[ewald]Formulating the decomposition problem.
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Marek Nečada
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notes/ewald.lyx
248
notes/ewald.lyx
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@ -152,6 +152,16 @@
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\ints}{\mathbb{Z}}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\reals}{\mathbb{R}}
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\end_inset
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\end_layout
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\begin_layout Title
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@ -166,6 +176,24 @@ Accelerating lattice mode calculations with
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Marek Nečada
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\end_layout
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\begin_layout Abstract
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The
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\begin_inset Formula $T$
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\end_inset
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-matrix approach is the method of choice for simulating optical response
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of a reasonably small system of compact linear scatterers on isotropic
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background.
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However, its direct utilisation for problems with infinite lattices is
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problematic due to slowly converging sums over the lattice.
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Here I develop a way to compute the problematic sums in the reciprocal
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space, making the
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\begin_inset Formula $T$
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\end_inset
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-matrix method very suitable for infinite periodic systems as well.
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\end_layout
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\begin_layout Section
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Formulation of the problem
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\end_layout
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@ -308,7 +336,7 @@ reference "eq:W definition"
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\end_inset
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-dimensional lattice.
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In electrostatics, one can solve this with problem with Ewald summation.
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In electrostatics, one can solve this problem with Ewald summation.
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Its basic idea is that if what asymptoticaly decays poorly in the direct
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space, will perhaps decay fast in the Fourier space.
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I use the same idea here, but everything will be somehow harder than in
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@ -404,13 +432,21 @@ where changed the sign of
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\end_inset
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.
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Fourier transform of product is convolution of Fourier transforms, so
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Fourier transform of product is convolution of Fourier transforms, so (using
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formula
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:Dirac comb uaFt"
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\end_inset
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for the Fourier transform of Dirac comb)
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\begin_inset Formula
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\begin{eqnarray*}
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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)\\
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& = & \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)}\\
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& = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}
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\end{eqnarray*}
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\begin{eqnarray}
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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 \\
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& = & \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 \\
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& = & \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}
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\end{eqnarray}
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\end_inset
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@ -434,7 +470,134 @@ Factor
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\end_inset
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As such, this is not extremely helpful because the the
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\emph on
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whole
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\emph default
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translation operator
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\begin_inset Formula $S$
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\end_inset
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has singularities in origin, hence its Fourier transform
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\begin_inset Formula $\uaft S$
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\end_inset
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will decay poorly.
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\end_layout
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\begin_layout Standard
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However, Fourier transform is linear, so we can in principle separate
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\begin_inset Formula $S$
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\end_inset
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in two parts,
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\begin_inset Formula $S=S^{\textup{L}}+S^{\textup{S}}$
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\end_inset
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.
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\begin_inset Formula $S^{\textup{S}}$
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\end_inset
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is a short-range part that decays sufficiently fast with distance so that
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its direct-space lattice sum converges well;
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\begin_inset Formula $S^{\textup{S}}$
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\end_inset
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must as well contain all the singularities of
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\begin_inset Formula $S$
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\end_inset
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in the origin.
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The other part,
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\begin_inset Formula $S^{\textup{L}}$
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\end_inset
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, will retain all the slowly decaying terms of
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\begin_inset Formula $S$
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\end_inset
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but it also has to be smooth enough in the origin, so that its Fourier
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transform
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\begin_inset Formula $\uaft{S^{\textup{L}}}$
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\end_inset
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decays fast enough.
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(The same idea lies behind the Ewald summation in electrostatics.) Using
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the linearity of Fourier transform and formulae
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:W definition"
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\end_inset
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and
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:W sum in reciprocal space"
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\end_inset
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, the operator
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\begin_inset Formula $W_{\alpha\beta}$
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\end_inset
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can then be re-expressed as
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\begin_inset Formula
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\begin{eqnarray}
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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 \\
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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}\\
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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}
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\end{eqnarray}
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\end_inset
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where both sums should converge nicely.
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\end_layout
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\begin_layout Section
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Finding a good decomposition
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\end_layout
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\begin_layout Standard
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The remaining challenge is therefore finding a suitable decomposition
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\begin_inset Formula $S^{\textup{L}}+S^{\textup{S}}$
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\end_inset
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such that both
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\begin_inset Formula $S^{\textup{S}}$
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\end_inset
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and
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\begin_inset Formula $\uaft{S^{\textup{L}}}$
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\end_inset
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decay fast enough with distance and are expressable analytically.
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With these requirements, I do not expect to find gaussian asymptotics as
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in the electrostatic Ewald formula—having
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\begin_inset Formula $\sim x^{-t}$
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\end_inset
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,
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\begin_inset Formula $t>d$
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\end_inset
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asymptotics would be nice, making the sums in
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:W Short definition"
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\end_inset
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,
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:W Long definition"
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\end_inset
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absolutely convergent.
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\end_layout
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\begin_layout Section
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@ -528,7 +691,7 @@ we have
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\end_inset
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we have
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we have (CHECK)
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\begin_inset Formula
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\[
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\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}
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@ -644,7 +807,8 @@ From the scaling property of delta function,
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\end_layout
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\begin_layout Standard
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From the book:
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From the Osgood's book (p.
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375):
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\end_layout
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\begin_layout Standard
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@ -722,22 +886,19 @@ reference "eq:Dirac comb factorisation"
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\end_layout
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\begin_layout Subsubsection
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Fourier transform
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Fourier transform (OK)
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\end_layout
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\begin_layout Standard
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From the book https://see.stanford.edu/materials/lsoftaee261/chap8.pdf, p.
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From the Osgood's book https://see.stanford.edu/materials/lsoftaee261/chap8.pdf,
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p.
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379
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\end_layout
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\begin_layout Standard
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(CHECK THIS)
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\[
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\uoft{\dc A}\left(\vect{\xi}\right)=\left|\det A^{-T}\right|\dc{}^{(d)}\left(A^{-T}\vect{\xi}\right).
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\uoft{\dc{\basis u}}\left(\vect{\xi}\right)=\left|\det\rec{\basis u}\right|\dc{\rec{\basis u}}^{(d)}\left(\vect{\xi}\right).
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\]
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\end_inset
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@ -746,29 +907,48 @@ And consequently, for unitary/angular frequency it is
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\end_layout
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\begin_layout Standard
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(CHECK THIS)
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\begin_inset Formula
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\begin{eqnarray}
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\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 \\
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& = & \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 \\
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& = & \left(2\pi\right)^{\frac{d}{2}}\left|\det\rec{\basis u}\right|\dc{\recb{\basis u}}\left(\vect k\right)\nonumber \\
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& = & \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}
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\end{eqnarray}
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset Note Note
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status open
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\begin_layout Plain Layout
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On the third line, we used the stretch theorem, getting
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\begin_inset Formula
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\[
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\dc{\recb{\basis u}}\left(\vect k\right)=\dc{2\pi\rec{\basis u}}\left(\vect k\right)=\left(2\pi\right)^{-d}\dc{\rec{\basis u}}\left(\frac{\vect k}{2\pi}\right)
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\]
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\end_inset
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\end_layout
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\end_inset
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\end_layout
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\begin_layout Subsubsection
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Convolution
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\[
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\uaft{\dc A}\left(\vect{\xi}\right)=\frac{\left|\det A^{-T}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\dc{}^{(d)}\left(\frac{1}{2\pi}A^{-T}\vect{\xi}\right).
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\]
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\end_inset
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Using my own
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\begin_inset Quotes eld
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\end_inset
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basis notation
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\begin_inset Quotes erd
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\end_inset
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TODO
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\begin_inset Formula
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\[
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\uoft{\dc{\basis u}}\left(\vect{\xi}\right)=\left|\det\rec{\basis u}\right|\dc{}^{(d)}\left(A^{-T}\vect{\xi}\right).
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\left(f\ast\dc{\basis u}\right)(\vect x)=\sum_{\vect t\in\basis u\ints^{d}}f(\vect x-\vect t)
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\]
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\end_inset
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