did some changes to include offdiag C in the perturbative part
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@ -100,6 +100,7 @@
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\newcommand{\bD}{\boldsymbol{D}}
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\newcommand{\bF}{\boldsymbol{F}}
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\newcommand{\bU}{\boldsymbol{U}}
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\newcommand{\bR}{\boldsymbol{R}}
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\newcommand{\bV}[2]{\boldsymbol{V}_{#1}^{#2}}
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\newcommand{\bW}{\boldsymbol{W}}
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\newcommand{\bX}{\boldsymbol{X}}
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@ -134,7 +135,7 @@
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\begin{document}
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\title{Notes on the project: Similarity Renormalization Group formalism applied to Green's function theory}
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\title{Notes on the project: Perturbative Analysis of the Similarity Renormalization Group formalism applied to the electronic Hamiltonian and Green's function theory}
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\author{Antoine \surname{Marie}}
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\email{amarie@irsamc.ups-tlse.fr}
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@ -392,23 +393,23 @@ Using SRG language, we define the diagonal and off-diagonal parts as
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H(0) =
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\begin{pmatrix}
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\bF & \bO \\
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\bO & \bC{}{}
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\bO & \bC{\text{d}}{}
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\end{pmatrix}
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+ \lambda
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\begin{pmatrix}
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\bO & \bV{}{} \\
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\bV{}{\dagger} & \bO
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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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which gives the following conditions
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\begin{align}
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\bHd{0}(0) &= \begin{pmatrix}
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\bF & \bO \\
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\bO & \bC{}{}
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\bO & \bC{\text{d}}{}
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\end{pmatrix} & \bHod{0}(0) &= \bO \\
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\bHd{1}(0) &= \bO & \bHod{1}(0) &= \begin{pmatrix}
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\bO & \bV{}{} \\
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\bV{}{\dagger} & \bO
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\bV{}{\dagger} & \bC{\text{od}}{}
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\end{pmatrix}
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\end{align}
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@ -440,8 +441,8 @@ Now turning to the first-order contribution to the MBPT matrix, we start by comp
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\begin{align}
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&\bEta{1} = \comm{\bHd{0}}{\bHod{1}} \\
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&= \begin{pmatrix}
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\bO & \bF^{(0)}\bV{}{(1)} - \bV{}{(1)}\bF^{(0)}\\
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\bC{}{(0)}\bV{}{(1),\dagger} - \bV{}{(1),\dagger}\bC{}{(0)} & \bO
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\bO & \bF^{(0)}\bV{}{(1)} - \bV{}{(1)}\bC{\text{d}}{(0)}\\
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\bC{\text{d}}{(0)}\bV{}{(1),\dagger} - \bV{}{(1),\dagger} \bF^{(0)} & \bC{\text{d}}{(0)} \bC{\text{od}}{(1)} - \bC{\text{od}}{(1)} \bC{\text{d}}{(0)}
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\end{pmatrix}
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\end{align}
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@ -451,16 +452,16 @@ Now turning to the first-order contribution to the MBPT matrix, we start by comp
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\dv{\bV{}{(1),\dagger}}{s} & \dv{\bC{}{(1)}}{s}
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\end{pmatrix} \\
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\dv{\bF^{(1)}}{s} &= \bO \Longleftrightarrow \color{red}{\boxed{\color{black}{\bF^{(1)}= \bO}}} \\
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\dv{\bC{}{(1)}}{s} &= \bO \Longleftrightarrow \color{red}{\boxed{\color{black}{\bC{}{(1)}= \bO}}} \\
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\dv{\bV{}{(1)}}{s} &= 2 \bF^{(0)}\bV{}{(1)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 \\
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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{\bV{}{(1)}}{s} &= 2 \bF^{(0)}\bV{}{(1)}\bC{\text{d}}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{\text{d}}{(0)})^2 \\
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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 two last equations can be solved differently depending on the form of $\bF$ and $\bC{}{}$.
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\subsubsection*{Diagonal $\bC{}{(0)}$}
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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 $\Delta\eps_R$ corresponds to the diagonal elements of the 2h1p and 2p1h sectors.
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\begin{align}
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(\dv{\bV{}{(1)}}{s})_{pQ} &= (2 \bF^{(0)}\bV{}{(1)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 )_{pQ}\\
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(\dv{\bV{}{(1)}}{s})_{pQ} &= (2 \bF^{(0)}\bV{}{(1)}\bC{\text{d}}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{\text{d}}{(0)})^2 )_{pQ}\\
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&= \sum_{rS} 2 f^{(0)}_{pr} v^{(1)}_{rS}c^{(0)}_{SQ} - \sum_{rs} f^{(0)}_{pr} f^{(0)}_{rs} v^{(1)}_{sQ} - \sum_{RS} v^{(1)}_{pR} c^{(0)}_{RS}c^{(0)}_{SQ} \\
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&= \sum_{rS} 2 \epsilon^{(0)}_p\delta_{pr} v^{(1)}_{rS}\Delta\epsilon^{(0)}_Q\delta_{SQ} \\
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&- \sum_{rs} \epsilon^{(0)}_p\delta_{pr} \epsilon^{(0)}_r\delta_{rs} v^{(1)}_{sQ} \\
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@ -470,28 +471,13 @@ In the following, upper case indices correspond to the 2h1p and 2p1h sectors whi
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&\color{red}{\boxed{\color{black}{v^{(1)}_{pQ}(s) = v^{(1)}_{pQ}(0) e^{-s(\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2} }}}
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\end{align}
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Note the close similarity with Evangelista's expressions for the off-diagonal part at first order!
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\subsubsection*{Non-diagonal $\bC{}{(0)}$}
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We follow the same development as before
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\begin{align}
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(\dv{\bV{}{(1)}}{s})_{pQ} &= (2 \bF^{(0)}\bV{}{(1)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 )_{pQ}\\
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&= \sum_{rS} 2 f^{(0)}_{pr} v^{(1)}_{rS}c^{(0)}_{SQ} - \sum_{rs} f^{(0)}_{pr} f^{(0)}_{rs} v^{(1)}_{sQ} - \sum_{RS} v^{(1)}_{pR} c^{(0)}_{RS}c^{(0)}_{SQ} \\
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&= \sum_{rS} 2 \epsilon^{(0)}_p\delta_{pr} v^{(1)}_{rS} c^{(0)}_{SQ} \\
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&- \sum_{rs} \epsilon^{(0)}_p\delta_{pr} \epsilon^{(0)}_r\delta_{rs} v^{(1)}_{sQ} \\
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&- \sum_{RS} v^{(1)}_{pR} c^{(0)}_{RS} c^{(0)}_{SQ} \\
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&= - (\epsilon^{(0)}_p)^2v^{(1)}_{pQ}+ \sum_{S} 2 \epsilon^{(0)}_p v^{(1)}_{pS} c^{(0)}_{SQ} - \sum_{RS} v^{(1)}_{pR} c^{(0)}_{RS} c^{(0)}_{SQ}
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(\dv{\bC{}{(1)}}{s})_{PQ} &= (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)_{PQ} \\
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&= \sum_{RS} 2 c^{(0)}_{PR} c^{(1)}_{RS} c^{(0)}_{SQ} - c^{(0)}_{PR} c^{(0)}_{RS} c^{(1)}_{SQ} - c^{(1)}_{PR} c^{(0)}_{RS} c^{(0)}_{SQ} \\
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&= 2 \Delta\epsilon^{(0)}_Pc^{(1)}_{PQ}\Delta\epsilon^{(0)}_Q - (\Delta\epsilon^{(0)}_P)^2 c^{(1)}_{PQ} - c^{(1)}_{PQ} (\Delta\epsilon^{(0)}_Q)^2 \\
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&= - (\Delta\epsilon^{(0)}_P - \Delta\epsilon^{(0)}_Q )^2 c^{(1)}_{PQ} \\
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&\color{red}{\boxed{\color{black}{c^{(1)}_{PQ}(s) = c^{(1)}_{PQ}(0) e^{-s(\Delta\epsilon^{(0)}_P - \Delta\epsilon^{(0)}_Q )^2} }}}
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\end{align}
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We obtain a set of coupled differential equations which seems far from being trivial to solve.
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In order to simplify the problem we consider the case when $\bF = \eps_p$.
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\begin{align}
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\dv{\bV{}{(1)}}{s} &= 2 \bF^{(0)}\bV{}{(1)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 \\
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&= 2 \eps_p\bV{}{(1)}\bC{}{(0)} - (\eps_p)^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 \\
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&= \bV{}{(1)} (\eps_p\mathbb{1} - \bC{}{(0)})^2
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\end{align}
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Now to solve this matrix differential equation, we just need to diagonalize $(\eps_p \mathbb{1} - \bC{}{(0)})^2$.
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Fortunately, this can be easily done because the eigenvalues of $\bC{}{(0)}$ are known to be the shifted RPA eigenvalues and the eigenvectors are given in Bintrim 2021.
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\textbf{\color{red}{IDEA: Can we put the non-diagonal part of C in the off-diag H?}}
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%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Second order Hamiltonian}
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@ -502,8 +488,8 @@ Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
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&\bEta{2} = \comm{\bHd{0}}{\bHod{2}} + \comm{\bHd{1}}{\bHod{1}} \\
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&= \comm{\bHd{0}}{\bHod{2}} \\
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&= \begin{pmatrix}
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\bO & \bF^{(0)}\bV{}{(2)} - \bV{}{(2)}\bF^{(0)}\\
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\bC{}{(0)}\bV{}{(2),\dagger} - \bV{}{(2),\dagger}\bC{}{(0)} & \bO
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\bO & \bF^{(0)}\bV{}{(2)} - \bV{}{(2)}\bC{\text{d}}{(0)}\\
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\bC{\text{d}}{(0)}\bV{}{(2),\dagger} - \bV{}{(2),\dagger}\bF^{(0)} & \bC{\text{d}}{(0)} \bC{\text{od}}{(2)} - \bC{\text{od}}{(2)} \bC{\text{d}}{(0)}
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\end{pmatrix}
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\end{align}
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@ -513,18 +499,15 @@ Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
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\dv{\bF^{(2)}}{s} & \dv{\bV{}{(2)}}{s} \\
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\dv{\bV{}{(2),\dagger}}{s} & \dv{\bC{}{(2)}}{s}
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\end{pmatrix} \\
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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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\dv{\bC{}{(2)}}{s} &= \bC{}{(0)}\bV{}{(1),\dagger }\bV{}{(1)} + \bV{}{(1),\dagger }\bV{}{(1)}\bC{}{(0)} - 2 \bV{}{(1)}\bF^{(0)}\bV{}{(1),\dagger}\\
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\dv{\bV{}{(2)}}{s} &= 2 \bF^{(0)}\bV{}{(2)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(2)} - \bV{}{(2)}(\bC{}{(0)})^2 \\
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\dv{\bV{}{(2),\dagger}}{s} &= 2 \bC{}{(0)}\bV{}{(2),\dagger}\bF^{(0)} - \bV{}{(2),\dagger}(\bF^{(0)})^2 - (\bC{}{(0)})^2\bV{}{(2),\dagger}
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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{\text{d}}{(0)}\bV{}{(1),\dagger}\\
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\dv{\bC{}{(2)}}{s} &= \\
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\dv{\bV{}{(2)}}{s} &= 2 \bF^{(0)}\bV{}{(2)}\bC{\text{d}}{(0)} - (\bF^{(0)})^2\bV{}{(2)} - \bV{}{(2)}(\bC{\text{d}}{(0)})^2 \\
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&- 2 \bV{}{(1)} \bC{\text{d}}{(0)} \bC{\text{od}}{(1)} + \bF^{(0)} \bV{}{(1)} \bC{\text{od}}{(1)} + \bV{}{(1)} \bC{\text{od}}{(1)} \bC{\text{d}}{(0)} \\
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\dv{\bV{}{(2),\dagger}}{s} &=
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\end{align}
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Once again the integration of these equations is much simpler if $\bC{}{(0)}$ is diagonal.
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\subsubsection*{Diagonal $\bC{}{(0)}$}
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\begin{align}
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&(\dv{\bF^{(2)}}{s})_{pq} = (\bF^{(0)}\bV{}{(1)}\bV{}{(1),\dagger} + \bV{}{(1)}\bV{}{(1),\dagger}\bF^{(0)} - 2 \bV{}{(1)}\bC{}{(0)}\bV{}{(1),\dagger})_{pq} \notag \\
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&(\dv{\bF^{(2)}}{s})_{pq} = (\bF^{(0)}\bV{}{(1)}\bV{}{(1),\dagger} + \bV{}{(1)}\bV{}{(1),\dagger}\bF^{(0)} - 2 \bV{}{(1)}\bC{\text{d}}{(0)}\bV{}{(1),\dagger})_{pq} \notag \\
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&= \sum_{rS} f^{(0)}_{pr} v^{(1)}_{rS} v^{(1),\dagger}_{Sq} + \sum_{Rs} v^{(1)}_{pR} v^{(1),\dagger}_{Rs} f^{(0)}_{sq} - 2\sum_{RS} v^{(1)}_{pR} c^{(0)}_{RS} v^{(1),\dagger}_{Sq} \notag \\
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&= \sum_{S} \eps^{(0)}_{p} v^{(1)}_{pS} v^{(1)}_{qS} + \sum_{R} \eps^{(0)}_{q} v^{(1)}_{pR} v^{(1)}_{qR} - 2\sum_{R} \Delta\eps^{(0)}_R v^{(1)}_{pR} v^{(1)}_{qR} \notag \\
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&= \sum_R (\eps^{(0)}_{p} + \eps^{(0)}_{q} - 2 \Delta\eps^{(0)}_R) v^{(1)}_{pR} v^{(1)}_{qR} \notag \\
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@ -533,21 +516,55 @@ Once again the integration of these equations is much simpler if $\bC{}{(0)}$ is
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&\color{red}{\boxed{\color{black}{- \sum_R \frac{\eps^{(0)}_{p} + \eps^{(0)}_{q} - 2 \Delta\eps^{(0)}_R}{(\eps^{(0)}_p - \Delta\eps^{(0)}_R)^2+ (\eps^{(0)}_q - \Delta\eps^{(0)}_R)^2}(1 - e^{-s [ (\eps^{(0)}_p - \Delta\eps^{(0)}_R)^2+ (\eps^{(0)}_q - \Delta\eps^{(0)}_R)^2]})}}} \notag
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\end{align}
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A similar derivation should give (\textbf{\textcolor{red}{TO CHECK}})
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%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Downfolding the SRG-transformed matrix}
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%%%%%%%%%%%%%%%%%%%%%%
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Now that we obtained the SRG-transformed Hamiltonian to a given order we can downfold it back to obtain a SRG-renormalized self-energy up to a given order.
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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{equation}
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\left\{
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\begin{aligned}
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(\bF^{(0)}(0) + \bF^{(2)}(s)) \bR^{1h/1p} + \bV{}{(1)}(s) \bR^{2h1p/2p1h} &= \omega \bR^{1h/1p} \\
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\bV{}{(1),\dagger}(s) \bR^{1h/1p} + (\bC{}{(0)}(0) +\bC{}{(2)}(s) ) \bR^{2h1p/2p1h}&= \omega \bR^{2h1p/2p1h}
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\end{aligned}
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\right.
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\end{equation}
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\begin{align}
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&c^{(2)}_{PQ}(s) = \notag \\
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&\color{red}{\boxed{\color{black}{- \sum_r \frac{\Delta\eps^{(0)}_{P} + \Delta\eps^{(0)}_{Q} - 2 \eps^{(0)}_r}{(\Delta\eps^{(0)}_P - \eps^{(0)}_r)^2+ (\Delta\eps^{(0)}_Q - \eps^{(0)}_r)^2}(1 - e^{-s [ (\Delta\eps^{(0)}_P - \eps^{(0)}_r)^2+ (\Delta\eps^{(0)}_P - \eps^{(0)}_r)^2]})}}} \notag
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&(\bF^{(0)}(0) + \bF^{(2)}(s)) + (\bV{}{(1)}(s) + \bV{}{(2)}(s))(\omega \mathbb{1} - \bC{}{(0)}(0) +\bC{}{(2)}(s) )^{-1} \notag \\
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&\dots (\bV{}{(1),\dagger}(s) + \bV{}{(2),\dagger}(s)) \bR^{1h/1p} = \omega \bR^{1h/1p}
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\end{align}
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\begin{align}
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&\dv{v^{(2)}_{pQ}}{s} = - (\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2 v^{(2)}_{pQ} \\
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&\color{red}{\boxed{\color{black}{v^{(2)}_{pQ}(s) = v^{(2)}_{pQ}(0) e^{-s(\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2} = 0 }}}
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&(\omega \mathbb{1} - \bC{}{(0)}(0) - \bC{}{(1)}(s) - \bC{}{(2)}(s) )^{-1} = (\omega \mathbb{1} - \bC{}{(0)}(0))^{-1} \\
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&+ (\omega \mathbb{1} - \bC{}{(0)}(0))^{-1}(\bC{}{(1)}(s) + \bC{}{(2)}(s) )(\omega \mathbb{1} - \bC{}{(0)}(0))^{-1} \notag \\
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&+ \dots \notag
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\end{align}
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Using this taylor expansion we can see that only the first term will contribute to second order in the self energy. Hence we have
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\begin{equation*}
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(\bF^{(0)}(0) + \bF^{(2)}(s) + \bV{}{(1)}(s)(\omega \mathbb{1} - \bC{}{(0)}(0))^{-1} \bV{}{(1),\dagger}(s))\bR^{1h/1p} = \omega \bR^{1h/1p}
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\end{equation*}
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Therefore we have to solve the following equation
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\begin{align}
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&(\tilde{\bF} + \tilde{\boldsymbol{\Sigma}}(\omega)) \bX{}{} = \omega \bX \\
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&\tilde{\bF} =\bF^{(0)}(0) + \bF^{(2)}(s) \\
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&\tilde{\boldsymbol{\Sigma}}(\omega) = \bV{}{(1)}(s)(\omega \mathbb{1} - \bC{}{(0)}(0))^{-1} \bV{}{(1),\dagger}(s)
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\end{align}
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%%%%%%%%%%%%%%%%%%%%%%
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\subsection{The SRG(2) quasi-particle equations}
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%%%%%%%%%%%%%%%%%%%%%%
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\subsubsection*{Non-diagonal $\bC{}{(0)}$}
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In this section, we report the GF(2), GW and GT quasi-particle equations.
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\appendix
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