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\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 
\[
I_{l}^{m}\equiv\int\ud\Omega_{\vect r}\,\ushD lm\left(\uvec r\right)e^{-\left(r_{\bot}^{2}+2s_{\bot}r_{\bot}\cos\left(\phi-\Phi\right)\right)^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}
\]

\end_inset


\end_layout

\begin_layout Standard
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


\begin_inset Formula 
\[
e^{-im\Phi}A_{l}^{m}\equiv\int_{0}^{2\pi}e^{-im\phi}e^{-w\cos\left(\phi-\Phi\right)}\ud\phi=e^{-im\Phi}\int_{0}^{2\pi}e^{-im\varphi}e^{-w\cos\varphi}\ud\varphi
\]

\end_inset


\begin_inset Formula 
\begin{align*}
A_{l}^{m} & =\int_{0}^{2\pi}e^{-im\varphi}\sum_{n=0}^{\infty}\frac{\left(-w\cos\varphi\right)^{n}}{n!}\ud\varphi\\
 & =\int_{0}^{2\pi}e^{-im\varphi}\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\left(e^{i\varphi}+e^{-i\varphi}\right)^{n}\ud\varphi\\
 & =\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\int_{0}^{2\pi}e^{-im\varphi}\sum_{k=0}^{n}\binom{n}{k}e^{ik\varphi}e^{-i\left(n-k\right)\varphi}\ud\varphi\\
 & =\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\sum_{k=0}^{n}\binom{n}{k}\int_{0}^{2\pi}e^{i\left(2k-n-m\right)\varphi}\ud\varphi\\
 & =2\pi\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\sum_{k=0}^{n}\binom{n}{k}\delta_{2k-n-m=0}\\
 & =2\pi\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\sum_{k=0}^{n}\binom{n}{k}\delta_{2k-n-m=0}\\
 & =2\pi\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\binom{n}{k}\delta_{2k-n-m=0}\\
 & =2\pi\sum_{k=0}^{\infty}\frac{\left(-w\right)^{2k-m}}{2^{2k-m}\left(2k-m\right)!}\binom{2k-m}{k}\delta_{2k-m\ge k}\\
 & =2\pi\sum_{k=0}^{\infty}\frac{\left(-w\right)^{2k-m}}{2^{2k-m}}\frac{1}{k!\left(k-m\right)!}\delta_{k-m\ge0}\\
 & =2\pi\sum_{k=\max\left(m,0\right)}^{\infty}\left(-\frac{w}{2}\right)^{2k-m}\frac{1}{k!\left(k-m\right)!}
\end{align*}

\end_inset


\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula 
\begin{align*}
A_{l}^{m} & =\int_{0}^{2\pi}e^{-im\varphi}\sum_{n=0}^{\infty}\frac{\left(-w\cos\varphi\right)^{n}}{n!}\ud\varphi\\
 & =\int_{0}^{2\pi}e^{-im\varphi}\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\left(e^{i\varphi}+e^{-i\varphi}\right)^{n}\ud\varphi\\
 & =\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\int_{0}^{2\pi}e^{-im\varphi}\sum_{k=0}^{n}\binom{n}{k}e^{i\left(n-k\right)\varphi}e^{-ik\varphi}\ud\varphi\\
 & =\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\sum_{k=0}^{n}\binom{n}{k}\int_{0}^{2\pi}e^{i\left(-2k+n-m\right)\varphi}\ud\varphi\\
 & =2\pi\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\sum_{k=0}^{n}\binom{n}{k}\delta_{-2k+n-m=0}\\
 & =2\pi\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\sum_{k=0}^{n}\binom{n}{k}\delta_{-2k+n-m=0}\\
 & =2\pi\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\binom{n}{k}\delta_{-2k+n-m=0}\\
 & =2\pi\sum_{k=0}^{\infty}\frac{\left(-w\right)^{2k+m}}{2^{2k+m}\left(2k+m\right)!}\binom{2k+m}{k}\delta_{2k+m\ge k}\\
 & =2\pi\sum_{k=0}^{\infty}\frac{\left(-w\right)^{2k+m}}{2^{2k+m}}\frac{1}{k!\left(k+m\right)!}\delta_{k+m\ge0}\\
 & =2\pi\sum_{k=\max\left(-m,0\right)}^{\infty}\frac{\left(-w\right)^{2k+m}}{2^{2k+m}}\frac{1}{k!\left(k+m\right)!}
\end{align*}

\end_inset


\end_layout

\end_inset

Althought it's not superobvious, this sum is symmetric w.r.t.
 sign change in 
\begin_inset Formula $m$
\end_inset

.
\end_layout

\begin_layout Standard
Let's do the polar integration next: 
\begin_inset Formula $r_{\bot}=r\sin\theta$
\end_inset


\begin_inset Formula 
\[
B_{l}^{m}\equiv\int_{0}^{\pi}\sin\theta\ud\theta\,P_{l}^{-m}\left(\cos\theta\right)P_{l}^{0}\left(\cos\theta\right)e^{-\left(\sin\theta\right)^{2}r^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\left(-\sin\theta\,rs_{\bot}\kappa^{2}\gamma_{\vect K}^{2}/4\tau\right)^{2k-m}
\]

\end_inset

Label 
\begin_inset Formula $u\equiv r^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau,v\equiv rs_{\bot}\kappa^{2}\gamma_{\vect K}^{2}/4\tau$
\end_inset

; then 
\begin_inset Formula 
\begin{align*}
B_{l}^{m} & =\int_{0}^{\pi}\sin\theta\ud\theta\,P_{l}^{-m}\left(\cos\theta\right)P_{l}^{0}\left(\cos\theta\right)e^{-u\left(\sin\theta\right)^{2}}\left(-v\sin\theta\right)^{2k-m}\\
 & =\int_{0}^{\pi}\sin\theta\ud\theta\,P_{l}^{-m}\left(\cos\theta\right)P_{l}^{0}\left(\cos\theta\right)\left(-v\sin\theta\right)^{2k-m}\sum_{a=0}^{\infty}\frac{\left(-u\right)^{a}}{a!}\left(\sin\theta\right)^{2a}\\
 & =\left(-v\right)^{2k-m}\sum_{a=0}^{\infty}\frac{\left(-u\right)^{a}}{a!}\int_{0}^{\pi}\sin\theta\ud\theta\,P_{l}^{-m}\left(\cos\theta\right)P_{l}^{0}\left(\cos\theta\right)\left(\sin\theta\right)^{2a+2k-m}
\end{align*}

\end_inset


\end_layout

\end_body
\end_document