3.5.5 $D_{2h}$ symmetry

Figure 3.10: Irreducible wedge of the first BZ of a diamond structure stressed along direction [110].

The Bravais cube of the crystal class $O_h$ can be converted to a parallelepiped of the orthorhombic system belonging to $D_{2h}$ in two ways [Bir74]:

1. Dilatation or compression along two of the three fourfold axes ${\ensuremath{\mathitbf{e}}}_i$. This results in a right parallelepiped with rectangular faces (cuboid). Of the five twofold axes ${\ensuremath{\mathitbf{e}}}_i$ and ${\ensuremath{\mathitbf{e}}}_s$ of $D_{4h}$, only the three ${\ensuremath{\mathitbf{e}}}_i$ along the edges of the parallelepiped remain. This symmetry reduction can be achieved by applying stress of different magnitude along two of the three equivalent $\langle 100 \rangle$ directions, simultaneously. In this case, the strain tensor is given by ${\ensuremath{\varepsilon_{11}}} \neq {\ensuremath{\varepsilon_{22}}} \neq {\ensuremath{\varepsilon_{33}}}$ and contains vanishing off-diagonal components.
2. The deformation originates from shearing the unit cube, thus altering the angles between the basis vectors. The result is a rectangular parallelepiped with rhombic base, which is also invariant under $D_{2h}$. Of the original five twofold axes ${\ensuremath{\mathitbf{e}}}_i$ and ${\ensuremath{\mathitbf{e}}}_s$ only two (diagonals of the base) remain. This type of lattice results, when uniaxial stress is applied along $\langle110\rangle$ or from biaxial strain in a {110} plane. The strain tensor has the form

   $$\ensuremath{{\underaccent{\bar}{\varepsilon}}} = \begin{pmatrix}{... ...varepsilon_{11}}} & 0\\ 0 & 0 & {\ensuremath{\varepsilon_{33}}}\end{pmatrix}\ ,$$ (3.44)

   
   
   where the components of the strain tensor can be related to stress according to (3.18).

This group has only eight symmetry elements (given in Table 3.2). The irreducible wedge with a volume of $\Omega_{\mathrm{BZ}}/8$ can be mapped onto six wedges of the unstrained lattice in the limit of vanishing strain.

When dilating or compressing along two of the three fourfold axes ${\ensuremath{\mathitbf{e}}}_i$ (case 1), any octant of the BZ can be chosen as the irreducible wedge. In the presence of uniaxial stress along [110] (case 2) a possible choice for the irreducible wedge is depicted in Figure 3.10. The six wedges labeled in Figure 3.10 can be transformed into the first wedge by the transformations

$$\ensuremath{{\underaccent{\bar}{T}}}_1 = \begin{pmatrix}1 & \hpha... ...\hphantom{-}0 & \hphantom{-}0\\ 0 & \hphantom{-}1 & \hphantom{-}0\end{pmatrix},$$

$$\ensuremath{{\underaccent{\bar}{T}}}_4 = \begin{pmatrix}1 & \hpha... ...{-}1\\ 1 & \hphantom{-}0 & \hphantom{-}0\\ 0 & -1 & \hphantom{-}0\end{pmatrix}.$$

E. Ungersboeck: Advanced Modelling Aspects of Modern Strained CMOS Technology