Manu: added some thoughts about the interacting case

Manuscript/FarDFT.tex

@@ -514,7 +514,36 @@ 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.   
+interesting. But I have no idea how meaningful such a solution is.\\   
+
+In the interacting case, the bi-ensemble (with the double excitation
+only) energy reads
+\beq
+%\begin{split}
+E^{\ew{}}\left(\Delta
+v\right)&=&(1-\ew{})E_0\left(\Delta
+v\right)+\ew{}E_2\left(\Delta
+v\right)
+\nonumber
+\\
+&=&(1-\ew{})E_0\left(\Delta
+v\right)+\ew{}\Big(2U-E_0\left(\Delta
+v\right)-E_1\left(\Delta
+v\right)\Big)
+\nonumber
+\\
+&=&(1-2\ew{})E_0\left(\Delta
+v\right)-\ew{}E_1\left(\Delta
+v\right)+2U\ew{}.
+%\end{split}
+\eeq
+In the vicinity of the symmetric regime ($\Delta
+v=0$), the excited-state energy is $E_1\left(\Delta
+v\right)\approx U$. In this case, the ensemble energy is concave if
+$\ew{}\leq 1/2$. One should check if $(1-2\ew{})E_0\left(\Delta
+v\right)-\ew{}E_1\left(\Delta
+v\right)$ remains concave away from this regime (I see no reason why it
+should be).
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