Manu: derived the hopefully proper expression for the individual Hx energy
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@ -92,7 +92,7 @@
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\newcommand{\manu}[1]{{\textcolor{blue}{ Manu: #1 }} }
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\newcommand{\manu}[1]{{\textcolor{blue}{ Manu: #1 }} }
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\newcommand{\beq}{\begin{eqnarray}}
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\newcommand{\beq}{\begin{eqnarray}}
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\newcommand{\eeq}{\nonumber\end{eqnarray}}
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\newcommand{\eeq}{\end{eqnarray}}
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\newcommand{\bmk}{\bm{\kappa}} % orbital rotation vector
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\newcommand{\bmk}{\bm{\kappa}} % orbital rotation vector
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\newcommand{\bmg}{\bm{\Gamma}} % orbital rotation vector
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\newcommand{\bmg}{\bm{\Gamma}} % orbital rotation vector
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\newcommand{\bfx}{\bf{x}}
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\newcommand{\bfx}{\bf{x}}
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@ -322,7 +322,11 @@ Tr}\left[{\bmg}^{(0)}{\bm h}\right]
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E^{(I)}&=&E^{{\bw}}+\sum_{K>0}\left(\delta_{IK}-w^{(K)}\right)\dfrac{\partial E^{{\bw}}}{\partial w^{(K)}}
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E^{(I)}&=&E^{{\bw}}+\sum_{K>0}\left(\delta_{IK}-w^{(K)}\right)\dfrac{\partial E^{{\bw}}}{\partial w^{(K)}}
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\nonumber\\
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\nonumber\\
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&=&
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&=&
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...+\dfrac{1}{2}\Tr(\bmg^{\bw} \, \bG \, \bmg^{\bw})
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...+\Tr(\bmg^{(I)} \, \bG \, \bmg^{\bw})
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-\dfrac{1}{2}\Tr(\bmg^{\bw} \, \bG \, \bmg^{\bw})+...
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\nonumber\\
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&=&...+
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\Tr\left[\left(\bmg^{(I)}-\dfrac{1}{2}\bmg^{\bw}\right) \, \bG \, \bmg^{\bw}\right]
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+...
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+...
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\eeq
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\eeq
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}
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}
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