messing around with orbital energies
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@ -220,7 +220,11 @@ In the KS formulation of eDFT, the universal ensemble functional (the weight-dep
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\begin{equation}
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\begin{equation}
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\F{}{\bw}[\n{}{}] = \Ts{\bw}[\n{}{}] + \E{\Hxc}{\bw}[\n{}{}],
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\F{}{\bw}[\n{}{}] = \Ts{\bw}[\n{}{}] + \E{\Hxc}{\bw}[\n{}{}],
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\end{equation}
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\end{equation}
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where $\Ts{\bw}[\n{}{}]$ and $\E{\Hxc}{}[\n{}{}]$ are the noninteracting ensemble kinetic energy functional and ensemble Hartree-exchange-correlation (Hxc) functional, respectively with
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where
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\begin{equation}
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\Ts{\bw}[\n{}{}] =
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\end{equation}
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and
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\begin{equation}
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\begin{equation}
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\label{eq:exc_def}
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\label{eq:exc_def}
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\begin{split}
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\begin{split}
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@ -230,6 +234,7 @@ where $\Ts{\bw}[\n{}{}]$ and $\E{\Hxc}{}[\n{}{}]$ are the noninteracting ensembl
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& = \frac{1}{2} \iint \frac{\n{}{}(\br{}) \n{}{}(\br{}')}{\abs{\br{}-\br{}'}} d\br{} d\br{}'+ \int \e{\xc}{\bw}[\n{}{}(\br{})] \n{}{}(\br{}) d\br{}.
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& = \frac{1}{2} \iint \frac{\n{}{}(\br{}) \n{}{}(\br{}')}{\abs{\br{}-\br{}'}} d\br{} d\br{}'+ \int \e{\xc}{\bw}[\n{}{}(\br{})] \n{}{}(\br{}) d\br{}.
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\end{split}
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\end{split}
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\end{equation}
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\end{equation}
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are the noninteracting ensemble kinetic energy functional and ensemble Hartree-exchange-correlation (Hxc) functional, respectively.
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Note that the weight-independent Hartree functional $\E{\Ha}{}[\n{}{}]$ causes the infamous ghost-interaction error (GIC) \cite{Gidopoulos_2002, Pastorczak_2014, Alam_2016, Alam_2017, Gould_2017} in eDFT, which is supposed to be cancelled by the weight-dependent xc functional $\E{\xc}{\bw}[\n{}{}]$.
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Note that the weight-independent Hartree functional $\E{\Ha}{}[\n{}{}]$ causes the infamous ghost-interaction error (GIC) \cite{Gidopoulos_2002, Pastorczak_2014, Alam_2016, Alam_2017, Gould_2017} in eDFT, which is supposed to be cancelled by the weight-dependent xc functional $\E{\xc}{\bw}[\n{}{}]$.
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From the GOK-DFT ensemble energy expression in Eq.~\eqref{eq:Ew-GOK}, we obtain \cite{Gross_1988b,Deur_2019}
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From the GOK-DFT ensemble energy expression in Eq.~\eqref{eq:Ew-GOK}, we obtain \cite{Gross_1988b,Deur_2019}
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@ -253,6 +258,7 @@ Equation \eqref{eq:dEdw} is our working equation for computing excitation energi
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\label{sec:func}
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\label{sec:func}
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The present work deals with the explicit construction of the (reduced) LDA xc functional $\e{\xc}{\bw}[\n{}{}]$ defined in Eq.~\eqref{eq:exc_def}.
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The present work deals with the explicit construction of the (reduced) LDA xc functional $\e{\xc}{\bw}[\n{}{}]$ defined in Eq.~\eqref{eq:exc_def}.
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Here, we restrict our study to the case of a two-state ensemble (\ie, $\Nens = 2$) where both the ground state ($I=0$) and the first doubly-excited state ($I=1$) are considered.
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Here, we restrict our study to the case of a two-state ensemble (\ie, $\Nens = 2$) where both the ground state ($I=0$) and the first doubly-excited state ($I=1$) are considered.
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Thus, we have $0 \le \ew{} \le 1/2$.
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The generalisation to a larger number of states (in particular the inclusion of the first singly-excited state) is trivial and left for future work.
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The generalisation to a larger number of states (in particular the inclusion of the first singly-excited state) is trivial and left for future work.
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We adopt the usual decomposition, and write down the weight-dependent xc functional as
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We adopt the usual decomposition, and write down the weight-dependent xc functional as
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@ -447,8 +453,12 @@ For the sake of clarity, the explicit expression of the VWN5 functional is not r
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Equation \eqref{eq:becw} can be recast
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Equation \eqref{eq:becw} can be recast
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\begin{equation}
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\begin{equation}
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\label{eq:eLDA}
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\label{eq:eLDA}
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\begin{split}
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\be{\xc}{\ew{}}(\n{}{})
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\be{\xc}{\ew{}}(\n{}{})
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= \e{\xc}{\LDA}(\n{}{}) + \ew{} \qty[\e{\xc}{(1)}(\n{}{}) - \e{\xc}{(0)}(\n{}{})],
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& = \e{\xc}{\LDA}(\n{}{}) + \ew{} \qty[\e{\xc}{(1)}(\n{}{}) - \e{\xc}{(0)}(\n{}{})]
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\\
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& = \e{\xc}{\LDA}(\n{}{}) + \ew{} \pdv{\e{\xc}{\ew{}}(\n{}{})}{\ew{}}
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\end{split}
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\end{equation}
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\end{equation}
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which nicely highlights the centrality of the LDA in the present eDFA.
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which nicely highlights the centrality of the LDA in the present eDFA.
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In particular, $\be{\xc}{(0)}(\n{}{}) = \e{\xc}{\LDA}(\n{}{})$.
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In particular, $\be{\xc}{(0)}(\n{}{}) = \e{\xc}{\LDA}(\n{}{})$.
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@ -456,7 +466,7 @@ Consequently, in the following, we name this correlation functional ``eLDA'' as
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Also, we note that, by construction,
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Also, we note that, by construction,
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\begin{equation}
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\begin{equation}
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\label{eq:dexcdw}
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\label{eq:dexcdw}
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\left. \pdv{\be{\xc}{\ew{}}[\n{}{}]}{\ew{I}}\right|_{\n{}{} = \n{}{\ew{}}(\br)} = \be{\xc}{(I)}[\n{}{\ew{}}(\br)] - \be{\xc}{(0)}[\n{}{\ew{}}(\br)].
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\left. \pdv{\be{\xc}{\ew{}}[\n{}{}]}{\ew{}}\right|_{\n{}{} = \n{}{\ew{}}(\br)} = \be{\xc}{(1)}[\n{}{\ew{}}(\br)] - \be{\xc}{(0)}[\n{}{\ew{}}(\br)].
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\end{equation}
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\end{equation}
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This embedding procedure can be theoretically justified by the generalised adiabatic connection formalism for ensembles (GACE)
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This embedding procedure can be theoretically justified by the generalised adiabatic connection formalism for ensembles (GACE)
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@ -487,17 +497,7 @@ The bonding and antibonding orbitals of the \ce{H2} molecule are given by
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\end{subequations}
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\end{subequations}
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where $\AO{A}$ and $\AO{B}$ are the two contracted Gaussian basis functions centred on each of the nucleus, and $S_{AB} = \braket{\AO{A}}{\AO{B}}$.
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where $\AO{A}$ and $\AO{B}$ are the two contracted Gaussian basis functions centred on each of the nucleus, and $S_{AB} = \braket{\AO{A}}{\AO{B}}$.
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As reference results, we consider CID (configuration interaction with doubles) computed in the same (minimal) basis set.
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The HF energies of the ground state and the doubly-excited states are
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The CID energies of the ground state and doubly-excited states are provided by the eigenvalues of the following CID matrix:
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\begin{equation}
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\bH_\CID =
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\begin{pmatrix}
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\E{\HF}{(0)} & \eK{12}
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\\
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\eK{12} & \E{\HF}{(1)}
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\end{pmatrix},
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\end{equation}
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with
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\begin{subequations}
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\begin{subequations}
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\begin{align}
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\begin{align}
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\label{eq:HF0}
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\label{eq:HF0}
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@ -507,7 +507,7 @@ with
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\E{\HF}{(1)} & = 2 \eHc{2} + 2 \eJ{22} - \eK{22},
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\E{\HF}{(1)} & = 2 \eHc{2} + 2 \eJ{22} - \eK{22},
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\end{align}
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\end{align}
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\end{subequations}
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\end{subequations}
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and
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with
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\begin{subequations}
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\begin{subequations}
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\begin{align}
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\begin{align}
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\eHc{p} & = \int \MO{p}{}(\br{}) \qty[-\frac{\nabla^2}{2} + \vext(\br{})] \MO{p}{}(\br{})d\br{},
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\eHc{p} & = \int \MO{p}{}(\br{}) \qty[-\frac{\nabla^2}{2} + \vext(\br{})] \MO{p}{}(\br{})d\br{},
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@ -518,8 +518,27 @@ and
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\end{align}
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\end{align}
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\end{subequations}
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\end{subequations}
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Note that, in the HF case, there is no self-interaction error as $\eJ{pp} = \eK{pp}$.
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Note that, in the HF case, there is no self-interaction error as $\eJ{pp} = \eK{pp}$.
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We also define the HF excitation energy as $\Ex{\HF}{(1)} = \E{\HF}{(1)} - \E{\HF}{(0)}$.
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The HF orbital energies are
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\begin{subequations}
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\begin{align}
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\eps{1}{\HF} & = \eHc{1} + 2\eJ{11} - \eK{11},
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\\
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\eps{2}{\HF} & = \eHc{2} + 2\eJ{12} - \eK{12}.
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\end{align}
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\end{subequations}
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The CID energies are explicitly given by
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As reference results, we consider CID (configuration interaction with doubles) computed in the same (minimal) basis set.
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The CID energies of the ground state and doubly-excited states are provided by the eigenvalues of the following CID matrix:
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\begin{equation}
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\bH_\CID =
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\begin{pmatrix}
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\E{\HF}{(0)} & \eK{12}
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\\
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\eK{12} & \E{\HF}{(1)}
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\end{pmatrix},
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\end{equation}
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These CID energies are explicitly given by
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\begin{subequations}
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\begin{subequations}
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\begin{align}
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\begin{align}
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\E{\CID}{(0)} & = \frac{\E{\HF}{(0)} + \E{\HF}{(1)}}{2} - \frac{1}{2} \sqrt{\qty(\E{\HF}{(1)} - \E{\HF}{(0)})^2 + 4 \eK{12}^2},
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\E{\CID}{(0)} & = \frac{\E{\HF}{(0)} + \E{\HF}{(1)}}{2} - \frac{1}{2} \sqrt{\qty(\E{\HF}{(1)} - \E{\HF}{(0)})^2 + 4 \eK{12}^2},
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@ -549,6 +568,14 @@ with
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\n{}{(1)}(\br{}) & = 2 \MO{2}{2}(\br{}),
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\n{}{(1)}(\br{}) & = 2 \MO{2}{2}(\br{}),
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\end{align}
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\end{align}
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Note that, contrary to the HF case, self-interaction is present in LDA.
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Note that, contrary to the HF case, self-interaction is present in LDA.
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The KS orbital energies are given by
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\begin{subequations}
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\begin{align}
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\eps{1}{\LDA} & = \eHc{1} + 2\eJ{11} + \ldots,
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\\
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\eps{2}{\LDA} & = \eHc{2} + 2\eJ{12} + \ldots.
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\end{align}
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\end{subequations}
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At the eLDA, we have
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At the eLDA, we have
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\begin{subequations}
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\begin{subequations}
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@ -562,12 +589,24 @@ At the eLDA, we have
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\end{subequations}
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\end{subequations}
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with $\be{\xc}{(0)}(\n{}{}) \equiv \e{\xc}{\LDA}(\n{}{})$ and $\be{\xc}{(1)}(\n{}{}) = \e{\xc}{\LDA}(\n{}{}) + \e{\xc}{(1)}(\n{}{}) - \e{\xc}{(0)}(\n{}{})$.
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with $\be{\xc}{(0)}(\n{}{}) \equiv \e{\xc}{\LDA}(\n{}{})$ and $\be{\xc}{(1)}(\n{}{}) = \e{\xc}{\LDA}(\n{}{}) + \e{\xc}{(1)}(\n{}{}) - \e{\xc}{(0)}(\n{}{})$.
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%\titou{Note that we do not consider symmetry-broken solutions.}
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Interestingly here, there is a strong connection between the LDA and eLDA excitation energies:
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Interestingly here, there is a strong connection between the LDA and eLDA excitation energies:
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\begin{equation}
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\begin{equation}
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\Ex{\eLDA}{(1)} = \Ex{\LDA}{(1)} + \int \qty( \e{\xc}{(1)} - \e{\xc}{(0)} )[\n{}{(1)}(\br{})] \n{}{(1)}(\br{}) d\br{}.
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\begin{split}
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\Ex{\eLDA}{(1)}
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& = \Ex{\LDA}{(1)} + \int \qty( \e{\xc}{(1)} - \e{\xc}{(0)} )[\n{}{(1)}(\br{})] \n{}{(1)}(\br{}) d\br{}.
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\\
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& = \Ex{\LDA}{(1)} + \int \left. \pdv{\e{\xc}{\ew{}}[\n{}{}]}{\ew{}} \right|_{\n{}{} = \n{}{(1)}(\br{})} \n{}{(1)}(\br{}) d\br{}.
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\end{split}
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\end{equation}
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\end{equation}
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The KS orbital energies are given by
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\begin{subequations}
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\begin{align}
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\eps{1}{\eLDA} & = \eHc{1} + 2\eJ{11} + \ldots,
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\\
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\eps{2}{\eLDA} & = \eHc{2} + 2\eJ{12} + \ldots.
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\end{align}
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\end{subequations}
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These equations can be combined to define three ensemble energies
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These equations can be combined to define three ensemble energies
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\begin{subequations}
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\begin{subequations}
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@ -583,16 +622,21 @@ These equations can be combined to define three ensemble energies
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\end{align}
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\end{align}
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\end{subequations}
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\end{subequations}
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which are all, by construction, linear with respect to $\ew{}$.
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which are all, by construction, linear with respect to $\ew{}$.
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Excitation energies can be easily extracted from these formulae via differenciation with respect to $\ew{}$.
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These energies given in Eqs.~\eqref{eq:EwHF}, \eqref{eq:EwLDA} and \eqref{eq:EweLDA} can also be obtained directly from the ensemble density $\n{}{\ew{}} = (1-\ew{}) \n{}{(0)} + \ew{} \n{}{(1)}$.
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Similar energies than the ones given in Eqs.~\eqref{eq:EwHF}, \eqref{eq:EwLDA} and \eqref{eq:EweLDA} can also be obtained directly from the ensemble density
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\begin{equation}
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\n{}{\ew{}} = (1-\ew{}) \n{}{(0)} + \ew{} \n{}{(1)}.
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\end{equation}
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(This is what one would do in practice, \ie, by performing a KS ensemble calculation.)
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(This is what one would do in practice, \ie, by performing a KS ensemble calculation.)
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We will label these energies as $\bE{}{\ew{}}$.
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We will label these energies as $\bE{}{\ew{}}$ to avoid confusion.
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\begin{widetext}
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\begin{widetext}
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For HF, we have
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For HF, we have
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\begin{equation}
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\begin{equation}
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\label{eq:bEwHF}
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\begin{split}
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\begin{split}
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\bE{\HF}{\ew{}}
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\bE{\HF}{\ew{}}
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& = \int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}
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& = \titou{\int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}}
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+ \frac{1}{2} \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
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+ \frac{1}{2} \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
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\\
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\\
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& = 2 (1-\ew{}) \eHc{1} + 2 \ew{} \eHc{2}
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& = 2 (1-\ew{}) \eHc{1} + 2 \ew{} \eHc{2}
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@ -602,9 +646,10 @@ For HF, we have
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which is clearly quadratic with respect to $\ew{}$ due to the ghost interaction error in the Hartree term.
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which is clearly quadratic with respect to $\ew{}$ due to the ghost interaction error in the Hartree term.
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In the case of the LDA, it reads
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In the case of the LDA, it reads
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\begin{equation}
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\begin{equation}
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\label{eq:bEwLDA}
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\begin{split}
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\begin{split}
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\bE{\LDA}{\ew{}}
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\bE{\LDA}{\ew{}}
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& = \int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}
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& = \titou{\int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}}
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+ \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
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+ \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
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+ \int \e{\xc}{\LDA}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{}
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+ \int \e{\xc}{\LDA}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{}
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\\
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\\
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@ -618,9 +663,10 @@ In the case of the LDA, it reads
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which is also clearly quadratic with respect to $\ew{}$ because the (weight-independent) LDA functional cannot compensate the ``quadraticity'' of the Hartree term.
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which is also clearly quadratic with respect to $\ew{}$ because the (weight-independent) LDA functional cannot compensate the ``quadraticity'' of the Hartree term.
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For eLDA, the ensemble energy can be decomposed as
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For eLDA, the ensemble energy can be decomposed as
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\begin{equation}
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\begin{equation}
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\label{eq:bEweLDA}
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\begin{split}
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\begin{split}
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\bE{\eLDA}{\ew{}}
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\bE{\eLDA}{\ew{}}
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& = \int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}
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& = \titou{\int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}}
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+ \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
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+ \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
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+ \int \be{\xc}{\ew{}}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{}
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+ \int \be{\xc}{\ew{}}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{}
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\\
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\\
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@ -646,12 +692,22 @@ which \textit{could} be linear with respect to $\ew{}$ if the weight-dependent x
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This would be, for example, the case with the exact xc functional.
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This would be, for example, the case with the exact xc functional.
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\end{widetext}
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\end{widetext}
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Extracting excitation energies from Eqs.~\eqref{eq:bEwHF}, \eqref{eq:bEwLDA} and \eqref{eq:bEweLDA} is more tricky.
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To do so, we will employ Eq.~\eqref{eq:dEdw}.
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The derivative discontinuity, modelled by the last term of the RHS of Eq.~\eqref{eq:dEdw} and only non-zero in the case of an explicitly weight-dependent functional, is straightforward to compute in our case [see Eq.~\eqref{eq:dexcdw}].
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\begin{align}
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\Eps{0}{\ew{}} & = 2 \eHc{1} + \ldots,
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\\
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\Eps{1}{\ew{}} & = 2 \eHc{2} + \ldots.
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\end{align}
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%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%
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%%% CONCLUSION %%%
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%%% CONCLUSION %%%
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%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%
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\section{Conclusion}
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\section{Conclusion}
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\label{sec:ccl}
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\label{sec:ccl}
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As concluding remarks, we would like to say that what we have done is awesome.
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As concluding remarks, we would like to say that what we have done, we think, is awesome.
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%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%
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%%% ACKNOWLEDGEMENTS %%%
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%%% ACKNOWLEDGEMENTS %%%
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