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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{\usht}[2]{\mathbb{S}_{#1}#2}
\end_inset


\begin_inset FormulaMacro
\newcommand{\bsht}[2]{\mathrm{S}_{#1}#2}
\end_inset


\begin_inset FormulaMacro
\newcommand{\pht}[2]{\mathfrak{\mathbb{H}}_{#1}#2}
\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


\begin_inset FormulaMacro
\newcommand{\ints}{\mathbb{Z}}
\end_inset


\begin_inset FormulaMacro
\newcommand{\reals}{\mathbb{R}}
\end_inset


\begin_inset FormulaMacro
\newcommand{\ush}[2]{Y_{#1,#2}}
\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 Abstract
The 
\begin_inset Formula $T$
\end_inset

-matrix approach is the method of choice for simulating optical response
 of a reasonably small system of compact linear scatterers on isotropic
 background.
 However, its direct utilisation for problems with infinite lattices is
 problematic due to slowly converging sums over the lattice.
 Here I develop a way to compute the problematic sums in the reciprocal
 space, making the 
\begin_inset Formula $T$
\end_inset

-matrix method very suitable for infinite periodic systems as well.
\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.
\begin_inset Foot
status open

\begin_layout Plain Layout
Note that 
\begin_inset Formula $d$
\end_inset

 here is dimensionality of the lattice, not the space it lies in, which
 I for certain reasons assume to be three.
 (TODO few notes on integration and reciprocal lattices in some appendix)
\end_layout

\end_inset

 In electrostatics, one can solve this 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 (using
 formula 
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:Dirac comb uaFt"

\end_inset

 for the Fourier transform of Dirac comb)
\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)\nonumber \\
 & = & \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 \\
 & = & \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}
\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

As such, this is not extremely helpful because the the 
\emph on
whole
\emph default
 translation operator 
\begin_inset Formula $S$
\end_inset

 has singularities in origin, hence its Fourier transform 
\begin_inset Formula $\uaft S$
\end_inset

 will decay poorly.
 
\end_layout

\begin_layout Standard
However, Fourier transform is linear, so we can in principle separate 
\begin_inset Formula $S$
\end_inset

 in two parts, 
\begin_inset Formula $S=S^{\textup{L}}+S^{\textup{S}}$
\end_inset

.
 
\begin_inset Formula $S^{\textup{S}}$
\end_inset

 is a short-range part that decays sufficiently fast with distance so that
 its direct-space lattice sum converges well; 
\begin_inset Formula $S^{\textup{S}}$
\end_inset

 must as well contain all the singularities of 
\begin_inset Formula $S$
\end_inset

 in the origin.
 The other part, 
\begin_inset Formula $S^{\textup{L}}$
\end_inset

, will retain all the slowly decaying terms of 
\begin_inset Formula $S$
\end_inset

 but it also has to be smooth enough in the origin, so that its Fourier
 transform 
\begin_inset Formula $\uaft{S^{\textup{L}}}$
\end_inset

 decays fast enough.
 (The same idea lies behind the Ewald summation in electrostatics.) Using
 the linearity of Fourier transform and formulae 
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:W definition"

\end_inset

 and 
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:W sum in reciprocal space"

\end_inset

, the operator 
\begin_inset Formula $W_{\alpha\beta}$
\end_inset

 can then be re-expressed as
\begin_inset Formula 
\begin{eqnarray}
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 \\
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}\\
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}
\end{eqnarray}

\end_inset

where both sums should converge nicely.
\end_layout

\begin_layout Section
Finding a good decomposition
\end_layout

\begin_layout Standard
The remaining challenge is therefore finding a suitable decomposition 
\begin_inset Formula $S^{\textup{L}}+S^{\textup{S}}$
\end_inset

 such that both 
\begin_inset Formula $S^{\textup{S}}$
\end_inset

 and 
\begin_inset Formula $\uaft{S^{\textup{L}}}$
\end_inset

 decay fast enough with distance and are expressable analytically.
 With these requirements, I do not expect to find gaussian asymptotics as
 in the electrostatic Ewald formula—having 
\begin_inset Formula $\sim x^{-t}$
\end_inset

, 
\begin_inset Formula $t>d$
\end_inset

 asymptotics would be nice, making the sums in 
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:W Short definition"

\end_inset

, 
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:W Long definition"

\end_inset

 absolutely convergent.
\end_layout

\begin_layout Standard
The translation operator 
\begin_inset Formula $S$
\end_inset

 for compact scatterers in 3d can be expressed as
\begin_inset Formula 
\[
S_{l',m',t'\leftarrow l,m,t}\left(\vect r\leftarrow\vect 0\right)=\sum_{p}c_{p}^{l',m',t'\leftarrow l,m,t}\ush p{m'-m}\left(\theta_{\vect r},\phi_{\vect r}\right)z_{p}^{(J)}\left(\left|\vect r\right|\right)
\]

\end_inset

where 
\begin_inset Formula $Y_{l,m}\left(\theta,\phi\right)$
\end_inset

 are the spherical harmonics, 
\begin_inset Formula $z_{p}^{(J)}\left(r\right)$
\end_inset

 some of the Bessel or Hankel functions (probably 
\begin_inset Formula $h_{p}^{(1)}$
\end_inset

 in the meaningful cases; TODO) and 
\begin_inset Formula $c_{p}^{l,m,t\leftarrow l',m',t'}$
\end_inset

 are some ugly but known coefficients (REF Xu 1996, eqs.
 76,77).
 
\end_layout

\begin_layout Standard
The spherical Hankel functions can be expressed analytically as (REF DLMF
 10.49.6, 10.49.1) 
\begin_inset Formula 
\[
h_{n}^{(1)}(r)=e^{ir}\sum_{k=0}^{n}\frac{i^{k-n-1}}{r^{k+1}}\frac{\left(n+k\right)!}{2^{k}k!\left(n-k\right)!},
\]

\end_inset

 so if we find a way to deal with the radial functions 
\begin_inset Formula $s_{q}(r)=e^{ir}r^{-q}$
\end_inset

, 
\begin_inset Formula $q=1,2$
\end_inset

 in 2d case or 
\begin_inset Formula $q=1,2,3$
\end_inset

 in 3d case, we get absolutely convergent summations in the direct space.
\end_layout

\begin_layout Subsection
2d
\end_layout

\begin_layout Standard

\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\uaft{S_{l',m',t'\leftarrow l,m,t}\left(\vect{\bullet}\leftarrow\vect 0\right)}(\vect k)=\sum_{p}c_{p}^{l',m',t'\leftarrow l,m,t}\ush p{m'-m}\left(\frac{\pi}{2},0\right)e^{i(m'-m)\phi}i^{m'-m}\pht{m'-m}{z_{p}^{(J)}}\left(\left|\vect k\right|\right)
\]

\end_inset


\end_layout

\begin_layout Subsection
3d
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\uaft{S_{l',m',t'\leftarrow l,m,t}\left(\vect{\bullet}\leftarrow\vect 0\right)}(\vect k)=\sum_{p}c_{p}^{l',m',t'\leftarrow l,m,t}\ush p{m'-m}\left(\theta_{\vect k},\phi_{\vect k}\right)\left(-i\right)^{p}\usht p{z_{p}^{(J)}}\left(\left|\vect k\right|\right)
\]

\end_inset


\end_layout

\begin_layout Section
(Appendix) Fourier vs.
 Hankel transform
\end_layout

\begin_layout Subsection
Three dimensions
\end_layout

\begin_layout Standard
Given a nice enough function 
\begin_inset Formula $f$
\end_inset

 of a real 3d variable, assume its factorisation into radial and angular
 parts 
\begin_inset Formula 
\[
f(\vect r)=\sum_{l,m}f_{l,m}(\left|\vect r\right|)\ush lm\left(\theta_{\vect r},\phi_{\vect r}\right).
\]

\end_inset

Acording to (REF Baddour 2010, eqs.
 13, 16), its Fourier transform can then be expressed in terms of Hankel
 transforms (CHECK normalisation of 
\begin_inset Formula $j_{n}$
\end_inset

, REF Baddour (1)) 
\begin_inset Formula 
\[
\uaft f(\vect k)=\frac{4\pi}{\left(2\pi\right)^{\frac{3}{2}}}\sum_{l,m}\left(-i\right)^{l}\left(\bsht{f_{l,m}}{}\right)\left(\left|\vect k\right|\right)\ush lm\left(\theta_{\vect k},\phi_{\vect k}\right)
\]

\end_inset

where the spherical Hankel transform 
\begin_inset Formula $\bsht l{}$
\end_inset

 of degree 
\begin_inset Formula $l$
\end_inset

 is defined as (REF Baddour eq.
 2)
\begin_inset Formula 
\[
\bsht lg(k)\equiv\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).
\]

\end_inset

Using this convention, the inverse spherical Hankel transform is given by
 (REF Baddour eq.
 3)
\begin_inset Formula 
\[
g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\, k^{2}\bsht lg(k)j_{l}(k),
\]

\end_inset

so it is not unitary.
 
\end_layout

\begin_layout Standard
An unitary convention would look like this:
\begin_inset Formula 
\begin{equation}
\usht lg(k)\equiv\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).\label{eq:unitary 3d Hankel tf definition}
\end{equation}

\end_inset

Then 
\begin_inset Formula $\usht l{}^{-1}=\usht l{}$
\end_inset

 and the unitary, angular-momentum Fourier transform reads
\begin_inset Formula 
\begin{eqnarray}
\uaft f(\vect k) & = & \frac{4\pi}{\left(2\pi\right)^{\frac{3}{2}}}\sqrt{\frac{\pi}{2}}\sum_{l,m}\left(-i\right)^{l}\left(\usht l{f_{l,m}}\right)\left(\left|\vect k\right|\right)\ush lm\left(\theta_{\vect k},\phi_{\vect k}\right)\nonumber \\
 & = & \sum_{l,m}\left(-i\right)^{l}\left(\usht l{f_{l,m}}\right)\left(\left|\vect k\right|\right)\ush lm\left(\theta_{\vect k},\phi_{\vect k}\right).\label{eq:Fourier v. Hankel tf 3d}
\end{eqnarray}

\end_inset

Cool.
\end_layout

\begin_layout Subsection
Two dimensions
\end_layout

\begin_layout Standard
Similarly in 2d, let the expansion of 
\begin_inset Formula $f$
\end_inset

 be 
\begin_inset Formula 
\[
f\left(\vect r\right)=\sum_{m}f_{m}\left(\left|\vect r\right|\right)e^{im\phi_{\vect r}},
\]

\end_inset

its Fourier transform is then (CHECK this, it is taken from the Wikipedia
 article on Hankel transform) 
\begin_inset Formula 
\begin{equation}
\uaft f\left(\vect k\right)=\sum_{m}i^{m}e^{im\phi_{\vect k}}\pht mf_{m}\left(\left|\vect k\right|\right)\label{eq:Fourier v. Hankel tf 2d}
\end{equation}

\end_inset

where the Hankel transform of order 
\begin_inset Formula $m$
\end_inset

 is defined as
\begin_inset Formula 
\begin{equation}
\pht mg\left(k\right)=\int_{0}^{\infty}\ud r\, g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition}
\end{equation}

\end_inset

which is already self-inverse, 
\begin_inset Formula $\pht m{}^{-1}=\pht m{}$
\end_inset

 (hence also unitary).
\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 (CHECK)
\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 Osgood's book (p.
 375):
\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 (OK)
\end_layout

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

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

\end_inset

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

\begin_layout Standard
\begin_inset Formula 
\begin{eqnarray}
\uaft{\dc{\basis u}}\left(\vect k\right) & = & \frac{1}{\left(2\pi\right)^{\frac{d}{2}}}\uoft{\dc{\basis u}}\left(\frac{\vect k}{2\pi}\right)\nonumber \\
 & = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\dc{\rec{\basis u}}^{(d)}\left(\frac{\vect k}{2\pi}\right)\nonumber \\
 & = & \left(2\pi\right)^{\frac{d}{2}}\left|\det\rec{\basis u}\right|\dc{\recb{\basis u}}\left(\vect k\right)\nonumber \\
 & = & \frac{\left|\det\recb{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\dc{\recb{\basis u}}\left(\vect k\right).\label{eq:Dirac comb uaFt}
\end{eqnarray}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
On the third line, we used the stretch theorem, getting
\begin_inset Formula 
\[
\dc{\recb{\basis u}}\left(\vect k\right)=\dc{2\pi\rec{\basis u}}\left(\vect k\right)=\left(2\pi\right)^{-d}\dc{\rec{\basis u}}\left(\frac{\vect k}{2\pi}\right)
\]

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsubsection
Convolution
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\left(f\ast\dc{\basis u}\right)(\vect x)=\sum_{\vect t\in\basis u\ints^{d}}f(\vect x-\vect t)
\]

\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