Manu: some thoughts illustrated with the Hubbard dimer model.

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Emmanuel Fromager 2020-04-25 13:26:28 +02:00
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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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