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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}
\end_inset


\begin_inset FormulaMacro
\newcommand{\uaft}[1]{\mathfrak{\mathbb{F}}#1}
\end_inset


\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

 and 
\begin_inset Formula $T_{\vect aα}=T_{\alpha}$
\end_inset

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

\end_inset

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

, yielding
\begin_inset Formula 
\begin{eqnarray*}
\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,\\
\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,\\
\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,\\
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.
\end{eqnarray*}

\end_inset

Therefore, in order to solve the modes, we need to compute the 
\begin_inset Quotes eld
\end_inset

lattice Fourier transform
\begin_inset Quotes erd
\end_inset

 of the translation operator,
\begin_inset Formula 
\begin{equation}
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}
\end{equation}

\end_inset


\end_layout

\begin_layout Section
Computing the Fourier sum of the translation operator
\end_layout

\begin_layout Standard
The problem evaluating 
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:W definition"

\end_inset

 is the asymptotic behaviour of the translation operator, 
\begin_inset Formula $S_{\vect 0α\leftarrow\vect bβ}\sim\left|\vect r_{\vect b}\right|^{-1}e^{ik_{0}\left|\vect r_{\vect b}\right|}$
\end_inset

 that makes the convergence of the sum quite problematic for any 
\begin_inset Formula $d>1$
\end_inset

-dimensional lattice.
 In electrostatics, one can solve this with problem with Ewald summation.
 Its basic idea is that if what asymptoticaly decays poorly in the direct
 space, will perhaps decay fast in the Fourier space.
 I use the same idea here, but everything will be somehow harder than in
 electrostatics.
\end_layout

\begin_layout Standard
Let us re-express the sum in 
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:W definition"

\end_inset

 in terms of integral with a delta comb
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
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}
\end{equation}

\end_inset

The translation operator 
\begin_inset Formula $S$
\end_inset

 is now a function defined in the whole 3D space; 
\begin_inset Formula $\vect r_{\alpha},\vect r_{\beta}$
\end_inset

 are the displacements of scatterers 
\begin_inset Formula $\alpha$
\end_inset

 and 
\begin_inset Formula $\beta$
\end_inset

 in a unit cell.
 The arrow notation 
\begin_inset Formula $S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})$
\end_inset

 means 
\begin_inset Quotes eld
\end_inset

translation operator for spherical waves originating in 
\begin_inset Formula $\vect r+\vect r_{\beta}$
\end_inset

 evaluated in 
\begin_inset Formula $\vect r_{\alpha}$
\end_inset


\begin_inset Quotes erd
\end_inset

 and obviously 
\begin_inset Formula $S$
\end_inset

 is in fact a function of a single 3d argument, 
\begin_inset Formula $S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})=S(\vect 0\leftarrow\vect r+\vect r_{\beta}-\vect r_{\alpha})=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha})$
\end_inset

.
 Expression 
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:W integral"

\end_inset

 can be rewritten as
\begin_inset Formula 
\[
W_{\alpha\beta}(\vect k)=\left(2\pi\right)^{\frac{d}{2}}\uaft{(\dc{\basis u}S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0))\left(\vect k\right)}
\]

\end_inset

where changed the sign of 
\begin_inset Formula $\vect r/\vect{\bullet}$
\end_inset

 has been swapped under integration, utilising evenness of 
\begin_inset Formula $\dc{\basis u}$
\end_inset

.
 Fourier transform of product is convolution of Fourier transforms, so
\begin_inset Formula 
\begin{eqnarray*}
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)\\
 & = & \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)}\\
 & = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}
\end{eqnarray*}

\end_inset


\begin_inset Note Note
status open

\begin_layout Plain Layout
Factor 
\begin_inset Formula $\left(2\pi\right)^{\frac{d}{2}}$
\end_inset

 cancels out with the 
\begin_inset Formula $\left(2\pi\right)^{-\frac{d}{2}}$
\end_inset

 factor appearing in the convolution/product formula in the unitary angular
 momentum convention.
 
\end_layout

\end_inset


\end_layout

\begin_layout Section
(Appendix) Hankel transform
\end_layout

\begin_layout Standard
Acording to Wikipedia page on Hankel transform, 
\begin_inset Formula 
\[
\uaft f(\vect k)=
\]

\end_inset


\end_layout

\begin_layout Section
(Appendix) 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 
\begin{equation}
Ш_{T}(t)=\sum_{n=-\infty}^{\infty}e^{2\pi int/T}\label{eq:1D Dirac comb Fourier series}
\end{equation}

\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 
\begin{equation}
\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}
\end{equation}

\end_inset

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

\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


\end_layout

\begin_layout Standard
From the book:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\dc A(\vect x)=\frac{1}{\left|\det A\right|}\dc{}^{(d)}\left(A^{-1}\vect x\right)
\]

\end_inset


\begin_inset Note Note
status open

\begin_layout Plain Layout
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

 SRSLY?, and hence
\begin_inset Formula 
\begin{equation}
\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}
\end{equation}

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsubsection
Fourier series representation
\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
Utilising the Fourier series for 1D Dirac comb 
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:1D Dirac comb Fourier series"

\end_inset

 and the factorisation 
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:Dirac comb factorisation"

\end_inset

, we get
\begin_inset Formula 
\begin{eqnarray*}
\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}}}\\
 & = & \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}}}.
\end{eqnarray*}

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsubsection
Fourier transform
\end_layout

\begin_layout Standard
From the book https://see.stanford.edu/materials/lsoftaee261/chap8.pdf, p.
 379
\end_layout

\begin_layout Standard
(CHECK THIS)
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\uoft{\dc A}\left(\vect{\xi}\right)=\left|\det A^{-T}\right|\dc{}^{(d)}\left(A^{-T}\vect{\xi}\right).
\]

\end_inset

And consequently, for unitary/angular frequency it is
\end_layout

\begin_layout Standard
(CHECK THIS)
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\uaft{\dc A}\left(\vect{\xi}\right)=\frac{\left|\det A^{-T}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\dc{}^{(d)}\left(\frac{1}{2\pi}A^{-T}\vect{\xi}\right).
\]

\end_inset

Using my own 
\begin_inset Quotes eld
\end_inset

basis notation
\begin_inset Quotes erd
\end_inset

 TODO 
\begin_inset Formula 
\[
\uoft{\dc{\basis u}}\left(\vect{\xi}\right)=\left|\det\rec{\basis u}\right|\dc{}^{(d)}\left(A^{-T}\vect{\xi}\right).
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
So, from the stretch theorem 
\begin_inset Formula $\uoft{(f(A\vect x))}=\frac{1}{\left|\det A\right|}\uoft{f\left(A^{-T}\vect{\xi}\right)}=\left|\det A^{-T}\right|\uoft{f\left(A^{-T}\vect{\xi}\right)}$
\end_inset


\end_layout

\begin_layout Plain Layout
From 
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:Dirac comb factorisation"

\end_inset

 and 
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:1D Dirac comb Ft ordinary freq"

\end_inset


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

\end_inset


\end_layout

\end_inset


\end_layout

\end_body
\end_document