Manu: done with II.A
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@ -336,27 +336,62 @@ we finally recover from Eqs.~\eqref{eq:KS_ens_density} and
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\right|_{\n{}{} = \n{\opGam{\bw}}{}}.
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\end{split}
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\eeq
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Note that, when $\bw=0$, the ensemble correlation functional becomes the
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conventional (ground-state) correlation functional $E_{\rm c}[n]$ and
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the ensemble density reduces to the ground-state one $n^{(0)}$. As a
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result,
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Eq.~(\ref{eq:exact_ener_level_dets}) can be simplified as follows:
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Note that, when $\bw=0$, the ensemble correlation functional reduces to the
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conventional (ground-state) correlation functional $E_{\rm c}[n]$. As a
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result, the regular KS-DFT expression is recovered from
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Eq.~(\ref{eq:exact_ener_level_dets}) for the ground-state energy:
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\beq
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\E{}{(0)}=\mel*{\Det{(0)}}{\hH}{\Det{(0)}} +
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\E{c}{}[\n{\Det{(0)}}{}],
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\eeq
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or, equivalently,
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\beq\label{eq:gs_ener_level_gs_lim}
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\begin{split}
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\E{}{(0)}&=\mel*{\Det{(0)}}{\hat{H}[\n{\Det{(0)}}{}]}{\Det{(0)}}
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\\
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&
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+
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\E{c}{}[\n{\Det{(0)}}{}]
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-\int \fdv{\E{c}{}[\n{\Det{(0)}}{}]}{\n{}{}(\br{})}\n{\Det{(0)}}{}(\br{})
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,
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\end{split}
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\eeq
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where the ground-state-density-functional Hamiltonian reads
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\beq
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\hat{H}[\n{\Det{(0)}}{}]=\hH+
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\sum^N_{i=1}\fdv{\E{c}{}[\n{\Det{(0)}}{}]}{\n{}{}(\br{i})}.
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\eeq
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Note that, when divided by the total number $N$ of electrons, the sum of
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the last two terms on the right-hand side of
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Eq.~(\ref{eq:gs_ener_level_gs_lim}) match the correlation component of
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Levy--Zahariev's shift in the KS potential~\cite{Levy_2014}.
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The excited-state ($I>0$) energy level expressions
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can also be simplified, thus leading to
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\beq\label{eq:excited_ener_level_gs_lim}
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\begin{split}
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\E{}{(I)}
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& = \mel*{\Det{(I)}}{\hH}{\Det{(I)}} + \E{c}{}[\n{}{(0)}]
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& = \mel*{\Det{(I)}}{\hat{H}[\n{\Det{(0)}}{}]}{\Det{(I)}}
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+
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\left.
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\pdv{\E{c}{\bw}[\n{\Det{(0)}}{}]}{\ew{I}}
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\right|_{\bw=0}.
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\\
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& + \int \fdv{\E{c}{}[\n{}{(0)}]}{\n{}{}(\br{})}
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\qty[ \n{\Det{(I)}}{}(\br{}) - \n{\opGam{\bw}}{}(\br{}) ] d\br{}
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\\
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&+
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\sum_{K>0} \qty(\delta_{IK} - \ew{K} )
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\left.
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\pdv{\E{c}{\bw}[\n{}{}]}{\ew{K}}
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\right|_{\n{}{} = \n{\opGam{\bw}}{}}.
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&
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+
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\E{c}{}[\n{\Det{(0)}}{}]
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-\int \fdv{\E{c}{}[\n{\Det{(0)}}{}]}{\n{}{}(\br{})}\n{\Det{(0)}}{}(\br{})
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\end{split}
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\eeq
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\titou{Manu, shall we mention that the last term in Eq.~\eqref{eq:exact_ener_level_dets} corresponds to the derivative discontinuity (DD)?}
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As readily seen from Eqs.~(\ref{eq:gs_ener_level_gs_lim}) and
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(\ref{eq:excited_ener_level_gs_lim}), any constant shift $\delta
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\E{c}{}[\n{\Det{(0)}}{}]/\delta n({\bf r})\rightarrow \delta
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\E{c}{}[\n{\Det{(0)}}{}]/\delta n({\bf r})+C$ in the correlation
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potential leaves the individual energy levels unchanged. As a result, in
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this context,
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the correlation derivative discontinuities that is expected to appear when the
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excitation process occurs~\cite{Levy_1995} are fully described by the ensemble
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correlation derivatives [second term on the right-hand side of
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Eq.~(\ref{eq:excited_ener_level_gs_lim})].
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%%%%%%%%%%%%%%%%
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\subsection{One-electron reduced density matrix formulation}
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