[ewald]Formulating the decomposition problem.

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Marek Nečada 2017-08-07 16:48:37 +03:00
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commit f3d27e74d8
1 changed files with 214 additions and 34 deletions

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@ -152,6 +152,16 @@
\end_inset
\begin_inset FormulaMacro
\newcommand{\ints}{\mathbb{Z}}
\end_inset
\begin_inset FormulaMacro
\newcommand{\reals}{\mathbb{R}}
\end_inset
\end_layout
\begin_layout Title
@ -166,6 +176,24 @@ Accelerating lattice mode calculations with
Marek Nečada
\end_layout
\begin_layout Abstract
The
\begin_inset Formula $T$
\end_inset
-matrix approach is the method of choice for simulating optical response
of a reasonably small system of compact linear scatterers on isotropic
background.
However, its direct utilisation for problems with infinite lattices is
problematic due to slowly converging sums over the lattice.
Here I develop a way to compute the problematic sums in the reciprocal
space, making the
\begin_inset Formula $T$
\end_inset
-matrix method very suitable for infinite periodic systems as well.
\end_layout
\begin_layout Section
Formulation of the problem
\end_layout
@ -308,7 +336,7 @@ reference "eq:W definition"
\end_inset
-dimensional lattice.
In electrostatics, one can solve this with problem with Ewald summation.
In electrostatics, one can solve this problem with Ewald summation.
Its basic idea is that if what asymptoticaly decays poorly in the direct
space, will perhaps decay fast in the Fourier space.
I use the same idea here, but everything will be somehow harder than in
@ -404,13 +432,21 @@ where changed the sign of
\end_inset
.
Fourier transform of product is convolution of Fourier transforms, so
Fourier transform of product is convolution of Fourier transforms, so (using
formula
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:Dirac comb uaFt"
\end_inset
for the Fourier transform of Dirac comb)
\begin_inset Formula
\begin{eqnarray*}
W_{\alpha\beta}(\vect k) & = & \left(\left(\uaft{\dc{\basis u}}\right)\ast\left(\uaft{S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\right)(\vect k)\\
& = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\left(\dc{\rec{\basis u}}^{(d)}\left(\frac{1}{2\pi}\vect{\circ}\right)\ast\left(\uaft{S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\right)\left(\vect k\right)\quad\mbox{(re-check facs)}\\
& = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}
\end{eqnarray*}
\begin{eqnarray}
W_{\alpha\beta}(\vect k) & = & \left(\left(\uaft{\dc{\basis u}}\right)\ast\left(\uaft{S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\right)(\vect k)\nonumber \\
& = & \frac{\left|\det\recb{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\left(\dc{\recb{\basis u}}^{(d)}\ast\left(\uaft{S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\right)\left(\vect k\right)\nonumber \\
& = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\sum_{\vect K\in\recb{\basis u}\ints^{d}}\left(\uaft{S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\left(\vect k-\vect K\right).\label{eq:W sum in reciprocal space}
\end{eqnarray}
\end_inset
@ -434,7 +470,134 @@ Factor
\end_inset
As such, this is not extremely helpful because the the
\emph on
whole
\emph default
translation operator
\begin_inset Formula $S$
\end_inset
has singularities in origin, hence its Fourier transform
\begin_inset Formula $\uaft S$
\end_inset
will decay poorly.
\end_layout
\begin_layout Standard
However, Fourier transform is linear, so we can in principle separate
\begin_inset Formula $S$
\end_inset
in two parts,
\begin_inset Formula $S=S^{\textup{L}}+S^{\textup{S}}$
\end_inset
.
\begin_inset Formula $S^{\textup{S}}$
\end_inset
is a short-range part that decays sufficiently fast with distance so that
its direct-space lattice sum converges well;
\begin_inset Formula $S^{\textup{S}}$
\end_inset
must as well contain all the singularities of
\begin_inset Formula $S$
\end_inset
in the origin.
The other part,
\begin_inset Formula $S^{\textup{L}}$
\end_inset
, will retain all the slowly decaying terms of
\begin_inset Formula $S$
\end_inset
but it also has to be smooth enough in the origin, so that its Fourier
transform
\begin_inset Formula $\uaft{S^{\textup{L}}}$
\end_inset
decays fast enough.
(The same idea lies behind the Ewald summation in electrostatics.) Using
the linearity of Fourier transform and formulae
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:W definition"
\end_inset
and
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:W sum in reciprocal space"
\end_inset
, the operator
\begin_inset Formula $W_{\alpha\beta}$
\end_inset
can then be re-expressed as
\begin_inset Formula
\begin{eqnarray}
W_{\alpha\beta}\left(\vect k\right) & = & W_{\alpha\beta}^{\textup{S}}\left(\vect k\right)+W_{\alpha\beta}^{\textup{L}}\left(\vect k\right)\nonumber \\
W_{\alpha\beta}^{\textup{S}}\left(\vect k\right) & = & \sum_{\vect R\in\basis u\ints^{d}}S^{\textup{S}}(\vect 0\leftarrow\vect R+\vect r_{\beta}-\vect r_{\alpha})e^{i\vect k\cdot\vect R}\label{eq:W Short definition}\\
W_{\alpha\beta}^{\textup{L}}\left(\vect k\right) & = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\sum_{\vect K\in\recb{\basis u}\ints^{d}}\left(\uaft{S^{\textup{L}}(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\left(\vect k-\vect K\right)\label{eq:W Long definition}
\end{eqnarray}
\end_inset
where both sums should converge nicely.
\end_layout
\begin_layout Section
Finding a good decomposition
\end_layout
\begin_layout Standard
The remaining challenge is therefore finding a suitable decomposition
\begin_inset Formula $S^{\textup{L}}+S^{\textup{S}}$
\end_inset
such that both
\begin_inset Formula $S^{\textup{S}}$
\end_inset
and
\begin_inset Formula $\uaft{S^{\textup{L}}}$
\end_inset
decay fast enough with distance and are expressable analytically.
With these requirements, I do not expect to find gaussian asymptotics as
in the electrostatic Ewald formula—having
\begin_inset Formula $\sim x^{-t}$
\end_inset
,
\begin_inset Formula $t>d$
\end_inset
asymptotics would be nice, making the sums in
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:W Short definition"
\end_inset
,
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:W Long definition"
\end_inset
absolutely convergent.
\end_layout
\begin_layout Section
@ -528,7 +691,7 @@ we have
\end_inset
we have
we have (CHECK)
\begin_inset Formula
\[
\uaft{Ш_{T}}(\omega)=\frac{\sqrt{2\pi}}{T}Ш_{\frac{2\pi}{T}}(\omega)=\frac{1}{\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}e^{-i\omega nT}
@ -644,7 +807,8 @@ From the scaling property of delta function,
\end_layout
\begin_layout Standard
From the book:
From the Osgood's book (p.
375):
\end_layout
\begin_layout Standard
@ -722,22 +886,19 @@ reference "eq:Dirac comb factorisation"
\end_layout
\begin_layout Subsubsection
Fourier transform
Fourier transform (OK)
\end_layout
\begin_layout Standard
From the book https://see.stanford.edu/materials/lsoftaee261/chap8.pdf, p.
From the Osgood's book https://see.stanford.edu/materials/lsoftaee261/chap8.pdf,
p.
379
\end_layout
\begin_layout Standard
(CHECK THIS)
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\uoft{\dc A}\left(\vect{\xi}\right)=\left|\det A^{-T}\right|\dc{}^{(d)}\left(A^{-T}\vect{\xi}\right).
\uoft{\dc{\basis u}}\left(\vect{\xi}\right)=\left|\det\rec{\basis u}\right|\dc{\rec{\basis u}}^{(d)}\left(\vect{\xi}\right).
\]
\end_inset
@ -746,29 +907,48 @@ And consequently, for unitary/angular frequency it is
\end_layout
\begin_layout Standard
(CHECK THIS)
\begin_inset Formula
\begin{eqnarray}
\uaft{\dc{\basis u}}\left(\vect k\right) & = & \frac{1}{\left(2\pi\right)^{\frac{d}{2}}}\uoft{\dc{\basis u}}\left(\frac{\vect k}{2\pi}\right)\nonumber \\
& = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\dc{\rec{\basis u}}^{(d)}\left(\frac{\vect k}{2\pi}\right)\nonumber \\
& = & \left(2\pi\right)^{\frac{d}{2}}\left|\det\rec{\basis u}\right|\dc{\recb{\basis u}}\left(\vect k\right)\nonumber \\
& = & \frac{\left|\det\recb{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\dc{\recb{\basis u}}\left(\vect k\right).\label{eq:Dirac comb uaFt}
\end{eqnarray}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Note Note
status open
\begin_layout Plain Layout
On the third line, we used the stretch theorem, getting
\begin_inset Formula
\[
\dc{\recb{\basis u}}\left(\vect k\right)=\dc{2\pi\rec{\basis u}}\left(\vect k\right)=\left(2\pi\right)^{-d}\dc{\rec{\basis u}}\left(\frac{\vect k}{2\pi}\right)
\]
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Subsubsection
Convolution
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\uaft{\dc A}\left(\vect{\xi}\right)=\frac{\left|\det A^{-T}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\dc{}^{(d)}\left(\frac{1}{2\pi}A^{-T}\vect{\xi}\right).
\]
\end_inset
Using my own
\begin_inset Quotes eld
\end_inset
basis notation
\begin_inset Quotes erd
\end_inset
TODO
\begin_inset Formula
\[
\uoft{\dc{\basis u}}\left(\vect{\xi}\right)=\left|\det\rec{\basis u}\right|\dc{}^{(d)}\left(A^{-T}\vect{\xi}\right).
\left(f\ast\dc{\basis u}\right)(\vect x)=\sum_{\vect t\in\basis u\ints^{d}}f(\vect x-\vect t)
\]
\end_inset