#LyX 2.4 created this file. For more info see https://www.lyx.org/ \lyxformat 584 \begin_document \begin_header \save_transient_properties true \origin unavailable \textclass article \use_default_options true \maintain_unincluded_children false \language finnish \language_package default \inputencoding utf8 \fontencoding auto \font_roman "default" "default" \font_sans "default" "default" \font_typewriter "default" "default" \font_math "auto" "auto" \font_default_family default \use_non_tex_fonts false \font_sc false \font_roman_osf false \font_sans_osf false \font_typewriter_osf false \font_sf_scale 100 100 \font_tt_scale 100 100 \use_microtype false \use_dash_ligatures true \graphics default \default_output_format default \output_sync 0 \bibtex_command default \index_command default \float_placement class \float_alignment class \paperfontsize 10 \spacing single \use_hyperref false \papersize a3paper \use_geometry true \use_package amsmath 1 \use_package amssymb 1 \use_package cancel 1 \use_package esint 1 \use_package mathdots 1 \use_package mathtools 1 \use_package mhchem 1 \use_package stackrel 1 \use_package stmaryrd 1 \use_package undertilde 1 \cite_engine basic \cite_engine_type default \biblio_style plain \use_bibtopic false \use_indices false \paperorientation portrait \suppress_date false \justification true \use_refstyle 1 \use_minted 0 \use_lineno 0 \index Index \shortcut idx \color #008000 \end_index \leftmargin 2cm \topmargin 2cm \rightmargin 2cm \bottommargin 2cm \secnumdepth 3 \tocdepth 3 \paragraph_separation indent \paragraph_indentation default \is_math_indent 0 \math_numbering_side default \quotes_style english \dynamic_quotes 0 \papercolumns 1 \papersides 1 \paperpagestyle default \tablestyle default \tracking_changes false \output_changes false \html_math_output 0 \html_css_as_file 0 \html_be_strict false \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 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 [Linton, (2.24)] with slightly modified notation and setting \begin_inset Formula $d_{c}=2$ \end_inset : \begin_inset Formula \[ G_{\Lambda}^{(1;\kappa)}\left(\vect r\right)=-\frac{1}{2\pi^{d_{c}/2}\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect r}\int_{1/\eta}^{\infty e^{i\pi/4}}e^{-\kappa^{2}\gamma^{2}t^{2}/4}e^{-\left|\vect r^{\bot}\right|^{2}/t^{2}}t^{1-d_{c}}\ud t \] \end_inset or, evaluated at point \begin_inset Formula $\vect s+\vect r$ \end_inset instead \begin_inset Formula \[ G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)=-\frac{1}{2\pi^{d_{c}/2}\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\left(\vect s+\vect r\right)}\int_{1/\eta}^{\infty e^{i\pi/4}}e^{-\kappa^{2}\gamma^{2}t^{2}/4}e^{-\left|\vect s^{\bot}+\vect r^{\bot}\right|^{2}/t^{2}}t^{1-d_{c}}\ud t \] \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 \[ G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)=-\frac{1}{2\pi^{d_{c}/2}\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\left(\vect s+\vect r\right)}\int_{\kappa^{2}\gamma_{m}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{m}^{2}/4\tau}\tau^{-\frac{d_{c}}{2}}\ud\tau \] \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 \[ \int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi^{d_{c}/2}\mathcal{A}}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\left(\vect s+\vect r\right)}\int_{\kappa^{2}\gamma_{\vect K}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\tau^{-\frac{d_{c}}{2}}\ud\tau \] \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 \[ \int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi^{d_{c}/2}\mathcal{A}}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ushD lm\left(\uvec r\right)\ush lm\left(\uvec K\right)\int_{\kappa^{2}\gamma_{\vect K}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\tau^{-\frac{d_{c}}{2}}\ud\tau \] \end_inset \end_layout \begin_layout Standard We also have \begin_inset Formula \begin{align*} e^{-\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau} & =e^{-\left(\left|\vect s_{\bot}\right|^{2}+\left|\vect r_{\bot}\right|^{2}+2\vect r_{\bot}\cdot\vect s_{\bot}\right)\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\\ & =e^{-\left|\vect s_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\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\tau}\right)^{n}, \end{align*} \end_inset 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^{d_{c}/2}\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^{-\frac{d_{c}}{2}-n}\ud\tau}_{\Delta_{n}^{\left(d_{\Lambda}\right)}}\\ & =-\frac{1}{2\pi^{d_{c}/2}\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}^{\left(d_{\Lambda}\right)}}{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^{d_{c}/2}\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}^{\left(d_{\Lambda}\right)}\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 \[ \int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi^{d_{c}/2}\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n}^{\left(d_{\Lambda}\right)}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left|\vect r_{\bot}\right|^{2n-k}\left(\cos\varphi\right)^{k} \] \end_inset and if we label \begin_inset Formula $\left|\vect r\right|\sin\vartheta\equiv\left|\vect r_{\bot}\right|$ \end_inset \begin_inset Formula \[ \int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi^{d_{c}/2}\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n}^{\left(d_{\Lambda}\right)}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\left|\vect r\right|^{2n-k}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\vartheta\right)^{2n-k}\left(\cos\varphi\right)^{k} \] \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 \[ \tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi^{d_{c}/2}\mathcal{A}\kappa^{1+l'}}\left(2l'+1\right)!!\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n}^{\left(d_{\Lambda}\right)}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\delta_{l'-l,2n-k}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\vartheta\right)^{l'-l}\left(\cos\varphi\right)^{k}. \] \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 \[ \tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi^{d_{c}/2}\mathcal{A}\kappa}\frac{\left(2l'+1\right)!!}{\kappa^{l'}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{n=0}^{l'}\frac{\left(-1\right)^{n}}{n!}\Delta_{n}^{\left(d_{\Lambda}\right)}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{l=\max\left(0,l'-2n\right)}^{l'-n}4\pi i^{l}\left(2\left|\vect s_{\bot}\right|\right)^{2n-l'+l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\sum_{m=-l}^{l}\ush lm\left(\uvec K\right)\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\vartheta\right)^{l'-l}\left(\cos\varphi\right)^{2n-l'+l}, \] \end_inset or replacing the anles with their original definition, \begin_inset Formula \[ \tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi^{d_{c}/2}\mathcal{A}\kappa}\frac{\left(2l'+1\right)!!}{\kappa^{l'}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{n=0}^{l'}\frac{\left(-1\right)^{n}}{n!}\Delta_{n}^{\left(d_{\Lambda}\right)}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{l=\max\left(0,l'-2n\right)}^{l'-n}4\pi i^{l}\left(2\left|\vect s_{\bot}\right|\right)^{2n-l'+l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\sum_{m=-l}^{l}\ush lm\left(\uvec K\right)\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\frac{\left|\vect r_{\bot}\right|}{\left|\vect r\right|}\right)^{l'-l}\left(\frac{\vect r_{\bot}\cdot\vect s_{\bot}}{\left|\vect r_{\bot}\right|\left|\vect s_{\bot}\right|}\right)^{2n-l'+l}, \] \end_inset and if we want a \begin_inset Formula $\sigma_{l'}^{m'}\left(\vect s,\vect k\right)$ \end_inset instead, we reverse the sign of \begin_inset Formula $\vect s$ \end_inset and replace all spherical harmonics with their dual counterparts: \begin_inset Formula \[ \sigma_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi^{d_{c}/2}\mathcal{A}\kappa}\frac{\left(2l'+1\right)!!}{\kappa^{l'}}\sum_{\vect K\in\Lambda^{*}}e^{-i\vect K\cdot\vect s}\sum_{n=0}^{l'}\frac{\left(-1\right)^{n}}{n!}\Delta_{n}^{\left(d_{\Lambda}\right)}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{l=\max\left(0,l'-2n\right)}^{l'-n}4\pi i^{l}\left(2\left|\vect s_{\bot}\right|\right)^{2n-l'+l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\sum_{m=-l}^{l}\ushD lm\left(\uvec K\right)\int\ud\Omega_{\vect r}\,\ush{l'}{m'}\left(\uvec r\right)\ush lm\left(\uvec r\right)\left(\frac{\left|\vect r_{\bot}\right|}{\left|\vect r\right|}\right)^{l'-l}\left(\frac{-\vect r_{\bot}\cdot\vect s_{\bot}}{\left|\vect r_{\bot}\right|\left|\vect s_{\bot}\right|}\right)^{2n-l'+l}, \] \end_inset and remembering that in the plane wave expansion the \begin_inset Quotes eld \end_inset duality \begin_inset Quotes erd \end_inset is interchangeable, \begin_inset Formula \[ \sigma_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi^{d_{c}/2}\mathcal{A}\kappa}\frac{\left(2l'+1\right)!!}{\kappa^{l'}}\sum_{\vect K\in\Lambda^{*}}e^{-i\vect K\cdot\vect s}\sum_{n=0}^{l'}\frac{\left(-1\right)^{n}}{n!}\Delta_{n}^{\left(d_{\Lambda}\right)}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{l=\max\left(0,l'-2n\right)}^{l'-n}4\pi i^{l}\left(2\left|\vect s_{\bot}\right|\right)^{2n-l'+l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\sum_{m=-l}^{l}\ush lm\left(\uvec K\right)\underbrace{\int\ud\Omega_{\vect r}\,\ush{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\frac{\left|\vect r_{\bot}\right|}{\left|\vect r\right|}\right)^{l'-l}\left(\frac{-\vect r_{\bot}\cdot\vect s_{\bot}}{\left|\vect r_{\bot}\right|\left|\vect s_{\bot}\right|}\right)^{2n-l'+l}}_{\equiv A_{l',l,m',m,n}^{\left(d_{\Lambda}\right)}}. \] \end_inset The angular integral is easier to evaluate when \begin_inset Formula $d_{\Lambda}=2$ \end_inset , because then \begin_inset Formula $\vect r_{\bot}$ \end_inset is parallel (or antiparallel) to \begin_inset Formula $\vect s_{\bot}$ \end_inset , which gives \begin_inset Formula \[ A_{l',l,m',m,n}^{\left(2\right)}=\left(-\frac{\vect r_{\bot}\cdot\vect s_{\bot}}{\left|\vect r_{\bot}\cdot\vect s_{\bot}\right|}\right)^{2n-l'+l}\int\ud\Omega_{\vect r}\,\ush{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\frac{\left|\vect r_{\bot}\right|}{\left|\vect r\right|}\right)^{2n} \] \end_inset and if we set the normal of the lattice correspond to the \begin_inset Formula $z$ \end_inset axis, the azimuthal part of the integral will become zero unless \begin_inset Formula $m'=m$ \end_inset for any meaningful spherical harmonics convention, and the polar part for the only nonzero case has a closed-form expression, see e.g. [Linton (A.15)], so one arrives at an expression similar to [Kambe II, (3.15)] \lang english \begin_inset Formula \begin{multline} \sigma_{l,m}^{\left(\mathrm{L},\eta\right)}\left(\vect k,\vect s\right)=-\frac{i^{l+1}}{\kappa^{2}\mathcal{A}}\pi^{3/2}2\left(\left(l-m\right)/2\right)!\left(\left(l+m\right)/2\right)!\times\\ \times\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\ush lm\left(\vect k+\vect K\right)\sum_{j=0}^{l-\left|m\right|}\left(-1\right)^{j}\gamma_{\vect K}^{2}^{2j+1}\times\\ \times\Delta_{j}\left(\frac{\kappa^{2}\gamma_{\vect K}^{2}}{4\eta^{2}},-i\kappa\gamma_{\vect K}^{2}s_{\perp}\right)\times\\ \times\sum_{\substack{s\\ j\le s\le\min\left(2j,l-\left|m\right|\right)\\ l-n+\left|m\right|\,\mathrm{even} } }\frac{1}{\left(2j-s\right)!\left(s-j\right)!}\frac{\left(-\kappa s_{\perp}\right)^{2j-s}\left(\left|\vect k+\vect K\right|/\kappa\right)^{l-s}}{\left(\frac{1}{2}\left(l-m-s\right)\right)!\left(\frac{1}{2}\left(l+m-s\right)\right)!}\label{eq:Ewald in 3D long-range part 1D 2D-1} \end{multline} \end_inset where \begin_inset Formula $s_{\perp}\equiv\vect s\cdot\uvec z=\vect s_{\bot}\cdot\uvec z$ \end_inset . If \begin_inset Formula $d_{\Lambda}=1$ \end_inset , the angular becomes more complicated to evaluate due to the different behaviour of the \begin_inset Formula $\vect r_{\bot}\cdot\vect s_{\bot}/\left|\vect r_{\bot}\right|\left|\vect s_{\bot}\right|$ \end_inset factor. The choice of coordinates can make most of the terms dissapear: if the lattice is set parallel to the \begin_inset Formula $z$ \end_inset axis, \begin_inset Formula $A_{l',l,m',m,n}^{\left(1\right)}$ \end_inset is zero unless \begin_inset Formula $m=0$ \end_inset , but one still has \begin_inset Formula \[ A_{l',l,m',0,n}^{\left(1\right)}=\pi\delta_{m',l'-l-2n}\lambda'_{l0}\lambda_{l'm'}\int_{-1}^{1}\ud x\,P_{l'}^{m'}\left(x\right)P_{l}^{0}\left(x\right)\left(1-x^{2}\right)^{\frac{l'-l}{2}} \] \end_inset where \begin_inset Formula $\lambda_{lm}$ \end_inset are constants depending on the conventions for spherical harmonics. This does not seem to have such a nice closed-form expression as in the 2D case, but it can be evaluated e.g. using the common recurrence relations for associated Legendre polynomials. \end_layout \end_body \end_document