485 lines
12 KiB
Plaintext
485 lines
12 KiB
Plaintext
#LyX 2.4 created this file. For more info see https://www.lyx.org/
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\pdf_author "Marek Nečada"
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\end_header
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\begin_body
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\begin_layout Section
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Infinite periodic systems
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\end_layout
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\begin_layout Subsection
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Formulation of the problem
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\end_layout
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\begin_layout Standard
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Assume a system of compact EM scatterers in otherwise homogeneous and isotropic
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medium, and assume that the system, i.e.
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both the medium and the scatterers, have linear response.
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A scattering problem in such system can be written as
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\begin_inset Formula
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\[
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A_{α}=T_{α}P_{α}=T_{α}(\sum_{β}S_{α\leftarrowβ}A_{β}+P_{0α})
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\]
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\end_inset
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where
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\begin_inset Formula $T_{α}$
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\end_inset
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is the
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\begin_inset Formula $T$
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\end_inset
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-matrix for scatterer α,
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\begin_inset Formula $A_{α}$
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\end_inset
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is its vector of the scattered wave expansion coefficient (the multipole
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indices are not explicitely indicated here) and
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\begin_inset Formula $P_{α}$
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\end_inset
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is the local expansion of the incoming sources.
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\begin_inset Formula $S_{α\leftarrowβ}$
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\end_inset
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is ...
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and ...
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is ...
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\end_layout
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\begin_layout Standard
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...
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\[
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\sum_{β}(\delta_{αβ}-T_{α}S_{α\leftarrowβ})A_{β}=T_{α}P_{0α}.
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\]
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\end_inset
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\end_layout
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\begin_layout Standard
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Now suppose that the scatterers constitute an infinite lattice
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\[
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\sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{\vect aα}S_{\vect aα\leftarrow\vect bβ})A_{\vect bβ}=T_{\vect aα}P_{0\vect aα}.
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\]
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\end_inset
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Due to the periodicity, we can write
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\begin_inset Formula $S_{\vect aα\leftarrow\vect bβ}=S_{α\leftarrowβ}(\vect b-\vect a)$
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\end_inset
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and
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\begin_inset Formula $T_{\vect aα}=T_{\alpha}$
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\end_inset
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.
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In order to find lattice modes, we search for solutions with zero RHS
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\begin_inset Formula
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\[
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\sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{α}S_{\vect aα\leftarrow\vect bβ})A_{\vect bβ}=0
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\]
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\end_inset
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and we assume periodic solution
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\begin_inset Formula $A_{\vect b\beta}(\vect k)=A_{\vect a\beta}e^{i\vect k\cdot\vect r_{\vect b-\vect a}}$
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\end_inset
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, yielding
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\begin_inset Formula
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\begin{eqnarray*}
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\sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{α}S_{\vect aα\leftarrow\vect bβ})A_{\vect a\beta}\left(\vect k\right)e^{i\vect k\cdot\vect r_{\vect b-\vect a}} & = & 0,\\
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\sum_{\vect bβ}(\delta_{\vect{0b}}\delta_{αβ}-T_{α}S_{\vect 0α\leftarrow\vect bβ})A_{\vect 0\beta}\left(\vect k\right)e^{i\vect k\cdot\vect r_{\vect b}} & = & 0,\\
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\sum_{β}(\delta_{αβ}-T_{α}\underbrace{\sum_{\vect b}S_{\vect 0α\leftarrow\vect bβ}e^{i\vect k\cdot\vect r_{\vect b}}}_{W_{\alpha\beta}(\vect k)})A_{\vect 0\beta}\left(\vect k\right) & = & 0,\\
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A_{\vect 0\alpha}\left(\vect k\right)-T_{α}\sum_{\beta}W_{\alpha\beta}\left(\vect k\right)A_{\vect 0\beta}\left(\vect k\right) & = & 0.
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\end{eqnarray*}
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\end_inset
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Therefore, in order to solve the modes, we need to compute the
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\begin_inset Quotes eld
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\end_inset
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lattice Fourier transform
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\begin_inset Quotes erd
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\end_inset
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of the translation operator,
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\begin_inset Formula
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\begin{equation}
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W_{\alpha\beta}(\vect k)\equiv\sum_{\vect b}S_{\vect 0α\leftarrow\vect bβ}e^{i\vect k\cdot\vect r_{\vect b}}.\label{eq:W definition}
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\end{equation}
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\end_inset
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\end_layout
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\begin_layout Subsection
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Computing the Fourier sum of the translation operator
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\end_layout
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\begin_layout Standard
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The problem evaluating
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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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is the asymptotic behaviour of the translation operator,
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\begin_inset Formula $S_{\vect 0α\leftarrow\vect bβ}\sim\left|\vect r_{\vect b}\right|^{-1}e^{ik_{0}\left|\vect r_{\vect b}\right|}$
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\end_inset
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that makes the convergence of the sum quite problematic for any
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\begin_inset Formula $d>1$
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\end_inset
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-dimensional lattice.
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\begin_inset Foot
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status open
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\begin_layout Plain Layout
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Note that
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\begin_inset Formula $d$
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\end_inset
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here is dimensionality of the lattice, not the space it lies in, which
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I for certain reasons assume to be three.
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(TODO few notes on integration and reciprocal lattices in some appendix)
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\end_layout
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\end_inset
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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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electrostatics.
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\end_layout
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\begin_layout Standard
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Let us re-express the sum in
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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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in terms of integral with a delta comb
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\begin_inset FormulaMacro
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\renewcommand{\basis}[1]{\mathfrak{#1}}
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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 Formula
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\begin{equation}
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W_{\alpha\beta}(\vect k)=\int\ud^{d}\vect r\dc{\basis u}(\vect r)S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})e^{i\vect k\cdot\vect r}.\label{eq:W integral}
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\end{equation}
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\end_inset
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The translation operator
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\begin_inset Formula $S$
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\end_inset
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is now a function defined in the whole 3d space;
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\begin_inset Formula $\vect r_{\alpha},\vect r_{\beta}$
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\end_inset
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are the displacements of scatterers
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\begin_inset Formula $\alpha$
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\end_inset
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and
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\begin_inset Formula $\beta$
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\end_inset
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in a unit cell.
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The arrow notation
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\begin_inset Formula $S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})$
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\end_inset
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means
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\begin_inset Quotes eld
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\end_inset
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translation operator for spherical waves originating in
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\begin_inset Formula $\vect r+\vect r_{\beta}$
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\end_inset
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evaluated in
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\begin_inset Formula $\vect r_{\alpha}$
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\end_inset
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\begin_inset Quotes erd
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\end_inset
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and obviously
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\begin_inset Formula $S$
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\end_inset
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is in fact a function of a single 3d argument,
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\begin_inset Formula $S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})=S(\vect 0\leftarrow\vect r+\vect r_{\beta}-\vect r_{\alpha})=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha})$
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\end_inset
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.
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Expression
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:W integral"
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\end_inset
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can be rewritten as
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\begin_inset Formula
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\[
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W_{\alpha\beta}(\vect k)=\left(2\pi\right)^{\frac{d}{2}}\uaft{(\dc{\basis u}S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0))\left(\vect k\right)}
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\]
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\end_inset
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where changed the sign of
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\begin_inset Formula $\vect r/\vect{\bullet}$
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\end_inset
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has been swapped under integration, utilising evenness of
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\begin_inset Formula $\dc{\basis u}$
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\end_inset
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.
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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)\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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& = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\sum_{\vect K\in\recb{\basis u}\ints^{d}}e^{i\left(\vect k-\vect K\right)\cdot\left(-\vect r_{\beta}+\vect r_{\alpha}\right)}\left(\uaft{S(\vect{\bullet}\leftarrow\vect 0)}\right)\left(\vect k-\vect K\right)\nonumber
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\end{eqnarray}
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\end_inset
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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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Factor
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\begin_inset Formula $\left(2\pi\right)^{\frac{d}{2}}$
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\end_inset
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cancels out with the
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\begin_inset Formula $\left(2\pi\right)^{-\frac{d}{2}}$
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\end_inset
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factor appearing in the convolution/product formula in the unitary angular
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momentum convention.
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\end_layout
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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 legendre
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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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\end_body
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\end_document
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