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BSEdyn.tex
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@ -362,12 +362,12 @@ In the optical limit of instantaneous electron-hole creation and destruction, im
where $\tau_{12} = t_1 - t_2$, $\theta$ is the Heaviside step function, and where $\tau_{12} = t_1 - t_2$, $\theta$ is the Heaviside step function, and
\begin{subequations} \begin{subequations}
\begin{align} \begin{align}
\chi_S(\bx_1,\bx_{2}) & = \mel{N}{T [\hpsi(\bx_1) \hpsi^{\dagger}(\bx_{2})] }{N,S}, \chi_S(\bx_1,\bx_{1'}) & = \mel{N}{T [\hpsi(\bx_1) \hpsi^{\dagger}(\bx_{1'})] }{N,S},
\\ \\
\tchi_S(\bx_1,\bx_{2}) & = \mel{N,S}{T [\hpsi(\bx_1) \hpsi^{\dagger}(\bx_{2})] }{N}. \tchi_S(\bx_1,\bx_{1'}) & = \mel{N,S}{T [\hpsi(\bx_1) \hpsi^{\dagger}(\bx_{1'})] }{N}.
\end{align} \end{align}
\end{subequations} \end{subequations}
The $\Om{S}{}$'s are the neutral excitation energies of interest. The $\Om{s}{}$'s are the neutral excitation energies of interest (with $\Om{s}{} = E^N_s - E^N_0$).
Picking up the $e^{+i \Om{S}{} t_2 }$ component of both $L(1,2; 1',2')$ and $L(6,2;5,2')$, simplifying further by $\tchi_S(\bx_2,\bx_{2'})$ on both sides of the BSE [see Eq.~\eqref{eq:BSE}], we seek the $e^{-i \Om{S}{} t_1 }$ Fourier component associated with the right-hand side of a modified dynamical BSE, which reads Picking up the $e^{+i \Om{S}{} t_2 }$ component of both $L(1,2; 1',2')$ and $L(6,2;5,2')$, simplifying further by $\tchi_S(\bx_2,\bx_{2'})$ on both sides of the BSE [see Eq.~\eqref{eq:BSE}], we seek the $e^{-i \Om{S}{} t_1 }$ Fourier component associated with the right-hand side of a modified dynamical BSE, which reads
\begin{multline} \label{eq:BSE_2} \begin{multline} \label{eq:BSE_2}
@ -534,6 +534,7 @@ One can verify that, in the static limit where $\Om{m}{\RPA} \to \infty$, the ma
evidencing that the standard static BSE problem is recovered from the present dynamical formalism in this limit. evidencing that the standard static BSE problem is recovered from the present dynamical formalism in this limit.
Due to excitonic effects, the lowest BSE excitation energy, $\Om{1}{}$, stands lower than the lowest RPA excitation energy, $\Om{1}{\RPA}$, so that, $\Om{ib}{S} - \Om{m}{\RPA} < 0 $ and $\widetilde{W}_{ij,ab}(\Om{S}{})$ has no resonances. Due to excitonic effects, the lowest BSE excitation energy, $\Om{1}{}$, stands lower than the lowest RPA excitation energy, $\Om{1}{\RPA}$, so that, $\Om{ib}{S} - \Om{m}{\RPA} < 0 $ and $\widetilde{W}_{ij,ab}(\Om{S}{})$ has no resonances.
This property holds for a few low lying $\Om{s}{}$ excitations but special care must be taken for higher ones.
Furthermore, $\Om{ib}{S}$ and $\Om{ja}{S}$ are necessarily negative quantities for in-gap low-lying BSE excitations. Furthermore, $\Om{ib}{S}$ and $\Om{ja}{S}$ are necessarily negative quantities for in-gap low-lying BSE excitations.
Thus, we have $\abs*{\Om{ib}{S} - \Om{m}{\RPA}} > \Om{m}{\RPA}$. Thus, we have $\abs*{\Om{ib}{S} - \Om{m}{\RPA}} > \Om{m}{\RPA}$.
As a consequence, we observe a reduction of the electron-hole screening, \ie, an enhancement of electron-hole binding energy, as compared to the standard static BSE, and consequently smaller (red-shifted) excitation energies. As a consequence, we observe a reduction of the electron-hole screening, \ie, an enhancement of electron-hole binding energy, as compared to the standard static BSE, and consequently smaller (red-shifted) excitation energies.
@ -1082,14 +1083,14 @@ The data that support the findings of this study are available within the articl
\appendix \appendix
\section{$L_0(1,3; 1',4)$ $(t_1)$-time Fourier transform} \section{$L_0(1,4; 1',3)$ $(t_1)$-time Fourier transform}
\label{app:A} \label{app:A}
In this Appendix, we derive Eqs.~\eqref{eq:iL0} to \eqref{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 Defining the $t_1$-time Fourier transform of $L_0(1,3;4,1')$ with
$(t_{1'} = t_1^{+})$ $(t_{1'} = t_1^{+})$
\begin{align} \begin{align}
[L_0](\bx_1,3;\bx_{1'},4 \; | \; \omega_1 ) = -i [L_0](\bx_1,4;\bx_{1'},3 \; | \; \omega_1 ) = -i
\int dt_1 e^{i \omega_1 t_1 } G(1,3)G(4,1') \int dt_1 e^{i \omega_1 t_1 } G(1,3)G(4,1')
\end{align} \end{align}
we plug-in the Fourier expansion of the Green's function, e.g. we plug-in the Fourier expansion of the Green's function, e.g.
@ -1098,14 +1099,14 @@ $(t_{1'} = t_1^{+})$
\end{align*} \end{align*}
with $\tau_{13} = (t_1-t_3)$ to obtain: with $\tau_{13} = (t_1-t_3)$ to obtain:
\begin{multline} \begin{multline}
[L_0](\bx_1,3;\bx_{1'},4 \;| \; \omega_1 ) = [L_0](\bx_1,4;\bx_{1'},3 \;| \; \omega_1 ) =
\\ \\
\int \frac{ d\omega }{ 2i\pi } \; G(\bx_1,\bx_3;\omega) \; G(\bx_4,\bx_{1'};\omega-\omega_1) \int \frac{ d\omega }{ 2i\pi } \; G(\bx_1,\bx_3;\omega) \; G(\bx_4,\bx_{1'};\omega-\omega_1)
e^{ i \omega t_3 } e^{-i (\omega-\omega_1) t_4 } \nonumber e^{ i \omega t_3 } e^{-i (\omega-\omega_1) t_4 } \nonumber
\end{multline} \end{multline}
With the change of variable $\omega \to \omega + {\omega_1}/2$ one obtains readily With the change of variable $\omega \to \omega + {\omega_1}/2$ one obtains readily
\begin{multline} \begin{multline}
[L_0](\bx_1,3;\bx_{1'},4 \; | \; \omega_1 ) = [L_0](\bx_1,4;\bx_{1'},3 \; | \; \omega_1 ) =
\\ \\
e^{ i \omega_1 t^{34} } e^{ i \omega_1 t^{34} }
\int \frac{ d\omega }{ 2i\pi } \; G\qty(\bx_1,\bx_3;\omega+ \frac{\omega_1}{2} ) G\qty(\bx_4,\bx_{1'};\omega-\frac{\omega_1}{2} ) \; \int \frac{ d\omega }{ 2i\pi } \; G\qty(\bx_1,\bx_3;\omega+ \frac{\omega_1}{2} ) G\qty(\bx_4,\bx_{1'};\omega-\frac{\omega_1}{2} ) \;