Manu: added some thoughts about the interacting case

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Emmanuel Fromager 2020-04-25 14:06:45 +02:00
parent 3941bb6ae2
commit 8095e7c9c5

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@ -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 considering $1/2\leq \ew{}\leq 1$ is meaningless. Of course, if we
employ approximate ground-state-based density-functional potentials and employ approximate ground-state-based density-functional potentials and
manage to converge the KS wavefunctions, one may obtain something 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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