mirror of https://gitlab.com/QEF/q-e.git
542 lines
21 KiB
TeX
542 lines
21 KiB
TeX
\documentclass[12pt,a4paper]{article}
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\def\version{7.3.1}
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\def\qe{{\sc Quantum ESPRESSO}}
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\textwidth = 17cm
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\textheight = 24cm
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\topmargin =-1 cm
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\oddsidemargin = 0 cm
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% BEWARE: don't revert from graphicx for epsfig, because latex2html
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% doesn't handle epsfig commands !!!
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\usepackage{graphicx}
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\usepackage{amssymb}
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\begin{document}
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\author{}
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\date{}
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\title{
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\vskip 1cm
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% title
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\Huge Points inside the Brillouin zone \\
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\Large Notes by Andrea Dal Corso (SISSA - Trieste)
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}
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\maketitle
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\newpage
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\section{Brillouin zone}
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\qe\ (QE) support for the definition of high symmetry lines inside the Brillouin
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zone (BZ) is still rather limited. However QE can calculate the coordinates
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of the vertexes of the BZ and of particular points inside the BZ. These
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notes show the shape and orientation of the BZ used by QE. The principal
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direct and reciprocal lattice vectors, as implemented
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in the routine \texttt{latgen}, are illustrated here together with the labels
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of each point. These labels can be given as input in a band or phonon
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calculation to define paths in the BZ. This feature is available with
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the option \texttt{tpiba\_b} or \texttt{crystal\_b} in a \texttt{'bands'}
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calculation or with the option \texttt{q\_in\_band\_form} in the input of the
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\texttt{matdyn.x} code.
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BEWARE: you need to explicitly specify \texttt{ibrav} to use this feature.
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Lines in reciprocal space are defined by giving the coordinates of the
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starting and ending points and the number of points of each line.
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The coordinates of the starting and ending points can be
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given explicitly with three real numbers or by giving the label of a
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point known to QE.
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For example:
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\begin{verbatim}
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X 10
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gG 25
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0.5 0.5 0.5 1
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\end{verbatim}
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indicate a path composed by two lines. The first line starts at point $X$,
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ends at point $\Gamma$, and has $10$ {\bf k} points. The second line starts
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at $\Gamma$, ends at the point of coordinates \texttt{(0.5,0.5,0.5)} and
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has $25$ {\bf k} points. Greek labels are prefixed by the letter
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\texttt{g}: \texttt{gG} indicates the $\Gamma$ point, \texttt{gS} the
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$\Sigma$ point etc. Subscripts are written after the label: the point $P_1$
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is indicated as \texttt{P1}. In the following section you can find the
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labels of the points defined in each BZ.
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There are many conventions to label high symmetry
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points inside the BZ. The variable \texttt{point\_label\_type} selects the
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set of labels used by QE. The default is \texttt{point\_label\_type='SC'} and
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the labels have been taken from W. Setyawan and S. Curtarolo, Comp. Mat. Sci.
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{\bf 49}, 299 (2010). Other choices can be more convenient in other situations.
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The names reported in the web pages
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\texttt{http://www.cryst.ehu.es/cryst/get\_kvec.html}
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are available for some BZ. You can use them by setting
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(\texttt{point\_label\_type='BI'}), others can be added in the future.
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This option is available only with \texttt{ibrav$\ne$0} and for
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all positive \texttt{ibrav} with the exception of the base centered
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monoclinic (\texttt{ibrav=13}), and triclinic (\texttt{ibrav=14}) lattices.
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In these cases you have to give all the coordinates of the {\bf k}-points.
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\subsection{\texttt{ibrav=1}, simple cubic lattice}
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The primitive vectors of the direct lattice are:
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\begin{eqnarray}
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{\bf a}_1 &=& a (1, 0, 0), \nonumber \\
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{\bf a}_2 &=& a (0, 1, 0), \nonumber \\
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{\bf a}_3 &=& a (0, 0, 1), \nonumber
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\nonumber
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\end{eqnarray}
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while the reciprocal lattice vectors are:
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\begin{eqnarray}
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{\bf b}_1 &=& {2\pi \over a} (1, 0, 0), \nonumber \\
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{\bf b}_2 &=& {2\pi \over a} (0, 1, 0), \nonumber \\
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{\bf b}_3 &=& {2\pi \over a} (0, 0, 1). \nonumber
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\nonumber
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\end{eqnarray}
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The Brilloin zone is:
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/cubic_bi.png}
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\end{center}
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\texttt{X$_1$} is available only with $\texttt{point\_label\_type='BI'}$.
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\subsection{\texttt{ibrav=2}, face centered cubic lattice}
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The primitive vectors of the direct lattice are:
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\begin{eqnarray}
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{\bf a}_1 &=& {a \over 2} (-1, 0, 1), \nonumber \\
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{\bf a}_2 &=& {a \over 2} (0, 1, 1), \nonumber \\
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{\bf a}_3 &=& {a \over 2} (-1, 1, 0), \nonumber
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\nonumber
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\end{eqnarray}
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while the reciprocal lattice vectors are:
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\begin{eqnarray}
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{\bf b}_1 &=& {2\pi \over a} (-1, -1, 1), \nonumber \\
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{\bf b}_2 &=& {2\pi \over a} (1, 1, 1), \nonumber \\
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{\bf b}_3 &=& {2\pi \over a} (-1, 1, -1). \nonumber
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\nonumber
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\end{eqnarray}
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The Brillouin zone is:
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/fcc_sc.png} \hspace{1.cm}
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\includegraphics[width=7.5cm,angle=0]{images/fcc_bi.png}
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\end{center}
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Labels corresponding to $\texttt{point\_label\_type='SC'}$ and to
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$\texttt{point\_label\_type='BI'}$ are shown on the left and on the right,
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respectively.
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\subsection{\texttt{ibrav=3}, body centered cubic lattice}
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The primitive vectors of the direct lattice are:
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\begin{eqnarray}
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{\bf a}_1 &=&{a \over 2} (1, 1, 1), \nonumber \\
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{\bf a}_2 &=&{a \over 2} (-1, 1, 1), \nonumber \\
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{\bf a}_3 &=&{a \over 2} (-1, -1, 1), \nonumber
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\nonumber
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\end{eqnarray}
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while the reciprocal lattice vectors are:
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\begin{eqnarray}
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{\bf b}_1 &=&{2\pi \over a} (1, 0, 1), \nonumber \\
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{\bf b}_2 &=&{2\pi \over a} (-1, 1, 0), \nonumber \\
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{\bf b}_3 &=&{2\pi \over a} (0, -1, 1). \nonumber
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\nonumber
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\end{eqnarray}
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/bcc_bi.png}
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\end{center}
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\texttt{H$_1$} is available only with $\texttt{point\_label\_type='BI'}$.
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\subsection{\texttt{ibrav=4}, hexagonal lattice}
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The primitive vectors of the direct lattice are:
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\begin{eqnarray}
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{\bf a}_1 &=& a (1, 0, 0), \nonumber \\
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{\bf a}_2 &=& a (-{1 \over 2}, {\sqrt{3} \over 2}, 0), \nonumber \\
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{\bf a}_3 &=& a (0, 0, {c\over a}), \nonumber
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\nonumber
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\end{eqnarray}
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while the reciprocal lattice vectors are:
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\begin{eqnarray}
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{\bf b}_1 &=& {2\pi \over a} (1, {1 \over \sqrt{3}}, 0), \nonumber \\
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{\bf b}_2 &=& {2\pi \over a} (0, {2 \over \sqrt{3}}, 0), \nonumber \\
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{\bf b}_3 &=& {2\pi \over a} (0, 0, {a\over c}). \nonumber
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\nonumber
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\end{eqnarray}
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The BZ is:
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/hex.png}
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\end{center}
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The figure has been obtained with ${c/a}=1.4$.
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\subsection{\texttt{ibrav=5}, trigonal lattice}
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The primitive vectors of the direct lattice are:
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\begin{eqnarray}
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{\bf a}_1 &=& a ({\sqrt{3}\over 2}\sin{\theta}, -{1\over 2} \sin{\theta},
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\cos{\theta}),
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\nonumber \\
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{\bf a}_2 &=& a (0, \sin{\theta}, \cos{\theta}),
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\nonumber \\
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{\bf a}_3 &=& a (-{\sqrt{3}\over 2} \sin{\theta}, -{1\over 2} \sin{\theta},
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\cos{\theta}),
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\nonumber \\
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\nonumber
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\end{eqnarray}
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while the reciprocal lattice vectors are:
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\begin{eqnarray}
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{\bf b}_1 &=& {2\pi \over a} ({1\over \sqrt{3} \sin{\theta}},
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-{1 \over 3 \sin{\theta}}, {1\over 3 \cos{\theta}}), \nonumber \\
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{\bf b}_2 &=& {2\pi \over a} (0,
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{2 \over 3 \sin{\theta}}, {1\over 3 \cos{\theta}}), \nonumber \\
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{\bf b}_3 &=& {2\pi \over a} (-{1\over \sqrt{3} \sin{\theta}},
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-{1 \over 3 \sin{\theta}}, {1\over 3 \cos{\theta}}), \nonumber \\
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\end{eqnarray}
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where $\sin{\theta}=\sqrt{2\over 3}\sqrt{1-\cos{\alpha}}$
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and $\cos{\theta}=\sqrt{1\over 3}\sqrt{1 + 2 \cos{\alpha}}$ and $\alpha$
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is the angle between any two primitive direct lattice vectors.
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There are two possible shapes of the BZ, depending on the
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value of the angle $\alpha$. For $\alpha < 90^\circ$ we
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have:
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/tri_1.png}
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\end{center}
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The figure has been obtained with $\alpha=70^\circ$.
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For $90^\circ < \alpha < 120^\circ$ we have:
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/tri_2.png}
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\end{center}
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The figure has been obtained with $\alpha=110^\circ$.
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\subsection{\texttt{ibrav=6}, simple tetragonal lattice}
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The primitive vectors of the direct lattice are:
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\begin{eqnarray}
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{\bf a}_1 &=& a (1, 0, 0), \nonumber \\
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{\bf a}_2 &=& a (0, 1, 0), \nonumber \\
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{\bf a}_3 &=& a (0, 0, {c\over a}), \nonumber
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\nonumber
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\end{eqnarray}
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while the reciprocal lattice vectors are:
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\begin{eqnarray}
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{\bf b}_1 &=& {2\pi \over a} (1, 0, 0), \nonumber \\
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{\bf b}_2 &=& {2\pi \over a} (0, 1, 0), \nonumber \\
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{\bf b}_3 &=& {2\pi \over a} (0, 0, {a\over c}). \nonumber
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\nonumber
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\end{eqnarray}
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/st.png}
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\end{center}
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The figure has been obtained with $c/a=1.4$.
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\subsection{\texttt{ibrav=7}, centered tetragonal lattice}
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The primitive vectors of the direct lattice are:
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\begin{eqnarray}
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{\bf a}_1 &=& {a \over 2} (1, -1, {c\over a}), \nonumber \\
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{\bf a}_2 &=& {a \over 2} (1, 1, {c\over a}), \nonumber \\
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{\bf a}_3 &=& {a \over 2} (-1, -1, {c\over a}), \nonumber
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\nonumber
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\end{eqnarray}
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while the reciprocal lattice vectors are:
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\begin{eqnarray}
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{\bf b}_1 &=& {2\pi \over a} (1, -1, 0), \nonumber \\
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{\bf b}_2 &=& {2\pi \over a} (0, 1, {a\over c}), \nonumber \\
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{\bf b}_3 &=& {2\pi \over a} (-1, 0, {a\over c}). \nonumber
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\nonumber
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\end{eqnarray}
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In this case there are two different shapes of the BZ depending on the
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$c/a$ ratio. For $c/a<1$ we have:
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/stc1.png}
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\end{center}
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The figure has been obtained with $c/a=0.5$ ($a>c$). For $c/a>1$ we have:
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/stc2_sc.png} \hspace{1cm}
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\includegraphics[width=7.5cm,angle=0]{images/stc2.png}
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\end{center}
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The figure has been obtained with $c/a=1.4$ ($a<c$).
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Labels corresponding to $\texttt{point\_label\_type='SC'}$ are shown on the left,
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those corresponding to $\texttt{point\_label\_type='BI'}$ on the right.
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\subsection{\texttt{ibrav=8}, simple orthorhombic lattice}
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The primitive vectors of the direct lattice are:
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\begin{eqnarray}
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{\bf a}_1 &=& a (1, 0, 0), \nonumber \\
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{\bf a}_2 &=& a (0, {b\over a}, 0), \nonumber \\
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{\bf a}_3 &=& a (0, 0, {c\over a}), \nonumber
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\nonumber
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\end{eqnarray}
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while the reciprocal lattice vectors are:
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\begin{eqnarray}
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{\bf b}_1 &=& {2\pi \over a} (1, 0, 0), \nonumber \\
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{\bf b}_2 &=& {2\pi \over a} (0, {a\over b}, 0), \nonumber \\
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{\bf b}_3 &=& {2\pi \over a} (0, 0, {a\over c}). \nonumber
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\nonumber
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\end{eqnarray}
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/so.png}
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\end{center}
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The figure has been obtained with $b/a=1.2$ and $c/a=1.5$.
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\subsection{\texttt{ibrav=9}, one-face centered orthorhombic lattice}
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The direct lattice vectors are:
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\begin{eqnarray}
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{\bf a}_1 &=& {a \over 2} (1, {b \over a}, 0), \nonumber \\
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{\bf a}_2 &=& {a \over 2} (-1, {b \over a}, 0), \nonumber \\
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{\bf a}_3 &=& a (0, 0, {c \over a}), \nonumber
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\nonumber
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\end{eqnarray}
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while the reciprocal lattice vectors are
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\begin{eqnarray}
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{\bf b}_1 &=& {2\pi \over a} (1, {a \over b}, 0), \nonumber \\
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{\bf b}_2 &=& {2\pi \over a} (-1, {a \over b}, 0), \nonumber \\
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{\bf b}_3 &=& {2\pi \over a} (0, 0, {a \over c}). \nonumber
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\nonumber
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\end{eqnarray}
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There is one shape that can have two orientations depending on the
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ratio between of $a$ and $b$:
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/ofco_2.png} \hspace{1cm}
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\includegraphics[width=6.5cm,angle=0]{images/ofco_1.png}
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\end{center}
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The figures have been obtained with $b/a=0.8$ and $c/a=1.4$ (left part $b<a$)
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and $b/a=1.2$ and $c/a=1.4$ (right part $b>a$).
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\subsection{\texttt{ibrav=10}, face centered orthorhombic lattice}
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The direct lattice vectors are:
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\begin{eqnarray}
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{\bf a}_1 &=& {a \over 2} (1, 0, {c \over a}), \nonumber \\
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{\bf a}_2 &=& {a \over 2} (1, {b \over a}, 0), \nonumber \\
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{\bf a}_3 &=& {a \over 2} (0, {b \over a}, {c \over a}). \nonumber
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\nonumber
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\end{eqnarray}
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while the reciprocal lattice vectors are
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\begin{eqnarray}
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{\bf b}_1 &=& {2\pi \over a} (1, -{a \over b}, {a \over c}), \nonumber \\
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{\bf b}_2 &=& {2\pi \over a} (1, {a \over b}, -{a \over c}), \nonumber \\
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{\bf b}_3 &=& {2\pi \over a} (-1, {a \over b}, {a \over c}). \nonumber
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\nonumber
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\end{eqnarray}
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In this case there are three different shapes
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that can be rotated in different ways depending on the relative
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sizes of $a$, $b$, and $c$.
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If $a$ is the shortest side, there are three different shapes according
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to
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\begin{equation}
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{1\over a^2} \lesseqqgtr
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{1\over b^2} + {1\over c^2},
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\label{uno}
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\end{equation}
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if $b$ is the shortest side there are three different shapes according to
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\begin{equation}
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{1\over b^2} \lesseqqgtr {1\over a^2} + {1\over c^2},
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\label{due}
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\end{equation}
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and if $c$ is the shortest side there are
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three different shapes according to
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\begin{equation}
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{1\over c^2} \lesseqqgtr {1\over a^2}
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+ {1\over b^2}.
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\label{tre}
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\end{equation}
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For each case there are two possibilities. If $a$
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is the shortest side, we can have $b<c$ or $b>c$, if $b$ is
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the shortest side, we can have $a<c$ or $a>c$, and finally
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if $c$ is the shortest side we can have $a<b$ or $a>b$. In total we
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have $18$ distinct cases. Not all cases give different BZ.
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All the cases with the $<$ sign in Eqs.~\ref{uno}, \ref{due}, \ref{tre} give
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the same shape of the BZ that differ for the relative sizes of the faces.
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All the cases with the $>$ sign in Eqs.~\ref{uno}, \ref{due}, \ref{tre} give
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the same shape with faces of different sizes and oriented in different ways.
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Finally the particular case with the $=$ sign in Eqs.~\ref{uno},
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\ref{due}, \ref{tre} give another shape
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with faces of different size and different orientations.
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We show all the 18 possibilities and the labels used in each case.
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We start with the case in which $a$ is the shortest side and show on the
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left the case $b<c$ and on the right the case $b>c$.
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The first possibility is that ${1\over a^2} <
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{1\over b^2} + {1\over c^2}$:
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/ofc_1.png} \hspace{1cm}
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\includegraphics[width=6.5cm,angle=0]{images/ofc_2.png}
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\end{center}
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The figures have been obtained with $b/a=1.2$ and $c/a=1.4$ (left part $b<c$),
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and with $b/a=1.4$ and $c/a=1.2$ (right part $b>c$).
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The second possibility is that ${1\over a^2} =
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{1\over b^2} + {1\over c^2}$:
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/ofc_13.png} \hspace{1cm}
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\includegraphics[width=5.5cm,angle=0]{images/ofc_14.png}
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\end{center}
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The figures have been obtained with $b/a=1.2$ and $c/a=1.80906807$
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(left part $b<c$) and with $b/a=1.80906807$ and $c/a=1.2$ (right part $b>c$).
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The third possibility is that ${1 \over a^2} > {1\over b^2} + {1\over c^2}$:
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\begin{center}
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\includegraphics[width=6.5cm,angle=0]{images/ofc_7.png} \hspace{1cm}
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\includegraphics[width=4.0cm,angle=0]{images/ofc_8.png}
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\end{center}
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The figures have been obtained with $b/a=1.2$ and $c/a=2.4$ (left part $b<c$),
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and with $b/a=2.4$ and $c/a=1.2$ (right part $b>c$).
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Then we consider the cases in which $b$ is the shortest side and show
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on the left the case in which $a<c$ and on the right the case $a>c$.
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We have the same three possibilities as before. The first possibility
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is that ${1 \over b^2} < {1\over a^2} + {1\over c^2}$:
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\begin{center}
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\includegraphics[width=7.5cm,angle=0]{images/ofc_3.png} \hspace{1cm}
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\includegraphics[width=7.5cm,angle=0]{images/ofc_4.png} \hspace{1cm}
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\end{center}
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The figures have been obtained with $b/a=0.9$ and $c/a=1.2$
|
|
(left part $a<c$) and $b/a=0.75$ and $c/a=0.95$ (right part $a>c$).
|
|
|
|
The second possibility is that ${1 \over b^2}={1\over a^2} + {1\over c^2}$:
|
|
\begin{center}
|
|
\includegraphics[width=7.5cm,angle=0]{images/ofc_15.png} \hspace{1cm}
|
|
\includegraphics[width=7.5cm,angle=0]{images/ofc_16.png}
|
|
\end{center}
|
|
The figures have been obtained with $b/a=0.8$ and $c/a=1.33333333333$ (left
|
|
part $a<c$), and $b/a=0.6$ and $c/a=0.75$ (right part $a>c$).
|
|
|
|
The third possibility is than ${1\over b^2}>{1\over a^2} + {1\over c^2}$:
|
|
\begin{center}
|
|
\includegraphics[width=7.5cm,angle=0]{images/ofc_9.png} \hspace{1cm}
|
|
\includegraphics[width=7.5cm,angle=0]{images/ofc_10.png}
|
|
\end{center}
|
|
The figures have been obtained with $b/a=0.8$ and $c/a=2.0$ (left part $a<c$),
|
|
and with $b/a=0.4$ and $c/a=0.5$ (right part $a>c$).
|
|
|
|
Finally we consider the case in which $c$ is the shortest side and show on
|
|
the left the case in which $a<b$ and on the right the case in which $a>b$.
|
|
|
|
The first possibility is that ${1\over c^2}<{1\over a^2} + {1\over b^2}$:
|
|
\begin{center}
|
|
\includegraphics[width=6.5cm,angle=0]{images/ofc_5.png} \hspace{1cm}
|
|
\includegraphics[width=7.5cm,angle=0]{images/ofc_6.png}
|
|
\end{center}
|
|
The figures have been obtained with $b/a=1.2$ and $c/a=0.85$ (left part $a<b$)
|
|
and $b/a=0.85$ and $c/a=0.75$ (right part $a>b$).
|
|
|
|
The second possibility is that ${1\over c^2}={1\over a^2} + {1\over b^2}$:
|
|
\begin{center}
|
|
\includegraphics[width=5.5cm,angle=0]{images/ofc_17.png} \hspace{1cm}
|
|
\includegraphics[width=7.5cm,angle=0]{images/ofc_18.png}
|
|
\end{center}
|
|
The figures have been obtained with $b/a=1.333333333$ and $c/a=0.8$
|
|
(left part $a<b$) and with $b/a=0.66$ and $c/a=0.5508422$ (right part
|
|
$a>b$).
|
|
|
|
Finally the third possibility is that
|
|
${1\over c^2} >{1\over a^2} + {1\over b^2}$:
|
|
\begin{center}
|
|
\includegraphics[width=5.0cm,angle=0]{images/ofc_11.png} \hspace{1cm}
|
|
\includegraphics[width=7.5cm,angle=0]{images/ofc_12.png}
|
|
\end{center}
|
|
The figures have been obtained with $b/a=2.0$ and $c/a=0.8$ (left part $a<b$),
|
|
and $b/a=0.5$ and $c/a=0.4$ (right part $a>b$).
|
|
|
|
\subsection{\texttt{ibrav=11}, body centered orthorhombic lattice}
|
|
The direct lattice vectors are:
|
|
\begin{eqnarray}
|
|
{\bf a}_1 &=& {a \over 2} (1, {b \over a}, {c \over a}), \nonumber \\
|
|
{\bf a}_2 &=& {a \over 2} (-1, {b \over a}, {c \over a}), \nonumber \\
|
|
{\bf a}_3 &=& {a \over 2} (-1, -{b \over a}, {c \over a}). \nonumber
|
|
\nonumber
|
|
\end{eqnarray}
|
|
while the reciprocal lattice vectors are:
|
|
\begin{eqnarray}
|
|
{\bf b}_1 &=& {2\pi \over a} (1, 0, {a \over c}), \nonumber \\
|
|
{\bf b}_2 &=& {2\pi \over a} (-1, {a \over b}, 0), \nonumber \\
|
|
{\bf b}_3 &=& {2\pi \over a} (0, -{a \over b}, {a \over c}). \nonumber
|
|
\nonumber
|
|
\end{eqnarray}
|
|
In this case the BZ has one shape that can be rotated in
|
|
different ways depending on the relative sizes of $a$, $b$, and $c$.
|
|
Similar orientations and BZ that differ only for the relative sizes of
|
|
the faces are obtained for the cases that have in common the longest side.
|
|
Therefore we distinguish the cases in which $a$ is the longest side
|
|
and $b<c$ or $b>c$, the cases in which $b$ is the longest side and
|
|
$a<c$ or $a>c$ and the cases in which $c$ is the longest side and $a<b$
|
|
or $a>b$. We have $6$ distinct cases.
|
|
|
|
First we take $a$ as the longest side and show
|
|
on the left the case $b<c$ and on the right the case $b>c$:
|
|
\begin{center}
|
|
\includegraphics[width=7.5cm,angle=0]{images/bco_4.png} \hspace{1.0cm}
|
|
\includegraphics[width=7.5cm,angle=0]{images/bco_5.png}
|
|
\end{center}
|
|
The figures have been obtained with $b/a=0.7$ and $c/a=0.85$ (left part
|
|
$b<c$) and $b/a=0.85$ and $c/a=0.7$ (right part $b>c$).
|
|
|
|
Then we take $b$ as the longest side and show on the left the case
|
|
in which $a<c$ and on the right the case in which $a>c$:
|
|
\begin{center}
|
|
\includegraphics[width=7.5cm,angle=0]{images/bco_2.png}\hspace{1cm}
|
|
\includegraphics[width=7.cm,angle=0]{images/bco_3.png}
|
|
\end{center}
|
|
The figures have been obtained with $b/a=1.4$ and $c/a=1.2$ (left part $a<c$)
|
|
and $b/a=1.2$ and $c/a=0.8$ (right part $a>c$).
|
|
|
|
Finally we take $c$ as the longest side and show on the left the case in
|
|
which $a<b$ and on the right the case in which $b<a$:
|
|
\begin{center}
|
|
\includegraphics[width=7.5cm,angle=0]{images/bco_1.png} \hspace{1.0 cm}
|
|
\includegraphics[width=7.5cm,angle=0]{images/bco_6.png}
|
|
\end{center}
|
|
The figures have been obtained with $b/a=1.2$ and $c/a=1.4$ (left part), and
|
|
$b/a=0.8$ and $c/a=1.2$ (right part).
|
|
|
|
\subsection{\texttt{ibrav=12}, simple monoclinic lattice, c unique}
|
|
The direct lattice vectors are:
|
|
\begin{eqnarray}
|
|
{\bf a}_1 &=& a (1, 0, 0), \nonumber \\
|
|
{\bf a}_2 &=& a ({b \over a} \cos{\gamma}, {b \over a}\sin{\gamma}, 0), \nonumber \\
|
|
{\bf a}_3 &=& a (0, 0, {c \over a}).
|
|
\nonumber
|
|
\end{eqnarray}
|
|
while the reciprocal lattice vectors are:
|
|
\begin{eqnarray}
|
|
{\bf b}_1 &=& {2\pi \over a} (1, -{\cos{\gamma}\over \sin{\gamma}}, 0), \nonumber \\
|
|
{\bf b}_2 &=& {2\pi \over a} (0, {a \over b \sin{\gamma}}, 0), \nonumber \\
|
|
{\bf b}_3 &=& {2\pi \over a} (0, 0, {a \over c}). \nonumber
|
|
\end{eqnarray}
|
|
The Brillouin zone is:
|
|
\begin{center}
|
|
\includegraphics[width=7.5cm,angle=0]{images/monoc.png}
|
|
\end{center}
|
|
The figure has been obtained with $b/a=0.8$, $c/a=1.4$ and $\cos{\gamma}=0.3$.
|
|
|
|
\subsection{\texttt{ibrav=-12}, simple monoclinic lattice, b unique}
|
|
The direct lattice vectors are:
|
|
\begin{eqnarray}
|
|
{\bf a}_1 &=& a (1, 0, 0), \nonumber \\
|
|
{\bf a}_2 &=& a (0, {b \over a}, 0), \nonumber \\
|
|
{\bf a}_3 &=& a ({c \over a} \cos{\beta}, 0, {c \over a}\sin{\beta}), \nonumber
|
|
\end{eqnarray}
|
|
while the reciprocal lattice vectors are:
|
|
\begin{eqnarray}
|
|
{\bf b}_1 &=& {2\pi \over a} (1, 0, -{\cos{\beta}\over \sin{\beta}}), \nonumber \\
|
|
{\bf b}_2 &=& {2\pi \over a} (0, {a \over b}, 0), \nonumber \\
|
|
{\bf b}_3 &=& {2\pi \over a} (0, 0, {a \over c \sin{\beta}}). \nonumber
|
|
\end{eqnarray}
|
|
The Brillouin zone is:
|
|
\begin{center}
|
|
\includegraphics[width=7.5cm,angle=0]{images/monob.png}
|
|
\end{center}
|
|
The figure has been obtained with $b/a=0.8$, $c/a=1.4$ and $\cos{\beta}=0.3$.
|
|
|
|
|
|
\subsection{\texttt{ibrav=13,14}, one-base centered monoclinic,
|
|
triclinic}
|
|
These lattices are not supported by this feature, you have to give
|
|
explicitly the coordinates of the path.
|
|
|
|
\section{Bibliography}
|
|
|
|
\noindent [1] G.F. Koster, Space groups and their representations, Academic press,
|
|
New York and London, (1957).
|
|
|
|
\noindent [2] C.J. Bradley and A.P. Cracknell, The mathematical theory of symmetry
|
|
in solids, Oxford University Press, (1972).
|
|
|
|
\noindent [3] W. Setyawan and S. Curtarolo, Comp. Mat. Sci. {\bf 49}, 299 (2010).
|
|
|
|
\noindent [4] E.S. Tasci, G. de la Flor, D. Orobengoa, C. Capillas,
|
|
J.M. Perez-Mato, M.I. Aroyo, ``An introduction to the tools hosted in the
|
|
Bilbao Crystallographic Server''. EPJ Web of Conferences 22 00009 (2012).
|
|
|
|
\end{document}
|