. -In anticipation, we swap $\nu$ and $\nu''$ in the second term, +$$\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} @@ -135,9 +135,9 @@ $$\begin{aligned} {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu} \end{aligned}$$ -All operators are in the Schrödinger picture from now on, hence we dropped the subscript $S$. +All operators are in the Schrödinger picture from now on, hence we dropped the subscript $$S$$. -To proceed, we need to rewrite $\hat{n}(\vb{q})$ somehow. +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: @@ -154,10 +154,10 @@ $$\begin{aligned} <label for="proof-density">Proof</label> <div class="hidden" markdown="1"> <label for="proof-density">Proof.</label> -Starting from the general definition of $\hat{n}$, -we write out the field operators $\hat{\Psi}(\vb{r})$, +Starting from the general definition of $$\hat{n}$$, +we write out the field operators $$\hat{\Psi}(\vb{r})$$, and insert the known non-interacting single-electron orbitals -$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$: +$$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$$: $$\begin{aligned} \hat{n}(\vb{r}) @@ -166,7 +166,7 @@ $$\begin{aligned} = \frac{1}{V} \sum_{\vb{k} \vb{k}'} e^{i (\vb{k}' - \vb{k}) \cdot \vb{r}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \end{aligned}$$ -Taking the Fourier transfom yields a Dirac delta function $\delta$: +Taking the Fourier transfom yields a Dirac delta function $$\delta$$: $$\begin{aligned} \hat{n}(\vb{q}) @@ -176,19 +176,19 @@ $$\begin{aligned} \end{aligned}$$ If we impose periodic boundary conditions -on our $D$-dimensional hypercube of volume $V$, -then $\vb{k}$ becomes discrete, -with per-value spacing $2 \pi / V^{1/D}$ along each axis. - -Consequently, each orbital $\psi_\vb{k}$ uniquely occupies -a volume $(2 \pi)^D / V$ in $\vb{k}$-space, so we make the approximation -$\sum_{\vb{k}} \approx V / (2 \pi)^D \int_{-\infty}^\infty \dd{\vb{k}}$. -This becomes exact for $V \to \infty$, -in which case $\vb{k}$ also becomes continuous again, +on our $$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}$$ also becomes continuous again, which is what we want for jellium. -We apply this standard trick from condensed matter physics to $\hat{n}$, -and $V$ cancels out: +We apply this standard trick from condensed matter physics to $$\hat{n}$$, +and $$V$$ cancels out: $$\begin{aligned} \hat{n}(\vb{q}) @@ -197,9 +197,9 @@ $$\begin{aligned} = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \end{aligned}$$ -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: +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: $$\begin{aligned} \hat{n}(-\vb{q}) @@ -208,13 +208,13 @@ $$\begin{aligned} = \hat{n}^\dagger(\vb{q}) \end{aligned}$$ -The summation variable $\vb{k}$ has an associated spin $\sigma$, -and $\hat{n}$ does not carry any spin. +The summation variable $$\vb{k}$$ has an associated spin $$\sigma$$, +and $$\hat{n}$$ does not carry any spin. </div> </div> -When neglecting interactions, it is tradition to rename $\chi$ to $\chi_0$. -We insert $\hat{n}$, suppressing spin: +When neglecting interactions, it is tradition to rename $$\chi$$ to $$\chi_0$$. +We insert $$\hat{n}$$, suppressing spin: $$\begin{aligned} \chi_0 @@ -227,16 +227,16 @@ $$\begin{aligned} {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu} \end{aligned}$$ -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 $\vb{k} \!+\! \vb{q}$, +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 $\sigma = \sigma'$). +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 -$\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}}$, -where $\xi_{\vb{k}}$ is the energy of a $\vb{k}$-orbital. +$$\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}}$$, +where $$\xi_{\vb{k}}$$ is the energy of a $$\vb{k}$$-orbital. Therefore: $$\begin{aligned} @@ -250,8 +250,8 @@ $$\begin{aligned} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu} \end{aligned}$$ -Notice that we have eliminated all dependence on $\Ket{\nu'}$, -so we remove it by $\sum_{\nu} \Ket{\nu} \Bra{\nu} = 1$: +Notice that we have eliminated all dependence on $$\Ket{\nu'}$$, +so we remove it by $$\sum_{\nu} \Ket{\nu} \Bra{\nu} = 1$$: $$\begin{aligned} \chi_0 @@ -268,9 +268,9 @@ $$\begin{aligned} \end{aligned}$$ 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 a thermal expectation $\Expval{}_0$: +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 $$\Expval{}_0$$: $$\begin{aligned} \chi_0 @@ -295,7 +295,7 @@ $$\begin{aligned} <label for="proof-commutator">Proof</label> <div class="hidden" markdown="1"> <label for="proof-commutator">Proof.</label> -In general, for any single-particle states labeled by $m$, $n$, $o$ and $p$, we have: +In general, for any single-particle states labeled by $$m$$, $$n$$, $$o$$ and $$p$$, we have: $$\begin{aligned} \comm{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p} &= \hat{c}_m^\dagger \hat{c}_n \hat{c}_o^\dagger \hat{c}_p - \hat{c}_o^\dagger \hat{c}_p \hat{c}_m^\dagger \hat{c}_n @@ -317,13 +317,13 @@ $$\begin{aligned} &= \hat{c}_m^\dagger \hat{c}_p \: \delta_{no} - \hat{c}_o^\dagger \hat{c}_n \: \delta_{pm} \end{aligned}$$ -In this case, $m = p = \vb{k}$ and $n = o = \vb{k} \!+\! \vb{q}$, +In this case, $$m = p = \vb{k}$$ and $$n = o = \vb{k} \!+\! \vb{q}$$, so the Kronecker deltas are unnecessary. </div> </div> -We substitute this result into $\chi_0$, -and reintroduce the spin index $\sigma$ associated with $\vb{k}$: +We substitute this result into $$\chi_0$$, +and reintroduce the spin index $$\sigma$$ associated with $$\vb{k}$$: $$\begin{aligned} \chi_0(\vb{q}, \omega) @@ -332,9 +332,9 @@ $$\begin{aligned} {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \end{aligned}$$ -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/) $n_F$. +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**: $$\begin{aligned} @@ -347,8 +347,8 @@ $$\begin{aligned} \end{aligned}$$ From this, we would like to get the -[dielectric function](/know/concept/dielectric-function/) $\varepsilon_r$. -Recall its definition, where $U_\mathrm{tot}$, $U_\mathrm{ext}$, and $U_\mathrm{ind}$ +[dielectric function](/know/concept/dielectric-function/) $$\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} @@ -361,17 +361,17 @@ 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 $U = U_\mathrm{tot}$. +$$\Phi_\mathrm{tot} = q_e U_\mathrm{tot}$$, +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: +$$\delta{\Expval{\hat{n}}} = \chi_0 U_\mathrm{tot}$$, +so the standard formula for $$\varepsilon_r$$ gives: $$\begin{aligned} \boxed{ @@ -381,7 +381,7 @@ $$\begin{aligned} } \end{aligned}$$ -Where $U_{ee}(\vb{q}) = q_e^2 / (\varepsilon_0 |\vb{q}|^2)$ +Where $$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, |