qpms/notes/ewald_1D_in_3D.lyx

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#LyX 2.4 created this file. For more info see https://www.lyx.org/
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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
\[
G_{\Lambda}^{(1;\kappa)}\left(\vect r\right)=-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect r}\int_{1/\eta}^{\infty e^{i\pi/4}}e^{-\kappa^{2}\gamma^{2}t^{2}/4}e^{-\left|\vect r^{\bot}\right|^{2}/t^{2}}t^{-1}\ud t
\]
\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
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\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
\[
\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left|\vect r_{\bot}\right|^{2n-k}\left(\cos\varphi\right)^{k}
\]
\end_inset
and if we label
\begin_inset Formula $\left|\vect r\right|\sin\vartheta\equiv\left|\vect r_{\bot}\right|$
\end_inset
\begin_inset Formula
\[
\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\left|\vect r\right|^{2n-k}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\vartheta\right)^{2n-k}\left(\cos\varphi\right)^{k}
\]
\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
\[
\tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi\mathcal{A}\kappa^{1+l'}}\left(2l'+1\right)!!\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\delta_{l'-l,2n-k}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\vartheta\right)^{l'-l}\left(\cos\varphi\right)^{k}.
\]
\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
\[
\tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi\mathcal{A}\kappa^{1+l'}}\left(2l'+1\right)!!\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{n=0}^{l'}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{l=\max\left(0,l'-2n\right)}^{l'-n}4\pi i^{l}\left(2\left|\vect s_{\bot}\right|\right)^{2n-l'+l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\sum_{m=-l}^{l}\ush lm\left(\uvec K\right)\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\vartheta\right)^{l'-l}\left(\cos\varphi\right)^{2n-l'+l}.
\]
\end_inset
\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
, and
\begin_inset Formula $\vartheta=\theta$
\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.
We are left with the polar integral
\begin_inset Formula
\[
\int_{0}^{\pi}\ud\theta\,\sin\theta P_{l'}^{-m'}\left(\cos\theta\right)P_{l}^{0}\left(\cos\theta\right)\left(\sin\theta\right)^{l'-l}
\]
\end_inset
for which I couldn't find an explicit form yet.
\end_layout
\begin_layout Section
X-aligned lattice
\end_layout
\begin_layout Standard
If we instead set
\begin_inset Formula $\vect s_{\bot}=\uvec z\left|\vect s_{\bot}\right|\cos\Theta+\uvec y\left|\vect s_{\bot}\right|\sin\Theta$
\end_inset
,
\begin_inset Formula $\vect r_{\bot}=\uvec z\left|\vect r_{\bot}\right|\cos\theta+\uvec y\left|\vect r_{\bot}\right|\sin\theta=\uvec z\left|\vect r\right|\cos\theta+\uvec y\left|\vect r\right|\sin\theta\sin\phi$
\end_inset
, we have
\begin_inset Formula $\vartheta=\Theta-\theta$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\vartheta\right)^{l'-l}\left(\cos\varphi\right)^{k}
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Note Note
status open
\begin_layout Plain Layout
BTW:
\begin_inset Formula
\begin{align*}
\left|\vect r_{\bot}\right|^{2} & =\left|\vect r\right|^{2}-\left|\vect r_{\parallel}\right|^{2}=\left|\vect r\right|^{2}-\left(\vect r\cdot\uvec K\right)^{2},\\
\vect r_{\bot}\cdot\vect s_{\bot} & =\vect r\cdot\vect s_{\bot}
\end{align*}
\end_inset
\end_layout
\end_inset
\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
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 $\vect r_{\bot}=\uvec xr_{\bot}\cos\phi+\uvec yr_{\bot}\sin\phi=\uvec xr\sin\theta\cos\phi+\uvec yr\sin\theta\sin\phi$
\end_inset
, we have
\begin_inset Formula
\[
\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}=s_{\bot}^{2}+r^{2}\left(\sin\theta\right)^{2}+2s_{\bot}r\sin\theta\cos\left(\phi-\Phi\right).
\]
\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
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 Plain Layout
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 Plain Layout
The angular integral (assuming it can be separated from the rest like this)
is
\begin_inset Formula
\[
I_{l'}^{m'}\equiv\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)e^{-\left(r_{\bot}^{2}+2s_{\bot}r_{\bot}\cos\left(\phi-\Phi\right)\right)^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}
\]
\end_inset
\end_layout
\begin_layout 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
\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
Using [DLMF 10.9.2],
\begin_inset Formula $\int_{0}^{2\pi}e^{-im'\varphi}e^{-w\cos\varphi}\ud\varphi=\int_{0}^{2\pi}\cos\left(m'\varphi\right)e^{i(iw)\cos\varphi}=2\pi i^{m'}J_{m'}\left(iw\right)$
\end_inset
we have
\begin_inset Formula
\[
e^{-m'\Phi}A_{l'}^{m'}=2\pi i^{m'}J_{m'}\left(iw\right),
\]
\end_inset
assuming that
\begin_inset Formula $w$
\end_inset
is real (which does not necessarily have to be true!); numerical experiments
in Sage show that the result is valid also for complex
\begin_inset Formula $w$
\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^{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
\end_layout
\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 Plain Layout
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
If we now perform the limit
\begin_inset Formula $r\to0$
\end_inset
and compare the radial parts (incl.
those in
\begin_inset Formula $u,v$
\end_inset
) powers, the leading term indices will have
\begin_inset Formula
\[
l'\sim l+2a+2k-m'
\]
\end_inset
so we can fix
\begin_inset Formula $2a+2k-m'=l'-l$
\end_inset
and get
\begin_inset Formula
\[
\int_{0}^{\pi}\sin\theta\ud\theta\,P_{l'}^{-m'}\left(\cos\theta\right)P_{l}^{0}\left(\cos\theta\right)\left(\sin\theta\right)^{l'-l}=\begin{cases}
0 & l'-l+m'\text{ odd}\\
? & l'-l+m'\text{ even}
\end{cases}
\]
\end_inset
\end_layout
\end_inset
\begin_inset Formula $ $
\end_inset
\end_layout
\end_body
\end_document