starting results
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@ -40,7 +40,7 @@
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\newcommand{\hH}{\Hat{H}}
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\newcommand{\hHc}{\Hat{h}}
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\newcommand{\hT}{\Hat{T}}
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\newcommand{\bH}{\Hat{T}}
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\newcommand{\bH}{\bm{H}}
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\newcommand{\hVext}{\Hat{V}_\text{ext}}
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\newcommand{\vext}{v_\text{ext}}
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\newcommand{\hWee}{\Hat{W}_\text{ee}}
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@ -68,6 +68,8 @@
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\newcommand{\EFCI}{E_\text{FCI}}
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\newcommand{\HF}{\text{HF}}
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\newcommand{\LDA}{\text{LDA}}
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\newcommand{\eLDA}{\text{eLDA}}
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\newcommand{\CID}{\text{CID}}
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\newcommand{\Hxc}{\text{Hxc}}
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\newcommand{\Ha}{\text{H}}
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\newcommand{\ex}{\text{x}}
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@ -82,7 +84,7 @@
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\newcommand{\bGamma}[1]{\bm{\Gamma}^{#1}}
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\newcommand{\bHc}{\bm{h}}
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\newcommand{\bF}[1]{\bm{F}^{#1}}
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\newcommand{\Ex}[1]{\Omega^{#1}}
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\newcommand{\Ex}[2]{\Omega_{#1}^{#2}}
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% elements
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\newcommand{\ew}[1]{w_{#1}}
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@ -91,6 +93,8 @@
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\newcommand{\eGamma}[2]{\Gamma_{#1}^{#2}}
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\newcommand{\hGamma}[2]{\Hat{\Gamma}_{#1}^{#2}}
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\newcommand{\eHc}[1]{h_{#1}}
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\newcommand{\eJ}[1]{J_{#1}}
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\newcommand{\eK}[1]{K_{#1}}
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\newcommand{\eF}[2]{F_{#1}^{#2}}
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\newcommand{\ON}[2]{f_{#1}^{#2}}
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@ -104,6 +108,8 @@
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\newcommand{\cMO}[2]{c_{#1}^{#2}}
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\newcommand{\AO}[1]{\chi_{#1}}
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\newcommand{\RHH}{R_{\ce{H-H}}}
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% units
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\newcommand{\IneV}[1]{#1 eV}
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\newcommand{\InAU}[1]{#1 a.u.}
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@ -200,7 +206,7 @@ Degeneracies can be easily handled.
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One of the key feature of eDFT in the present context is that one can easily extract individual excitation energies from the ensemble energy via differentiation with respect to individual weights:
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\begin{equation}
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\pdv{\E{}{\bw}}{\ew{I}} = \E{}{(I)} - \E{}{(0)},
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\pdv{\E{}{\bw}}{\ew{I}} = \E{}{(I)} - \E{}{(0)} = \Ex{}{(I)},
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\end{equation}
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where the weights are normalised by setting $\ew{0} = 1 - \sum_{I \ne 0} \ew{I}$.
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@ -469,7 +475,159 @@ In the case of a homogeneous system (or equivalently within the LDA), substituti
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%%%%%%%%%%%%%%%
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\section{Results}
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\label{sec:res}
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Here, we do \ce{H2} because \ce{H2} is very interesting.
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Here, we consider as testing ground the minimal-basis \ce{H2} molecule.
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We select STO-3G as minimal basis, and study the behaviour of the total energy of \ce{H2} as a function of the internuclear distance $\RHH$ (in bohr).
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The bonding and antibonding orbitals of the \ce{H2} molecule are given by
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\begin{subequations}
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\begin{align}
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\MO{1}{}(\br{}) & = \qty[ \AO{A}(\br{}) + \AO{B}(\br{}) ]/\sqrt{2 + S_{AB}},
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\\
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\MO{2}{}(\br{}) & = \qty[ \AO{A}(\br{}) - \AO{B}(\br{}) ]/\sqrt{2 - S_{AB}},
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\end{align}
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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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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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with
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\begin{subequations}
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\begin{align}
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\label{eq:HF0}
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\E{\HF}{(0)} & = \eHc{1} + 2 \eJ{11} - \eK{11},
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\\
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\label{eq:HF1}
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\E{\HF}{(1)} & = \eHc{2} + 2 \eJ{22} - \eK{22},
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\end{align}
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\end{subequations}
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and
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\begin{subequations}
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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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\\
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\eJ{pq} & = \iint \frac{\MO{p}{}(\br{})\MO{p}{}(\br{}) \MO{q}{}(\br{}')\MO{q}{}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}',
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\\
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\eK{pq} & = \iint \frac{\MO{p}{}(\br{})\MO{q}{}(\br{}) \MO{q}{}(\br{}')\MO{p}{}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'.
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\end{align}
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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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The CID energies are explicitly given by
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\begin{subequations}
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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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\\
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\E{\CID}{(1)} & = \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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\end{align}
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\end{subequations}
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and the CID excitation energy reads
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\begin{equation}
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\Ex{\CID}{(1)} = \sqrt{\qty(\E{\HF}{(1)} - \E{\HF}{(0)})^2 + 4 \eK{12}^2} \ge \Ex{\HF}{(1)}.
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\end{equation}
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At the (ground-state) LDA level (\ie, we only consider ground-state functionals), these energies reads
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\begin{subequations}
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\begin{align}
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\label{eq:LDA0}
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\E{\LDA}{(0)} & = \eHc{1} + 2 \eJ{11} + \int \e{\xc}{\LDA}[\n{}{(0)}(\br{})] \n{}{(0)}(\br{}) d\br{},
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\\
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\label{eq:LDA1}
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\E{\LDA}{(1)} & = \eHc{2} + 2 \eJ{22} + \int \e{\xc}{\LDA}[\n{}{(1)}(\br{})] \n{}{(1)}(\br{}) d\br{},
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\end{align}
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\end{subequations}
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with
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\begin{align}
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\n{}{(0)}(\br{}) & = 2 \MO{1}{2}(\br{}),
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&
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\n{}{(1)}(\br{}) & = 2 \MO{2}{2}(\br{}),
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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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At the eLDA, we have
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\begin{subequations}
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\begin{align}
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\label{eq:eLDA0}
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\E{\eLDA}{(0)} & = \eHc{1} + 2 \eJ{11} + \int \be{\xc}{(0)}[\n{}{(0)}(\br{})] \n{}{(0)}(\br{}) d\br{},
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\\
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\label{eq:eLDA1}
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\E{\eLDA}{(1)} & = \eHc{2} + 2 \eJ{22} + \int \be{\xc}{(1)}[\n{}{(1)}(\br{})] \n{}{(1)}(\br{}) d\br{},
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\end{align}
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\end{subequations}
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with $\be{\xc}{(0)} \equiv \e{\xc}{\LDA}$ 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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\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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\end{equation}
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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{align}
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\label{eq:EwHF}
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\E{\HF}{\ew{}} & = (1-\ew{}) \E{\HF}{(0)} + \ew{} \E{\HF}{(1)},
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\\
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\label{eq:EwLDA}
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\E{\LDA}{\ew{}} & = (1-\ew{}) \E{\LDA}{(0)} + \ew{} \E{\LDA}{(1)},
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\\
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\label{eq:EweLDA}
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\E{\eLDA}{\ew{}} & = (1-\ew{}) \E{\eLDA}{(0)} + \ew{} \E{\eLDA}{(1)},
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\end{align}
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\end{subequations}
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which are all, by construction, linear 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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(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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For HF, we have
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\begin{equation}
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\begin{split}
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\bE{\HF}{\ew{}}
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& = (1-\ew{}) \eHc{1} + \ew{} \eHc{2}
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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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& = \ldots
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\end{split}
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\end{equation}
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which is clearly quadratic with respect to $\ew{}$.
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For LDA, we have
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\begin{equation}
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\begin{split}
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\bE{\LDA}{\ew{}}
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& = (1-\ew{}) \eHc{1} + \ew{} \eHc{2}
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+ \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
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\\
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& + \int \e{\xc}{\LDA}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{}
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\\
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& = \ldots ,
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\end{split}
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\end{equation}
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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, we have
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\begin{equation}
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\begin{split}
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\bE{\eLDA}{\ew{}}
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& = (1-\ew{}) \eHc{1} + \ew{} \eHc{2}
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+ \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
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\\
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& + \int \be{\xc}{\ew{}}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{}
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\\
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& = \ldots ,
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\end{split}
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\end{equation}
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which *could* be linear with respect to the weight if the weight-dependent xc functional is very well constructed.
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This would be, for example, the case with the exact xc functional.
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%%%%%%%%%%%%%%%%%%
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%%% CONCLUSION %%%
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