Start writing notes on dirac combs
Former-commit-id: f91e766e82dc8b84246a895bf95fc9d59afa996d
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Marek Nečada
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\pdf_title "Accelerating lattice mode calculations with T-matrix method"
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\pdf_author "Marek Nečada"
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\begin_body
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\begin_layout Standard
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\begin_inset FormulaMacro
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\newcommand{\uoft}[1]{\mathfrak{F}#1}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\uaft}[1]{\mathfrak{\mathbb{F}}#1}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\vect}[1]{\mathbf{#1}}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\ud}{\mathrm{d}}
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\end_inset
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\newcommand{\basis}[1]{\mathfrak{#1}}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\dc}[1]{Ш_{#1}}
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\end_layout
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\begin_layout Title
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Accelerating lattice mode calculations with
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\begin_inset Formula $T$
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\end_inset
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-matrix method
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\end_layout
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\begin_layout Author
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Marek Nečada
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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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\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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.
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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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\[
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\sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{\vect aα}S_{\vect aα\leftarrow\vect bβ})A_{\vect bβ}=0
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\]
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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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\end_inset
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.
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\end_layout
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\begin_layout Section
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Multidimensional Dirac comb
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\end_layout
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\begin_layout Subsection
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1D
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\end_layout
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\begin_layout Standard
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This is all from Wikipedia
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\end_layout
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\begin_layout Subsubsection
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Definitions
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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{eqnarray*}
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Ш(t) & \equiv & \sum_{k=-\infty}^{\infty}\delta(t-k)\\
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Ш_{T}(t) & \equiv & \sum_{k=-\infty}^{\infty}\delta(t-kT)=\frac{1}{T}Ш\left(\frac{t}{T}\right)
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\end{eqnarray*}
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\end_inset
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\end_layout
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\begin_layout Subsubsection
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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 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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\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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With unitary ordinary frequency Ft., i.e.
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\[
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\uoft f(\vect{\xi})\equiv\int_{\mathbb{R}^{n}}f(\vect x)e^{-2\pi i\vect x\cdot\vect{\xi}}\ud^{n}\vect x
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\]
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\end_inset
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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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\end_inset
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and with unitary angular frequency Ft., i.e.
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\begin_inset Formula
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\[
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\uaft f(\vect k)\equiv\frac{1}{\left(2\pi\right)^{n}}\int_{\mathbb{R}^{n}}f(\vect x)e^{-2\pi i\vect x\cdot\vect k}\ud^{n}\vect x
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\]
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\end_inset
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we have
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\begin_inset Formula
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\[
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\uaft{Ш_{T}}(\omega)=\frac{\sqrt{2\pi}}{T}Ш_{\frac{2\pi}{T}}(\omega)=\frac{1}{\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}e^{-i\omega nT}
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\]
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\end_inset
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\end_layout
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\begin_layout Subsection
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Dirac comb for multidimensional lattices
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\end_layout
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\begin_layout Subsubsection
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Definitions
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\end_layout
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\begin_layout Standard
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Let
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\begin_inset Formula $d$
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\end_inset
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be the dimensionality of the real vector space in question, and let
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\begin_inset Formula $\basis u\equiv\left\{ \vect u_{i}\right\} _{i=1}^{d}$
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\end_inset
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denote a basis for some lattice in that space.
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Let the corresponding lattice delta comb be
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\begin_inset Formula
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\[
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\dc{\basis u}\left(\vect x\right)\equiv\sum_{n_{1}=-\infty}^{\infty}\ldots\sum_{n_{d}=-\infty}^{\infty}\delta\left(\vect x-\sum_{i=1}^{d}n_{i}\vect u_{i}\right).
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\]
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\end_inset
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\end_layout
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\begin_layout Standard
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Furthemore, let
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\begin_inset Formula $\rec{\basis u}\equiv\left\{ \rec{\vect u}_{i}\right\} _{i=1}^{d}$
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\end_inset
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be the reciprocal lattice basis, that is the basis satisfying
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\begin_inset Formula $\vect u_{i}\cdot\rec{\vect u_{j}}=\delta_{ij}$
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\end_inset
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.
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This slightly differs from the usual definition of a reciprocal basis,
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here denoted
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\begin_inset Formula $\recb{\basis u}\equiv\left\{ \recb{\vect u_{i}}\right\} _{i=1}^{d}$
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\end_inset
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, which satisfies
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\begin_inset Formula $\vect u_{i}\cdot\recb{\vect u_{j}}=2\pi\delta_{ij}$
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\end_inset
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instead.
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\end_layout
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\begin_layout Subsubsection
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Factorisation of a multidimensional lattice delta comb
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\end_layout
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\begin_layout Standard
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By simple drawing, it can be seen that
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\begin_inset Formula
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\[
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\dc{\basis u}(\vect x)=c_{\basis u}\prod_{i=1}^{d}\dc{}\left(\vect x\cdot\rec{\vect u_{i}}\right)
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\]
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\end_inset
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where
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\begin_inset Formula $c_{\basis u}$
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\end_inset
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is some numerical volume factor.
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In order to determine
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\begin_inset Formula $c_{\basis u}$
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\end_inset
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, let us consider only the
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\begin_inset Quotes eld
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\end_inset
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zero tooth
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\begin_inset Quotes erd
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\end_inset
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of the comb, leading to
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\begin_inset Formula
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\[
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\delta^{d}(\vect x)=c_{\basis u}\prod_{i=1}^{d}\delta\left(\vect x\cdot\rec{\vect u_{i}}\right).
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\]
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\end_inset
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From the scaling property of delta function,
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\begin_inset Formula $\delta(ax)=\left|a\right|^{-1}\delta(x)$
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\end_inset
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, we get
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\begin_inset Formula
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\[
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\delta^{d}(\vect x)=c_{\basis u}\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert ^{-1}\delta\left(\vect x\cdot\frac{\rec{\vect u_{i}}}{\left\Vert \rec{\vect u_{i}}\right\Vert }\right).
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\]
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\end_inset
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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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\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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\end_inset
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\end_layout
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\begin_layout Subsubsection
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Fourier series representation
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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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\end_body
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\end_document
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