Equations for third order
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Notes/Notes.tex
130
Notes/Notes.tex
@ -415,7 +415,7 @@ A general upfolded MBPT matrix can be written as
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Using SRG language, we define the diagonal and off-diagonal parts as
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\begin{equation}
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\label{eq:H_MBPT_partitioning}
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H(0) =
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\bH(0) =
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\begin{pmatrix}
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\bF{}{} & \bO \\
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\bO & \bC{}{}
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@ -440,7 +440,7 @@ we could also have defined it like this
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\bV{}{\dagger} & \bC{\text{od}}{}
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\end{pmatrix}
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\end{equation}
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However, in the end this choice was not wise and the reason is explained why in Appendix~\ref{sec:partitioning}.
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However, in the end this alternative choice was not judicious and the reason is explained why in Appendix~\ref{sec:partitioning}.
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The initial conditions corresponding to the first partitioning are
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\begin{align}
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@ -521,8 +521,8 @@ Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
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\end{align}
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\begin{align}
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\dv{\bH^{(2)}}{s} &= \comm{\bEta{2}}{\bHd{0}} + \comm{\bEta{1}}{\bHd{1}} + \comm{\bEta{0}}{\bHd{2}} \\
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&= \comm{\bEta{2}}{\bHd{0}} \notag
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\dv{\bH^{(2)}}{s} &= \comm{\bEta{2}}{\bH^{(0)}} + \comm{\bEta{1}}{\bH^{(1)}} + \comm{\bEta{0}}{\bH^{(2)}} \\
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&= \comm{\bEta{2}}{\bHd{0}} + \comm{\bEta{1}}{\bHod{1}} \notag
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\end{align}
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\begin{align}
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@ -540,6 +540,25 @@ Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
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\bEta{3} = \comm{\bHd{0}}{\bHod{3}} + \comm{\bHd{2}}{\bHod{1}}
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\end{equation}
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We will show after that $\bV{}{(2)} = \bO$, and hence $\bEta{2} = 0$, but we already use this result to simplify derivations
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\begin{equation}
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\dv{\bH^{(3)}}{s} = \comm{\bEta{3}}{\bHd{0}} + \comm{\bEta{1}}{\bHd{2}}
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\end{equation}
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\begin{align}
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\dv{\bF{}{(3)}}{s} &= \bO \\
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\dv{\bC{}{(3)}}{s} &= \bO\\
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\dv{\bV{}{(3)}}{s} &= 2 \bF{}{(0)}\bV{}{(3)}\bC{}{(0)} - (\bF{}{(0)})^2\bV{}{(3)} - \bV{}{(3)}(\bC{}{(0)})^2 \\
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&+ 2 \bF{}{(0)}\bV{}{(1)}\bC{}{(2)} + 2 \bF{}{(2)}\bV{}{(1)}\bC{}{(0)} \notag \\
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&- \left( \bF{}{(0)}\bF{}{(2)} + \bF{}{(2)}\bF{}{(0)} \right)\bV{}{(1)} \notag \\
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&- \bV{}{(1)}\left( \bC{}{(0)}\bC{}{(2)} + \bC{}{(2)}\bC{}{(0)} \right) \notag \\
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\dv{\bV{}{(3),\dagger}}{s} &= 2 \bC{}{(0)}\bV{}{(3),\dagger}\bF{}{(0)} - \bV{}{(3),\dagger}(\bF{}{(0)})^2 - (\bC{}{(0)})^2\bV{}{(3),\dagger} \\
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&+ 2 \bC{}{(0)}\bV{}{(1),\dagger}\bF{}{(2)} + 2 \bC{}{(2)}\bV{}{(1),\dagger}\bF{}{(0)} \notag \\
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&- \bV{}{(1),\dagger}\left( \bF{}{(0)}\bF{}{(2)} + \bF{}{(2)}\bF{}{(0)} \right) \notag \\
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&-\left( \bC{}{(0)}\bC{}{(2)} + \bC{}{(2)}\bC{}{(0)} \right)\bV{}{(1),\dagger} \notag
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\end{align}
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%///////////////////////////%
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\subsubsection{Forth order differential equations}
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% ///////////////////////////%
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@ -548,13 +567,17 @@ Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
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\bEta{4} = \comm{\bHd{0}}{\bHod{4}} + \comm{\bHd{2}}{\bHod{2}} + \comm{\bHd{3}}{\bHod{1}}
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\end{equation}
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\begin{equation}
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\dv{\bH^{(4)}}{s} = \comm{\bEta{4}}{\bHd{0}} + \comm{\bEta{2}}{\bHd{2}} + \comm{\bEta{1}}{\bHd{3}}
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\end{equation}
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%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Integrating order by order}
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%%%%%%%%%%%%%%%%%%%%%%
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In the following, upper case indices correspond to the 2h1p and 2p1h sectors while lower case indices correspond to the 1h and 1p sectors. Also the $\Delta\epsilon_R$ corresponds to the diagonal elements of the 2h1p and 2p1h sectors.
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%///////////////////////////%
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\subsubsection{First order Hamiltonian elements}
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%///////////////////////////%
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\begin{align}
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\dv{\bF{}{(1)}}{s} &= \bO \Longleftrightarrow \bF{}{(1)}(s) = \bF{}{(1)}(0) \Longleftrightarrow \color{red}{\boxed{\color{black}{\bF{}{(1)}(s)= \bO}}} \\
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@ -562,16 +585,16 @@ In the following, upper case indices correspond to the 2h1p and 2p1h sectors whi
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\dv{\bV{}{(1),\dagger}}{s} &= 2 \bC{}{(0)}\bV{}{(1),\dagger}\bF{}{(0)} - \bV{}{(1),\dagger}(\bF{}{(0)})^2 - (\bC{}{(0)})^2\bV{}{(1),\dagger} \\
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\dv{\bC{}{(1)}}{s} &= \bO \Longleftrightarrow \bC{}{(1)}(s) = \bC{}{(1)}(0) \Longleftrightarrow \color{red}{\boxed{\color{black}{\bC{}{(1)}(s)= \bO}}}
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\end{align}
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The differential equation for the coupling blocks can be solved in the GF(2) case because in this case $\bC{}{}(0)$ is diagonal (see Appendix~\ref{sec:diagC}).
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Inspired by this remark we transform $\bC{}{(0)}$ to its diagonal representation using
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The differential equation for the coupling blocks can be solved in the GF(2) case because $\bC{}{\GF}(0)$ is diagonal.
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Inspired by this fact we transform $\bC{}{(0)}$ to its diagonal representation using
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$\bC{}{(0)} = \bU \bD^{(0)} \bU^{-1}$ and insert it in the differential equation for $\bV{}{(1)}$
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\begin{align}
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\dv{\bV{}{(1)}}{s} &= 2 \bF{}{(0)}\bV{}{(1)}\bU \bD^{(0)} \bU^{-1}- (\bF{}{(0)})^2\bV{}{(1)} - \bV{}{(1)}\bU (\bD^{(0)})^2 \bU^{-1}\\
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\dv{\bV{}{(1)}}{s}\bU &= 2 \bF{}{(0)}\bV{}{(1)}\bU \bD^{(0)} - (\bF{}{(0)})^2\bV{}{(1)} \bU - \bV{}{(1)}\bU (\bD^{(0)})^2 \\
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\dv{\bW^{(1)}}{s} &= 2 \bF{}{(0)}\bW^{(1)} \bD^{(0)} - (\bF{}{(0)})^2\bW^{(1)} - \bW^{(1)} (\bD^{(0)})^2
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\end{align}
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where in the last line we have defined the matrix of screened integral.
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Note that in the GF(2) case $\bU = \mathbb{1}$ and thus $\bW = \bV{}{}$.
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where in the last line we have defined the matrix of screened integral $\bW^{(1)}=\bV{}{(1)} \bU$.
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Note that in the GF(2) case $\bU = \mathbb{1}$ and thus $\bW^{(1)}=\bV{}{(1)}$.
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\begin{align}
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&(\dv{\bW^{(1)}}{s})_{p,(q,v)} = (2 \bF{}{(0)}\bW^{(1)} \bD^{(0)} - (\bF{}{(0)})^2\bW^{(1)} - \bW^{(1)} (\bD^{(0)})^2)_{p,(q,v)} \notag \\
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&= \sum_{r,(s,x)} 2 F_{pr}^{(0)}W_{r,(s,x)}^{(1)}D_{(s,x),(q,v)}^{(0)} - \sum_{r,s} F_{pr}^{(0)} F_{rs}^{(0)}W_{s,(q,v)}^{(1)} \notag \\
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@ -581,10 +604,12 @@ Note that in the GF(2) case $\bU = \mathbb{1}$ and thus $\bW = \bV{}{}$.
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\end{align}
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This equation can be integrated to give
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\begin{equation}
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W_{p,(q,v)}^{(1)}(s) = W_{p,(q,v)}^{(1)}(0) e^{- (F_{pp}^{(0)} - D_{(q,v),(q,v)}^{(0)})^2 s}
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\color{red}{\boxed{\color{black}{W_{p,(q,v)}^{(1)}(s) = W_{p,(q,v)}^{(1)}(0) e^{- (F_{pp}^{(0)} - D_{(q,v),(q,v)}^{(0)})^2 s}}}}
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\end{equation}
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%///////////////////////////%
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\subsubsection{Second order Hamiltonian elements}
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%///////////////////////////%
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\begin{align}
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\dv{\bF{}{(2)}}{s} &= \bF{}{(0)}\bV{}{(1)}\bV{}{(1),\dagger} + \bV{}{(1)}\bV{}{(1),\dagger}\bF{}{(0)} - 2 \bV{}{(1)}\bC{}{(0)}\bV{}{(1),\dagger} \\
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@ -646,6 +671,43 @@ Which finally gives
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\end{itemize}
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The equation for $\bV{}{(2)}$ and $\bV{}{(2),\dagger}$ are the same that for their first order counterpart.
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So we define $\bW^{(2)} = \bV{}{(2)} \bU$ and we obtain
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\begin{equation}
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W_{p,(q,v)}^{(2)}(s) = W_{p,(q,v)}^{(2)}(0) e^{- (F_{pp}^{(0)} - D_{(q,v),(q,v)}^{(0)})^2 s} = 0
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\end{equation}
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Therefore, we have
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\begin{align}
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&\color{red}{\boxed{\color{black}{\bV{}{(2)}(s)= \bO}}} & &\color{red}{\boxed{\color{black}{\bV{}{(2),\dagger}(s)= \bO}}}
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\end{align}
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Finally, the elements of $\bC{}{(2)}$ can be obtained exactly like their $\bF{}{(2)}$ counterpart just by swapping the role of $\bF{}{(0)}$ and $\bC{}{(0)}$ in the formula.
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%///////////////////////////%
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\subsubsection{Third order Hamiltonian elements}
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%///////////////////////////%
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\begin{align}
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\dv{\bF{}{(3)}}{s} &= \bO \Longleftrightarrow \bF{}{(3)}(s) = \bF{}{(3)}(0) \Longleftrightarrow \color{red}{\boxed{\color{black}{\bF{}{(3)}(s)= \bO}}} \\
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\dv{\bC{}{(3)}}{s} &= \bO \Longleftrightarrow \bC{}{(1)}(s) = \bC{}{(1)}(0) \Longleftrightarrow \color{red}{\boxed{\color{black}{\bC{}{(3)}(s)= \bO}}}
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\end{align}
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The homogeneous equations for $\bV{}{(3)}$ and $\bV{}{(3), \dagger}$ give homogeneous solutions equal to zero for the same reasons as $\bV{}{(2)}$ and $\bV{}{(2), \dagger}$.
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Therefore the differential equations can be simplified to
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\begin{align}
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\dv{\bV{}{(3)}}{s} &= 2 \bF{}{(0)}\bV{}{(1)}\bC{}{(2)} + 2 \bF{}{(2)}\bV{}{(1)}\bC{}{(0)} \notag \\
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&- \left( \bF{}{(0)}\bF{}{(2)} + \bF{}{(2)}\bF{}{(0)} \right)\bV{}{(1)} \notag \\
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&- \bV{}{(1)}\left( \bC{}{(0)}\bC{}{(2)} + \bC{}{(2)}\bC{}{(0)} \right) \notag \\
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\dv{\bV{}{(3),\dagger}}{s} &= 2 \bC{}{(0)}\bV{}{(1),\dagger}\bF{}{(2)} + 2 \bC{}{(2)}\bV{}{(1),\dagger}\bF{}{(0)} \notag \\
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&- \bV{}{(1),\dagger}\left( \bF{}{(0)}\bF{}{(2)} + \bF{}{(2)}\bF{}{(0)} \right) \notag \\
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&-\left( \bC{}{(0)}\bC{}{(2)} + \bC{}{(2)}\bC{}{(0)} \right)\bV{}{(1),\dagger} \notag
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\end{align}
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and can be solved by integration of the right side using the previous analytic formula for Hamiltonian elements.
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%///////////////////////////%
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\subsubsection{Forth order Hamiltonian elements}
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%///////////////////////////%
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%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Downfolding the SRG-transformed matrix}
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%%%%%%%%%%%%%%%%%%%%%%
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@ -688,7 +750,7 @@ To zeroth order in the coupling we recover the HF quasiparticle energies which m
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Now turning to the first non-zero correction to the quasiparticle equation which is the second order equation
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\begin{align}
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\bSig^{(2)} (\omega) &= \bV{}{\hhp,(1)} \bU^{\hhp} \text{diag}(\frac{1}{\omega - D_{(i,v),(i,v)}}) (\bU^{\hhp})^{-1} (\bV{}{\hhp,(1)})^{\mathsf{T}} \notag \\
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\bSig_c^{(2)} (\omega) &= \bV{}{\hhp,(1)} \bU^{\hhp} \text{diag}(\frac{1}{\omega - D_{(i,v),(i,v)}}) (\bU^{\hhp})^{-1} (\bV{}{\hhp,(1)})^{\mathsf{T}} \notag \\
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&+ \bV{}{\pph,(1)} \bU^{\pph} \text{diag}(\frac{1}{\omega - D_{(a,v),(a,v)}}) (\bU^{\pph})^{-1} (\bV{}{\pph,(1)})^{\mathsf{T}} \notag \\
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&= \bW^{\hhp,(1)} \text{diag}(\frac{1}{\omega - D_{(i,v),(i,v)}}) (\bW^{\hhp,(1)})^{\mathsf{T}} \notag \\
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&+ \bW^{\pph,(1)} \text{diag}(\frac{1}{\omega - D_{(a,v),(a,v)}}) (\bW^{\pph,(1)})^{\mathsf{T}} \notag
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@ -700,12 +762,12 @@ Now turning to the first non-zero correction to the quasiparticle equation which
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In the GF(2) case we have $D_{(i,v),(i,v)} = D_{ija,ija} = \epsilon_i + \epsilon_j - \epsilon_a$ and $D_{(a,v),(a,v)} = D_{iab,iab} = \epsilon_a + \epsilon_b - \epsilon_i $ and the $W$ matrix elements have been defined in Eq.~(\ref{eq:GF2_sERI}).
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\begin{align}
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(\bSig^{(2)} (\omega,s))_{pq} &= \sum_{ija} W_{p,i[ja]}^{(1)} \frac{1}{\omega - D_{ija,ija}}(W^{\mathsf{T}})_{i[ja],q}^{(1)} \notag \\
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(\bSig_c^{(2)} (\omega,s))_{pq} &= \sum_{ija} W_{p,i[ja]}^{(1)} \frac{1}{\omega - D_{ija,ija}}(W^{\mathsf{T}})_{i[ja],q}^{(1)} \notag \\
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&+ \sum_{iab} W_{p,[ia]b}^{(1)} \frac{1}{\omega - D_{iab,iab}}(W^{\mathsf{T}})_{[ia]b,q}^{(1)} \notag \\
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&= \sum_{ija} \frac{W_{pa,ij}^{(1)} W_{qa,ij}^{(1)} }{\omega - (\epsilon_i + \epsilon_j - \epsilon_a)} \notag \\
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&+ \sum_{iab} \frac{W_{pi,ab}^{(1)} W_{qi,ab}^{(1)}}{\omega - (\epsilon_a + \epsilon_b - \epsilon_i)} \notag \\
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&= \sum_{ija} \frac{W_{pa,ij}^{(0)} W_{qa,ij}^{(0)} }{\omega - (\epsilon_i + \epsilon_j - \epsilon_a)} e^{-(\epsilon_p - \Delta_{ij}^a)^2s}e^{-(\epsilon_q - \Delta_{ij}^a)^2s} \notag \\
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&+ \sum_{iab} \frac{W_{pi,ab}^{(0)} W_{qi,ab}^{(0)} }{\omega - (\epsilon_a + \epsilon_b - \epsilon_i)} e^{-(\epsilon_p - \Delta_{i}^{ab})^2s}e^{-(\epsilon_q - \Delta_{i}^{ab})^2s} \notag
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&= \sum_{ija} \frac{W_{pa,ij}^{(1)}(0) W_{qa,ij}^{(1)}(0) }{\omega - (\epsilon_i + \epsilon_j - \epsilon_a)} e^{-(\epsilon_p - \Delta_{ij}^a)^2s}e^{-(\epsilon_q - \Delta_{ij}^a)^2s} \notag \\
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&+ \sum_{iab} \frac{W_{pi,ab}^{(1)}(0) W_{qi,ab}^{(1)}(0) }{\omega - (\epsilon_a + \epsilon_b - \epsilon_i)} e^{-(\epsilon_p - \Delta_{i}^{ab})^2s}e^{-(\epsilon_q - \Delta_{i}^{ab})^2s} \notag
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\end{align}
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\item \textbf{GW}
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@ -714,8 +776,8 @@ Now turning to the first non-zero correction to the quasiparticle equation which
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\begin{align}
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\label{eq:SRGGW_selfenergy}
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\Sigma_{pq}^{\GW}(\omega) &= \sum_{iv} \frac{W_{pi,v}^{(0)} W_{qi,v}^{(0)}}{\omega - \epsilon_i + \Omega_{v}^{\dRPA} - \ii \eta}e^{-(\epsilon_p - \epsilon_i + \Omega_v)^2s} e^{-(\epsilon_q - \epsilon_i + \Omega_v)^2s} \notag \\
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&+ \sum_{av} \frac{W_{pa,v}^{(0)} W_{qa,v}^{(0)} }{\omega - \epsilon_a - \Omega_{v}^{\dRPA} + \ii \eta} e^{-(\epsilon_p - \epsilon_a - \Omega_v)^2s}e^{-(\epsilon_q - \epsilon_a - \Omega_v)^2s} \notag
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(\Sigma_c^{\GW}(\omega,s))_{pq} &= \sum_{iv} \frac{W_{pi,v}^{(1)}(0) W_{qi,v}^{(1)}(0)}{\omega - \epsilon_i + \Omega_{v}^{\dRPA} - \ii \eta}e^{-(\epsilon_p - \epsilon_i + \Omega_v)^2s} e^{-(\epsilon_q - \epsilon_i + \Omega_v)^2s} \notag \\
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&+ \sum_{av} \frac{W_{pa,v}^{(1)}(0) W_{qa,v}^{(1)}(0)}{\omega - \epsilon_a - \Omega_{v}^{\dRPA} + \ii \eta} e^{-(\epsilon_p - \epsilon_a - \Omega_v)^2s}e^{-(\epsilon_q - \epsilon_a - \Omega_v)^2s} \notag
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\end{align}
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\item \textbf{GT}
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@ -724,17 +786,18 @@ Now turning to the first non-zero correction to the quasiparticle equation which
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\end{itemize}
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Now that we have all first order blocks and $\bF{}{(2)}$ in analytical form we have every ingredients for the second order quasi-particle equation.
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Now that we have analytical expression for the second order correlation part of the self-energy as well as for $\bF{}{(2)}$, we have every ingredients for the second order quasi-particle equation.
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In the previous formula we can see that the diagonal elements at $s \to \infty$ correspond to the same values as in the usual diagonal static approximation.
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Therefore, the SRG as a renormalization group method removes the coupling $\bV{}{}$ while incorporating some of its physics in the non-coupled problem $\bF{}{}$.
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Therefore, the SRG as a true renormalization group method removes the coupling $\bV{}{}$ while incorporating some of its physics in the non-coupled problem $\bF{}{}$.
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This formalism gives us a rationalization of the diagonal static approximation from a RG perspective.
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In addition, this gives us a way to define a non-diagonal static approximation which is not straightforward to define by simply looking at Eq.~(\ref{eq:GW_selfenergy}).
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The usual non-diagonal static approximation used in qsGW is
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In addition, this gives us an unambiguous way to define a non-diagonal static approximation.
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On the other hand the usual non-diagonal static self-energy used in qsGW,
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\begin{equation}
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(\Sigma_c^{qsGW})_{pq} = \frac{\Sigma_c(\epsilon_p)_{pq} + \Sigma_c(\epsilon_q)_{qp}}{2}
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\end{equation}
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is not unambiguously defined because this is not the only possible choice starting from the $GW$ self-energy.
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If we define $x=\epsilon_p^{(0)} - \epsilon_r^{(0)} \pm \Omega_v$ and $y=\epsilon_q^{(0)} - \epsilon_r^{(0)} \pm \Omega_v$ then the SRG qsGW is of the form $(x+y)/(x^2 + y^2)$ while the usual qsGW is of the form $(x+y)/2xy$.
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Note that both diagonal are of the form $1/x$ which is consistent with the fact that the SRG diagonal correspond to the usual static diagonal.
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@ -929,8 +992,7 @@ We have
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\label{sec:partitioning}
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%=================================================================%
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At this point, we aren't sure if the off-diagonal part of $\bC{}{}$ should be included or not in the off-diagonal part of the Hamiltonian $\bH_\text{od}$.
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In the following derivation, we choose to do so because the other case can be obtained simply by taking $\bC{\text{od}}{} = \boldsymbol{0}$ and $\bC{\text{d}}{} = \bC{}{}$.
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In this section we discuss the alternative partitioning of the Hamiltonian mentioned previously.
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%///////////////////////////%
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\subsubsection{First order Hamiltonian}
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@ -956,7 +1018,6 @@ Now turning to the first-order contribution to the MBPT matrix, we start by comp
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\dv{\bV{}{(1),\dagger}}{s} &= 2 \bC{\text{d}}{(0)}\bV{}{(1),\dagger}\bF{}{(0)} - \bV{}{(1),\dagger}(\bF{}{(0)})^2 - (\bC{\text{d}}{(0)})^2\bV{}{(1),\dagger} \\
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\dv{\bC{}{(1)}}{s} &= 2 \bC{\text{d}}{(0)}\bC{\text{od}}{(1)}\bC{\text{d}}{(0)}- (\bC{\text{d}}{(0)})^2\bC{\text{od}}{(1)} - \bC{\text{od}}{(1)}(\bC{\text{d}}{(0)})^2
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\end{align}
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The last two equations can be solved differently depending on the form of $\bF{}{}$ and $\bC{}{}$.
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%///////////////////////////%
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\subsubsection{Second order Hamiltonian}
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@ -994,24 +1055,7 @@ Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
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In order to choose what to do with $\bC{\text{od}}{}$ we look at the downfolded SRG quasiparticle equation.
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\begin{equation}
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\label{eq:H_SRGMBPT}
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H(s) =
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\begin{pmatrix}
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\bF{}{(0)}(0) + \bF{}{(2)}(s) & \bV{}{(1)}(s) + \bV{}{(2)}(s) \\
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\bV{}{(1),\dagger}(s) + \bV{}{(2),\dagger}(s) & \bC{}{(0)}(0) +\bC{}{(2)}(s)
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\end{pmatrix}
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\end{equation}
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\begin{widetext}
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\begin{equation}
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\left\{
|
||||
\begin{aligned}
|
||||
(\bF{}{(0)}(0) + \bF{}{(2)}(s)) \bR^{1h/1p} + (\bV{}{(1)}(s) + \bV{}{(2)}(s)) \bR^{2h1p/2p1h} &= \omega \bR^{1h/1p} \\
|
||||
(\bV{}{(1),\dagger}(s) + \bV{}{(2),\dagger}(s)) \bR^{1h/1p} + (\bC{}{(0)}(0) +\bC{}{(2)}(s) ) \bR^{2h1p/2p1h}&= \omega \bR^{2h1p/2p1h}
|
||||
\end{aligned}
|
||||
\right.
|
||||
\end{equation}
|
||||
|
||||
\begin{equation}
|
||||
(\bF{}{(0)}(0) + \bF{}{(2)}(s)) + (\bV{}{(1)}(s) + \bV{}{(2)}(s)) (\omega \mathbb{1} - \bC{\text{d}}{(0)}(0) - \bC{\text{od}}{(1)}(s) -\bC{}{(2)}(s) )^{-1} (\bV{}{(1),\dagger}(s) + \bV{}{(2),\dagger}(s)) \bR^{1h/1p} = \omega \bR^{1h/1p}
|
||||
@ -1023,10 +1067,6 @@ In order to choose what to do with $\bC{\text{od}}{}$ we look at the downfolded
|
||||
(\bF{}{(0)}(0) + \bF{}{(2)}(s)) + \bV{}{(1)}(s)(\omega \mathbb{1} - \bC{\text{d}}{(0)}(0))^{-1} \bV{}{(1),\dagger}(s) \bR^{1h/1p} = \omega \bR^{1h/1p}
|
||||
\end{equation}
|
||||
\end{widetext}
|
||||
So if we choose to put the off-diagonal part of $\bC{}{}$ in the off-diagonal $\bH{}{}$ we see that the off diagonal part of $\bC{}{}$ is not present in the second order quasi-particle equation.
|
||||
We believe that this is not desirable.
|
||||
In the following, we will integrate order by order the differential equations obtained above in the case $\bC{\text{od}}{} = \boldsymbol{0}$ and $\bC{\text{d}}{} = \bC{}{}$.
|
||||
The expression in the other case are given in Appendix~\ref{sec:diagC}.
|
||||
|
||||
% \section{Old stuff}
|
||||
|
||||
|
||||
Loading…
Reference in New Issue
Block a user