messing around with orbital energies

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Pierre-Francois Loos 2019-11-26 22:50:38 +01:00
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@ -220,7 +220,11 @@ In the KS formulation of eDFT, the universal ensemble functional (the weight-dep
\begin{equation}
\F{}{\bw}[\n{}{}] = \Ts{\bw}[\n{}{}] + \E{\Hxc}{\bw}[\n{}{}],
\end{equation}
where $\Ts{\bw}[\n{}{}]$ and $\E{\Hxc}{}[\n{}{}]$ are the noninteracting ensemble kinetic energy functional and ensemble Hartree-exchange-correlation (Hxc) functional, respectively with
where
\begin{equation}
\Ts{\bw}[\n{}{}] =
\end{equation}
and
\begin{equation}
\label{eq:exc_def}
\begin{split}
@ -230,6 +234,7 @@ where $\Ts{\bw}[\n{}{}]$ and $\E{\Hxc}{}[\n{}{}]$ are the noninteracting ensembl
& = \frac{1}{2} \iint \frac{\n{}{}(\br{}) \n{}{}(\br{}')}{\abs{\br{}-\br{}'}} d\br{} d\br{}'+ \int \e{\xc}{\bw}[\n{}{}(\br{})] \n{}{}(\br{}) d\br{}.
\end{split}
\end{equation}
are the noninteracting ensemble kinetic energy functional and ensemble Hartree-exchange-correlation (Hxc) functional, respectively.
Note that the weight-independent Hartree functional $\E{\Ha}{}[\n{}{}]$ causes the infamous ghost-interaction error (GIC) \cite{Gidopoulos_2002, Pastorczak_2014, Alam_2016, Alam_2017, Gould_2017} in eDFT, which is supposed to be cancelled by the weight-dependent xc functional $\E{\xc}{\bw}[\n{}{}]$.
From the GOK-DFT ensemble energy expression in Eq.~\eqref{eq:Ew-GOK}, we obtain \cite{Gross_1988b,Deur_2019}
@ -245,14 +250,15 @@ where $\Eps{I}{\bw} = \sum_{p}^{\Norb} \ON{p}{(I)} \eps{p}{\bw}$, $\eps{p}{\bw}$
\end{equation}
where $\hHc(\br{}) = -\frac{\nabla^2}{2} + \vext(\br{})$, $\MO{p}{\bw}(\br{})$ is a KS orbital, $\ON{p}{(I)}$ its occupancy for the state $I$, and $\n{}{\bw} = \sum_{I=0}^{\Nens-1} \ew{I} \n{}{(I)}$ is the ensemble density.
Equation \eqref{eq:dEdw} is our working equation for computing excitation energies.
%%%%%%%%%%%%%%%%%%
%%% FUNCTIONAL %%%
%%%%%%%%%%%%%%%%%%
\section{Functional}
\label{sec:func}
The present work deals with the explicit construction of the (reduced) LDA xc functional $\e{\xc}{\bw}[\n{}{}]$ defined in Eq.~\eqref{eq:exc_def}.
Here, we restrict our study to the case of a two-state ensemble (\ie, $\Nens = 2$) where both the ground state ($I=0$) and the first doubly-excited state ($I=1$) are considered.
Here, we restrict our study to the case of a two-state ensemble (\ie, $\Nens = 2$) where both the ground state ($I=0$) and the first doubly-excited state ($I=1$) are considered.
Thus, we have $0 \le \ew{} \le 1/2$.
The generalisation to a larger number of states (in particular the inclusion of the first singly-excited state) is trivial and left for future work.
We adopt the usual decomposition, and write down the weight-dependent xc functional as
@ -447,8 +453,12 @@ For the sake of clarity, the explicit expression of the VWN5 functional is not r
Equation \eqref{eq:becw} can be recast
\begin{equation}
\label{eq:eLDA}
\begin{split}
\be{\xc}{\ew{}}(\n{}{})
= \e{\xc}{\LDA}(\n{}{}) + \ew{} \qty[\e{\xc}{(1)}(\n{}{}) - \e{\xc}{(0)}(\n{}{})],
& = \e{\xc}{\LDA}(\n{}{}) + \ew{} \qty[\e{\xc}{(1)}(\n{}{}) - \e{\xc}{(0)}(\n{}{})]
\\
& = \e{\xc}{\LDA}(\n{}{}) + \ew{} \pdv{\e{\xc}{\ew{}}(\n{}{})}{\ew{}}
\end{split}
\end{equation}
which nicely highlights the centrality of the LDA in the present eDFA.
In particular, $\be{\xc}{(0)}(\n{}{}) = \e{\xc}{\LDA}(\n{}{})$.
@ -456,7 +466,7 @@ Consequently, in the following, we name this correlation functional ``eLDA'' as
Also, we note that, by construction,
\begin{equation}
\label{eq:dexcdw}
\left. \pdv{\be{\xc}{\ew{}}[\n{}{}]}{\ew{I}}\right|_{\n{}{} = \n{}{\ew{}}(\br)} = \be{\xc}{(I)}[\n{}{\ew{}}(\br)] - \be{\xc}{(0)}[\n{}{\ew{}}(\br)].
\left. \pdv{\be{\xc}{\ew{}}[\n{}{}]}{\ew{}}\right|_{\n{}{} = \n{}{\ew{}}(\br)} = \be{\xc}{(1)}[\n{}{\ew{}}(\br)] - \be{\xc}{(0)}[\n{}{\ew{}}(\br)].
\end{equation}
This embedding procedure can be theoretically justified by the generalised adiabatic connection formalism for ensembles (GACE)
@ -487,17 +497,7 @@ The bonding and antibonding orbitals of the \ce{H2} molecule are given by
\end{subequations}
where $\AO{A}$ and $\AO{B}$ are the two contracted Gaussian basis functions centred on each of the nucleus, and $S_{AB} = \braket{\AO{A}}{\AO{B}}$.
As reference results, we consider CID (configuration interaction with doubles) computed in the same (minimal) basis set.
The CID energies of the ground state and doubly-excited states are provided by the eigenvalues of the following CID matrix:
\begin{equation}
\bH_\CID =
\begin{pmatrix}
\E{\HF}{(0)} & \eK{12}
\\
\eK{12} & \E{\HF}{(1)}
\end{pmatrix},
\end{equation}
with
The HF energies of the ground state and the doubly-excited states are
\begin{subequations}
\begin{align}
\label{eq:HF0}
@ -507,7 +507,7 @@ with
\E{\HF}{(1)} & = 2 \eHc{2} + 2 \eJ{22} - \eK{22},
\end{align}
\end{subequations}
and
with
\begin{subequations}
\begin{align}
\eHc{p} & = \int \MO{p}{}(\br{}) \qty[-\frac{\nabla^2}{2} + \vext(\br{})] \MO{p}{}(\br{})d\br{},
@ -518,8 +518,27 @@ and
\end{align}
\end{subequations}
Note that, in the HF case, there is no self-interaction error as $\eJ{pp} = \eK{pp}$.
We also define the HF excitation energy as $\Ex{\HF}{(1)} = \E{\HF}{(1)} - \E{\HF}{(0)}$.
The HF orbital energies are
\begin{subequations}
\begin{align}
\eps{1}{\HF} & = \eHc{1} + 2\eJ{11} - \eK{11},
\\
\eps{2}{\HF} & = \eHc{2} + 2\eJ{12} - \eK{12}.
\end{align}
\end{subequations}
The CID energies are explicitly given by
As reference results, we consider CID (configuration interaction with doubles) computed in the same (minimal) basis set.
The CID energies of the ground state and doubly-excited states are provided by the eigenvalues of the following CID matrix:
\begin{equation}
\bH_\CID =
\begin{pmatrix}
\E{\HF}{(0)} & \eK{12}
\\
\eK{12} & \E{\HF}{(1)}
\end{pmatrix},
\end{equation}
These CID energies are explicitly given by
\begin{subequations}
\begin{align}
\E{\CID}{(0)} & = \frac{\E{\HF}{(0)} + \E{\HF}{(1)}}{2} - \frac{1}{2} \sqrt{\qty(\E{\HF}{(1)} - \E{\HF}{(0)})^2 + 4 \eK{12}^2},
@ -549,6 +568,14 @@ with
\n{}{(1)}(\br{}) & = 2 \MO{2}{2}(\br{}),
\end{align}
Note that, contrary to the HF case, self-interaction is present in LDA.
The KS orbital energies are given by
\begin{subequations}
\begin{align}
\eps{1}{\LDA} & = \eHc{1} + 2\eJ{11} + \ldots,
\\
\eps{2}{\LDA} & = \eHc{2} + 2\eJ{12} + \ldots.
\end{align}
\end{subequations}
At the eLDA, we have
\begin{subequations}
@ -562,12 +589,24 @@ At the eLDA, we have
\end{subequations}
with $\be{\xc}{(0)}(\n{}{}) \equiv \e{\xc}{\LDA}(\n{}{})$ and $\be{\xc}{(1)}(\n{}{}) = \e{\xc}{\LDA}(\n{}{}) + \e{\xc}{(1)}(\n{}{}) - \e{\xc}{(0)}(\n{}{})$.
%\titou{Note that we do not consider symmetry-broken solutions.}
Interestingly here, there is a strong connection between the LDA and eLDA excitation energies:
\begin{equation}
\Ex{\eLDA}{(1)} = \Ex{\LDA}{(1)} + \int \qty( \e{\xc}{(1)} - \e{\xc}{(0)} )[\n{}{(1)}(\br{})] \n{}{(1)}(\br{}) d\br{}.
\begin{split}
\Ex{\eLDA}{(1)}
& = \Ex{\LDA}{(1)} + \int \qty( \e{\xc}{(1)} - \e{\xc}{(0)} )[\n{}{(1)}(\br{})] \n{}{(1)}(\br{}) d\br{}.
\\
& = \Ex{\LDA}{(1)} + \int \left. \pdv{\e{\xc}{\ew{}}[\n{}{}]}{\ew{}} \right|_{\n{}{} = \n{}{(1)}(\br{})} \n{}{(1)}(\br{}) d\br{}.
\end{split}
\end{equation}
The KS orbital energies are given by
\begin{subequations}
\begin{align}
\eps{1}{\eLDA} & = \eHc{1} + 2\eJ{11} + \ldots,
\\
\eps{2}{\eLDA} & = \eHc{2} + 2\eJ{12} + \ldots.
\end{align}
\end{subequations}
These equations can be combined to define three ensemble energies
\begin{subequations}
@ -583,16 +622,21 @@ These equations can be combined to define three ensemble energies
\end{align}
\end{subequations}
which are all, by construction, linear with respect to $\ew{}$.
Excitation energies can be easily extracted from these formulae via differenciation with respect to $\ew{}$.
These energies given in Eqs.~\eqref{eq:EwHF}, \eqref{eq:EwLDA} and \eqref{eq:EweLDA} can also be obtained directly from the ensemble density $\n{}{\ew{}} = (1-\ew{}) \n{}{(0)} + \ew{} \n{}{(1)}$.
Similar energies than the ones given in Eqs.~\eqref{eq:EwHF}, \eqref{eq:EwLDA} and \eqref{eq:EweLDA} can also be obtained directly from the ensemble density
\begin{equation}
\n{}{\ew{}} = (1-\ew{}) \n{}{(0)} + \ew{} \n{}{(1)}.
\end{equation}
(This is what one would do in practice, \ie, by performing a KS ensemble calculation.)
We will label these energies as $\bE{}{\ew{}}$.
We will label these energies as $\bE{}{\ew{}}$ to avoid confusion.
\begin{widetext}
For HF, we have
\begin{equation}
\label{eq:bEwHF}
\begin{split}
\bE{\HF}{\ew{}}
& = \int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}
& = \titou{\int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}}
+ \frac{1}{2} \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
\\
& = 2 (1-\ew{}) \eHc{1} + 2 \ew{} \eHc{2}
@ -602,9 +646,10 @@ For HF, we have
which is clearly quadratic with respect to $\ew{}$ due to the ghost interaction error in the Hartree term.
In the case of the LDA, it reads
\begin{equation}
\label{eq:bEwLDA}
\begin{split}
\bE{\LDA}{\ew{}}
& = \int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}
& = \titou{\int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}}
+ \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
+ \int \e{\xc}{\LDA}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{}
\\
@ -618,9 +663,10 @@ In the case of the LDA, it reads
which is also clearly quadratic with respect to $\ew{}$ because the (weight-independent) LDA functional cannot compensate the ``quadraticity'' of the Hartree term.
For eLDA, the ensemble energy can be decomposed as
\begin{equation}
\label{eq:bEweLDA}
\begin{split}
\bE{\eLDA}{\ew{}}
& = \int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}
& = \titou{\int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}}
+ \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
+ \int \be{\xc}{\ew{}}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{}
\\
@ -646,12 +692,22 @@ which \textit{could} be linear with respect to $\ew{}$ if the weight-dependent x
This would be, for example, the case with the exact xc functional.
\end{widetext}
Extracting excitation energies from Eqs.~\eqref{eq:bEwHF}, \eqref{eq:bEwLDA} and \eqref{eq:bEweLDA} is more tricky.
To do so, we will employ Eq.~\eqref{eq:dEdw}.
The derivative discontinuity, modelled by the last term of the RHS of Eq.~\eqref{eq:dEdw} and only non-zero in the case of an explicitly weight-dependent functional, is straightforward to compute in our case [see Eq.~\eqref{eq:dexcdw}].
\begin{align}
\Eps{0}{\ew{}} & = 2 \eHc{1} + \ldots,
\\
\Eps{1}{\ew{}} & = 2 \eHc{2} + \ldots.
\end{align}
%%%%%%%%%%%%%%%%%%
%%% CONCLUSION %%%
%%%%%%%%%%%%%%%%%%
\section{Conclusion}
\label{sec:ccl}
As concluding remarks, we would like to say that what we have done is awesome.
As concluding remarks, we would like to say that what we have done, we think, is awesome.
%%%%%%%%%%%%%%%%%%%%%%%%
%%% ACKNOWLEDGEMENTS %%%