Continue writing on Ewald summation.
Former-commit-id: a01eb920f0e49ce30fd29b8694a614a4af9993b1
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Marek Nečada
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216
ewald.lyx
216
ewald.lyx
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@ -51,7 +51,7 @@
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\output_sync 0
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\bibtex_command default
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\index_command default
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\paperfontsize default
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\paperfontsize 10
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\spacing single
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\use_hyperref true
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\pdf_title "Accelerating lattice mode calculations with T-matrix method"
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@ -65,8 +65,8 @@
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\pdf_colorlinks false
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\pdf_backref false
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\pdf_pdfusetitle true
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\papersize default
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\use_geometry false
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\papersize a5paper
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\use_geometry true
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\use_package amsmath 1
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\use_package amssymb 1
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\use_package cancel 1
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@ -90,6 +90,10 @@
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\shortcut idx
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\color #008000
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\end_index
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\leftmargin 2cm
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\topmargin 2cm
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\rightmargin 2cm
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\bottommargin 2cm
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\secnumdepth 3
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\tocdepth 3
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\paragraph_separation indent
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@ -234,26 +238,85 @@ Now suppose that the scatterers constitute an infinite lattice
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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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\[
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\sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{\vect aα}S_{\vect aα\leftarrow\vect bβ})A_{\vect bβ}=0
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\sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{α}S_{\vect aα\leftarrow\vect bβ})A_{\vect bβ}=0
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\]
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\end_inset
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and we assume periodic solution
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\begin_inset Formula $A_{\vect b\alpha}(\vect k)=A_{\vect a\alpha}e^{i\vect k\cdot\vect r_{\vect b-\vect a}}$
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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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.
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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 Section
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Multidimensional Dirac comb
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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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In electrostatics, one can solve this with 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 things will be somehow harder than in electrostat
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ics.
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\end_layout
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\begin_layout Section
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(Appendix) Multidimensional Dirac comb
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\end_layout
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\begin_layout Subsection
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@ -286,9 +349,9 @@ Fourier series representation
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\begin_layout Standard
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\begin_inset Formula
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\[
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Ш_{T}(t)=\sum_{n=-\infty}^{\infty}e^{2\pi int/T}
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\]
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\begin{equation}
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Ш_{T}(t)=\sum_{n=-\infty}^{\infty}e^{2\pi int/T}\label{eq:1D Dirac comb Fourier series}
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\end{equation}
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\end_inset
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@ -313,9 +376,9 @@ With unitary ordinary frequency Ft., i.e.
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we have
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\begin_inset Formula
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\[
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\uoft{Ш_{T}}(f)=\frac{1}{T}Ш_{\frac{1}{T}}(f)=\sum_{n=-\infty}^{\infty}e^{-i2\pi fnT}
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\]
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\begin{equation}
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\uoft{Ш_{T}}(f)=\frac{1}{T}Ш_{\frac{1}{T}}(f)=\sum_{n=-\infty}^{\infty}e^{-i2\pi fnT}\label{eq:1D Dirac comb Ft ordinary freq}
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\end{equation}
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\end_inset
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@ -439,16 +502,41 @@ From the scaling property of delta function,
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\end_inset
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\end_layout
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\begin_layout Standard
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From the book:
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\[
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\dc A(\vect x)=\frac{1}{\left|\det A\right|}\dc{}^{(d)}\left(A^{-1}\vect x\right)
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\]
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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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Applying both sides to a test function that is one at the origin, we get
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\begin_inset Formula $c_{\basis u}=\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert $
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\end_inset
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, and hence
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SRSLY?, and hence
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\begin_inset Formula
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\[
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\dc{\basis u}(\vect x)=\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert \dc{}\left(\vect x\cdot\rec{\vect u_{i}}\right).
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\]
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\begin{equation}
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\dc{\basis u}(\vect x)=\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert \dc{}\left(\vect x\cdot\rec{\vect u_{i}}\right).\label{eq:Dirac comb factorisation}
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\end{equation}
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\end_inset
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\end_layout
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\end_inset
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@ -459,9 +547,103 @@ Applying both sides to a test function that is one at the origin, we get
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Fourier series representation
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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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Utilising the Fourier series for 1D Dirac comb
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:1D Dirac comb Fourier series"
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\end_inset
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and the factorisation
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:Dirac comb factorisation"
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\end_inset
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, we get
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\begin_inset Formula
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\begin{eqnarray*}
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\dc{\basis u}(\vect x) & = & \prod_{j=1}^{d}\left\Vert \rec{\vect u_{j}}\right\Vert \sum_{n_{j}=-\infty}^{\infty}e^{2\pi in_{i}\vect x\cdot\rec{\vect u_{i}}}\\
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& = & \left(\prod_{j=1}^{d}\left\Vert \rec{\vect u_{j}}\right\Vert \right)\sum_{\vect n\in\mathbb{Z}^{d}}e^{2\pi i\vect x\cdot\sum_{k=1}^{d}n_{k}\rec{\vect u_{k}}}.
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\end{eqnarray*}
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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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Fourier transform
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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
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\[
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\uoft{\dc A}\left(\vect{\xi}\right)=\dc{}^{(d)}\left(A^{T}\vect{\xi}\right).
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\]
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset Note Note
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status open
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\begin_layout Plain Layout
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So, from the stretch theorem
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\begin_inset Formula $\uoft{(f(A\vect x))}=\frac{1}{\left|\det A\right|}\uoft{f\left(A^{-T}\vect{\xi}\right)}=\left|\det A^{-T}\right|\uoft{f\left(A^{-T}\vect{\xi}\right)}$
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\end_inset
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\end_layout
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\begin_layout Plain Layout
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From
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:Dirac comb factorisation"
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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:1D Dirac comb Ft ordinary freq"
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\end_inset
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\begin_inset Formula
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\[
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\uoft{\dc{\basis u}}(\vect{\xi})=\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert \dc{}\left(\vect x\cdot\rec{\vect u_{i}}\right).
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\]
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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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\end_body
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\end_document
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