Translational domain structures (translation twins)

Wondratschek, H.

doi:10.1107/97809553602060000538

International
Tables for
Crystallography
Volume A1
Symmetry relations between space groups
Edited by Hans Wondratschek and Ulrich Müller

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International Tables for Crystallography (2006). Vol. A1. ch. 1.2, pp. 20-21 | 1 | 2 |

Section 1.2.7.3. Translational domain structures (translation twins)

Hans Wondratschek^a ^*

^a Institut für Kristallographie, Universität, D-76128 Karlsruhe, Germany
Correspondence e-mail: wondra@physik.uni-karlsruhe.de

1.2.7.3. Translational domain structures (translation twins)

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In Example 1.2.7.2.4, a phase transition was discussed which involves only translationengleiche group–subgroup relations and, hence, only orientational relations between the domains. The following two examples treat klassengleiche transitions, i.e. $[{\cal H}]$ is a klassengleiche subgroup of $[{\cal G}]$ , and translational domain structures, also called translation twins, may appear.

Translational domain structures consist of domains which are parallel, i.e. have the same orientation of their structures (and thus of their lattices) but differ in their location because of the loss of translations of the parent phase in the phase transition. The origins of the larger unit cells of the phase B with subgroup $[{\cal H}]$ may coincide with any of the origins of the smaller unit cells of the parent structure A with space group $[{\cal G}]$ . Again the number of such domain states is equal to the index of $[{\cal H}]$ in $[{\cal G}]$ , $[[\,i\,]=|{\cal G}:{\cal H}|]$ ; the number of symmetry states is $[[\,i_N\,]=|{\cal G}:{\cal N}_{{\cal G}}({\cal H})|]$ .

Example 1.2.7.3.1.

Let $[{\cal G}=Fm\overline{3}m]$ , No. 225, with lattice parameter a and $[{\cal H}=Pm\overline{3}m]$ , No. 221, with the same lattice parameter a. The relation $[{\cal H} \,\lt\, {\cal G}]$ is of index 4 and is found between the disordered and ordered modifications of the alloy AuCu₃. In the disordered state, one Au and three Cu atoms occupy the positions of a cubic F-lattice statistically; in the ordered compound the Au atoms occupy the positions of a cubic P-lattice whereas the Cu atoms occupy the centres of all faces of this cube. According to IT A, Table 15.2.1.4 , the Euclidean normalizer of $[{\cal H}]$ is $[{\cal N}_{{\cal E}}({\cal H})=Im\overline{3}m]$ with lattice parameter a. The additional I centring translations of $[{\cal N}_{{\cal E}}({\cal H})]$ are not translations of $[{\cal G}]$ and thus $[{\cal N}_{{\cal G}}({\cal H})={\cal H}]$ . There are four domain states, each one with its own distinct space group and symmetry state $[{\cal H}_j, \ j=1,\ \ldots\ 4,]$ and consequently its own conventional origin relative to the origin of the disordered crystal A with the space group $[{\cal G}]$ . The origin shifts of $[{\bf B}_j]$ relative to the origin of $[{\bf A}]$ are [0,0,0;] $[{{1}\over{2}},{{1}\over{2}},0;]$ $[{{1}\over{2}},0,{{1}\over{2}}]$ and $[0,{{1}\over{2}},{{1}\over{2}}]$ .

These shifts do not show up in the macroscopic properties of the domains. Indeed, one is normally neither interested in those translations of $[{\cal G}]$ which are lost in the transition to the subgroup $[{\cal H}]$ , nor in the position of the conventional origin of $[{\bf B}]$ relative to that of $[{\bf A}]$ but only in the orientation of the domain states $[{\bf B}_k]$ . If so, the observed relations are not governed by the space groups $[{\cal G}]$ and $[{\cal H}]$ but by $[{\cal G}]$ and Hermann's group $[{\cal M}]$ , $[{\cal G}\geq{\cal M}\geq{\cal H}]$ , cf. Lemma 1.2.8.1.2. The group $[{\cal M}]$ is uniquely determined as the space group with the translations of $[{\cal G}]$ and the point-group operations of $[{\cal H}]$ . The group $[{\cal M}]$ can thus be characterized as that translationengleiche subgroup of $[{\cal G}]$ which is at the same time a klassengleiche supergroup of $[{\cal H}]$ . This group $[{\cal M}]$ plays a role in the practical treatment of domains. It was applied to domain structures first by Janovec (1976).

In the current literature, the following considerations are mostly restricted to the point groups of the phases involved. In the following, the use of Hermann's group $[{\cal M}]$ is discussed in parallel with the normal use of the point groups. The (admittedly rather abstract) discussion may thus be unfamiliar to the reader. Nevertheless, it is offered here because it opens up the possibility of treating phase transitions on a microscopic or atomistic level, whereas the point-group approach can only deal with the continuum or macroscopic aspect. The microscopic approach is necessary in particular when discussing domain boundaries, which will not be done here.

Definition 1.2.7.3.2. Two domain states $[{\bf B}_1]$ and $[{\bf B}_k]$ with space groups $[{\cal H}_1]$ and $[{\cal H}_k]$ and point groups $[{\cal P}_{{\cal H}_1}]$ and $[{\cal P}_{{\cal H}_k}]$ have the same orientation state if their orientation is identical, i.e. if the linear part of the operation $[{\sf g}_k\in{\cal G}]$ of Lemma 1.2.7.2.3 is the identity operation. This means that $[{\sf g}_k\in{\cal G}]$ is a translation $[{\sf t}\in{\cal T}]$ and implies that the point groups of $[{\bf B}_1]$ and $[{\bf B}_k]$ are the same: $[{\cal P}_{{\cal H}_1}={\cal P}_{{\cal H}_k}]$ . Thus the space groups $[{\cal H}_1]$ and $[{\cal H}_k]$ are subgroups of the same space group $[{\cal M}]$ .

Lemma 1.2.7.3.3. The number of orientation states in the transition A $[\longrightarrow]$ B with space groups $[{\cal G}]$ $[\longrightarrow]$ $[{\cal H}_m]$ is $[|{\cal G}:{\cal M}_m|]$ , i.e. the index of $[{\cal M}_m]$ in $[{\cal G}]$ , where $[{\cal M}_m]$ is Hermann's group in the sequence $[{\cal G}\geq{\cal M}_m\geq{\cal H}_m]$ . These orientation states belong to $[|{\cal G}:{\cal N}_{{\cal G}}({\cal M}_m)|]$ space groups. The number of domain states which belong to the same orientation state is $[|{\cal M}_m:{\cal H}_m|]$ , i.e. the index of $[{\cal H}_m]$ in $[{\cal M}_m]$ .

In Example 1.2.7.3.1 of AuCu₃, $[{\cal G} = {\cal M}_1]$ because $[{\cal H}_1]$ is a klassengleiche subgroup of $[{\cal G}]$ . Therefore, $[|{\cal G}:{\cal M}_1|=1]$ and all four domain states belong to the same orientation state. This is obvious visually, because, as stated above, all four domain states are parallel and only shifted against each other.

Lemma 1.2.7.3.4. Because of the isomorphism $[({\cal G}:{\cal M})\cong({\cal P}_{{\cal G}}:{\cal P}_{{\cal H}})]$ between the factor groups $[({\cal G}:{\cal M})]$ and $[({\cal P}_{{\cal G}}:{\cal P}_{{\cal H}})]$ , the results of the application of the groups $[{\cal G}]$ and $[{\cal M}]$ are the same as the results of the application of the groups $[{\cal P}_{{\cal G}}]$ and $[{\cal P}_{{\cal H}}]$ . The latter application is called the `continuum approach to phase transitions' which is nearly always applied in practice.

Lemma 1.2.7.3.3 is the microscopic formulation of the (macroscopic) continuum treatment of phase transitions and forms the bridge from the (macroscopic) continuum to the (microscopic) atomistic approach to phase transitions.

Example 1.2.7.3.5.

There is an order–disorder transition of the alloy β-brass, CuZn. In the disordered state the Cu and Zn atoms occupy statistically the positions of a cubic I lattice with space group $[{\cal G}= Im\overline{3}m]$ , No. 229. In the ordered state, both kinds of atoms form a cubic primitive lattice P each, and one kind of atom occupies the centres of the cubes of the other, such that a space group $[Pm\overline{3}m]$ , No. 221, is formed, see also Example 1.3.3.1 . For the space groups the relation $[{\cal G}=Im\overline{3}m>{\cal H}=Pm\overline{3}m]$ of index 2 holds with the same cubic lattice parameter a. In this case, $[{\cal G}={\cal N}_{{\cal E}}({\cal H})]$ , see IT A, Table 15.2.1.4 . As the index $[|{\cal G}:{\cal H}|=2]$ , there are two domain states with their crystal structures shifted relative to each other by $[{{1}\over{2}}({\bf a}+{\bf b}+{\bf c})]$ . Thus, both domain states belong to the same orientation state. This also follows from $[{\cal G} = {\cal M}]$ . Because $[{\cal G}={\cal N}_{{\cal G}}({\cal H})]$ and thus $[|{\cal G}:{\cal N}_{{\cal G}}({\cal H})|=1]$ , there is one symmetry state, and both domain states belong to the same space group.

References

Janovec, V. (1976). A symmetry approach to domain structures. Ferroelectrics, 12, 43–53.Google Scholar

International Tables for Crystallography (2006). Vol. A1. ch. 1.2, pp. 20-21