Hexagonal (HEX)#
Pearson symbol: hP
Constructor: HEX()
It is defined by two parameters \(a\) and \(c\). Standardized primitive and conventional cells in the default orientation are
Transformation matrix from standardized primitive cell to standardized conventional cell is
K-path#
\(\mathrm{\Gamma-M-K-\Gamma-A-L-H-A\vert L-M\vert K-H}\)
Point |
\(\times\boldsymbol{b}_1^s\) |
\(\times\boldsymbol{b}_2^s\) |
\(\times\boldsymbol{b}_3^s\) |
|---|---|---|---|
\(\mathrm{\Gamma}\) |
\(0\) |
\(0\) |
\(0\) |
\(\mathrm{A}\) |
\(0\) |
\(0\) |
\(1/2\) |
\(\mathrm{H}\) |
\(1/3\) |
\(1/3\) |
\(1/2\) |
\(\mathrm{K}\) |
\(1/3\) |
\(1/3\) |
\(0\) |
\(\mathrm{L}\) |
\(1/2\) |
\(0\) |
\(1/2\) |
\(\mathrm{M}\) |
\(1/2\) |
\(0\) |
\(0\) |
Variations#
There are no variations for hexagonal lattice.
One example is predefined: hex with \(a = \pi\) and \(c = 2\pi\).
Examples#
Brillouin zone and default kpath#
import wulfric as wulf
cell = wulf.cell.get_cell_example("HEX")
backend = wulf.visualization.PlotlyBackend()
backend.plot(cell, kind="brillouin-kpath")
# Save an image:
backend.save("hex_reciprocal.png")
# Interactive plot:
backend.show()
Primitive, Wigner-Seitz and conventional cells#
Click on the legend to hide a cell.
import wulfric as wulf
cell = wulf.cell.get_cell_example("HEX")
backend = wulf.visualization.PlotlyBackend()
backend.plot(cell, kind="primitive", label="primitive", color="black")
backend.plot(cell, kind="wigner-seitz", label="wigner-seitz", color="green")
# Save an image:
backend.save("hex_real.png")
# Interactive plot:
backend.show()
Cell standardization#
Since parameters \(a\) and \(c\) are not restricted (i.e. \(a = c\) is allowed), wulfric uses angles \(\alpha\), \(\beta\) and \(\gamma\) to determine the standard form of the cell. For the primitive cell in a standard form \(\alpha^s = \beta^s = 90^{\circ}\) and \(\gamma^s = 120^{\circ}\).
Matrix \(\boldsymbol{S}\) is constructed in two steps.
Step 1#
If \(\alpha = \beta = \dfrac{\pi}{2}\) then
\[(\boldsymbol{a}_1^{(1)}, \boldsymbol{a}_2^{(1)}, \boldsymbol{a}_3^{(1)}) = (\boldsymbol{a}_1, \boldsymbol{a}_2, \boldsymbol{a}_3)\]and
\[\begin{split}\boldsymbol{S}_1 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad \boldsymbol{S}_1^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\end{split}\]If \(\beta = \gamma = \dfrac{\pi}{2}\) then
\[(\boldsymbol{a}_1^{(1)}, \boldsymbol{a}_2^{(1)}, \boldsymbol{a}_3^{(1)}) = (\boldsymbol{a}_2, \boldsymbol{a}_3, \boldsymbol{a}_1)\]and
\[\begin{split}\boldsymbol{S}_1 = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \qquad \boldsymbol{S}_1^{-1} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\end{split}\]If \(\alpha = \gamma = \dfrac{\pi}{2}\) then
\[(\boldsymbol{a}_1^{(1)}, \boldsymbol{a}_2^{(1)}, \boldsymbol{a}_3^{(1)}) = (\boldsymbol{a}_3, \boldsymbol{a}_1, \boldsymbol{a}_2)\]and
\[\begin{split}\boldsymbol{S}_1 = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} \qquad \boldsymbol{S}_1^{-1} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\end{split}\]
Step 2#
If \(\gamma^{(1)} = \dfrac{2}{3}\pi\) then
\[(\boldsymbol{a}_1^s, \boldsymbol{a}_2^s, \boldsymbol{a}_3^s) = (\boldsymbol{a}_1^{(1)}, \boldsymbol{a}_2^{(1)}, \boldsymbol{a}_3^{(1)})\]and
\[\begin{split}\boldsymbol{S}_2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad \boldsymbol{S}_2^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\end{split}\]If \(\gamma^{(1)} = \dfrac{\pi}{3}\) then
\[(\boldsymbol{a}_1^s, \boldsymbol{a}_2^s, \boldsymbol{a}_3^s) = (\boldsymbol{a}_2^{(1)}, -\boldsymbol{a}_1^{(1)}, \boldsymbol{a}_3^{(1)})\]and
\[\begin{split}\boldsymbol{S}_2 = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad \boldsymbol{S}_2^{-1} = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\end{split}\]