From 299bb6fc0308e26352ba655e114ce3c523764111 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Marek=20Ne=C4=8Dada?= Date: Sun, 21 Jun 2020 18:59:32 +0300 Subject: [PATCH] Kinda reasonable form of 1D in 3D Ewald. Former-commit-id: eb5c54026cff0c2889d32cd2c6288e7be2cbb3e9 --- notes/ewald_1D_in_3D.lyx | 316 ++++++++++++++++++++++++++++++++++++++- 1 file changed, 311 insertions(+), 5 deletions(-) diff --git a/notes/ewald_1D_in_3D.lyx b/notes/ewald_1D_in_3D.lyx index b1bfad3..75ca080 100644 --- a/notes/ewald_1D_in_3D.lyx +++ b/notes/ewald_1D_in_3D.lyx @@ -162,6 +162,25 @@ \end_inset +\end_layout + +\begin_layout Section +General formula +\end_layout + +\begin_layout Standard +We need to find the expansion coefficient +\end_layout + +\begin_layout Standard +\begin_inset Formula +\begin{equation} +\tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{i}{\kappa j_{l'}\left(\kappa\left|\vect r\right|\right)}\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(\kappa)}\left(\vect s+\vect r,\vect k\right)\ushD{l'}{m'}\left(\uvec r\right).\label{eq:tau extraction formula} +\end{equation} + +\end_inset + + \end_layout \begin_layout Standard @@ -354,8 +373,153 @@ e^{-\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2} hence \begin_inset Formula +\begin{align*} +\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)\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)\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{\left(\left|\vect r_{\bot}\right|^{2}+2\vect r_{\bot}\cdot\vect s_{\bot}\right)\kappa^{2}\gamma_{\vect K}^{2}}{4}\right)^{n}\underbrace{\int_{\kappa^{2}\gamma_{\vect K}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect s_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\tau^{-1-n}\ud\tau}_{\Delta_{n+1/2}}\\ + & =-\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)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\Delta_{n+1/2}}{n!}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(-\frac{\left(\left|\vect r_{\bot}\right|^{2}+2\vect r_{\bot}\cdot\vect s_{\bot}\right)\kappa^{2}\gamma_{\vect K}^{2}}{4}\right)^{n}\\ + & =-\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)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left|\vect r_{\bot}\right|^{2(n-k)}\left(2\vect r_{\bot}\cdot\vect s_{\bot}\right)^{k} +\end{align*} + +\end_inset + +If we label +\begin_inset Formula $\left|\vect r_{\bot}\right|\left|\vect s_{\bot}\right|\cos\varphi\equiv\vect r_{\bot}\cdot\vect s_{\bot}$ +\end_inset + +, we have +\begin_inset Formula \[ -\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ushD lm\left(\uvec r\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{\left(\left|\vect r_{\bot}\right|^{2}+2\vect r_{\bot}\cdot\vect s_{\bot}\right)\kappa^{2}\gamma_{\vect K}^{2}}{4}\right)^{n}\underbrace{\int_{\kappa^{2}\gamma_{\vect K}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect s_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\tau^{-1-n}\ud\tau}_{\Delta_{n+1/2}} +\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left|\vect r_{\bot}\right|^{2n-k}\left(\cos\varphi\right)^{k} +\] + +\end_inset + +and if we label +\begin_inset Formula $\left|\vect r\right|\sin\theta\equiv\left|\vect r_{\bot}\right|$ +\end_inset + + +\begin_inset Formula +\[ +\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\left|\vect r\right|^{2n-k}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\theta\right)^{2n-k}\left(\cos\varphi\right)^{k} +\] + +\end_inset + +Now let's put the RHS into +\begin_inset CommandInset ref +LatexCommand eqref +reference "eq:tau extraction formula" +plural "false" +caps "false" +noprefix "false" + +\end_inset + + and try eliminating some sum by taking the limit +\begin_inset Formula $\left|\vect r\right|\to0$ +\end_inset + +. + We have +\begin_inset Formula $j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\sim\left(\left|\vect K\right|\left|\vect r\right|\right)^{l}/\left(2l+1\right)!!$ +\end_inset + +; the denominator from +\begin_inset CommandInset ref +LatexCommand eqref +reference "eq:tau extraction formula" +plural "false" +caps "false" +noprefix "false" + +\end_inset + + behaves like +\begin_inset Formula $j_{l'}\left(\kappa\left|\vect r\right|\right)\sim\left(\kappa\left|\vect r\right|\right)^{l'}/\left(2l'+1\right)!!.$ +\end_inset + + The leading terms are hence those with +\begin_inset Formula $\left|\vect r\right|^{l-l'+2n-k}$ +\end_inset + +. + So +\begin_inset Formula +\[ +\tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi\mathcal{A}\kappa^{1+l'}}\left(2l'+1\right)!!\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\delta_{l'-l,2n-k}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\theta\right)^{l'-l}\left(\cos\varphi\right)^{k}. +\] + +\end_inset + +Let's now focus on rearranging the sums; we have +\begin_inset Formula +\[ +S(l')\equiv\sum_{l=0}^{\infty}\sum_{n=0}^{\infty}\sum_{k=0}^{n}\delta_{l'-l,2n-k}f(l',l,n,k)=\sum_{l=0}^{\infty}\sum_{n=0}^{\infty}\sum_{k=0}^{n}\delta_{l'-l,2n-k}f(l',l,n,2n-l'+l) +\] + +\end_inset + +We have +\begin_inset Formula $0\le k\le n$ +\end_inset + +, hence +\begin_inset Formula $0\le2n-l'+l\le n$ +\end_inset + +, hence +\begin_inset Formula $-2n\le-l'+l\le-n$ +\end_inset + +, hence also +\begin_inset Formula $l'-2n\le l\le l'-n$ +\end_inset + +, which gives the opportunity to swap the +\begin_inset Formula $l,n$ +\end_inset + + sums and the +\begin_inset Formula $l$ +\end_inset + +-sum becomes finite; so also consuming +\begin_inset Formula $\sum_{k=0}^{n}\delta_{l'-l,2n-k}$ +\end_inset + + we get +\begin_inset Formula +\[ +S(l')=\sum_{n=0}^{\infty}\sum_{l=\max(0,l'-2n)}^{l'-n}f(l',l,n,2n-l'+l). +\] + +\end_inset + +Finally, we see that the interval of valid +\begin_inset Formula $l$ +\end_inset + + becomes empty when +\begin_inset Formula $l'-n<0$ +\end_inset + +, i.e. + +\begin_inset Formula $n>l'$ +\end_inset + +; so we get a finite sum +\begin_inset Formula +\[ +S(l')=\sum_{n=0}^{l'}\sum_{l=\max(0,l'-2n)}^{l'-n}f(l',l,n,2n-l'+l). +\] + +\end_inset + +Applying rearrangement, +\begin_inset Formula +\[ +\tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi\mathcal{A}\kappa^{1+l'}}\left(2l'+1\right)!!\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{n=0}^{l'}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{l=\max\left(0,l'-2n\right)}^{l'-n}4\pi i^{l}\left(2\left|\vect s_{\bot}\right|\right)^{2n-l'+l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\sum_{m=-l}^{l}\ush lm\left(\uvec K\right)\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\theta\right)^{l'-l}\left(\cos\varphi\right)^{2n-l'+l}. \] \end_inset @@ -363,6 +527,107 @@ hence \end_layout +\begin_layout Section +Z-aligned lattice +\end_layout + +\begin_layout Standard +Now we set some 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. +\begin_inset Note Note +status open + +\begin_layout Plain Layout +(TODO check the meaning of +\begin_inset Formula $\vect k$ +\end_inset + + and possible additional phase factor.) +\end_layout + +\end_inset + + If we write +\begin_inset Formula $\vect s_{\bot}=\uvec x\left|\vect s_{\bot}\right|\cos\Phi+\uvec y\left|\vect s_{\bot}\right|\sin\Phi$ +\end_inset + +, +\begin_inset Formula $\vect r_{\bot}=\uvec x\left|\vect r_{\bot}\right|\cos\phi+\uvec y\left|\vect r_{\bot}\right|\sin\phi=\uvec x\left|\vect r\right|\sin\theta\cos\phi+\uvec y\left|\vect r\right|\sin\theta\sin\phi$ +\end_inset + +, we have +\begin_inset Formula $\varphi=\phi-\Phi$ +\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 +\[ +\tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi\mathcal{A}\kappa^{1+l'}}\left(2l'+1\right)!!\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{n=0}^{l'}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{l=\max\left(0,l'-2n\right)}^{l'-n}4\pi i^{l}\left(2\left|\vect s_{\bot}\right|\right)^{2n-l'+l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\ush l0\left(\uvec K\right)\underbrace{\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD l0\left(\uvec r\right)\left(\sin\theta\right)^{l'-l}\left(\cos\varphi\right)^{2n-l'+l}}_{\equiv A_{l',l,n,m'}}. +\] + +\end_inset + +Let's also fix the (dual) 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 + +the angular integral then becomes (we also use +\begin_inset Formula $e^{-im'\phi}=e^{im'\Phi}e^{-im'\varphi}$ +\end_inset + +) +\begin_inset Formula +\begin{align*} +A_{l',l,n,m'} & \equiv\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD l0\left(\uvec r\right)\left(\sin\theta\right)^{l'-l}\left(\cos\varphi\right)^{2n-l'+l}\\ + & =\lambda'_{l'm'}\lambda'_{l0}e^{im'\Phi}\int_{0}^{\pi}\ud\theta\,\sin\theta P_{l'}^{-m'}\left(\cos\theta\right)P_{l}^{0}\left(\cos\theta\right)\left(\sin\theta\right)^{l'-l}\int_{0}^{2\pi}\ud\varphi\,e^{-im'\varphi}\left(\cos\varphi\right)^{2n-l'+l}. +\end{align*} + +\end_inset + +The asimuthal integral evaluates to +\begin_inset Formula +\[ +\int_{0}^{2\pi}\ud\varphi\,e^{-im'\varphi}\left(\cos\varphi\right)^{2n-l'+l}=\pi\delta_{\left|m'\right|,2n-l'+l} +\] + +\end_inset + + (note that +\begin_inset Formula $2n-l'+l\ge0$ +\end_inset + + as it's the former index +\begin_inset Formula $k$ +\end_inset + +). + That eliminates one of the two remaining (finite) sums. +\end_layout + \begin_layout Standard \begin_inset Note Note status open @@ -386,6 +651,10 @@ BTW: \end_layout \begin_layout Standard +\begin_inset Note Note +status open + +\begin_layout Plain Layout Now we set the conventions: let the lattice lie on the \begin_inset Formula $z$ \end_inset @@ -418,6 +687,38 @@ Now we set the conventions: let the lattice lie on the \end_inset + +\end_layout + +\begin_layout Plain Layout +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 +\end_layout + +\begin_layout Plain Layout +\begin_inset Formula +\begin{align*} +\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}}\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)\ush l0\left(\uvec K\right)\times\\ + & \quad\times\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD l0\left(\uvec r\right)\sum_{n=0}^{\infty}\Delta_{n+1/2}\frac{1}{n!}\left(-\frac{\left(\left|\vect r_{\bot}\right|^{2}+2\vect r_{\bot}\cdot\vect s_{\bot}\right)\kappa^{2}\gamma_{\vect K}^{2}}{4}\right)^{n}. +\end{align*} + +\end_inset + +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 + Also, in this convention \begin_inset Formula $\ush lm\left(\uvec K\right)=0$ \end_inset @@ -437,7 +738,7 @@ Also, in this convention \end_layout -\begin_layout Standard +\begin_layout Plain Layout Let's also fix the spherical harmonics for now, \begin_inset Formula \[ @@ -449,7 +750,7 @@ Let's also fix the spherical harmonics for now, \end_layout -\begin_layout Standard +\begin_layout Plain Layout The angular integral (assuming it can be separated from the rest like this) is \begin_inset Formula @@ -462,7 +763,7 @@ I_{l'}^{m'}\equiv\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)e^{-\ \end_layout -\begin_layout Standard +\begin_layout Plain Layout Let's further extract the azimuthal part \begin_inset Formula $\left(w\equiv2r_{\bot}s_{\bot}\kappa^{2}\gamma_{\vect K}^{2}/4\tau\right)$ \end_inset @@ -556,7 +857,7 @@ Althought it's not superobvious, this sum is symmetric w.r.t. . \end_layout -\begin_layout Standard +\begin_layout Plain Layout Let's do the polar integration next: \begin_inset Formula $r_{\bot}=r\sin\theta$ \end_inset @@ -616,6 +917,11 @@ so we can fix \end_inset +\end_layout + +\end_inset + + \begin_inset Formula $ $ \end_inset