Example of symmetry action in lattices, fig draft.

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Marek Nečada 2019-08-03 12:06:54 +03:00
parent 64fdeb893e
commit 705e61053f
3 changed files with 236 additions and 12 deletions

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@ -712,15 +712,15 @@ This means that the field expansion coefficients
transform as transform as
\begin_inset Formula \begin_inset Formula
\begin{align} \begin{align}
\rcoeffptlm p{\tau}lm & \mapsto\rcoeffptlm{\pi_{g}^{-1}(p)}{\tau}lmD_{m,\mu'}^{\tau l}\left(g\right),\nonumber \\ \rcoeffptlm p{\tau}lm & \overset{g}{\longmapsto}\rcoeffptlm{\pi_{g}^{-1}(p)}{\tau}lmD_{m,\mu'}^{\tau l}\left(g\right),\nonumber \\
\outcoeffptlm p{\tau}lm & \mapsto\outcoeffptlm{\pi_{g}^{-1}(p)}{\tau}lmD_{m,\mu'}^{\tau l}\left(g\right).\label{eq:excitation coefficient under symmetry operation} \outcoeffptlm p{\tau}lm & \overset{g}{\longmapsto}\outcoeffptlm{\pi_{g}^{-1}(p)}{\tau}lmD_{m,\mu'}^{\tau l}\left(g\right).\label{eq:excitation coefficient under symmetry operation}
\end{align} \end{align}
\end_inset \end_inset
Obviously, the expansion coefficients belonging to particles in different Obviously, the expansion coefficients belonging to particles in different
orbits do not mix together. orbits do not mix together.
As before, we introduce a short-hand block-matrix notation for As before, we introduce a short-hand pairwise matrix notation for
\begin_inset CommandInset ref \begin_inset CommandInset ref
LatexCommand eqref LatexCommand eqref
reference "eq:excitation coefficient under symmetry operation" reference "eq:excitation coefficient under symmetry operation"
@ -730,14 +730,23 @@ noprefix "false"
\end_inset \end_inset
(TODO avoid notation clash here in a more consistent and readable way!) (TODO avoid notation clash here in a more consistent and readable way!
\begin_inset Formula
\begin{align}
\rcoeffp p & \overset{g}{\longmapsto}\tilde{J}\left(g\right)\rcoeffp{\pi_{g}^{-1}(p)},\nonumber \\
\outcoeffp p & \overset{g}{\longmapsto}\tilde{J}\left(g\right)\outcoeffp{\pi_{g}^{-1}(p)},\label{eq:excitation coefficient under symmetry operation matrix form}
\end{align}
\end_inset
and also a global block-matrix form
\end_layout \end_layout
\begin_layout Standard \begin_layout Standard
\begin_inset Formula \begin_inset Formula
\begin{align} \begin{align}
\rcoeff & \mapsto J\left(g\right)a,\nonumber \\ \rcoeff & \overset{g}{\longmapsto}J\left(g\right)a,\nonumber \\
\outcoeff & \mapsto J\left(g\right)\outcoeff.\label{eq:excitation coefficient under symmetry operation block form} \outcoeff & \overset{g}{\longmapsto}J\left(g\right)\outcoeff.\label{eq:excitation coefficient under symmetry operation global block form}
\end{align} \end{align}
\end_inset \end_inset
@ -1134,7 +1143,8 @@ The transformation to the symmetry adapted basis
\begin_inset Formula $J(g)$ \begin_inset Formula $J(g)$
\end_inset \end_inset
. : this can happen if the point group symmetry maps some of the scatterers
from the reference unit cell to scatterers belonging to other unit cells.
This is illustrated in Fig. This is illustrated in Fig.
\begin_inset CommandInset ref \begin_inset CommandInset ref
@ -1147,14 +1157,232 @@ noprefix "false"
\end_inset \end_inset
. .
Fig.
\begin_inset CommandInset ref
LatexCommand ref
reference "Phase factor illustration"
plural "false"
caps "false"
noprefix "false"
\end_inset
a shows a hexagonal periodic array with
\begin_inset Formula $p6m$
\end_inset
wallpaper group symmetry, with lattice vectors
\begin_inset Formula $\vect a_{1}=\left(a,0\right)$
\end_inset
and
\begin_inset Formula $\vect a_{2}=\left(a/2,\sqrt{3}a/2\right)$
\end_inset
.
If we delimit our representative unit cell as the Wigner-Seitz cell with
origin in a
\begin_inset Formula $D_{6}$
\end_inset
point group symmetry center (there is one per each unit cell).
Per unit cell, there are five different particles placed on the unit cell
boundary, and we need to make a choice to which unit cell the particles
on the boundary belong; in our case, we choose that a unit cell includes
the particles on the left as denoted by different colors.
If the Bloch vector is at the upper
\begin_inset Formula $M$
\end_inset
point,
\begin_inset Formula $\vect k=\vect M_{1}=\left(0,2\pi/\sqrt{3}a\right)$
\end_inset
, it creates a relative phase of
\begin_inset Formula $\pi$
\end_inset
between the unit cell rows, and the original
\begin_inset Formula $D_{6}$
\end_inset
symmetry is reduced to
\begin_inset Formula $D_{2}$
\end_inset
.
The
\begin_inset Quotes eld
\end_inset
horizontal
\begin_inset Quotes erd
\end_inset
mirror operation
\begin_inset Formula $\sigma_{xz}$
\end_inset
maps, acording to our boundary division, all the particles only inside
the same unit cell, e.g.
\begin_inset Formula
\begin{align*}
\outcoeffp{\vect 0A} & \overset{\sigma_{xz}}{\longmapsto}\tilde{J}\left(\sigma_{xz}\right)\outcoeffp{\vect 0E},\\
\outcoeff_{\vect 0C} & \overset{\sigma_{xz}}{\longmapsto}\tilde{J}\left(\sigma_{xz}\right)\outcoeffp{\vect 0C},
\end{align*}
\end_inset
as in eq.
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:excitation coefficient under symmetry operation"
plural "false"
caps "false"
noprefix "false"
\end_inset
.
However, both the
\begin_inset Quotes eld
\end_inset
vertical
\begin_inset Quotes erd
\end_inset
mirroring
\begin_inset Formula $\sigma_{yz}$
\end_inset
and the
\begin_inset Formula $C_{2}$
\end_inset
rotation map the boundary particles onto the boundaries that do not belong
to the reference unit cell with
\begin_inset Formula $\vect n=\left(0,0\right)$
\end_inset
, so we have, explicitly writing down also the lattice point indices
\begin_inset Formula $\vect n$
\end_inset
,
\begin_inset Formula
\begin{align*}
\outcoeffp{\vect 0A} & \overset{\sigma_{yz}}{\longmapsto}\tilde{J}\left(\sigma_{yz}\right)\outcoeffp{\left(0,1\right)E},\\
\outcoeff_{\vect 0C} & \overset{\sigma_{yz}}{\longmapsto}\tilde{J}\left(\sigma_{yz}\right)\outcoeffp{\left(1,0\right)C},
\end{align*}
\end_inset
but we want
\begin_inset Formula $J(g)$
\end_inset
to operate only inside one unit cell, so we use the Bloch condition
\begin_inset Formula $\outcoeffp{\vect n,\alpha}=\outcoeffp{\vect 0,\alpha}\left(\vect k\right)e^{i\vect k\cdot\vect R_{\vect n}}$
\end_inset
: in this case, we have
\begin_inset Formula $\outcoeffp{\left(0,1\right)\alpha}=\outcoeffp{\vect 0\alpha}e^{i\vect M_{1}\cdot\vect a_{2}}=\outcoeffp{\vect 0\alpha}e^{i0}=\outcoeffp{\vect 0\alpha}$
\end_inset
,
\begin_inset Formula $\outcoeffp{\left(1,0\right)\alpha}=e^{i\vect M_{1}\cdot\vect a_{2}}\outcoeffp{\vect 0\alpha}=e^{i\pi}\outcoeffp{\vect 0\alpha}=-\outcoeffp{\vect 0\alpha},$
\end_inset
so
\begin_inset Formula
\begin{align*}
\outcoeffp{\vect 0A} & \overset{\sigma_{yz}}{\longmapsto}-\tilde{J}\left(\sigma_{yz}\right)\outcoeffp{\vect 0E},\\
\outcoeff_{\vect 0C} & \overset{\sigma_{yz}}{\longmapsto}\tilde{J}\left(\sigma_{yz}\right)\outcoeffp{\vect 0C}.
\end{align*}
\end_inset
If we set instead
\begin_inset Formula $\vect k=\vect K=\left(4\pi/3a,0\right),$
\end_inset
the original
\begin_inset Formula $D_{6}$
\end_inset
point group symmetry reduces to
\begin_inset Formula $D_{3}$
\end_inset
and the unit cells can obtain a relative phase factor of
\begin_inset Formula $e^{-2\pi i/3}$
\end_inset
(blue) or
\begin_inset Formula $e^{2\pi i/3}$
\end_inset
(red).
The
\begin_inset Formula $\sigma_{xz}$
\end_inset
mirror symmetry, as in the previous case, acts purely inside the reference
unit cell with our boundary division.
However, for a counterclockwise
\begin_inset Formula $C_{3}$
\end_inset
rotation, as an example we have
\begin_inset Formula
\begin{align*}
\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},\\
\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},\\
\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},
\end{align*}
\end_inset
because in this case, the Bloch condition gives
\begin_inset Formula $\outcoeffp{\left(0,-1\right)\alpha}=\outcoeffp{\vect 0\alpha}e^{i\vect K\cdot\left(-\vect a_{2}\right)}=\outcoeffp{\vect 0\alpha}e^{-4\pi i/3}=\outcoeffp{\vect 0\alpha}e^{2\pi i/3}=\outcoeffp{\vect 0\alpha}$
\end_inset
,
\begin_inset Formula $\outcoeffp{\left(1,-1\right)\alpha}=\outcoeffp{\vect 0\alpha}e^{i\vect K\cdot\left(\vect a_{1}-\vect a_{2}\right)}=e^{-2\pi i/3}\outcoeffp{\vect 0\alpha}.$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Float figure \begin_inset Float figure
placement document placement document
alignment document alignment document
wide false wide false
sideways false sideways false
status open status collapsed
\begin_layout Plain Layout \begin_layout Plain Layout
\align center
\begin_inset Graphics
filename p6m_mpoint.png
width 100col%
\end_inset
\end_layout
\begin_layout Plain Layout
\begin_inset Graphics
filename p6m_kpoint.png
width 100col%
\end_inset
\end_layout \end_layout
@ -1174,10 +1402,6 @@ name "Phase factor illustration"
\end_inset \end_inset
\end_layout
\begin_layout Plain Layout
\end_layout \end_layout
\end_inset \end_inset