done with removing full

This commit is contained in:
Pierre-Francois Loos 2020-07-22 15:43:27 +02:00
parent bad32fe5c9
commit 561c8b6f84

View File

@ -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
\begin{subequations}
\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{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
\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.
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.
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.
@ -1082,14 +1083,14 @@ The data that support the findings of this study are available within the articl
\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}
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}
[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')
\end{align}
we plug-in the Fourier expansion of the Green's function, e.g.
@ -1098,14 +1099,14 @@ $(t_{1'} = t_1^{+})$
\end{align*}
with $\tau_{13} = (t_1-t_3)$ to obtain:
\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)
e^{ i \omega t_3 } e^{-i (\omega-\omega_1) t_4 } \nonumber
\end{multline}
With the change of variable $\omega \to \omega + {\omega_1}/2$ one obtains readily
\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} }
\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} ) \;