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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 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*}
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\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}
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\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
\[
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\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi^{d_{c}/2}\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n}^{\left(d_{\Lambda}\right)}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left|\vect r_{\bot}\right|^{2n-k}\left(\cos\varphi\right)^{k}
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\]
\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
\[
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\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi^{d_{c}/2}\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n}^{\left(d_{\Lambda}\right)}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\left|\vect r\right|^{2n-k}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\vartheta\right)^{2n-k}\left(\cos\varphi\right)^{k}
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\]
\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
\[
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\tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi^{d_{c}/2}\mathcal{A}\kappa^{1+l'}}\left(2l'+1\right)!!\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n}^{\left(d_{\Lambda}\right)}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\delta_{l'-l,2n-k}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\vartheta\right)^{l'-l}\left(\cos\varphi\right)^{k}.
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\]
\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
\[
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\tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi^{d_{c}/2}\mathcal{A}\kappa}\frac{\left(2l'+1\right)!!}{\kappa^{l'}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{n=0}^{l'}\frac{\left(-1\right)^{n}}{n!}\Delta_{n}^{\left(d_{\Lambda}\right)}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{l=\max\left(0,l'-2n\right)}^{l'-n}4\pi i^{l}\left(2\left|\vect s_{\bot}\right|\right)^{2n-l'+l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\sum_{m=-l}^{l}\ush lm\left(\uvec K\right)\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\vartheta\right)^{l'-l}\left(\cos\varphi\right)^{2n-l'+l},
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\]
\end_inset
or replacing the anles with their original definition,
\begin_inset Formula
\[
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\tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi^{d_{c}/2}\mathcal{A}\kappa}\frac{\left(2l'+1\right)!!}{\kappa^{l'}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{n=0}^{l'}\frac{\left(-1\right)^{n}}{n!}\Delta_{n}^{\left(d_{\Lambda}\right)}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{l=\max\left(0,l'-2n\right)}^{l'-n}4\pi i^{l}\left(2\left|\vect s_{\bot}\right|\right)^{2n-l'+l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\sum_{m=-l}^{l}\ush lm\left(\uvec K\right)\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\frac{\left|\vect r_{\bot}\right|}{\left|\vect r\right|}\right)^{l'-l}\left(\frac{\vect r_{\bot}\cdot\vect s_{\bot}}{\left|\vect r_{\bot}\right|\left|\vect s_{\bot}\right|}\right)^{2n-l'+l},
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\]
\end_inset
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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
2020-06-21 21:47:34 +03:00
2020-06-22 05:46:35 +03:00
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.
2020-06-21 21:47:34 +03:00
\end_layout
\end_body
\end_document
