diff --git a/Manuscript/FarDFT.tex b/Manuscript/FarDFT.tex
index eb5532f..d5f1d6f 100644
--- a/Manuscript/FarDFT.tex
+++ b/Manuscript/FarDFT.tex
@@ -241,9 +241,9 @@ From the GOK-DFT ensemble energy expression in Eq.~\eqref{eq:Ew-GOK}, we obtain
 \end{equation}
 where $\Eps{I}{\bw} = \sum_{p}^{\Norb} \ON{p}{(I)} \eps{p}{\bw}$, $\eps{p}{\bw}$ is the $p$th KS orbital energy given by the ensemble KS equation
 \begin{equation}
-	\qty( -\frac{\nabla^2}{2}  + \vext(\br{}) + \fdv{\E{\Hxc}{\bw}[\n{}{}]}{\n{}{}(\br{})}) \MO{p}{\bw}(\br{}) = \eps{p}{\bw} \MO{p}{\bw}(\br{}),
+	\qty( \hHc(\br{}) + \fdv{\E{\Hxc}{\bw}[\n{}{}]}{\n{}{}(\br{})}) \MO{p}{\bw}(\br{}) = \eps{p}{\bw} \MO{p}{\bw}(\br{}),
 \end{equation}
-(where $\MO{p}{\bw}(\br{})$ 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.
+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.
 
 %%%%%%%%%%%%%%%%%%
@@ -587,15 +587,16 @@ which are all, by construction, linear 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 $\n{}{\ew{}} = (1-\ew{}) \n{}{(0)} + \ew{} \n{}{(1)}$.
 (This is what one would do in practice, \ie, by performing a KS ensemble calculation.)
 We will label these energies as $\bE{}{\ew{}}$.
-
+\begin{widetext}
 For HF, we have
 \begin{equation}
 \begin{split}
 	\bE{\HF}{\ew{}} 
-	& = (1-\ew{}) 2 \eHc{1} + \ew{} 2 \eHc{2} 
+	& = \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{}'
 	\\
-	& = \ldots
+	& = 2 (1-\ew{}) \eHc{1} + 2 \ew{} \eHc{2} 
+	+ (1-\ew{})^2 (2\eJ{11}- \eK{11}) + \ew{}^2 (2\eJ{22}- \eK{22}) + 2 (1-\ew{})\ew{} (2 \eJ{12} - \eK{12}),
 \end{split}
 \end{equation}
 which is clearly quadratic with respect to $\ew{}$ due to the ghost interaction error in the Hartree term.
@@ -604,12 +605,15 @@ For LDA, we have
 \begin{equation}
 \begin{split}
 	\bE{\LDA}{\ew{}} 
-	& = (1-\ew{}) 2 \eHc{1} + \ew{} 2 \eHc{2} 
+	& = \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{}
 	\\
-	& + \int \e{\xc}{\LDA}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{}
+	& = 2 (1-\ew{}) \eHc{1} + 2 \ew{} \eHc{2} 
+	+ 2(1-\ew{})^2 \eJ{11} + 2\ew{}^2 \eJ{22} + 4 (1-\ew{})\ew{} \eJ{12}
 	\\
-	& = \ldots ,
+	& + (1-\ew{}) \int \e{\xc}{\LDA}[\n{}{\ew{}}(\br{})] \n{}{(0)}(\br{}) d\br{}
+	+ \ew{} \int \e{\xc}{\LDA}[\n{}{\ew{}}(\br{})] \n{}{(1)}(\br{}) d\br{}
 \end{split}
 \end{equation}
 which is also clearly quadratic with respect to $\ew{}$ because the (weight-independent) LDA functional cannot compensate the ``quadraticity'' of the Hartree term.
@@ -618,16 +622,23 @@ For eLDA, we have
 \begin{equation}
 \begin{split}
 	\bE{\eLDA}{\ew{}} 
-	& = (1-\ew{}) 2 \eHc{1} + \ew{} 2 \eHc{2} 
+	& = \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{}
 	\\
-	& + \int \be{\xc}{\ew{}}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{}
+	& = 2 (1-\ew{}) \eHc{1} + 2 \ew{} \eHc{2} 
+	+ 2(1-\ew{})^2 \eJ{11} + 2\ew{}^2 \eJ{22} + 4 (1-\ew{})\ew{} \eJ{12}
 	\\
-	& = \ldots ,
+	& + (1-\ew{})^2 \int \be{\xc}{(0)}[\n{}{\ew{}}(\br{})] \n{}{(0)}(\br{}) d\br{}
+	+ \ew{}^2 \int \be{\xc}{(1)}[\n{}{\ew{}}(\br{})] \n{}{(1)}(\br{}) d\br{}
+	\\
+	& + (1-\ew{})\ew{} \int \be{\xc}{(0)}[\n{}{\ew{}}(\br{})] \n{}{(1)}(\br{}) d\br{}
+	+ \ew{}(1-\ew{}) \int \be{\xc}{(1)}[\n{}{\ew{}}(\br{})] \n{}{(0)}(\br{}) d\br{}
 \end{split}
 \end{equation}
-which *could* be linear with respect to the weight if the weight-dependent xc functional is very well constructed.
+which \textit{could} be linear with respect to the weight if the weight-dependent xc functional is very well constructed.
 This would be, for example, the case with the exact xc functional.
+\end{widetext}
 
 %%%%%%%%%%%%%%%%%%
 %%% CONCLUSION %%%