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@@ -459,6 +459,62 @@ is the Hxc potential, with
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+\section{Some thougths illustrated with the Hubbard dimer model}
+
+The definition of an ensemble density functional relies on the concavity
+of the ensemble energy with respect to the external potential. In the
+case of the Hubbard dimer, the singlet triensemble non-interacting
+energy (which contains both singly- and doubly-excited states) reads
+\beq
+\begin{split}
+\mathcal{E}_{\rm KS}^{\bw}\left(\Delta
+v\right)=&(1-\ew{1}-\ew{2})\mathcal{E}_0\left(\Delta 
+v\right)+\ew{1}\mathcal{E}_1\left(\Delta 
+v\right)
+\\
+&+\ew{2}\mathcal{E}_2\left(\Delta 
+v\right),
+\end{split}
+\eeq 
+where $\mathcal{E}_0\left(\Delta
+v\right)=2\varepsilon_0\left(\Delta
+v\right)$, $\mathcal{E}_1\left(\Delta
+v\right)=0$, $\mathcal{E}_2\left(\Delta
+v\right)=-2\varepsilon_0\left(\Delta
+v\right)$, and
+\beq
+\varepsilon_0\left(\Delta
+v\right)=-\sqrt{t^2+\dfrac{\Delta v^2}{4}},
+\eeq
+thus leading to
+\beq
+\mathcal{E}_{\rm KS}^{\bw}\left(\Delta
+v\right)=-2\left(1-\ew{1}-2\ew{2}\right)\sqrt{t^2+\dfrac{\Delta
+v^2}{4}}.
+\eeq
+If we ignore the single excitation ($\ew{1}=0$) and denote
+$\ew{}=\ew{2}$, the ensemble energy becomes 
+\beq
+\mathcal{E}_{\rm KS}^{\ew{}}\left(\Delta
+v\right)=-2(1-2\ew{})\sqrt{t^2+\dfrac{\Delta
+v^2}{4}}.
+\eeq
+As readily seen, it is concave only if $\ew{}\leq 1/2$. Outside the
+usual range of weight values, it is convex, thus preventing any density
+to be ensemble non-interacting $v$-representable. This statement is
+based on the Legendre--Fenchel transform expression of the
+non-interacting ensemble kinetic energy functional:
+\beq
+T^{\ew{}}_{\rm s}(n)=\sup_{\Delta
+v}\left\{\mathcal{E}_{\rm KS}^{\ew{}}\left(\Delta
+v\right)+\Delta
+v\times(n-1)\right\}.
+\eeq  
+In this simple example, ignoring the single excitation is fine. However,
+considering $1/2\leq \ew{}\leq 1$ is meaningless. Of course, if we
+employ approximate ground-state-based density-functional potentials and
+manage to converge the KS wavefunctions, one may obtain something
+interesting. But I have no idea how meaningful such a solution is.   
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