diff --git a/Manuscript/FarDFT.tex b/Manuscript/FarDFT.tex
index d15e516..eb5532f 100644
--- a/Manuscript/FarDFT.tex
+++ b/Manuscript/FarDFT.tex
@@ -480,9 +480,9 @@ We select STO-3G as minimal basis, and study the behaviour of the total energy o
 The bonding and antibonding orbitals of the \ce{H2} molecule are given by
 \begin{subequations}
 \begin{align}
-	\MO{1}{}(\br{}) & = \qty[ \AO{A}(\br{}) + \AO{B}(\br{}) ]/\sqrt{2 + S_{AB}},
+	\MO{1}{}(\br{}) & = \qty[ \AO{A}(\br{}) + \AO{B}(\br{}) ]/\sqrt{2(1 + S_{AB})},
 	\\
-	\MO{2}{}(\br{}) & = \qty[ \AO{A}(\br{}) - \AO{B}(\br{}) ]/\sqrt{2 - S_{AB}},
+	\MO{2}{}(\br{}) & = \qty[ \AO{A}(\br{}) - \AO{B}(\br{}) ]/\sqrt{2(1 - S_{AB})},
 \end{align}
 \end{subequations}
 where $\AO{A}$ and $\AO{B}$ are the two contracted Gaussian basis functions centred on each of the nucleus, and $S_{AB} = \braket{\AO{A}}{\AO{B}}$.
@@ -501,10 +501,10 @@ with
 \begin{subequations}
 \begin{align}
 	\label{eq:HF0}
-	\E{\HF}{(0)} & = \eHc{1} + 2 \eJ{11} - \eK{11},
+	\E{\HF}{(0)} & = 2 \eHc{1} + 2 \eJ{11} - \eK{11},
 	\\
 	\label{eq:HF1}
-	\E{\HF}{(1)} & = \eHc{2} + 2 \eJ{22} - \eK{22},
+	\E{\HF}{(1)} & = 2 \eHc{2} + 2 \eJ{22} - \eK{22},
 \end{align}
 \end{subequations}
 and 
@@ -536,10 +536,10 @@ At the (ground-state) LDA level (\ie, we only consider ground-state functionals)
 \begin{subequations}
 \begin{align}
 	\label{eq:LDA0}
-	\E{\LDA}{(0)} & = \eHc{1} + 2 \eJ{11} + \int \e{\xc}{\LDA}[\n{}{(0)}(\br{})] \n{}{(0)}(\br{}) d\br{},
+	\E{\LDA}{(0)} & = 2 \eHc{1} + 2 \eJ{11} + \int \e{\xc}{\LDA}[\n{}{(0)}(\br{})] \n{}{(0)}(\br{}) d\br{},
 	\\
 	\label{eq:LDA1}
-	\E{\LDA}{(1)} & = \eHc{2} + 2 \eJ{22} + \int \e{\xc}{\LDA}[\n{}{(1)}(\br{})] \n{}{(1)}(\br{}) d\br{},
+	\E{\LDA}{(1)} & = 2 \eHc{2} + 2 \eJ{22} + \int \e{\xc}{\LDA}[\n{}{(1)}(\br{})] \n{}{(1)}(\br{}) d\br{},
 \end{align}
 \end{subequations}
 with
@@ -554,13 +554,13 @@ At the eLDA, we have
 \begin{subequations}
 \begin{align}
 	\label{eq:eLDA0}
-	\E{\eLDA}{(0)} & = \eHc{1} + 2 \eJ{11} + \int \be{\xc}{(0)}[\n{}{(0)}(\br{})] \n{}{(0)}(\br{}) d\br{},
+	\E{\eLDA}{(0)} & = 2 \eHc{1} + 2 \eJ{11} + \int \be{\xc}{(0)}[\n{}{(0)}(\br{})] \n{}{(0)}(\br{}) d\br{},
 	\\
 	\label{eq:eLDA1}
-	\E{\eLDA}{(1)} & = \eHc{2} + 2 \eJ{22} + \int \be{\xc}{(1)}[\n{}{(1)}(\br{})] \n{}{(1)}(\br{}) d\br{},
+	\E{\eLDA}{(1)} & = 2 \eHc{2} + 2 \eJ{22} + \int \be{\xc}{(1)}[\n{}{(1)}(\br{})] \n{}{(1)}(\br{}) d\br{},
 \end{align}
 \end{subequations}
-with $\be{\xc}{(0)} \equiv \e{\xc}{\LDA}$ and $\be{\xc}{(1)}(\n{}{}) = \e{\xc}{\LDA}(\n{}{}) + \e{\xc}{(1)}(\n{}{}) - \e{\xc}{(0)}(\n{}{})$.
+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.}
 
@@ -592,19 +592,19 @@ For HF, we have
 \begin{equation}
 \begin{split}
 	\bE{\HF}{\ew{}} 
-	& = (1-\ew{}) \eHc{1} + \ew{} \eHc{2} 
+	& = (1-\ew{}) 2 \eHc{1} + \ew{} 2 \eHc{2} 
 	+ \frac{1}{2} \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
 	\\
 	& = \ldots
 \end{split}
 \end{equation}
-which is clearly quadratic with respect to $\ew{}$.
+which is clearly quadratic with respect to $\ew{}$ due to the ghost interaction error in the Hartree term.
 
 For LDA, we have
 \begin{equation}
 \begin{split}
 	\bE{\LDA}{\ew{}} 
-	& = (1-\ew{}) \eHc{1} + \ew{} \eHc{2} 
+	& = (1-\ew{}) 2 \eHc{1} + \ew{} 2 \eHc{2} 
 	+ \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{}
@@ -618,7 +618,7 @@ For eLDA, we have
 \begin{equation}
 \begin{split}
 	\bE{\eLDA}{\ew{}} 
-	& = (1-\ew{}) \eHc{1} + \ew{} \eHc{2} 
+	& = (1-\ew{}) 2 \eHc{1} + \ew{} 2 \eHc{2} 
 	+ \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{}