INTERNATIONAL SCHOOL ON FUNDAMENTAL CRYSTALLOGRAPHY AND WORKSHOP ON STRUCTURAL PHASE TRANSITIONS. 30 August - 4 September 2017
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1 INTERNATIONAL SCHOOL ON FUNDAMENTAL CRYSTALLOGRAPHY AND WORKSHOP ON STRUCTURAL PHASE TRANSITIONS 30 August - 4 September 2017
2 ROURKELA INTERNATIONAL CRYSTALLOGRAPHY SCHOOL BILBAO CRYSTALLOGRAPHIC SERVER PRACTICAL EXERCISES SYMMETRY CONSIDERATIONS IN STRUCTURAL PHASE TRANSITIONS Mois I. Aroyo Universidad del Pais Vasco, Bilbao, Spain
3 PHASE TRANSITIONS ACTIVE REPRESENTATIONS, ORDER PARAMETERS AND ISOTROPY SUBGROUPS
4 Landau theory of phase transitions Active irrep the symmetry break taking place in structural phase transitions is due to the condensation (i.e. the change from zero to a non-zero amplitude) of one or a set of collective degrees of freedom that transform according to a single irreducible representation of the space group of the high-symmetry phase (the so-called active irrep). Order parameter The amplitudes {Qi, i=1,...,n} that become spontaneously nonzero in the low-symmetry phase, constitute the so-called order parameter, and the n- dimensional irrep describing its transformation properties is the active irrep of the transition. Isotropy subgroups The symmetry group of the low-symmetry phase for which the amplitudes {Qi, i=1,...,n} that become spontaneously non-zero in the low-symmetry phase, is called an isotropy subgroup of the corresponding active irrep
5 Problem: We know the high symmetry and the active irrep or order parameter and we want to know the possible symmetries of the distorted phase G, D(g)? possible isotropy subgroups for a given active irrep? G: space group of the high-symmetry phase D={D(g), g G}: matrix irrep of G Q=(Q1, Q2,,Qn): order parameter transforming according to D Transformation of the order-parameter: Isotropy subgroup H of D: Domain-related structures: (lost operations) D(g)Q=Q, g G D(g)Q=Q, g H<G D(g)Q=Q Q, g G
6 EXAMPLE Isotropy subgroup Possible phase transitions/symmetry breaks of material of symmetry I4/mmm without changing the unit cell (i.e. all translations are kept in the isotropy subgroups, t-subgroups) k*={k, k, k,..., k n } D k* (I,t)= exp-ikt exp-ik t 0exp-ik t exp-ik n t D k* (I,t)Q=Q, g H<G k 0 Irreps of I4/mmm for k=0?
7 Bilbao Crystallographic Server: POINT Isotropy subgroup 2h=2100 2h =2110 mv=m100 md=m110 Determine the isotropy subgroups for the 1-dim irreps
8 Isotropy subgroup One-dimensional irreps of I4/mmm, k=0 What about the isotropy subgroup of A1u?
9 Isotropy subgroup Two-dimensional irreps of I4/mmm, k=0 Matrices of Eg Isotropy subgroups of Eg of I4/mmm, depending on the order-parameter direction bm II at bm II bt bm II at+bt bm II -at+bt
10 Isotropy subgroup Non-equivalent isotropy subgroups of I4/mmm k=0 CELLSUB 10
11 Subgroups of C2/m type Isotropy subgroup direct calculations bm II at bm II bt bm II at+bt bm II -at+bt Bilbao Crystallographic Server CELLSUB
12 EXERCISES Problem Determine the isotropy subgroups of the irreps of P4mm with k=0 E(4)= E(m-)=
13 Problem SOLUTION Irreps of P4mm at k=0 (Γ point) Bilbao Crystallographic Server CELLSUB P4mm P4 for 1-dim irreps rather trivial, for n-dim one must apply the matrix equations or use some group theoretical "tricks" Pmm2 Cmm2 Pm Cm P1 isotropy subgroup depends on the "direction" of the 2-dim order parameter. E[g] Q=Q {g}=h
14 PROBLEM Subgroups compatible with a given supercell or some propagation vector SUBGROUPS The program SUBGROUPS provides the possible subgroups of a space group which are possible for a given supercell. The program provides a list of the set of space groups or a graph showing the group-subgroup hierarchy, grouped into conjugacy classes. Other alternatives for the input of the program: Instead of the whole set of subgroups, the output can be limited to subgroups having a chosen common subgroup of lowest symmetry, common point group of lowest symmetry, or groups which belong to a specific crystal class. Instead of a supercell, a set of modulation wave vectors can be given, including complete or partial wave-vectors stars. When a set of wave-vectors is used as input, the output can be further refined introducing the Wyckoff positions of the atoms and/or a set of irreducible representations. 14
15 SUBGROUPS INPUT form 15
16 EXERCISES Problem (cont) Problem (cont) With the help of the program SUBGROUPS: (i) Determine the isotropy subgroups of the irreps of I4/mmm with k=0 (ii) Determine the isotropy subgroups of the irreps of P4mm with k=0 16
17 Isotropy subgroup Problem: Prediction of probable symmetries for compounds of a family, or for the same compound at different conditions due to a common active irrep, with the order parameter taking different directions Perovskites are known to have systematically a soft or unstable mode with irrep R4+ of their symmetry group Pm-3m (221): Isotropy subgroups of R4+: Example: SrZrO 3 SrZrO 3 CeAlO 3 BaPbO 3 LaCoO 3 CeAlO3: sequence with of phase transitions with the temperature I4/mcm Imma R-3c Pm-3m
18 EXERCISES Problem (cont) With the help of the program SUBGROUPS: Determine the isotropy subgroups of the irrep R4+ of Pm-3m (221), with R=1/2, 1/2,1/2. 18
19 Problem: Determine the active irrep if the symmetry break G > H is known (inverse Landau problem) G active irrep? H comparison of the unit cells k-vector(s) of the active irrep: star of the active irrep irreps of P4mm at k=0 (Γ point) e ik.t =1, t TH k.t=2πn G P4mm P4 Pmm2 Cmm2 Pm Cm P1 from the possible isotropy subgroups, identify possible active irreps with the known k-vector via computer: COPL (isotropy) SYMMODES (Bilbao server) --- as by-product AMPLIMODES (Bilbao server) --- as by product
20 Inverse Landau Secondary spontaneous variables In the low symmetry phase, apart from the order parameter and quantities transforming according to the active irrep, variables or degrees of freedom transforming according to other irreps can also condense or become spontaneous. The only requirement is that they are compatible with the low-symmetry space group. This is a realization of the Von Neumann principle: any variable/degree of freedom compatible with the symmetry of the crystal is allowed and will therefore in general have a non- zero value. Using the concept of isotropy subgroup, we can say that any quantity with transformation properties given by an irrep having an isotropy subgroup containing the group H of the distorted phase will be spontaneous in the transition (i.e. it will change from zero to non-zero values in the distorted phase). Therefore, while the active irrep must have H as an isotropy subgroup, the irreps associated to spontaneous secondary variables have in general an isotropy subgroup which is a supergroup of H.
21 EXAMPLE Bilbao Crystallographic Server Phase transitions of YMnO 3 P6 3 /mmc P6 3 cm (a-b, a+2b,c; 0 0 0) active irrep? Inverse Landau SYMMODES and SUBGROUPGRAPH Irrep Dir Subgr Size GM1+ (a) oup P63/mmc 1 GM2- (a) P63mc 1 K1 (a,0) P63/mcm 3 K3 (a,0) P63cm 3 secondary irreps/modes spontaneous crystal tensor quantities that transform according to GM2-: Polarization along z active irrep C. Fennie and K. Rabe, cond-mat/ FIG. 1: Group-subgroup sequence of allowed phase transitions connecting the high-temperature prototypic phase, P6 3 /mmc, to the low-temperature ferroelectric phase, P6 3 cm. Improper ferroelectric coupling of secondary variables/modes with order parameter (faintness index)? : program INVARIANTS (Isotropy)
22 Inverse Landau SYMMODES - Symmetry modes INPUT: G > H, [i], W G Group-subgroup graph Classes of conjugate subgroups Wyckoff position splittings Primary and secondary modes Isotropy subgroups and irreps of G SUBGROUPGRAPH WYCKSPLIT SUBGROUPS ISOTROPY
23 EXAMPLE Phase transitions of YMnO 3 P6 3 /mmc P6 3 cm (a-b, a+2b,c; 0 0 0) Inverse Landau SYMMODES - Symmetry modes
24 Symmetry modes for P63/mmc >P63cm symmetry break SYMMODES
25 EXERCISES Problem Monoclinic phase of the system PbZr1-xTixO3 Consider the perovskite-like ferroelectric system PbZr1-xTixO3 (PZT). Some measurements have revealed a monoclinic phase (with no cell multiplication) between the previously established tetragonal (P4mm) and rhombohedral (R3m) regions in its phase diagram as a function of x. Both phases, P4mm and R3m, are ferroelectric distorted phases of the perovskite, due to the condensation of a polar mode of symmetry at k=0. The perfect perovskite structure PbBO3 is cubic Pm-3m (Z=1) with positions: Pb 1b, B 1a, O 3d. (i) Using SYMMODES obtain a valid transtormation matrix for the pairs Pm-3m -- P4mm, and Pm-3m -- R3m, and check that indeed the two phases, P4mm and R3m, can be assigned to the same active irrep, for two different directions of the order parameter. Take notice of the active irrep and these directions. (ii) A reasonable assumption about the detected monoclinic structure is that it must be some bridging phase with the order parameter changing between the two special directions obtained in (i). Its symmetry would then be given by a common subgroup of the tetragonal and rhomboedral space groups. Use COMMONSUBS to predict under this assumption the space group of the monoclinic phase. Take notice of the transformation matrix relating it with the space group P4mm. (iv) From the transformation matrices for the pairs Pm-3m --- P4mm and P4mm --- monoclinic space, obtain the transformation matrix relating Pm-3m and the monoclinic space group. Using SYMMODES again demonstrate that the active irrep of the postulated monoclinic space group is indeed the same as for the other two phases. Compare the order parameter direction with those obtained in (i). (v) Use TRANSTRU to derive a starting structural model of the monoclinic phase (with a single mixed site for the Zr/Ti atoms), which you could use as the starting point for a refinement of the structure.
26 EXERCISES Problem (cont) Consider the sequence with of phase transitions with the temperature of CeAlO3: [6] [12] [8] I4/mcm (140) Imma (74) R-3c (167) Pm-3m (221) Using SYMMODES show that the transitions can be assigned to the same active irrep, but different directions of the order parameter. The occupied atomic positions are the following: Ce: 1b, Al: 1a, O: 3d 26
27 Symmetry breaks with several active irreps Inverse Landau P P2 1 (c,-a-2b,a; 3/4 0 3/4) active irreps? H is not an isotropy subgroup: two active irreps are necessary... Irrep Dir Subgr Size GM1 (a) P GM4 (a) P312 1 GM5 (a,1.732a) C GM6 (a,1.732a) C2 1 M2 (0,0,a) P M3 (0,0,a) P from SYMMODES and SUBGROUPGRAPH (Bilbao server): P Γ 4 P312 C222 1 Γ 5 Γ 6 C2 M 3 P P M 2 P2 1 is not an isotropy subgroup: more than one active irrep necessary. P2 1 probable intermediate phase! In this case, two active irreps are necessary. which ones? GM6: responsible of polarization along the monoclinic axis: it can be a primary or secondary effect. Can be indicated only by experiment or simulations
28 Symmetry breaks with several active irreps Pseudo-proper ferroelasticity of SrAl 2 O 4 (Larsson et al. 2008) P P2 1 P two unstable irrep distortions: Γ 4 P312 C222 1 Γ (b) M 3-1q 0-10 Γ 6 Γ 6 C2 M 3 P P M 2 E(mRy) M 2-1q -40 P
29 Example: SrBi2Ta2O9 (SBT) Initial data: G I4/mmm (139) [8] Sr: 2a Ta: 4e Bi: 4e O1: 2b O2: 4e O3: 4d O4: 8g H Cmc2 1 (36) Basis transformation: c, a-b, a+b (a, b, c) H =(a, b, c) G
30 Example: Symmetry Modes I 4/mmm E u X + 2 F mmm X 3 Three contributions for the global distortion E u X 3 X + 2 F mm2 C mcm C mca ISOTROPY SUBGROUPS C mc2 1 It is necessary at least two modes for breaking the symmetry until Cmc2 1 Check the results by the program SYMMODES G=I4/mmm (139), H=Cmc21(36), [i]=8, WP: 8g
31 Example: Modes for 8g position X 3 E u X + 2
32 EXERCISES Problem A compound has Pnma (62) symmetry at high temperatures and has space group P (4) at low temperatures, keeping essentially the same lattice, except for some strain. Using SYMMODES obtain the graph of maximal subgroups relating both symmetries. Check that at least two irreps must be active to explain the symmetry of the distorted structure. Indicate the possible pairs of active irreps in the distorted phase. Indicate the possible (alternative) symmetries of a probable intermediate phase.
33 Further problems We know the symmetry break and active irrep and want to derive further "consequences" Other possible phases in the phase diagram coming from the same order parameter INVARIANTS - ISODISPLACE spontaneous ferroic (switchable) quantities only ferroic species needed POINT primary and secondary spontaneous degrees of freedom/modes: transition mechanism. COPL (Isotropy) SYMMODES separation of structural parameters into collective modes with very different weight in the distorted structure. Mode decomposition ISODISPLACE (Isotropy) COSETS AMPLIMODES Amplimodes+ FullProf= direct struct. refinement temperature/pressure dependence of variables/modes: Landau analysis INVARIANTS (isotropy) Domain structure: orientational relations, domain walls, domain related equivalent structures COSETS NORMALIZER
34 PHASE TRANSITIONS SYMMETRY MODE ANALYSIS
35 Symmetry-mode analysis of distorted structures Modes are collective correlated atomic displacements fulfilling certain symmetry properties. Structural distortions in distorted structures can be decomposed into contributions of modes with different symmetries given by irreducible representations of the parent space group. One can then distinguish primary and secondary (induced) distortions, which will have in general quite different weights in the structure, and will respond differently to external perturbations. The use of symmetry-adapted modes in the description of distorted structures introduces a natural physical hierarchy among the structural parameters. This can be useful not only for investigating the physical mechanisms that stabilize these phases, but also for pure crystallographic purposes. The set of structural parameters used in a mode description of a distorted phase will in general be better adapted for a controlled refinement of the structure, for comparative studies between different materials or for ab-initio calculations. AMPLIMODES is a computer program available on the Bilbao Crystallographic Server that can perform the symmetry-mode analysis of any distorted structure of displacive type. The analysis consists in decomposing the symmetry-breaking distortion present in the distorted structure into contributions from different symmetry-adapted modes. Given the high- and the low-symmetry structures, AMPLIMODES determines the atomic displacements that relate them, defines a basis of symmetry-adapted modes, and calculates the amplitudes and polarization vectors of the distortion modes of different symmetry frozen in the structure.
36 Problem: SYMMETRY-MODE ANALYSIS AMPLIMODES Distorted Structure H = High-Symmetry Structure G + distortion mode = Amplitude x polarization vector Description of a mode : frozen modes e 3 u(atoms) = Q e e 1 amplitude polarization vector e 4 e 2 e = ( e 1,e 2,e 3,e 4 ) normalization: e e e e 4 2 =1 (within a unit cell) comparison of amplitudes of different frozen distortion modes AMPLIMODES calculates the amplitudes and polarization vectors of all distortion modes with different symmetries (irreps) frozen in a distorted structure.
37 Example: Phase transitions in SrBi2Ta2O9 (SBT) Initial data: Hightemperature prototype G I4/mmm (139) Sr: 2a Ta: 4e Bi: 4e Group-theoretical analysis [8] O1: 2b O2: 4e O3: 4d O4: 8g Lowtemperature ferroelectric H Cmc2 1 (36) Basis transformation: c, a-b, a+b (a, b, c) H =(a, b, c) G
38 Example: SrBi2Ta2O9 (SBT) I4/mmm (139) Cmc2 1 (36) Sr: 2a Bi: 4e Ta: 4e O1: 2b O2: 4e O3: 4d O4: 8g Sr: 2a Ta: 4e Bi: 4e O1: 2b O2: 4e O3: 4d O4: 8g Sr: 4a Ta: 8b Bi: 8b O1: 4a O2: 8b O3: 8b O4: 8b O5: 8b Structural data Global distortion
39 I 4/mmm Example: SrBi2Ta2O9 (SBT) E u X + 2 F mmm X 3 At least two modes for breaking the symmetry to Cmc2 1 Coupled order parameters E u X 3 X + 2 X + 2 Three types of contributions to the global distortion F mm2 C mcm C mca C mc2 1 cb a {Distortion} = Q Eu (E u )+Q X + 2 (X + 2 )+Q X 3 (X 3 ) Å Å Å
40 EXERCISES Problem J. Appl.Cryst. (2009), 42, Pm-3m P4mm Symmetry-mode analysis of the ferroelectric phase transition to the orthorhombic Amm2 structure of BaTiO 3 Amm2 Q(100) R3m Q(1/ 3,1/ 3,1/ 3) Q(1/ 2,1/ 2,0) Amm2 R3m Amm2 P4mm b c Li et al. J. Appl. Phys. (2005) Polarization
41 Problem SOLUTION Input of AMPLIMODES: Pm-3m Amm2 phase of BaTiO 3 b Amm2 4 parameters c Transf.
42 Reference Structure Problem SOLUTION -lattice parameters calculations b=c= 2a -splitting of some atomic orbits
43 Atom Mappings Problem SOLUTION -displacements smaller than a given tolerance -displacements of atoms in the asymmetric unit: ucell=ruasym -amplitude of the total distortion: (atomic displacements) 2 unit cell
44 Polar Structures Problem SOLUTION -global translation of the structure as a whole due to the arbitrary choice of the origin along z of Amm2 number of free structural parameters?
45 Summary Output Problem SOLUTION basis of the symmetry-adapted modes according to the atomic orbits -symmetry modes restricted to single atomic orbit and characterised by the irreps of the high-symmetry group -number of symmetry modes of a given type -total number of modes= number of free structural parameters
46 Summary Output Problem SOLUTION Summary of the Amplitudes of irrep distortions =( ) 1/2 characteristics of the irrep distortions: k-vector, irrep, direction in the irrep space, isotropy subgroup, dimension(number of modes), amplitude primary and secondary distortion
47 Problem SOLUTION Irrep Distortions: Amplitudes and Polarization Vectors Irrep distortion characteristics: wave vector direction isotropy subgroup transformation matrix Amplitude of the irrep distortion
48 Problem SOLUTION Irrep Distortions: Amplitudes and Polarization Vectors Polarization vector of irrep distortion: Set of normalized displacements multiplied by the amplitude result in the irrep distortion Polarization vectors (algebraic presentation) Invariance of the polarization vector with temperature, composition, etc. Polarization vectors (crystallographic presentation)
49 Basis of Symmetry-adapted modes Problem SOLUTION -normalization of modes: U(i) 2 = 1 A 2 unit cell º -orthogonality of modes: Umode1(i).Umode2(i) = 0 unit cell
50 Summary: The orthorhombic Amm2 structure of BaTiO 3 ( Kwei et al. (1993) b neutron-powder 190 K ) c max. atomic displ. : 0.13 Ǻ Mode decomposition of distortion: Q T1u x + Q T2u x Å Å T 1u - mode T 2u - mode polar ferroelectric mode Q T1u >> Q T2u
51 Summary: The orthorhombic Amm2 structure of BaTiO 3 Mode decomposition of structure distortion: b primary mode-order parameter: c Q T1u x + Q T2u x T 1u - mode T 2u - mode Q T1u >> Q T2u secondary mode 3 faintness index Q T2u ~ Q T1u
52 EXERCISES Problem S 2 Sn 2 P 6 has a monoclinic P12 1 /c1 phase at high temperatures and a ferroelectric non-centrosymmetric phase of symmetry P1c1 at low temperature (the structures are given in the Exercise Data file). 1.Obtain with AMPLIMODES that the ferroelectric structure has two dirtortion modes, a primary one that yields the P1c1 space group, i.e. the order parameter distortion, and a secondary one compatible with the parent symmetry. 2.Change arbitrarily some of the coordinates of P121/c1, but displacing the positions only a small amount (below 1Å). Check that AMPLIMODES gives the same results (amplitude and polarization vector) for the symmetrybreaking mode, changing only the form of the fully symmetric mode GM1. 3.Which atoms are moving more strongly in the transition? Derive the atomic positions of the Sn atoms in two virtual P1c1 structures having only the experimental primary ferroelectric mode with an amplitude of 0.1 Å and of 0.2 Å. One could use such virtual structures in an ab-initio calculation to characterize the energy variation of the system as a function of the ferroelectric mode.
53 EXERCISES Problem Carry out a symmetry mode analysis of the room-temperature phase P63cm of KNiCl3, considering this phase as a distortion from a virtual parent phase of symmetry P63/mmc (the structures are given in the Structure data file). There are different types of symmetry-adapted modes contributing to the global distortion. What symmetry modes could be related to the primary order parameter, and which are related to the secondary effects? Is it correct to classify the room-temperature form of KNiCl3 as proper ferroelectric? Hint: Use the databases and computer tools on the Bilbao Crystallographic Server to determine the transformation matrix for the symmetry break: P63/mmc P63cm.
54 Problem SOLUTION programs: INDEX, SUBGROUPGRAPH Group-subgroup symmetry break P63/mmc > P63cm, index 6
55 Problem SOLUTION Group-subgroup symmetry break P63/mmc > P63cm, index 6
56 Problem SOLUTION programs: AMPLIMODES GM2- K1 K3
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