qpms/lepaper/infinite.lyx

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\pdf_author "Marek Nečada"
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\begin_body

\begin_layout Section
Infinite periodic systems
\end_layout

\begin_layout Subsection

\lang english
Formulation of the problem
\end_layout

\begin_layout Standard

\lang english
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

\lang english
...
\end_layout

\begin_layout Standard

\lang english
\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

\lang english
Now suppose that the scatterers constitute an infinite lattice
\end_layout

\begin_layout Standard

\lang english
\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 Subsection

\lang english
Computing the Fourier sum of the translation operator
\end_layout

\begin_layout Standard

\lang english
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 $S_{\vect0α\leftarrow\vect bβ}\sim\left|\vect r_{\vect b}\right|^{-1}e^{ik_{0}\left|\vect r_{\vect b}\right|}$
\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

\lang english
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

\lang english
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

\lang english
\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 $S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})=S(\vect0\leftarrow\vect r+\vect r_{\beta}-\vect r_{\alpha})=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect0)=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha})$
\end_inset

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

\end_inset

 can be rewritten as
\begin_inset Formula 
\[
W_{\alpha\beta}(\vect k)=\left(2\pi\right)^{\frac{d}{2}}\uaft{(\dc{\basis u}S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect0))\left(\vect k\right)}
\]

\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\vect0)}\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\vect0)}\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\vect0)}\right)\left(\vect k-\vect K\right)\label{eq:W sum in reciprocal space}\\
 & = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\sum_{\vect K\in\recb{\basis u}\ints^{d}}e^{i\left(\vect k-\vect K\right)\cdot\left(-\vect r_{\beta}+\vect r_{\alpha}\right)}\left(\uaft{S(\vect{\bullet}\leftarrow\vect0)}\right)\left(\vect k-\vect K\right)\nonumber 
\end{eqnarray}

\end_inset


\begin_inset Note Note
status open

\begin_layout Plain Layout

\lang english
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

\lang english
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 legendre
\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}}(\vect0\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\vect0)}\right)\left(\vect k-\vect K\right)\label{eq:W Long definition}
\end{eqnarray}

\end_inset

where both sums should converge nicely.
\end_layout

\end_body
\end_document