Little Orthogonality Theorem (LOT)

Transcription

1 Little Orthogonality Theorem (LOT) Take diagonal elements of D matrices in RG * D R D R i j G ij mi N * D R D R N i j G G ij ij RG mi mi ( ) By definition, D j j j R TrD R ( R). Sum GOT over β: * * ( ) D R D R D R R i j i j G G ij RG RG mi mi N N N ij RG * ( ) D R R N i j G mi m i. Now yields: ij * ( ) ( ) * ( ) ( ) i j i j G D R R R R ij NGij RG RG mi ( ) * ( ) : i j LOT ( R) R NG RG ij N Orthogonality of characters

2 ( ) * ( ) : i j LOT ( R) R NG RG Orthogonality of characters can be seen and could help to complete them ij C I C g 6 v v E 0 x, y z R z C I C C g 8 4v 4 v d B x y B E z xy ( x, y) z R T E 8C C 6 6S N 4 d d 4 G E z r x y 0 0 (, ) T 0 ( R, R, R ) T 0 ( x, y, z) r x y z

3 ( ) * ( ) : i j LOT ( R) R NG RG ( ) i j : i ( R) NG RG ij Consequences: Used to certify irreps LOT : is the same R C, so one can sum over classes: C n ( i) * ( j) C ( C) C NGij Orthogonal vectors have component for each class and there is one of them for each irrep. Thus, the number of irreps does not exceed the number of classes.

4 Dimension m i of irrep i LOT: n ( C) C N CClasses Recall: in irrep i () i ( i) () i ( i) i C somma tracce in classe C: m i C nc ( C) character nc nc () i where C * Imxm represents Dirac's character R. R m in irrep i, n ( ) N m CClasses ( i) * ( j) C G ij set i=j: for each irrep i, ( ) CClasses ( i) nc C NG m () i i C G C G i ( i) nc ( C ) CClasses in Cv irrep E, I, R, 0, nr me 4 4 N n C 6

5 Practical use: one starts with random basis reducible representations. We know that U exists such that = U U + Tr is invariant, so D( R )= trace of reducible representation: ( ) ( R) n j j ( R) j 5

6 trace of reducible representation: ( ) ( R) n j j ( R) j One can find n using the LOT : ( R) R N ( i) * ( j) j G ij RG ( R) ( R) n ( R) ( R) n N ( i) * ( j) ( i) * j j G ij RG RG j j ni ( R) ( R) N G RG ( i) * number of occurrences of irrep i in arbitrary basis = size of determinants for each irrep 6

7 Remark on the number of orthogonal vectors GOT: RG * D R D R i j G ij mi N For each i, D (i) is m i x m i and for each of the m i entries there is a N G component vector ; these m i vectors from all irreps are all orthogonal! The number of orthogonal vectors cannot exceed number of components. i m i N G 7

8 Burnside theorem mi NG i Regular representation G elements -> vectors: E=R (,0,0,0,0,0) C =R (0,,0,0,0,0)... E C C C C E C E C a b c b c a c a b a b c c a b b c a E C C C E C C C E C : EC, C C, C E, a c b a, c b that is: C :,,, 46, 54, 65 8

9 G elements -> vectors: E=R (,0,0,0,0,0) C =R (0,,0,0,0,0)... Rearrangement theorem: multiplication by R i permutation matrix E C C C C E C E C a b c b c a c a b a b c c a b b c a E C C C E C C C E C : EC, C C, C E, a c b a, c b that is: C :,,, 46, 54, 65 DC ( ) D( E) Diag(,,,,,) Tr D( C) 0 TrD( R) 0 R except TrD( E) 6 9

10 Groups, Tr D( R) 0 except Tr D( E) NG The regular representation i s reducible. How? i ni R R E E N N G R * i * i * i n * i E NG E mi. N G m m bloch occurs m times! i i i Burnside theorem mi NG i G C I C g 6 v v E 0 x, y z R z 0

11 There is a Second character orthogonality theorem (orthogonality of columns): irreps ( i i) ( i) * NG ( C) ( C'), C and C' classes CC' n C Consider the matrix whose entries are those of the character table: Q () i T { (C )} Q ( j ) (C ) j ij i Multiply each entry of Q T by the number of elements of each class divided by the order of the Group, making the new matrix: (j) n (C ) Ci i Q' { } N G

12 Q (j) n (C ) () i Ci i { (C )} Q' { } j N G The product is given by (j) ( ) * ( ') ' (C ) n (C ) i Ck k QQ Q Q ij ik kj k N k kclasses G ( i) * ( j) LOT : one can sum over classes: nc ( C) C C N ( QQ') { } QQ' I Q' Q I. ij ij G ij N G kirreps (k) * (k) n (C ) (C ) C i j ij i Therefore the columns are also orthogonal.

13 kirreps (k) * (k) NG (C ) (C ) i j ij n C i There is one vector for each class There is one component for each irrep The number of orthogonal vectors cannot exceede the number of components Hence the number of classes the number of irreps. Since the other orthogonality theorem for rows implies that the number of irreps the number of classes, the number of classes=the number of irreps.

14 How to generalize the elementary trick to symmetrize/antisymmetrize functions: j D i * j ( j) multiply the GOT : D R D R O( i) by RG ij N ( j) i * G ( j) G ( j) D R R ij ij RG mi mi j ( ) ( ) ( ) ( ) ( ) R ecall : j D R j j R j R j. The result i G ( ) ( ) f x f x f x i ( j) * G () i D R R ij RG mi i * ( j) ( j) D R R ij RG N N, that is, the action i i i * i ( j) ( i) R is P ij NG RG i m N m P D R P Projection Operators is projection operator and j i.. of 4

15 m i i P D R R NG RG i * generalized projection operator Practical use with any function to project into an irrep: Start from any basis function: i j i cj (, ) Then, if P c( i, ) 0 try another ψ, otherwise gives symmetry-adapted functions i component. There is no need to diagonalize any matrix to get the symmetry-adapted basis. The projection operator is a Group property, like the D. j 5

16 There is no need for the D matrices: Start from any basis function: c( j, ) P c( j, ) j j j j i i P c i, irrep i ( i) i ( i) * P P ( R) R R () i projects on irrep i; we only need ( R) To make basis take several and Gram-Schmidt orthogonalization. 6

17 i Summary of main Group integration formulas * GOT : D R D R dd RG N i j G ij mi ( j) 8 ( ) D ( ) j * : i j LOT ( R) ( R) NG RG Number of times irrep i is present in basis: i * n i R R NG R Projector : ( ) N ( i) ( i) * G ( C) ( C') CC' nc ( J) * ab a' b' jj' aa' bb' ( i) i * P R R R ij Burnside theorem mi NG i 7 7

18 Simpler Uses of Group Theory N C I C g 6 v v E 0 x, y z R z minimal basis set : s, p x, p y, p z for N and the s orbitals s, s, s for the H atoms. N atom is not shifted by any R, therefore s and pz are assigned to while (px, py) is a basis for E. Hydrogen Basis: 0 0 s 0 s s

19 possible choice of axes: a b alias c Reducible representation: hydrogen basis 0 0 s 0 s s E 0 0 C 0 0 C a 0 0 b 0 0 c

20 a b alias s s s c E 0 0 C 0 0 C a 0 0 b 0 0 c Traces : I, C 0, i G R * ni R R N C I C N 6 v E ( H ) 0 v G z R 0 ( x, y) z **0 * n R 6 R 6 ne (* **0 *0*) 6 n (* **0 **( )) 0 6 ( H ) E 0 0

21 Projection operators as matrices on Hydrogen Basis to get the symmetry-adapted functions Simplest method: projection on irrep 0 0 s 0 s s E 0 0 C 0 0 C 0 0 a 0 0 b 0 0 c projectors on irreps: ( ) ( i) i * P R R R P ( ) R R P ( ) R R cting with P ( ) on any Hydrogen one finds the obvious s s s C I C g 6 v v E. 0 x, y z R z

22 Projection operators on irrep E 0 0 s 0 s s E 0 0 C 0 0 C 0 0 a 0 0 b 0 0 c projectors on irreps: ( ) ( i) i * P R R R C I C g 6 v v E 0 x, y z R z P ( E)

23 P ( E) Start projecting from s: 0 0 C I C g 6 v v E 0 x, y z R z The particular one obtains depends on s orbital: from s, which is v () invariant, one gets the v () invariant result s s s 6 s s s 6 x if axes are taken is like x (even for )

24 0 0 s 0 s s While from s, which is v() invariant, one got the v() invariant result E s s s P s 6 ( ) Orthonormalizing from s, one gets E s s s P s ' 6 ( ) s lternative method: projection on irrep and component s transforms like y (, ) ( x, y) E E * P xx Dxx ( R) R 6 R We need appropriate D matrices for that 4 4

25 bove, we introduced the following matrices for irrep E: irrep E of C : 0 c s c s D( E) D( C ) D( C ) 0 s c s c 0 c s c s D( ) D( ) D( ) a c b 0 s c s c c cos( ), s sin( ) v a :x changes sign y goes to y c a b x 5 5

26 However now the geometry is: a b D( C ), D( C ) remain 0 a( x, y) ( x, y), that is, a 0 c C a b C a c New matrices for irrep E D( C ), D( C ) a 0 0 b c 6

27 Now we can proceed with the calculation of E E * P xx Dxx ( R) R 6 R Operators R on Hydrogen basis E 0 0 C 0 0 C a 0 0 b 0 0 c Recall: ( H ) E 7 7

28 ROperators R on Hydrogen basis E 0 0 C 0 0 C 0 0 a 0 0 b 0 0 c D( C ), D( C ) P E xx Input data: E E * P xx Dxx ( R) R 6 R 0 a 0 b c E P xx [ E ( C C) a b c ] E P yy

29 original basis: 7 functions s,p x,p y,p z,s,s,s 77 Determinant is broken: basis: s, p z, E bases: (p x,p y ),(, ) Hamiltonian matrices: (p x, ),(py, ) The two E determinants are identical (elements cannot depend on component, so there can be no difference between x and y). i j Recall O O( i) ij 9 9

30 Molecular Orbitals of CH 4 CH 4 belongs to the tetrahedral symmetry Group Td; we may put the C atom at the origin and H atoms at (-,,),(,-,),(-,-,-),(,,-) in appropriate units. There are N G =4 operations. There are C and C rotations around CH bonds,c and S 4 rotations around the x,y,z axes, and 6 σd reflexions in the planes that contain two CH bonds. The C orbitals (s,px, py, pz) are classified at once under and T. T E 8C C 6 6S N 4 d d 4 G E z r x y 0 0 (, ) T 0 ( R, R, R ) T 0 ( x, y, z) r x y z 0 0

31 4 Characters of H : build 4x4 representation like above (slower method) 4 T E 8C C 6 6S N 4 d d 4 G E z r x y 0 0 (, ) T 0 ( R, R, R ) T 0 ( x, y, z) r x y z Faster method: for each operation each moved atom adds 0 to each unmoved atom adds each orbital which changes sign adds - (not in present problem)... Class moved unmoved E C C 6 6S d 4 Next, examine all other R

32 Tetrahedron: T d 8C Class moved unmoved E C C 6 6S d E 8C

33 C Class moved unmoved E C C S d E 8C C

34 6 d one for each pair of H Class moved unmoved E C C S d E 8C C 6 d 4 4

35 6S 4 Class moved unmoved E C C d 6S E 8C C 6 6S d 4 5 5

36 Characters of H 4 n n T 4 E 8C C 6 6S d 4 ( H ) i 4 * ni R R N T E 8C C 6 6S N 4 d d 4 G E z r x y 0 0 (, ) T 0 ( R, R, R ) T 0 ( x, y, z) Class moved unmoved 4 8 *0 6* 6*0 G R n R 4 R 4 4**8 0 *( )*6 0 4* *( )* n E * 0 0 ( )*6 0 4* 0 0 **6 0 0 n T 4 4 H T 4 E C C d 6S r x y z 6 6

37 Characters of H : (still faster method) 4 4 T E 8C C 6 6S N 4 d d 4 G E z r x y 0 0 (, ) T 0 ( R, R, R ) T 0 ( x, y, z) r x y z subtract the totalsymmetric combination 4 4 E 8C C 6 6S d 4 ( H ) ( H ) 0 H T 4 7 7

38 How to build Symmetry adapted orbitals from H states using Projectors We can build the actual states very simply if we know the destinations of orbitals Td E 8C C 6 d 6S4 NG 4 r E z r x y Class moved unmoved destinations atom E 0 4 8C,,,,,,4,4 C 4 0,,4 6,,,,,4 d 6S 4 0,,,,4, (, ) T 0 ( R, R, R ) T 0 ( x, y, z) Projector : ( ) P ( ) x y z ( i) i * P R R R R R 4 (obviously) 8 8

39 T E 8C C 6 6S N 4 d d 4 G E z r x y 0 0 (, ) T 0 ( R, R, R ) T 0 ( x, y, z) r x y z Class moved unmoved destinations atom E 0 4 8C,,,,,,4,4 C 4 0,,4 6,,,,,4 d 6S 4 0,,,,4,4 4 Symmetry adapted orbitals Projector : ( ) ( i) i * P R R ( T ) P E C S d 4 C d S4 Start for instance from atom we obtain an obital of T R ( T ) P [ 4 ] 4 4 ( T T ) [ 4 hus, P 4] to be normalized 9 9 4

40 The above methods grants orbitals in T ( x, y, z): 4 How can we find the components, i.e. orbitals x, or y, z? T T i P D R R xx G RT d T T i P D R R yy m N m N G RT d xx yy * * Requires D matrices : action of the Group operations on (x,y,z)linear combinations Faster way One can always choose at least one R to diagonalize simultaneously with Dirac s characters, like C around z. Then the orbital z which transforms like z has eigenvalue under C rotation, like >+ > >+ >. To obtain z in a smart way symmetrize on 40 40

41 ( T ) Start from the action of P C S on, P ( T ) C d 4 S d 4 [ 4 ] [ 4 ] [ 4 ] ( T ) Simplify and get P [ 4 ] ( T ) Do the same on : P [ 4 ] P z P P ( T ) ( T ) [ 4 ] [ 4 ] ( T ) ( ) ( ) [( ) 4 ] 4

42 P z ( T ) ( T ) ( ) ( ) [( ) 4 ] 4 T 4 simplify and norm z T 4 T 4 x y In a similar way, P P ( T ) [ 4 ] [ 4 ] Without symmetry, 8*8 determinant, With group thy 4 determinants * T T T C, C, C, C, s p x x p y y p z z but x,y,z are identical. Since H elements cannot depend on components. Indeed, only occupied levels are seen in XPS, the upper one is times degenerate. i j (recall O O( i) ) ij 4 4

43 Broken representations! C v : (x,y) basis for E has subgroup Z C I C g 6 v v E 0 x, y z R z Z I C C * E * z x, y However Z is abelian and degeneracy is broken. Basis for conjugate representations: i m x iy e cos i sin i m x iy e cos i sin are not mixed by Z 4 i i i i i i i i : : : C e e e C e e e e e 4 i i i i i i : : C e e e C e e e Rule: H G subgroup irreps of G are still representations of H, since h H does not mix different representations of G However: some representations become reducible in the subgroup. Example: (x,y) is an irreducible basis for E in C but is broken in Z. v ctually, ( x iy) and (x-iy) are not mixed by rotations, but by reflc e tions. 4 4

44 Important example: angular momentum eigenstates Characters and crystal field splitting j, m irrep of O(), with j m j j j Point Groups are subgroups. How does irrep break in subgroups? Consider first rotations around z axis. ll rotations by same angle belong to same class, so the character is the same independently of the axis of rotation. J j, m m j, m z j j j ij z im j R j, m exp j, m e j, m z j j j ( j) im j matrix of j representation is D ( Rz ) diag( e ) j ( j) ( j) z m j j ( ) TrD ( R ) e. j im 44 44

45 Trick to evaluate the sum: i i j im j i i ij i( j) ij ij i( j) ij sin e e ( e e e ) e ( e e e ) m j j j ( j) ( j) z m j j ( ) TrD ( R ) e. j ( ) ( im ) j i j i j sin e ( e e ) sin[ j ] m j i j ll terms cancel except two: for instance for J= i i im j i i i i i i i i i i sin e e ( e e e e ) e ( e e e e ) m j i i 5 i/ 5 i/ 5 ( e e ) sin( ). i 5 i / i / i / i / i / i / i / i / i / 5 i / ( e e e e e ) ( e e e e e ) j im 45 45

46 ( j) ( ) m j j j sin[ j ] im j e sin The jm, j basis of O() is reducible in all point Groups for J>. The orbitals of free atoms belong to irreps of O(). In molecules, atomic orbitals that are displaced by symmetry operations mix in the basis of a representation of the Group. Instead, the orbitals of atoms that are not displaced by symmetry operations make reducible transformations that can be analyzed in Irreps of the point Group. Number of times irrep i is present in basis: i ni R R N * 46 46

47 The Cubic group O h Is often realized by octahedra To draw an octahedron start by a cube centered at (0,0,0) E C F B 48 operations (twice as many as in T d ) D 47 47

48 The Cubic group O h The O h Group is the point Group of many interesting solids, including complexes like CuSO 4 5H O and FeCl where a transition metal ion at the center of an octahedron; most often the octahedral symmetry is only approximate, as in CuSO 4 5H O where the central Cu ++ is bound to 4 H O molecules and SO 4 ions, and some superconducting cuprates.. CuSO 4 5H O Iron(III) chloride. By contrast, Iron(II) choride is green 48 48

49 Cube: O h Group 6C 4 coordinate axes, clockwise or counterclockwise E 6C

50 Cube: O h Group C E 6C 4 C 50 50

51 Cube: O h Group axes in xy plane, axes in xz plane, axes in yz plane 6 C ' E C 4 6C 6 C ' 5 5

52 Cube: O h Group 8C E C 4 6C 6 C ' 8C 5 5

53 Cube: O h Group i : ( x, y, z) ( x, y, z) i * R E C 4 6C 6 C ' 8C i 5 5

54 6S 4 E C 4 6C 6 C ' 8C i 6S

55 h E C 4 6C 6 C ' 8C i 6S 4 h 55 55

56 6 d x=y and x=-y x=z and x=-z, z=y and z=-y E C 4 6C 6 C ' 8C i 6S 4 h 6 d 56 56

57 8S 6 rotation around a diagonal of cube followed by a reflection in plane through the cube center. S 6 Front view of octahedron faces: C 6 E C 4 6C 6 C ' 8C i 6S 4 h 6 8S d