Implement some of Javier's notes.
Former-commit-id: 3881eccd2bbca4975d50c4a749751b7c134d6698
This commit is contained in:
Marek Nečada
parent
c70317dc25
commit
2a890c56ac
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@ -481,7 +481,11 @@ The single-particle scattering problem at frequency
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\end_inset
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.
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Inside this volume, the electric field can be expanded as
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Inside
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\begin_inset Formula $\openball 0{R^{>}}\backslash B_{0}\left(R\right)$
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\end_inset
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, the electric field can be expanded as
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\begin_inset Note Note
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status open
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@ -770,7 +774,7 @@ literal "false"
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its (maximum) refractive index.
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\begin_inset Note Note
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status open
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status collapsed
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\begin_layout Plain Layout
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\begin_inset Formula
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@ -1281,7 +1285,7 @@ In practice, the multiple-scattering problem is solved in its truncated
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\begin_inset Formula $l\le L_{p}$
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\end_inset
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, laeving only
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, leaving only
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\begin_inset Formula $N_{p}=2L_{p}\left(L_{p}+2\right)$
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\end_inset
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@ -1428,11 +1432,7 @@ Let
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\end_inset
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where an explicit formula for the (regular)
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\emph on
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translation operator
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\emph default
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where an explicit formula for the regular translation operator
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\begin_inset Formula $\tropr$
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\end_inset
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@ -1547,7 +1547,7 @@ reference "eq:regular vswf translation"
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,
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\begin_inset Formula
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\[
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\vect E\left(\vect r,\omega\right)=\sum_{\tau,l,m}\rcoeffptlm p{\tau}lm\sum_{\tau'l'm'}\tropr_{\tau lm;\tau'l'm'}\left(k\left(\vect r_{q}-\vect r_{p}\right)\right)\vswfrtlm{\tau'}{l'}{m'}\left(\vect r-\vect r_{q}\right)
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\vect E\left(\vect r,\omega\right)=\sum_{\tau,l,m}\rcoeffptlm p{\tau}lm\sum_{\tau'l'm'}\tropr_{\tau lm;\tau'l'm'}\left(k\left(\vect r_{q}-\vect r_{p}\right)\right)\vswfrtlm{\tau'}{l'}{m'}\left(k\left(\vect r-\vect r_{q}\right)\right)
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\]
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\end_inset
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@ -1579,7 +1579,12 @@ reference "eq:regular vswf coefficient translation"
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\end_inset
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(note the reversed indices; TODO redefine them in
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(note the reversed indices
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\begin_inset Note Note
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status open
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\begin_layout Plain Layout
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; TODO redefine them in
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:regular vswf translation"
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@ -1593,7 +1598,12 @@ reference "eq:singular vswf translation"
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\end_inset
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? Similarly, if we had only outgoing waves in the original expansion around
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?
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\end_layout
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\end_inset
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) Similarly, if we had only outgoing waves in the original expansion around
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\begin_inset Formula $\vect r_{p}$
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\end_inset
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@ -329,6 +329,16 @@ noprefix "false"
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\begin_layout Standard
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As in the case of a finite system, eq.
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:Multiple-scattering problem unit cell"
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plural "false"
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caps "false"
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noprefix "false"
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\end_inset
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can be written in a shorter block-matrix form,
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\begin_inset Formula
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\begin{equation}
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@ -526,7 +536,17 @@ noprefix "false"
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\begin_inset Formula $\left|\vect k\right|=\sqrt{\epsilon\mu}\omega/c_{0}$
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\end_inset
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(modulo lattice points; TODO write this a clean way).
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(modulo reciprocal lattice points
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\begin_inset Note Note
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status open
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\begin_layout Plain Layout
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TODO write this in a clean way
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\end_layout
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\end_inset
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).
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A somehow challenging step is to distinguish the different bands that can
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all be very close to the empty lattice approximation, especially if the
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particles in the systems are small.
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@ -687,7 +707,7 @@ translation operator for spherical waves originating in
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\end_inset
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is in fact a function of a single 3d argument,
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\begin_inset Formula $S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})=S(\vect 0\leftarrow\vect r+\vect r_{\beta}-\vect r_{\alpha})=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha})$
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\begin_inset Formula $S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})=S(\vect0\leftarrow\vect r+\vect r_{\beta}-\vect r_{\alpha})=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect0)=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha})$
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\end_inset
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.
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@ -701,7 +721,7 @@ reference "eq:W integral"
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can be rewritten as
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\begin_inset Formula
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\[
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W_{\alpha\beta}(\vect k)=\left(2\pi\right)^{\frac{d}{2}}\uaft{(\dc{\basis u}S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0))\left(\vect k\right)}
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W_{\alpha\beta}(\vect k)=\left(2\pi\right)^{\frac{d}{2}}\uaft{(\dc{\basis u}S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect0))\left(\vect k\right)}
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\]
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\end_inset
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@ -735,10 +755,10 @@ reference "eq:Dirac comb uaFt"
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(REF?) for the Fourier transform of Dirac comb)
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\begin_inset Formula
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\begin{eqnarray}
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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 \\
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& = & \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 \\
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& = & \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}\\
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& = & \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\vect 0)}\right)\left(\vect k-\vect K\right)\nonumber
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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 \\
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& = & \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 \\
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& = & \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}\\
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& = & \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
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\end{eqnarray}
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\end_inset
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@ -840,8 +860,8 @@ reference "eq:W sum in reciprocal space"
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\begin_inset Formula
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\begin{eqnarray}
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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 \\
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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}\\
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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}
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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}\\
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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}
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\end{eqnarray}
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\end_inset
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@ -879,7 +899,7 @@ CHECK THE FOLLOWING EXPRESSION FOR CORRECT FUNCTION ARGUMENTS
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\begin_inset Formula
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\begin{equation}
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\sigma_{\nu}^{\mu}\left(\vect k\right)=\sum_{\vect n\in\ints^{d}\backslash\left\{ \vect 0\right\} }e^{i\vect{\vect k}\cdot\vect R_{\vect n}}\ush{\nu}{\mu}\left(\uvec{R_{n}}\right)h_{n}^{(1)}\left(R_{n}\right),\label{eq:sigma lattice sums}
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\sigma_{\nu}^{\mu}\left(\vect k\right)=\sum_{\vect n\in\ints^{d}\backslash\left\{ \vect0\right\} }e^{i\vect{\vect k}\cdot\vect R_{\vect n}}\ush{\nu}{\mu}\left(\uvec{R_{n}}\right)h_{n}^{(1)}\left(R_{n}\right),\label{eq:sigma lattice sums}
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\end{equation}
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\end_inset
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@ -98,11 +98,11 @@ If the system has nontrivial point group symmetries, group theory gives
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\end_layout
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\begin_layout Standard
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As an example, if our system has a
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As an example, if the system has a
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\begin_inset Formula $D_{2h}$
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\end_inset
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symmetry and our truncated
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symmetry and the corresponding truncated
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\begin_inset Formula $\left(I-T\trops\right)$
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\end_inset
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@ -961,7 +961,7 @@ where
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\begin_inset Formula $1$
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\end_inset
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through
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to
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\begin_inset Formula $d_{n}$
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\end_inset
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@ -969,7 +969,7 @@ where
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\begin_inset Formula $i$
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\end_inset
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goes from 1 through the multiplicity of irreducible representation
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goes from 1 to the multiplicity of irreducible representation
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\begin_inset Formula $\Gamma_{n}$
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\end_inset
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@ -1328,8 +1328,8 @@ horizontal
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the same unit cell, e.g.
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\begin_inset Formula
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\begin{align*}
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\outcoeffp{\vect 0A} & \overset{\sigma_{xz}}{\longmapsto}\tilde{J}\left(\sigma_{xz}\right)\outcoeffp{\vect 0E},\\
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\outcoeff_{\vect 0C} & \overset{\sigma_{xz}}{\longmapsto}\tilde{J}\left(\sigma_{xz}\right)\outcoeffp{\vect 0C},
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\outcoeffp{\vect0A} & \overset{\sigma_{xz}}{\longmapsto}\tilde{J}\left(\sigma_{xz}\right)\outcoeffp{\vect0E},\\
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\outcoeff_{\vect0C} & \overset{\sigma_{xz}}{\longmapsto}\tilde{J}\left(\sigma_{xz}\right)\outcoeffp{\vect0C},
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\end{align*}
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\end_inset
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@ -1374,8 +1374,8 @@ vertical
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,
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\begin_inset Formula
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\begin{align*}
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\outcoeffp{\vect 0A} & \overset{\sigma_{yz}}{\longmapsto}\tilde{J}\left(\sigma_{yz}\right)\outcoeffp{\left(0,1\right)E},\\
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\outcoeff_{\vect 0C} & \overset{\sigma_{yz}}{\longmapsto}\tilde{J}\left(\sigma_{yz}\right)\outcoeffp{\left(1,0\right)C},
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\outcoeffp{\vect0A} & \overset{\sigma_{yz}}{\longmapsto}\tilde{J}\left(\sigma_{yz}\right)\outcoeffp{\left(0,1\right)E},\\
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\outcoeff_{\vect0C} & \overset{\sigma_{yz}}{\longmapsto}\tilde{J}\left(\sigma_{yz}\right)\outcoeffp{\left(1,0\right)C},
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\end{align*}
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\end_inset
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@ -1385,22 +1385,22 @@ but we want
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\end_inset
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to operate only inside one unit cell, so we use the Bloch condition
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\begin_inset Formula $\outcoeffp{\vect n,\alpha}=\outcoeffp{\vect 0,\alpha}\left(\vect k\right)e^{i\vect k\cdot\vect R_{\vect n}}$
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\begin_inset Formula $\outcoeffp{\vect n,\alpha}=\outcoeffp{\vect0,\alpha}\left(\vect k\right)e^{i\vect k\cdot\vect R_{\vect n}}$
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\end_inset
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: in this case, we have
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\begin_inset Formula $\outcoeffp{\left(0,1\right)\alpha}=\outcoeffp{\vect 0\alpha}e^{i\vect M_{1}\cdot\vect a_{2}}=\outcoeffp{\vect 0\alpha}e^{i0}=\outcoeffp{\vect 0\alpha}$
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\begin_inset Formula $\outcoeffp{\left(0,1\right)\alpha}=\outcoeffp{\vect0\alpha}e^{i\vect M_{1}\cdot\vect a_{2}}=\outcoeffp{\vect0\alpha}e^{i0}=\outcoeffp{\vect0\alpha}$
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\end_inset
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,
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\begin_inset Formula $\outcoeffp{\left(1,0\right)\alpha}=e^{i\vect M_{1}\cdot\vect a_{2}}\outcoeffp{\vect 0\alpha}=e^{i\pi}\outcoeffp{\vect 0\alpha}=-\outcoeffp{\vect 0\alpha},$
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\begin_inset Formula $\outcoeffp{\left(1,0\right)\alpha}=e^{i\vect M_{1}\cdot\vect a_{2}}\outcoeffp{\vect0\alpha}=e^{i\pi}\outcoeffp{\vect0\alpha}=-\outcoeffp{\vect0\alpha},$
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\end_inset
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so
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\begin_inset Formula
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\begin{align*}
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\outcoeffp{\vect 0A} & \overset{\sigma_{yz}}{\longmapsto}-\tilde{J}\left(\sigma_{yz}\right)\outcoeffp{\vect 0E},\\
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\outcoeff_{\vect 0C} & \overset{\sigma_{yz}}{\longmapsto}\tilde{J}\left(\sigma_{yz}\right)\outcoeffp{\vect 0C}.
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\outcoeffp{\vect0A} & \overset{\sigma_{yz}}{\longmapsto}-\tilde{J}\left(\sigma_{yz}\right)\outcoeffp{\vect0E},\\
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\outcoeff_{\vect0C} & \overset{\sigma_{yz}}{\longmapsto}\tilde{J}\left(\sigma_{yz}\right)\outcoeffp{\vect0C}.
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\end{align*}
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\end_inset
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@ -1439,19 +1439,19 @@ the original
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rotation, as an example we have
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\begin_inset Formula
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\begin{align*}
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\outcoeffp{\vect 0A} & \overset{C_{3}}{\longmapsto}\tilde{J}\left(C_{3}\right)\outcoeffp{\left(0,-1\right)E}=e^{2\pi i/3}\tilde{J}\left(C_{3}\right)\outcoeffp{\vect 0E},\\
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\outcoeff_{\vect 0C} & \overset{C_{3}}{\longmapsto}\tilde{J}\left(C_{3}\right)\outcoeffp{\left(1,-1\right)A}=e^{-2\pi i/3}\tilde{J}\left(C_{3}\right)\outcoeffp{\vect 0A},\\
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\outcoeff_{\vect 0B} & \overset{C_{3}}{\longmapsto}\tilde{J}\left(C_{3}\right)\outcoeffp{\left(1,-1\right)B}=e^{-2\pi i/3}\tilde{J}\left(C_{3}\right)\outcoeffp{\vect 0B},
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\outcoeffp{\vect0A} & \overset{C_{3}}{\longmapsto}\tilde{J}\left(C_{3}\right)\outcoeffp{\left(0,-1\right)E}=e^{2\pi i/3}\tilde{J}\left(C_{3}\right)\outcoeffp{\vect0E},\\
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\outcoeff_{\vect0C} & \overset{C_{3}}{\longmapsto}\tilde{J}\left(C_{3}\right)\outcoeffp{\left(1,-1\right)A}=e^{-2\pi i/3}\tilde{J}\left(C_{3}\right)\outcoeffp{\vect0A},\\
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\outcoeff_{\vect0B} & \overset{C_{3}}{\longmapsto}\tilde{J}\left(C_{3}\right)\outcoeffp{\left(1,-1\right)B}=e^{-2\pi i/3}\tilde{J}\left(C_{3}\right)\outcoeffp{\vect0B},
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\end{align*}
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\end_inset
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because in this case, the Bloch condition gives
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\begin_inset Formula $\outcoeffp{\left(0,-1\right)\alpha}=\outcoeffp{\vect 0\alpha}e^{i\vect K\cdot\left(-\vect a_{2}\right)}=\outcoeffp{\vect 0\alpha}e^{-4\pi i/3}=\outcoeffp{\vect 0\alpha}e^{2\pi i/3}=\outcoeffp{\vect 0\alpha}$
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\begin_inset Formula $\outcoeffp{\left(0,-1\right)\alpha}=\outcoeffp{\vect0\alpha}e^{i\vect K\cdot\left(-\vect a_{2}\right)}=\outcoeffp{\vect0\alpha}e^{-4\pi i/3}=\outcoeffp{\vect0\alpha}e^{2\pi i/3}=\outcoeffp{\vect0\alpha}$
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\end_inset
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,
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\begin_inset Formula $\outcoeffp{\left(1,-1\right)\alpha}=\outcoeffp{\vect 0\alpha}e^{i\vect K\cdot\left(\vect a_{1}-\vect a_{2}\right)}=e^{-2\pi i/3}\outcoeffp{\vect 0\alpha}.$
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\begin_inset Formula $\outcoeffp{\left(1,-1\right)\alpha}=\outcoeffp{\vect0\alpha}e^{i\vect K\cdot\left(\vect a_{1}-\vect a_{2}\right)}=e^{-2\pi i/3}\outcoeffp{\vect0\alpha}.$
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\end_inset
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