Ewald 1D in 3D notes

Former-commit-id: 771165b42dc07d0681588cf1ee4e047d938c1b45

notes/ewald_1D_in_3D.lyx

@@ -0,0 +1,444 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 584
+\begin_document
+\begin_header
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+\index Index
+\shortcut idx
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+\end_index
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+\end_header
+
+\begin_body
+
+\begin_layout Title
+1D in 3D Ewald sum
+\end_layout
+
+\begin_layout Standard
+\begin_inset FormulaMacro
+\newcommand{\ud}{\mathrm{d}}
+\end_inset
+
+
+\begin_inset FormulaMacro
+\newcommand{\abs}[1]{\left|#1\right|}
+\end_inset
+
+ 
+\begin_inset FormulaMacro
+\newcommand{\vect}[1]{\mathbf{#1}}
+\end_inset
+
+
+\begin_inset FormulaMacro
+\newcommand{\uvec}[1]{\hat{\mathbf{#1}}}
+\end_inset
+
+
+\lang english
+
+\begin_inset FormulaMacro
+\newcommand{\ush}[2]{Y_{#1}^{#2}}
+\end_inset
+
+
+\begin_inset FormulaMacro
+\newcommand{\ushD}[2]{Y'_{#1}^{#2}}
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset FormulaMacro
+\newcommand{\vsh}{\vect A}
+\end_inset
+
+
+\begin_inset FormulaMacro
+\newcommand{\vshD}{\vect{A'}}
+\end_inset
+
+
+\begin_inset FormulaMacro
+\newcommand{\wfkc}{\vect y}
+\end_inset
+
+
+\begin_inset FormulaMacro
+\newcommand{\wfkcout}{\vect u}
+\end_inset
+
+
+\begin_inset FormulaMacro
+\newcommand{\wfkcreg}{\vect v}
+\end_inset
+
+
+\begin_inset FormulaMacro
+\newcommand{\wckcreg}{a}
+\end_inset
+
+
+\begin_inset FormulaMacro
+\newcommand{\wckcout}{f}
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+[Linton, (2.24)] with slightly modified notation and setting 
+\begin_inset Formula $d_{c}=2$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+G_{\Lambda}^{(1;\kappa)}\left(\vect r\right)=-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect r}\int_{1/\eta}^{\infty e^{i\pi/4}}e^{-\kappa^{2}\gamma^{2}t^{2}/4}e^{-\left|\vect r^{\bot}\right|^{2}/t^{2}}t^{-1}\ud t
+\]
+
+\end_inset
+
+or, evaluated at point 
+\begin_inset Formula $\vect s+\vect r$
+\end_inset
+
+ instead
+\begin_inset Formula 
+\[
+G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)=-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\left(\vect s+\vect r\right)}\int_{1/\eta}^{\infty e^{i\pi/4}}e^{-\kappa^{2}\gamma^{2}t^{2}/4}e^{-\left|\vect s^{\bot}+\vect r^{\bot}\right|^{2}/t^{2}}t^{-1}\ud t
+\]
+
+\end_inset
+
+The integral can be by substitutions taken into the form 
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\lang english
+\begin_inset Formula 
+\[
+G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{2\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}e^{i\vect K_{m}\cdot\vect r}\int_{1/\eta}^{\infty\exp\left(i\pi/4\right)}e^{-\kappa^{2}\gamma_{m}^{2}\zeta^{2}/4}e^{-\left|\vect r_{\bot}\right|^{2}/\zeta^{2}}\zeta^{1-d_{c}}\ud\zeta
+\]
+
+\end_inset
+
+Try substitution 
+\begin_inset Formula $t=\zeta^{2}$
+\end_inset
+
+: then 
+\begin_inset Formula $\ud t=2\zeta\,\ud\zeta$
+\end_inset
+
+ (
+\begin_inset Formula $\ud\zeta=\ud t/2t^{1/2}$
+\end_inset
+
+) and
+\begin_inset Formula 
+\[
+G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{4\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}e^{i\vect K_{m}\cdot\vect r}\int_{1/\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\kappa^{2}\gamma_{m}^{2}t/4}e^{-\left|\vect r_{\bot}\right|^{2}/t}t^{\frac{-d_{c}}{2}}\ud t
+\]
+
+\end_inset
+
+Try subst.
+ 
+\begin_inset Formula $\tau=k^{2}\gamma_{m}^{2}/4$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\lang english
+\begin_inset Formula 
+\[
+G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{4\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}e^{i\vect K_{m}\cdot\vect r}\left(\frac{\kappa\gamma_{m}}{2}\right)^{d_{c}}\int_{\kappa^{2}\gamma_{m}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{m}^{2}/4\tau}\tau^{\frac{-d_{c}}{2}}\ud\tau
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)=-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\left(\vect s+\vect r\right)}\int_{\kappa^{2}\gamma_{m}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{m}^{2}/4\tau}\tau^{-1}\ud\tau
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+[Linton, (2.25)] with slightly modified notation:
+\begin_inset Formula 
+\[
+G_{\Lambda}^{(1;\kappa)}\left(\vect r\right)=-\frac{1}{\sqrt{4\pi}\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect r}\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}\left|\vect r^{\bot}\right|^{2j}}{j!}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2j-1}\Gamma_{j\vect K}
+\]
+
+\end_inset
+
+We want to express an expansion in a shifted point, so let's substitute
+ 
+\begin_inset Formula $\vect r\to\vect s+\vect r$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)=-\frac{1}{\sqrt{4\pi}\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\left(\vect s+\vect r\right)}\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}\left|\vect s^{\bot}+\vect r^{\bot}\right|^{2j}}{j!}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2j-1}\Gamma_{j\vect K}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Let's do the integration to get 
+\begin_inset Formula $\tau_{l}^{m}\left(\vect s,\vect k\right)$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\left(\vect s+\vect r\right)}\int_{\kappa^{2}\gamma_{\vect K}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\tau^{-1}\ud\tau
+\]
+
+\end_inset
+
+The 
+\begin_inset Formula $\vect r$
+\end_inset
+
+-dependent plane wave factor can be also written as
+\begin_inset Formula 
+\begin{align*}
+e^{i\vect K\cdot\vect r} & =e^{i\left|\vect K\right|\vect r\cdot\uvec K}=4\pi\sum_{lm}i^{l}\mathcal{J}'_{l}^{m}\left(\left|\vect K\right|\vect r\right)\ush lm\left(\uvec K\right)\\
+ & =4\pi\sum_{lm}i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ushD lm\left(\uvec{\vect r}\right)\ush lm\left(\uvec K\right)
+\end{align*}
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+or the other way around
+\begin_inset Formula 
+\[
+e^{i\vect K\cdot\vect r}=4\pi\sum_{lm}i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec{\vect r}\right)\ushD lm\left(\uvec K\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+so
+\begin_inset Formula 
+\[
+\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ushD lm\left(\uvec r\right)\ush lm\left(\uvec K\right)\int_{\kappa^{2}\gamma_{\vect K}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\tau^{-1}\ud\tau
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Now we set the conventions: let the lattice lie on the 
+\begin_inset Formula $z$
+\end_inset
+
+ axis, so that 
+\begin_inset Formula $\vect s_{\bot},\vect r_{\bot}$
+\end_inset
+
+ lie in the 
+\begin_inset Formula $xy$
+\end_inset
+
+-plane, (TODO check the meaning of 
+\begin_inset Formula $\vect k$
+\end_inset
+
+ and possible additional phase factor.) If we write
+\begin_inset Formula $\vect s_{\bot}=\uvec xs_{\bot}\cos\Phi+\uvec ys_{\bot}\sin\Phi$
+\end_inset
+
+, 
+\begin_inset Formula $\vect r_{\bot}=\uvec xr_{\bot}\cos\phi+\uvec yr_{\bot}\sin\phi$
+\end_inset
+
+, we have 
+\begin_inset Formula 
+\[
+\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}=s_{\bot}^{2}+r_{\bot}^{2}+2s_{\bot}r_{\bot}\cos\left(\phi-\Phi\right).
+\]
+
+\end_inset
+
+Also, in this convention 
+\begin_inset Formula $\ush lm\left(\uvec K\right)=0$
+\end_inset
+
+ for 
+\begin_inset Formula $m\ne0$
+\end_inset
+
+, so
+\begin_inset Formula 
+\[
+\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{l}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ushD l0\left(\uvec r\right)\ush l0\left(\uvec K\right)\int_{\kappa^{2}\gamma_{\vect K}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left(s_{\bot}^{2}+r_{\bot}^{2}+2s_{\bot}r_{\bot}\cos\left(\phi-\Phi\right)\right)^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\tau^{-1}\ud\tau
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Let's also fix the spherical harmonics for now, 
+\begin_inset Formula 
+\[
+\ushD lm\left(\uvec r\right)=\lambda'_{lm}e^{-im\phi}P_{l}^{-m}\left(\cos\theta\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+The angular integral (assuming it can be separated from the rest like this)
+ is
+\begin_inset Formula 
+\[
+I_{l}^{m}\equiv\int\ud\Omega_{\vect r}\,\ushD lm\left(\uvec r\right)e^{i\vect K\cdot\vect r_{\parallel}}e^{-2s_{\bot}r_{\bot}\cos\left(\phi-\Phi\right)}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+which can be separated even more into two integrals
+\begin_inset Formula 
+\[
+I_{l}^{m}=\lambda'_{lm}\left(\int_{0}^{2\pi}e^{-im\phi}e^{-2s_{\bot}r_{\bot}\cos\left(\phi-\Phi\right)}\ud\phi\right)\left(\int_{0}^{\pi}P_{l}^{-m}\left(\cos\theta\right)e^{i\left|\vect K\right|\left|\vect r\right|\cos\theta}\sin\theta\,\ud\theta\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document

notes/ewald_23_z_nonzero.lyx

@@ -255,7 +255,7 @@ Ewald long range integral
 Linton has (2.24):
 \begin_inset Formula 
 \[
-G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{2\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}\int_{1/\eta}^{\infty\exp\left(i\pi/4\right)}e^{-\kappa^{2}\gamma_{m}^{2}\zeta^{2}/4}e^{-\left|\vect r_{\bot}\right|^{2}/\zeta^{2}}\zeta^{1-d_{c}}\ud\zeta
+G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{2\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}e^{i\vect K_{m}\cdot\vect r}\int_{1/\eta}^{\infty\exp\left(i\pi/4\right)}e^{-\kappa^{2}\gamma_{m}^{2}\zeta^{2}/4}e^{-\left|\vect r_{\bot}\right|^{2}/\zeta^{2}}\zeta^{1-d_{c}}\ud\zeta
 \]
 
 \end_inset
@@ -275,7 +275,7 @@ Try substitution
 ) and
 \begin_inset Formula 
 \[
-G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{4\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}\int_{1/\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\kappa^{2}\gamma_{m}^{2}t/4}e^{-\left|\vect r_{\bot}\right|^{2}/t}t^{\frac{-d_{c}}{2}}\ud t
+G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{4\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}e^{i\vect K_{m}\cdot\vect r}\int_{1/\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\kappa^{2}\gamma_{m}^{2}t/4}e^{-\left|\vect r_{\bot}\right|^{2}/t}t^{\frac{-d_{c}}{2}}\ud t
 \]
 
 \end_inset
@@ -293,7 +293,7 @@ Try subst.
 \lang english
 \begin_inset Formula 
 \[
-G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{4\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}\left(\frac{\kappa\gamma_{m}}{2}\right)^{d_{c}}\int_{\kappa^{2}\gamma_{m}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{m}^{2}/4\tau}\tau^{\frac{-d_{c}}{2}}\ud\tau
+G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{4\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}e^{i\vect K_{m}\cdot\vect r}\left(\frac{\kappa\gamma_{m}}{2}\right)^{d_{c}}\int_{\kappa^{2}\gamma_{m}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{m}^{2}/4\tau}\tau^{\frac{-d_{c}}{2}}\ud\tau
 \]
 
 \end_inset