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\pdf_title "Accelerating lattice mode calculations with T-matrix method"
\pdf_author "Marek Nečada"
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\begin_body

\begin_layout Standard
\begin_inset FormulaMacro
\newcommand{\uoft}[1]{\mathfrak{F}#1}
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\begin_inset FormulaMacro
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\begin_inset FormulaMacro
\newcommand{\vect}[1]{\mathbf{#1}}
\end_inset


\begin_inset FormulaMacro
\newcommand{\ud}{\mathrm{d}}
\end_inset


\begin_inset FormulaMacro
\newcommand{\basis}[1]{\mathfrak{#1}}
\end_inset


\begin_inset FormulaMacro
\newcommand{\dc}[1]{Ш_{#1}}
\end_inset


\begin_inset FormulaMacro
\newcommand{\rec}[1]{#1^{-1}}
\end_inset


\begin_inset FormulaMacro
\newcommand{\recb}[1]{#1^{\widehat{-1}}}
\end_inset


\end_layout

\begin_layout Title
Accelerating lattice mode calculations with 
\begin_inset Formula $T$
\end_inset

-matrix method
\end_layout

\begin_layout Author
Marek Nečada
\end_layout

\begin_layout Section
Formulation of the problem
\end_layout

\begin_layout Standard
Assume a system of compact EM scatterers in otherwise homogeneous and isotropic
 medium, and assume that the system, i.e.
 both the medium and the scatterers, have linear response.
 A scattering problem in such system can be written as
\begin_inset Formula 
\[
A_{α}=T_{α}P_{α}=T_{α}(\sum_{β}S_{α\leftarrowβ}A_{β}+P_{0α})
\]

\end_inset

where 
\begin_inset Formula $T_{α}$
\end_inset

 is the 
\begin_inset Formula $T$
\end_inset

-matrix for scatterer α, 
\begin_inset Formula $A_{α}$
\end_inset

 is its vector of the scattered wave expansion coefficient (the multipole
 indices are not explicitely indicated here) and 
\begin_inset Formula $P_{α}$
\end_inset

 is the local expansion of the incoming sources.
 
\begin_inset Formula $S_{α\leftarrowβ}$
\end_inset

 is ...
 and ...
 is ...
\end_layout

\begin_layout Standard
...
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{β}(\delta_{αβ}-T_{α}S_{α\leftarrowβ})A_{β}=T_{α}P_{0α}.
\]

\end_inset


\end_layout

\begin_layout Standard
Now suppose that the scatterers constitute an infinite lattice
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{\vect aα}S_{\vect aα\leftarrow\vect bβ})A_{\vect bβ}=T_{\vect aα}P_{0\vect aα}.
\]

\end_inset

Due to the periodicity, we can write 
\begin_inset Formula $S_{\vect aα\leftarrow\vect bβ}=S_{α\leftarrowβ}(\vect b-\vect a)$
\end_inset

.
 In order to find lattice modes, we search for solutions with zero RHS
\begin_inset Formula 
\[
\sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{\vect aα}S_{\vect aα\leftarrow\vect bβ})A_{\vect bβ}=0
\]

\end_inset

and we assume periodic solution 
\begin_inset Formula $A_{\vect b\alpha}(\vect k)=A_{\vect a\alpha}e^{i\vect k\cdot\vect r_{\vect b-\vect a}}$
\end_inset

.
\end_layout

\begin_layout Section
Multidimensional Dirac comb
\end_layout

\begin_layout Subsection
1D
\end_layout

\begin_layout Standard
This is all from Wikipedia
\end_layout

\begin_layout Subsubsection
Definitions
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{eqnarray*}
Ш(t) & \equiv & \sum_{k=-\infty}^{\infty}\delta(t-k)\\
Ш_{T}(t) & \equiv & \sum_{k=-\infty}^{\infty}\delta(t-kT)=\frac{1}{T}Ш\left(\frac{t}{T}\right)
\end{eqnarray*}

\end_inset


\end_layout

\begin_layout Subsubsection
Fourier series representation
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
Ш_{T}(t)=\sum_{n=-\infty}^{\infty}e^{2\pi int/T}
\]

\end_inset


\end_layout

\begin_layout Subsubsection
Fourier transform
\end_layout

\begin_layout Standard
With unitary ordinary frequency Ft., i.e.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\uoft f(\vect{\xi})\equiv\int_{\mathbb{R}^{n}}f(\vect x)e^{-2\pi i\vect x\cdot\vect{\xi}}\ud^{n}\vect x
\]

\end_inset

we have 
\begin_inset Formula 
\[
\uoft{Ш_{T}}(f)=\frac{1}{T}Ш_{\frac{1}{T}}(f)=\sum_{n=-\infty}^{\infty}e^{-i2\pi fnT}
\]

\end_inset

 and with unitary angular frequency Ft., i.e.
\begin_inset Formula 
\[
\uaft f(\vect k)\equiv\frac{1}{\left(2\pi\right)^{n}}\int_{\mathbb{R}^{n}}f(\vect x)e^{-2\pi i\vect x\cdot\vect k}\ud^{n}\vect x
\]

\end_inset

we have
\begin_inset Formula 
\[
\uaft{Ш_{T}}(\omega)=\frac{\sqrt{2\pi}}{T}Ш_{\frac{2\pi}{T}}(\omega)=\frac{1}{\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}e^{-i\omega nT}
\]

\end_inset


\end_layout

\begin_layout Subsection
Dirac comb for multidimensional lattices
\end_layout

\begin_layout Subsubsection
Definitions
\end_layout

\begin_layout Standard
Let 
\begin_inset Formula $d$
\end_inset

 be the dimensionality of the real vector space in question, and let 
\begin_inset Formula $\basis u\equiv\left\{ \vect u_{i}\right\} _{i=1}^{d}$
\end_inset

 denote a basis for some lattice in that space.
 Let the corresponding lattice delta comb be
\begin_inset Formula 
\[
\dc{\basis u}\left(\vect x\right)\equiv\sum_{n_{1}=-\infty}^{\infty}\ldots\sum_{n_{d}=-\infty}^{\infty}\delta\left(\vect x-\sum_{i=1}^{d}n_{i}\vect u_{i}\right).
\]

\end_inset


\end_layout

\begin_layout Standard
Furthemore, let 
\begin_inset Formula $\rec{\basis u}\equiv\left\{ \rec{\vect u}_{i}\right\} _{i=1}^{d}$
\end_inset

 be the reciprocal lattice basis, that is the basis satisfying 
\begin_inset Formula $\vect u_{i}\cdot\rec{\vect u_{j}}=\delta_{ij}$
\end_inset

.
 This slightly differs from the usual definition of a reciprocal basis,
 here denoted 
\begin_inset Formula $\recb{\basis u}\equiv\left\{ \recb{\vect u_{i}}\right\} _{i=1}^{d}$
\end_inset

, which satisfies 
\begin_inset Formula $\vect u_{i}\cdot\recb{\vect u_{j}}=2\pi\delta_{ij}$
\end_inset

 instead.
\end_layout

\begin_layout Subsubsection
Factorisation of a multidimensional lattice delta comb
\end_layout

\begin_layout Standard
By simple drawing, it can be seen that 
\begin_inset Formula 
\[
\dc{\basis u}(\vect x)=c_{\basis u}\prod_{i=1}^{d}\dc{}\left(\vect x\cdot\rec{\vect u_{i}}\right)
\]

\end_inset

where 
\begin_inset Formula $c_{\basis u}$
\end_inset

 is some numerical volume factor.
 In order to determine 
\begin_inset Formula $c_{\basis u}$
\end_inset

, let us consider only the 
\begin_inset Quotes eld
\end_inset

zero tooth
\begin_inset Quotes erd
\end_inset

 of the comb, leading to
\begin_inset Formula 
\[
\delta^{d}(\vect x)=c_{\basis u}\prod_{i=1}^{d}\delta\left(\vect x\cdot\rec{\vect u_{i}}\right).
\]

\end_inset

From the scaling property of delta function, 
\begin_inset Formula $\delta(ax)=\left|a\right|^{-1}\delta(x)$
\end_inset

, we get
\begin_inset Formula 
\[
\delta^{d}(\vect x)=c_{\basis u}\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert ^{-1}\delta\left(\vect x\cdot\frac{\rec{\vect u_{i}}}{\left\Vert \rec{\vect u_{i}}\right\Vert }\right).
\]

\end_inset

Applying both sides to a test function that is one at the origin, we get
 
\begin_inset Formula $c_{\basis u}=\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert $
\end_inset

, and hence
\begin_inset Formula 
\[
\dc{\basis u}(\vect x)=\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert \dc{}\left(\vect x\cdot\rec{\vect u_{i}}\right).
\]

\end_inset


\end_layout

\begin_layout Subsubsection
Fourier series representation
\end_layout

\begin_layout Subsubsection
Fourier transform
\end_layout

\end_body
\end_document