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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 \[ S_{l',m',t'\leftarrow l,m,t}\left(\vect r\leftarrow\vect 0\right)=\sum_{p}c_{p}^{l',m',t'\leftarrow l,m,t}Y_{p,m'-m}\left(\theta_{\vect r},\phi_{\vect r}\right)z_{p}^{(J)}\left(\left|\vect r\right|\right) \] \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 (TODO) and \begin_inset Formula $c_{p}^{l,m,t\leftarrow l',m',t'}$ \end_inset are some ugly but known coefficients (Xu 1996, eqs. 76,77). \end_layout \begin_layout Section (Appendix) Hankel transform \end_layout \begin_layout Standard Acording to (Baddour 2010, eq. 13) (CHECK FACTORS) \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 (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