Sangalli again

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Pierre-Francois Loos 2020-06-17 17:32:21 +02:00
parent bb02d36819
commit abaae0f8c0

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@ -281,19 +281,19 @@ One can now build the dynamical Bethe-Salpeter equation (dBSE) Hamiltonian, whic
\end{equation} \end{equation}
with with
\begin{subequations} \begin{subequations}
\begin{align} \begin{gather}
R(\omega) & = \Delta\eGW{} + 2 \sigma \ERI{vc}{cv} - W_R(\omega) R(\omega) = \Delta\eGW{} + 2 \sigma \ERI{vc}{cv} - W_R(\omega)
\\ \\
C(\omega) & = 2 \sigma \ERI{vc}{cv} - W_C(\omega) C(\omega) = 2 \sigma \ERI{vc}{cv} - W_C(\omega)
\end{align} \end{gather}
\end{subequations} \end{subequations}
($\sigma = 1$ for singlets and $\sigma = 0$ for triplets) and ($\sigma = 1$ for singlets and $\sigma = 0$ for triplets) and
\begin{subequations} \begin{subequations}
\begin{align} \begin{gather}
W_R(\omega) & = \ERI{vv}{cc} + \frac{4 \ERI{vv}{vc} \ERI{vc}{cc}}{\omega - \Omega - \Delta\eGW{}} W_R(\omega) = \ERI{vv}{cc} + \frac{4 \ERI{vv}{vc} \ERI{vc}{cc}}{\omega - \Omega - \Delta\eGW{}}
\\ \\
W_C(\omega) & = \ERI{vc}{cv} + \frac{4 \ERI{vc}{cv}^2}{\omega - \Omega} W_C(\omega) = \ERI{vc}{cv} + \frac{4 \ERI{vc}{cv}^2}{\omega - \Omega}
\end{align} \end{gather}
\end{subequations} \end{subequations}
are the elements of the dynamically-screened Coulomb potential for the resonant and coupling blocks of the dBSE Hamiltonian. are the elements of the dynamically-screened Coulomb potential for the resonant and coupling blocks of the dBSE Hamiltonian.
It can be easily shown that solving the equation It can be easily shown that solving the equation
@ -315,11 +315,13 @@ Within the static approximation, the BSE Hamiltonian is
\end{pmatrix} \end{pmatrix}
\end{equation} \end{equation}
with with
\begin{align} \begin{subequations}
R^{\stat} & = R(\omega = \Delta\eGW{}) = \Delta\eGW{} + 2 \sigma \ERI{vc}{vc} - W_R(\omega = \Delta\eGW{}) \begin{gather}
R^{\stat} = \Delta\eGW{} + 2 \sigma \ERI{vc}{vc} - W_R(\omega = \Delta\eGW{})
\\ \\
C^{\stat} & = C(\omega = 0) = 2 \sigma \ERI{vc}{vc} - W_C(\omega = 0) C^{\stat} = 2 \sigma \ERI{vc}{vc} - W_C(\omega = 0)
\end{align} \end{gather}
\end{subequations}
In the static approximation, only one pair of solutions (per spin manifold) is obtained by diagonalizing $\bH^{\BSE}$. In the static approximation, only one pair of solutions (per spin manifold) is obtained by diagonalizing $\bH^{\BSE}$.
There are, like in the dynamical case, opposite in sign. There are, like in the dynamical case, opposite in sign.
Therefore, the static BSE Hamiltonian does not produce spurious excitations but misses the (singlet) double excitation. Therefore, the static BSE Hamiltonian does not produce spurious excitations but misses the (singlet) double excitation.
@ -516,11 +518,13 @@ The static Hamiltonian for this theory is just the usual RPAx (or TDHF) Hamilton
\end{pmatrix} \end{pmatrix}
\end{equation} \end{equation}
with with
\begin{align} \begin{subequations}
A^{\stat} & = \Delta\eGF{} + 2 \sigma \ERI{vc}{vc} - \ERI{vv}{cc} \begin{gather}
A^{\stat} = \Delta\eGF{} + 2 \sigma \ERI{vc}{vc} - \ERI{vv}{cc}
\\ \\
B^{\stat} & = 2 \sigma \ERI{vc}{vc} - \ERI{vc}{cv} B^{\stat} = 2 \sigma \ERI{vc}{vc} - \ERI{vc}{cv}
\end{align} \end{gather}
\end{subequations}
The dynamical part of the kernel for BSE2 (that we will call dRPAx for notational consistency) is a bit ugly but it simplifies greatly in the case of the present model to yield The dynamical part of the kernel for BSE2 (that we will call dRPAx for notational consistency) is a bit ugly but it simplifies greatly in the case of the present model to yield
\begin{equation} \begin{equation}
\bH^{\dRPAx} = \bH^{\RPAx} + \bH^{\dRPAx} = \bH^{\RPAx} +
@ -531,17 +535,21 @@ The dynamical part of the kernel for BSE2 (that we will call dRPAx for notationa
\end{pmatrix} \end{pmatrix}
\end{equation} \end{equation}
with with
\begin{align} \begin{subequations}
A^{\updw}(\omega) & = - \frac{4 \ERI{cv}{vv} \ERI{vc}{cc} - \ERI{vc}{cc}^2 - \ERI{cv}{vv}^2 }{\omega - 2 \Delta\eGF{}} \begin{gather}
A^{\updw}(\omega) = - \frac{4 \ERI{cv}{vv} \ERI{vc}{cc} - \ERI{vc}{cc}^2 - \ERI{cv}{vv}^2 }{\omega - 2 \Delta\eGF{}}
\\ \\
B^{\updw} & = - \frac{4 \ERI{vc}{cv}^2 - \ERI{cc}{cc} \ERI{vc}{cv} - \ERI{vv}{vv} \ERI{vc}{cv} }{2 \Delta\eGF{}} B^{\updw} = - \frac{4 \ERI{vc}{cv}^2 - \ERI{cc}{cc} \ERI{vc}{cv} - \ERI{vv}{vv} \ERI{vc}{cv} }{2 \Delta\eGF{}}
\end{align} \end{gather}
\end{subequations}
and and
\begin{align} \begin{subequations}
A^{\upup}(\omega) & = - \frac{ \ERI{vc}{cc}^2 + \ERI{cv}{vv}^2 }{\omega - 2 \Delta\eGF{}} \begin{gather}
A^{\upup}(\omega) = - \frac{ \ERI{vc}{cc}^2 + \ERI{cv}{vv}^2 }{\omega - 2 \Delta\eGF{}}
\\ \\
B^{\upup} & = - \frac{\ERI{cc}{cc} \ERI{vc}{cv} + \ERI{vv}{vv} \ERI{vc}{cv} }{2 \Delta\eGF{}} B^{\upup} = - \frac{\ERI{cc}{cc} \ERI{vc}{cv} + \ERI{vv}{vv} \ERI{vc}{cv} }{2 \Delta\eGF{}}
\end{align} \end{gather}
\end{subequations}
Note that the coupling blocks $B$ are frequency independent, as they should. Note that the coupling blocks $B$ are frequency independent, as they should.
This has an important consequence as this lack of frequency dependence removes one of the spurious pole. This has an important consequence as this lack of frequency dependence removes one of the spurious pole.
The singlet manifold has then the right number of excitations. The singlet manifold has then the right number of excitations.
@ -602,18 +610,22 @@ The dynamical BSE Hamiltonian with Sangalli's kernel is
\end{pmatrix} \end{pmatrix}
\end{equation} \end{equation}
with with
\begin{align} \begin{subequations}
H_{ia,jb}(\omega) & = \delta_{ij} \delta_{ab} (\eGW{a} - \eGW{i}) + \Xi_{ia,jb} (\omega)
\\
K_{ia,jb}(\omega) & = \Xi_{ia,bj} (\omega)
\end{align}
and
\begin{gather} \begin{gather}
\Xi_{ia,jb} (\omega) = \sum_{m \neq n} \frac{ C_{ia,mn} C_{jb,mn} }{\omega - ( \omega_{m} + \omega_{n} + 2i\eta)} H_{ia,jb}(\omega) = \delta_{ij} \delta_{ab} (\eGW{a} - \eGW{i}) + \Xi_{ia,jb} (\omega)
\\
K_{ia,jb}(\omega) = \Xi_{ia,bj} (\omega)
\end{gather}
\end{subequations}
and
\begin{subequations}
\begin{gather}
\Xi_{ia,jb} (\omega) = \sum_{m \neq n} \frac{ C_{ia,mn} C_{jb,mn} }{\omega - ( \omega_{m} + \omega_{n})}
\\ \\
C_{ia,mn} = \frac{1}{2} \sum_{jb,kc} \qty{ \qty[ \ERI{ij}{kc} \delta_{ab} + \ERI{kc}{ab} \delta_{ij} ] \qty[ R_{m,jc} R_{n,kb} C_{ia,mn} = \frac{1}{2} \sum_{jb,kc} \qty{ \qty[ \ERI{ij}{kc} \delta_{ab} + \ERI{kc}{ab} \delta_{ij} ] \qty[ R_{m,jc} R_{n,kb}
+ R_{m,kb} R_{n,jc} ] } + R_{m,kb} R_{n,jc} ] }
\end{gather} \end{gather}
\end{subequations}
where $R_{m,ia}$ are the elements of the RPA eigenvectors. where $R_{m,ia}$ are the elements of the RPA eigenvectors.
Here $i$, $j$, and $k$ are occupied orbitals, $a$, $b$, and $c$ are unoccupied orbitals, and $m$ and $n$ label single excitations. Here $i$, $j$, and $k$ are occupied orbitals, $a$, $b$, and $c$ are unoccupied orbitals, and $m$ and $n$ label single excitations.
@ -624,10 +636,11 @@ For the two-level model, Sangalli's kernel reads
K(\omega) & = \Xi_K (\omega) K(\omega) & = \Xi_K (\omega)
\end{align} \end{align}
\begin{align} \begin{gather}
\Xi_H (\omega) = \sum_{m \neq n} \frac{ C_{ia,mn} C_{jb,mn} }{\omega - ( \omega_{m} + \omega_{n} + 2i\eta)} \Xi_H (\omega) = 2 \frac{ [\ERI{vv}{vc} + \ERI{vc}{cc}]^2 }{\omega - 2\omega_1}
\end{align} \\
\Xi_C (\omega) = 0
\end{gather}
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\section{Take-home messages} \section{Take-home messages}