diff --git a/notes/ewald.lyx b/notes/ewald.lyx index b41e741..337dba8 100644 --- a/notes/ewald.lyx +++ b/notes/ewald.lyx @@ -152,6 +152,16 @@ \end_inset +\begin_inset FormulaMacro +\newcommand{\ints}{\mathbb{Z}} +\end_inset + + +\begin_inset FormulaMacro +\newcommand{\reals}{\mathbb{R}} +\end_inset + + \end_layout \begin_layout Title @@ -166,6 +176,24 @@ Accelerating lattice mode calculations with 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 @@ -308,7 +336,7 @@ reference "eq:W definition" \end_inset -dimensional lattice. - In electrostatics, one can solve this with problem with Ewald summation. + 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 @@ -404,13 +432,21 @@ where changed the sign of \end_inset . - Fourier transform of product is convolution of Fourier transforms, so + 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)\\ - & = & \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*} +\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 @@ -434,7 +470,134 @@ Factor \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 Section @@ -528,7 +691,7 @@ we have \end_inset -we have +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} @@ -644,7 +807,8 @@ From the scaling property of delta function, \end_layout \begin_layout Standard -From the book: +From the Osgood's book (p. + 375): \end_layout \begin_layout Standard @@ -722,22 +886,19 @@ reference "eq:Dirac comb factorisation" \end_layout \begin_layout Subsubsection -Fourier transform +Fourier transform (OK) \end_layout \begin_layout Standard -From the book https://see.stanford.edu/materials/lsoftaee261/chap8.pdf, p. +From the Osgood's 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 \[ -\uoft{\dc A}\left(\vect{\xi}\right)=\left|\det A^{-T}\right|\dc{}^{(d)}\left(A^{-T}\vect{\xi}\right). +\uoft{\dc{\basis u}}\left(\vect{\xi}\right)=\left|\det\rec{\basis u}\right|\dc{\rec{\basis u}}^{(d)}\left(\vect{\xi}\right). \] \end_inset @@ -746,29 +907,48 @@ And consequently, for unitary/angular frequency it is \end_layout \begin_layout Standard -(CHECK THIS) +\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 \[ -\uaft{\dc A}\left(\vect{\xi}\right)=\frac{\left|\det A^{-T}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\dc{}^{(d)}\left(\frac{1}{2\pi}A^{-T}\vect{\xi}\right). -\] - -\end_inset - -Using my own -\begin_inset Quotes eld -\end_inset - -basis notation -\begin_inset Quotes erd -\end_inset - - TODO -\begin_inset Formula -\[ -\uoft{\dc{\basis u}}\left(\vect{\xi}\right)=\left|\det\rec{\basis u}\right|\dc{}^{(d)}\left(A^{-T}\vect{\xi}\right). +\left(f\ast\dc{\basis u}\right)(\vect x)=\sum_{\vect t\in\basis u\ints^{d}}f(\vect x-\vect t) \] \end_inset