clean up appendix

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Pierre-Francois Loos 2020-07-01 22:20:06 +02:00
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commit 9309438271

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@ -59,6 +59,7 @@
% operators
\newcommand{\hH}{\Hat{H}}
\newcommand{\ha}{\Hat{a}}
% methods
\newcommand{\KS}{\text{KS}}
@ -185,7 +186,7 @@
\newcommand{\ra}{\rightarrow}
\newcommand{\hpsi}{\Hat{\psi}}
\newcommand{\ha}{\Hat{a}}
\renewcommand{\ha}{\Hat{a}}
\newcommand{\tchi}{\Tilde{\chi}}
\newcommand{\bx}{\mathbf{x}}
@ -1084,7 +1085,7 @@ The data that support the findings of this study are available within the articl
\section{$L_0(1,3; 1',4)$ $(t_1)$-time Fourier transform}
\label{appendixA}
We derive in this Appendix Eqs.~\ref{eq:iL0} to ~\ref{eq:iL0bis}.
In this Appendix, we derive Eqs.~\eqref{eq:iL0} to \eqref{eq:iL0bis}.
Defining the $t_1$-time Fourier transform of $L_0(1,3;4,1')$ with
$(t_{1'} = t_1^{+})$
\begin{align}
@ -1107,68 +1108,70 @@ With the change of variable $\omega \to \omega + {\omega_1}/2$ one obtains readi
\int \frac{ d\omega }{ 2i\pi } \; G\qty(x_1,x_3;\omega+ \frac{\omega_1}{2} ) G\qty(x_4,x_{1'};\omega-\frac{\omega_1}{2} ) \;
e^{ i \omega \tau_{34} }
\end{equation}
with $\tau_{34} = ( t_3 - t_4 )$ and $t^{34}= (t_3+t_4)/2$.
Using now the Lehman representation of the Green's functions (Eq.~\ref{eq:G-Lehman}), and picking up the poles associated with the occupied (virtual) states in the upper (lower) half-plane for $\tau_{34} > 0$ ($\tau_{34} < 0$), one obtains using the residue theorem (with $\tau = \tau_{34})$
with $\tau_{34} = t_3 - t_4$ and $t^{34}= (t_3+t_4)/2$.
Using now the Lehman representation of the Green's functions (Eq.~\eqref{eq:G-Lehman}), and picking up the poles associated with the occupied (virtual) states in the upper (lower) half-plane for $\tau_{34} > 0$ ($\tau_{34} < 0$), one obtains using the residue theorem (with $\tau = \tau_{34})$
\begin{equation}
\begin{split}
\int \frac{ d \omega }{2i\pi} \; G\qty(x_1,x_3; \omega + \homu ) G\qty(x_4,x_{1'}; \omega - \homu ) e^{ i \omega \tau }
& = \sum_{bj}
\frac{ \phi_b(x_1) \phi_b^*(x_3) \phi_j(x_4) \phi_j^*(x_{1'})} { \omega_1 - (\varepsilon_b - \varepsilon_j) + i\eta }
\qty[ \theta(\tau) e^{i ( \varepsilon_j + \homu ) \tau } + \theta(-\tau) e^{i ( \varepsilon_b - \homu ) \tau } ]
\frac{ \phi_b(x_1) \phi_b^*(x_3) \phi_j(x_4) \phi_j^*(x_{1'})} { \omega_1 - (\e{b} - \e{j}) + i\eta }
\qty[ \theta(\tau) e^{i ( \e{j} + \homu ) \tau } + \theta(-\tau) e^{i ( \e{b} - \homu ) \tau } ]
\\
& - \sum_{bj} \frac{ \phi_j(x_1) \phi_j^*(x_3) \phi_b(x_4) \phi_b^*(x_{1'})} { \omega_1 + (\varepsilon_b - \varepsilon_j ) -i\eta }
\qty[ \theta(\tau) e^{i ( \varepsilon_j - \homu ) \tau } + \theta(-\tau) e^{i ( \varepsilon_b + \homu ) \tau } ]
& - \sum_{bj} \frac{ \phi_j(x_1) \phi_j^*(x_3) \phi_b(x_4) \phi_b^*(x_{1'})} { \omega_1 + (\e{b} - \e{j} ) -i\eta }
\qty[ \theta(\tau) e^{i ( \e{j} - \homu ) \tau } + \theta(-\tau) e^{i ( \e{b} + \homu ) \tau } ]
\\
& + \sum_{ab} \text{ pp terms } + \sum_{ij} \text{ hh terms }
\end{split}
\end{equation}
where (pp) and (hh) labels particle-particle and hole-hole channels neglected here.
Projecting onto $\phi_a^*(x_1) \phi_i(x_{1'})$ selects the first line of the RHS, leading to Eq.~\ref{eq:iL0bis}
with $ (\omega_1 \rightarrow \Omega_s )$.
Projecting onto $\phi_a^*(x_1) \phi_i(x_{1'})$ selects the first line of the RHS, leading to Eq.~\eqref{eq:iL0bis}
with $ (\omega_1 \to \Omega_s )$.
\section{ $\mel{N}{T [\hpsi(6) \hpsi^{\dagger}(5)] }{N,s}$ in the electron/hole product basis }
\section{ $\mel{N}{T [\hpsi(6) \hpsi^{\dagger}(5)] }{N,s}$ in the electron/hole product basis }
We now derive in some more details Eq.~\ref{eq:spectral65}.
Starting with:
We now derive in some more details Eq.~\eqref{eq:spectral65}.
Starting with
\begin{equation}
\mel{N}{T [\hpsi(6) \hpsi^{\dagger}(5)] }{N,s}
= \theta(\tau_{65}) \mel{N}{ \hpsi(6) \hpsi^{\dagger}(5) }{N,s}
- \theta(-\tau_{65}) \mel{N}{ \hpsi^{\dagger}(5) \hpsi(6) }{N,s}
\end{equation}
we use the relation between operators in their Heisenberg and Schr\"{o}dinger representations (Eq.~\ref{Eisenberg}) to obtain:
we use the relation between operators in their Heisenberg and Schr\"{o}dinger representations [see Eq.~\eqref{Eisenberg}] to obtain
\begin{equation}
\langle N | T [\hpsi(6) \hpsi^{\dagger}(5)] | N,s \rangle = \\
+ \theta(\tau_{65}) \mel{N}{ \hpsi(x_6) e^{-i{\hat H} \tau_{65}} \hpsi^{\dagger}(x_5) }{N,s} e^{ i E^N_0 t_6 } e^{ - i E^N_s t_5 }
- \theta(-\tau_{65}) \mel{N}{ \hpsi^{\dagger}(x_5) e^{ i{\hat H} \tau_{65}} \hpsi(x_6) }{N,s} e^{ i E^N_0 t_5 } e^{ - i E^N_s t_6 }
\mel{N}{T [\hpsi(6) \hpsi^{\dagger}(5)]}{N,s} = \\
+ \theta(\tau_{65}) \mel{N}{ \hpsi(x_6) e^{-i\hH \tau_{65}} \hpsi^{\dagger}(x_5) }{N,s} e^{ i E^N_0 t_6 } e^{ - i E^N_s t_5 }
- \theta(-\tau_{65}) \mel{N}{ \hpsi^{\dagger}(x_5) e^{ i\hH \tau_{65}} \hpsi(x_6) }{N,s} e^{ i E^N_0 t_5 } e^{ - i E^N_s t_6 }
\end{equation}
with $E^N_0$ the N-electron ground-state energy and $E^N_s$ the enrgy of the s-th excited state $| N,s \rangle$. Expanding now the field operators with creation/destruction operators in the MO basis
with $E^N_0$ the $N$-electron ground-state energy and $E^N_s$ the energy of the $s$th excited state $\ket{N,s}$.
Expanding now the field operators with creation/destruction operators in the orbital basis
\begin{align*}
\hpsi(x_6) & = \sum_p \phi_p(x_6) {\hat a}_p
\hpsi(x_6) & = \sum_p \phi_p(x_6) \ha_p
&
\hpsi^{\dagger}(x_5) & = \sum_q \phi_q^{*}(x_5) {\hat a}^{\dagger}_q
\hpsi^{\dagger}(x_5) & = \sum_q \phi_q^{*}(x_5) \ha^{\dagger}_q
\end{align*}
one obtains
one gets
\begin{equation}
\langle N | T [\hpsi(6) \hpsi^{\dagger}(5)] | N,s \rangle =
\sum_{pq} \phi_p(x_6) \phi_q^{*}(x_5) \;
\big[ \; \theta(\tau_{65}) \mel{N}{ {\hat a}_p e^{-i{\hat H} \tau_{65}} {\hat a}^{\dagger}_q }{N,s} e^{ i E^N_0 t_6 } e^{ - i E^N_s t_5 } \\
- \theta(-\tau_{65}) \mel{N}{ {\hat a}^{\dagger}_q e^{ i{\hat H} \tau_{65}} {\hat a}_p }{N,s} e^{ i E^N_0 t_5 } e^{ - i E^N_s t_6 } \; \big]
\mel{N}{T [\hpsi(6) \hpsi^{\dagger}(5)]}{N,s} =
\sum_{pq} \phi_p(x_6) \phi_q^{*}(x_5)
\qty[ \theta(\tau_{65}) \mel{N}{ \ha_p e^{-i \hH \tau_{65}} \ha^{\dagger}_q }{N,s} e^{ i E^N_0 t_6 } e^{ - i E^N_s t_5 }
- \theta(-\tau_{65}) \mel{N}{ \ha^{\dagger}_q e^{ i \hH \tau_{65}} \ha_p }{N,s} e^{ i E^N_0 t_5 } e^{ - i E^N_s t_6 } ]
\end{equation}
We now act on the $N$-electron ground-state with
\begin{align*}
e^{i{\hat H} \tau_{65} } {\hat a}^{\dagger}_p | N \rangle &=
e^{i ( E^N_0 + \varepsilon_p ) \tau_{65} } | N \rangle &
e^{ -i{\hat H} \tau_{65} } {\hat a}_q | N \rangle &=
e^{-i ( E^N_0 - \varepsilon_q ) \tau_{65} } | N \rangle
e^{i\hH \tau_{65} } \ha^{\dagger}_p \ket{N} &=
e^{i ( E^N_0 + \e{p} ) \tau_{65} } \ket{N} &
e^{ -i\hH \tau_{65} } \ha_q \ket{N} &=
e^{-i ( E^N_0 - \e{q} ) \tau_{65} } \ket{N}
\end{align*}
where $\lbrace \varepsilon_{p/q} \rbrace$ are quasiparticle energies, such as the $GW$ ones, namely proper addition/removal energies. Taking the associated bras that we plug into the MOs product basis expansion of $\langle N | T [\hpsi(6) \hpsi^{\dagger}(5)] | N,s \rangle $ one obtains:
where $\lbrace \e{p/q} \rbrace$ are quasiparticle energies, such as the $GW$ ones, namely proper addition/removal energies.
Taking the associated bras that we plug into the orbital product basis expansion of $\mel{N}{T [\hpsi(6) \hpsi^{\dagger}(5)]}{N,s}$ one obtains:
\begin{equation}
\langle N | T [\hpsi(6) \hpsi^{\dagger}(5)] | N,s \rangle =
\sum_{pq} \phi_p(x_6) \phi_q^{*}(x_5) \;
\big[ \; \theta(\tau_{65}) \mel{N}{ {\hat a}_p {\hat a}^{\dagger}_q }{N,s} e^{ -i \varepsilon_p \tau_{65} } e^{ - i \Omega_s t_5 }
- \theta(-\tau_{65}) \mel{N}{ {\hat a}^{\dagger}_q {\hat a}_p }{N,s} e^{ -i \varepsilon_q \tau_{65} } e^{ - i \Omega_s t_6 } \; \big]
\mel{N}{T [\hpsi(6) \hpsi^{\dagger}(5)]}{N,s} =
\sum_{pq} \phi_p(x_6) \phi_q^{*}(x_5)
\qty[ \theta(\tau_{65}) \mel{N}{ \ha_p \ha^{\dagger}_q }{N,s} e^{ -i \e{p} \tau_{65} } e^{ - i \Omega_s t_5 }
- \theta(-\tau_{65}) \mel{N}{ \ha^{\dagger}_q \ha_p }{N,s} e^{ -i \e{q} \tau_{65} } e^{ - i \Omega_s t_6 } ]
\end{equation}
leading to Eq.~\ref{eq:spectral65} with $\Omega_s = (E^N_s - E^N_0)$, $t_6 = \tau_{65}/2 + t^{65}$ and $t_5 = - \tau_{65}/2 + t^{65}$. \\
leading to Eq.~\eqref{eq:spectral65} with $\Omega_s = (E^N_s - E^N_0)$, $t_6 = \tau_{65}/2 + t^{65}$ and $t_5 = - \tau_{65}/2 + t^{65}$.
\end{widetext}