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Crystallography II

Crystallography II. Group Theory I-32 point group of the crystal (the first classification) 1.Introduction Crystal = Lattice structure basis 1) Point symmetry 7 C.S. primitive 1801—1807 lattice

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Crystallography II

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  1. Crystallography II Group Theory I-32 point group of the crystal (the first classification) 1.Introduction Crystal = Lattice structure basis 1)Point symmetry 7 C.S. primitive 1801—1807 lattice 2) Point symmetry Geometric 14B.L. Crystallo- graphy Translational symmetry 1818----1839

  2. Crystal = Lattice structure basis 1801—1830 3)Point symmetry 32 point group of crystals To describe the physical properties of the crystals in quantity and quality

  3. . 1Monoclinic 1 1 Lattice2 2 2 basic structure unit The point group can be regarded as the first classification of the crystal

  4. Same crystal system Same Bravis lattice Different basis unit content Different point symmetry collections Different crystals This means we can use the point symmetry collection to classify the crystals. By this way, we get 32 point groups Other reasons for the point group study To obtain 32 point system patterns. If adding them to 14B.L., we can obtain some of 230 space groups( the general structure of the crystals)

  5. To symbolize all symmetry operation possessed by crystals. • To describe the physical properties of crystal. The essential characteristics of point symmetry collections • The production of any two operation is also a member of the collection ( closure). • The collection must include an identity 1 (E) • For each operation R, there is an inverse R-1, which is included in the collection to meet R.R-1=1 • The multiplication of operations are associative, (AB)C=A(BC) In mathematics, this specific collection is called a group （群）.

  6. The order （阶） of group : the number of the elements in the group. Point group (点群) of crystals: { 4, 42=2,43, 1…} {2, 1…} {1}

  7. { 4, 42=2,43, 1…} {2, 1…} {1}

  8. In each crystal, there are different point symmetry collections at different point positions. The highest point symmetry collection is called the point group of the crystal. 2. the procedure of 32 point group of the crystal From the pure math combination theory point of view , all possibilities of the combination from 1. 2, 3, 4, 6, 1, 2, 3, 4, 6 are C101+C102+…….+C1010

  9. Then check each possibility Meet the crystal system condition? Meet the four basic requirements of the group? If so, we get one point group of the crystal, or vice versa. Some disadvantages of combination method : have enormous possibilities make some confuses need a lot of combination laws of the symmetry operation In order to overcome these defects, it’s better for us to discuss the point group for each crystal system individually.

  10. The procedure is that adding more symmetries in a certain crystal system, if these new symmetry elements additions still remain the crystal condition, and also meet the four basic requirements of the group . We get one point group which below to this crystal system. 3. The crystallographic point group for each crystal system 3.1 Triclinic In this system, we have 1(E) or/ and 1(i) point symmetries. Any other point symmetry elements addition is not allowed since this addition will change the crystal Condition. So we have two point group in Triclinic They are {1}, {1,1}

  11. Just the same as the symmetry description, we have 4 ways to describe the point group. Symbology Geometry Mathematics Language International Schoenflies 1 C1 general equivalent positions {E} ………. 1 S2 symmetry element distribution {E,i} ……….

  12. 1Triclinic 1 1 Lattice different basic structure unit in the same C.S. , , , , , , , , ,

  13. Background knowledge: Stereographic projection 极射赤面投影 N pole A 2-D method to describe the relationship of points, lines, and planes in 3-D. general position in N hemisphere and projection general position in S hemisphere and projection S pole the enantiomorphical position and it’s projection 1(C1) 1(S2) , , , , ,

  14. The concept of Holosymmetric point group. （全对称点群） the point group in the crystal system with the largest number of symmetry operation. The holosymmetric point group of Triclinic is 1 (S2). 3.2 Monoclinic According to the definition of this C.S. , we can discuss the possibilities of the point group in monoclinic . 2 2 Five possibilities 1, 1, 2, 2

  15. 1 2 combination {1, 2} a point group in Monoclinic,Called 2 (C2) b) 1 2 combination {1,2},……, called m (C1h)

  16. c) 1 2 combination {1[00]}{2[001]}= -1 0 0 -1 0 0 = 1 0 0 0 -1 0 0 -1 0 0 1 0 0 0 -1 0 0 1 0 0 -1 ={2[001]} A mirror plane to be produced which is perpendicular to the 2-fold axis. Add 1, we have a collection {1, 2, 1, m} or {E, C2, i, h }…. ,called 2/m (C2h)

  17. d) 1 2 combination {1[000]}{2[001]} = {2[001]} All the symmetry elements are included in c) case. e) 2 2 combination {2[001]}{2[001]}={1[000]}{2[001]}{2[001]} = {1[000]} {1[001]}= {1[000]} Not a new point group 2 (C2), m (C1h), 2/m (C2h)

  18. 3.3 Orthorhombic 2, 2, 2 2, 2, 6 possibilities 1, 1; 2, 2 • 1 2, 2, 2 {1, 2[001], 2[010], 2[100]} A point of group of Orthorhombic, called 222 (D2)

  19. b) 12[100], 2[010}, combination {1[000]}{2[100]}{2[010]} ={2[001]} Get a symmetry set{ 1[000]}, 2[100], 2[010], 2[001]}. It ‘s a point group in Orthorhombic, called mm2 (C2v). c)1 2, 2, 2 combination {1[000]]{2[001]}={2[001]} c 2/m Along the b, c directions, we have the same results , ,

  20. a b 2/m; 2/m We get a symmetry set,{ 1[000], 1[000]; 2[001], 2[100], 2[010]; 2[001], 2[010],2[100]}. This collection from a point group which is below to Orthorhombic, called 2/m2/m2/m or mmm (D2h) , ,

  21. d) 1 2, 2 combination The result was included in the point group of mmm case c) e) 2 2, 2 combination The results were included in case b) and c). f) 2 2, 2, 2 combination The result was include in the case c). 3.4 Tetragonal According to the limitation of the crystal condition, we have 222 (D2) ,mm2 (C2v), 2/m2/m2/m or mmm (D2h)

  22. 2/m2/m2/m or mmm (D2h) { 1[000];2[100], 2[010] 2[001]; 1[000],2[001], 2[010],2[100]}. 222 (D2), { 1[000];2[100], 2[010] 2[001]} mm2 (C2v), { 1[000];2[001];2[001], 2[010]}

  23. C.S………. • B.L. oP,oC(oA,oB),oI,oF • Point Group (x,y,z) • B.L. Point Groups Point Space Group Structure of Crystals 222 (D2) , mm2 (C2v) , , mmm (D2h) ,

  24. pO 222 (D2)

  25. Totally we have 66 symmorphic space group. They are CS Symmorphic space group Triclinic P1,P1 Monoclinic P2,Pm, P2/m; B2, Bm, B2/m Orthorhombic P222, Pmm2, Pmmm; C222, Cmm2,Cmmm;Ammm; I222, Imm2, Immm; F222,Fmm2, Fmmm Tetragonal P4,P4,P4/m,P422,P4mm,P42m,(P4m2),P4/mmm; I4,I4,I4/m,I422,I4mm,I42m,( I4m2), I4/mmm; Cubic P23,Pm3,P432,P43m,Pm3m; I23, Im3,I432,I43m,Im3m; F23, Fm3,F432,F43m,Fm3m Trigonal P3,P3,P312,(P321),P3m1,(P31m), P3m1,(P31m) R3,R3, R32,R3m,R3m Hexagonal P 6,P6,P6/m,P622, P6mm,P6m2,(P62m),P6/mm

  26. 4 4 about 9 possibilities 1, 1, 2, 2, 4 • 1 4 combination {1[000]}.{4[001]}={4[001]} The collection is { 1[000], 41[001], 42[001], 43[001]}. …. formed a point group of this crystal system and called 4 (C4)

  27. b)1 4 combination {1[000]}.{4[001]}={4[001]} The collection of { 1, 41, 42=2, 43} was obtained. ……, A point group was formed, called 4(S4) c) 1 4 combination {1[000]}{42{001]}. ={2[001]} {1[000]}{41{001]}. ={ 41[001]} A mirror plane and 4-fold symmetry axes were produced. The symmetry collection was {E,41, 2, 43, i, S41,S43, h}, … , ,

  28. It formed a point of group in this crystal system, called 4/m (C4h) point group. d)1 4 combination The result is the same as the case c) e)2 4 combination We have two cases

  29. The first case was include in case a), and the second case may result in a new point group in this crystal system. Suppose the 2 along a direction. The 4-fold symmetry will ask for another 2-fold axes along b direction. According to {41[001]}{2[100]}={2[110]}, two 2-fold axes are produced along [110] and [110]. Finally by this combination, we get a symmetry elements Collection {E; C41,C42=C2, C43;2C2’;2C2’’}…… It forms a point group , called 422 (D4)

  30. f)2 4 combination We also meet two choices, which are similar to case e). The first case is include in case b) (1 4 combination). The second one is that the 2-fold axis is perpendicular to 4 axis. Since {42[001]}{2[100]}={2[010]} and {4[001]}{2[100]}={2[110]}

  31. By other combination, no new symmetry operations are obtained. From the above pattern, we get a symmetry collection{E; S43, S42=C2, S41;2C2’;2d}…….. We get a point group in this crystal system, called 42m (D2d) g)2 4 combination The result is the same as c)case , , , ,

  32. asked for by 4-fold axis. Allowed and still not to be discussed. May form a new point group. According to the above basic symmetry elements, we can obtain the following patterns. , , , ,

  33. From the above patterns, we can find other two mirrors Out to form a symmetry collection, {E;C41, C42=C2,C43; 2v; 2d}, which meets the all needs for the point group. It is a point group , called 4mm (C4v). h addition will not change the crystal condition. The symmetry collection is different from 4mm point of group. It is quite possible to form a new point group. , , , ,

  34. {4[001]}{2[001]}={43[001]} means 4 axis was produced along c direction. {42[001]}{2[001]}={1[001]} means 1 axis was produced . Reference the point group of 4mm, the other two mirrors in [110], [1 10] can be founded ( 2d). The collection may be {E; i ;C41,C42=C2,C43;S41,S43;h ;2v; 2d? } Totally there are 16 positions. But in the above collection, , , , ,

  35. So the collection needs other 4 symmetry elements to form a closure set. Where are they? {2[100]}{1[000]}={2[100]} {2[010]}{1[000]}={2[010]} {2[110]}{1[000]}={2[110]} {2[110]}{1[000]}={2[110]} {E; i ;C41,C42=C2,C43;S41,S43;h ;2v; 2d ,2C2’;2C2’’}

  36. It meets the all needs for the point group in this crystal system, called 4/m2/m2/m (D4h). h)2 4 combination included in the case c) (4/m ) included in the case f) (42m)

  37. included in the case g) 4/m2/m2/m. i)4 4 combination has been discussed in c case. 4/m

  38. 4/m2/m2/m (D4h) 4mm(C4v) {E; i ;C41,C42=C2,C43;S41,S43;h ;2v; 2d ,2C2’;2C2’’} {E;C41, C42=C2,C43; 2v;2d} 42m(D2d) {E;S43, S42=C2,S41;2C2’;2d} 422(D4) {E;C41, C42=C2,C43; 2C2’,2C2’’} 4/m(C4h) {E; i ,C41, C42=C2,C43; S41 ,S43,h} 4(S4) {E;S41,S42=C2,S43} 4(C4) {E;C41, C42=C2,C43}

  39. , , 4(S4) 4/m (C4h) 422(D4) 42m (D2d) 4(C4) , , , , , , , , 4mm (C4v) 4/m2/m 2/m(D4h)

  40. Please give a brief summary for tetragonal 7 point groups by yourselves. 3.5 The point groups in Trigonal 3 3 1, 1, 2, 2, 3 about 9 possibilities a) 1 3 combination b) 1 3 combination {1, 31,32} ,….. {1;S65,S64=C32, 3 ( C3 ) point group S63=i,S62=C3,S6}, ….. 3 (S6 or C3i) point group , , ,

  41. Combination point groupGeometric patterns a)1 3 3 C3 b) 1 3 3 S6 c) 2 3 32 D3

  42. , , d) 2 3 32/m D3d e) 2 3 3m D3v , , , ,

  43. So far, we get all point groups in Trigonal crystal system. They are 3 (C3), 3(S6 or C3i) , 32(D3), 3m(C3v), 32/m or 3m(D3d). 3.6 Hexgonal This crystal system is with the main axis, e. g. 6 and or 6, and very similar to Trigonal system. Here we just give a list.

  44. Combination point groupGeometric patterns a)1 6 6 C6 b) 1 6 6 C3h c) 2 6 622 D6

  45. d) 2 6 6/m C6h 2 6 6mm C6v e) 2 6 6m2 D3h

  46. 2 6 6/m2/m2/m D6h Six 2-fold axes, seven mirrors, one E , one i , plus C61, C62=C3, C63=C2, C64=C32, C65; S35, S31; S61, S65.

  47. 3.7 Cubic crystal system This is a little more difficult to treat than the above Systems since it is without the principle axis. a) The first question is that we can obtain a point group just using 1(E) and eight 3-fold axes? The answer is no, since {3[111]}{3[111]}= 0 -1 0 0 0 1 0 0 1 1 0 0 -1 0 0 0 1 0 -1 0 0 = 0 1 0 ={2[010]} 0 0 -1

  48. By repeating mutiplcations, no other new symmetry elements will be produced. {1, 31[111], 32[111]; 31[111], 32[111] ,31[111], 32[111] ,31[111], 32[111], 2[100], 2[010], 2[001]}…. This is a point group , called 23 (T).

  49. 3 , 3 , 3 , 3 23(T) 1 2 4 4 b) Now we try to add 1 at the origin or 2 which is perpendicular to the 2-fold axes,which does not break the crystal condition, and meets……The most important results are the production of 3(S65) along body diagonal directions111. The collection is {1, 1; 31[111], 32[111],3[111], 35[111]; 2[100], 2[010], 2[001] 31[111], 32[111],3[111], 35[111]; 2[100], 2[010], 2[001] 31[111], 32[111],3[111], 35[111]; 31[111], 32[111],3[111], 35[111];}……

  50. It forms a point group of the cubic system, called 2/m3 or m3 (Th). • The mirrors contain both the 3-fold axes and 2-fold axes in 23 point group. This addition still remains the crystal condition, and maybe form a point group. , , , , , ,

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