Little Orthogonality Theorem (LOT)

Transcription

1 Little Orthogonality Theorem (LOT) Take diagonal elements of D matrices in R G i * j G αµ βν = δδ ij µνδαβ mi D R D R N N i * j G G Dµµ R Dββ R = δijδ µ β δ µβ = δijδµβ R G mi mi j j ( j) By definition, Dββ R = TrD R = χ ( R). Sum GOT over β: β N N = χ = δδ = δ * ( ) ( ) * D R D R D ( R) ( R) β i j i j G G µµ ββ µµ ij µβ R G R G β mi mi N ij R G * χ D R R = N i j G µµ mi m i δ. Now yields: * * µµ i j i j G D R R R R ij NG ij µ R G R G µ mi ij N χ = χ χ = δ = δ () * : i j LOT ( R) R = NG R G χ χ δ ij µ Orthogonality of characters

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

3 : χ χ δ () i * ( j) LOT R R = NG R G = = () i i j: χ ( R) NG R G ij Consequences: Used to certify irreps LOT : χ is the same R C, so one can sum over classes: C n ( C) ( C) N () i * ( j) Cχ χ = Gδij 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 χ () i miω ( C) = character = n Recall: in irrep i () i C () i where ω C * Imxm represents Dirac's character mω in irrep i, n N m C Classes C LOT: n χ ( C) χ C = N δ C Classes () i * ( j) C G ij () i nc χ C = NG set i=j: for each irrep i, C Classes n C Ω= () i i C G C = G i = () i nc ( ωc ) R R C Classes in Cv irrep E, ωi =, ωr =, ω σ = 0, nr = me = = 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: χ( R) n j jχ ( R) j = 5

6 χ R = n χ R ( j) trace of reducible representation: j j One can find n using the LOT : χ ( R) χ R = N δ () i * ( j) j G ij R G χ( R) χ ( R) = n χ ( R) χ ( R) = n N δ () i * ( j) () i * j j G ij R G R G j j n = i χ( R) χ ( R) N G R G () 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: R G i * j G αµ βν = δδ ij µνδαβ mi D R D R 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 σ σ σ a b c σ σ σ b c a σ σ σ c a b σ σ σ a b c σ σ σ c a b σ σ σ b c a E C C C E C C C E C : E C, C C, C E, σ a σ c σ b σ a, σ c σ b that is: C :,,, 6, 5, 6 5 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 σ σ σ a b c σ σ σ b c a σ σ σ c a b σ σ σ a b c σ σ σ c a b σ σ σ b c a E C C C E C C C E C : E C, C C, C E, σ a σ c σ b σ a, σ c σ b that is: C :,,, 6, 5, 6 5 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 = χ R χ R = χ E χ E N N G R * i * i * i n = * i E NG E mi. N χ = χ = G m m bloch occurs m times! i i i Burnside theorem mi = NG i G C I C σ g = 6 v v E ( xy) 0, z R z 0

11 There is a Second character orthogonality theorem (orthogonality of columns): irreps () i i () i * NG χ ( C) χ ( C') = δ, C and C' classes CC' n C Consider the matrix whose entries are those of the character table: Q = χ () i T { (C )} Q = χ ( 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 χ (C ) Ci i Q' = { } N G

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

13 k irreps (k) * (k) NG χ (C ) χ (C ) = δ 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 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: β ψ ( i ) ( j ) * N ( j ) ( j) multiply the GOT : D R D R = Oi δδ δ by ψ R G αµ βν ij µν αβ β ( j) i * G ( j) G ( j) β Dαµ R Dβν R = ψ β δδ ij αβ δµν = ψα δijδ µν R G mi β mi ( j ) ψ D R = ψ ψ Rψ = Rψ R ecall : j j j j j β βν β β ν ν. β β The result i G ( ± ) f x = f x ± f x ( i ) ( j) * G () i D αµ R Rψ ν = ψ α δδ ij µν R G mi ( i ) * ( j) ( j) Dαµ ( R) Rψν = ψα δijδ µν R G N N, that is, the action i i i * i ( j) () i αµ = αµ R is Pαµ ψν = ψ α δδ ij µν NG R G i αµ m N m P D R P Projection Operators is projection operator ν α and j = i.. of

15 m = i i Pαν Dαµ R R NG R G ( i ) * generalized projection operator Practical use with any function to project into an irrep: Start from any basis function: ψ= jν i i µµ µ cj (, νψ ) Then, if P ψ = ci (, µψ ) = 0 try another ψ, otherwise gives symmetry-adapted functions i component µ. 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 ν 5

16 There is no need for the D matrices: Start from any basis function: ψ = cj (, νψ ) Pψ = cj (, µψ ) jν j j j ν µµ µ i i µµ ψ = µψµ P ci, irrep i µ µ = = () i i () i * P Pµµ χ ( R) R µ R () i projects on irrep i; we only need To make basis take several and Gram-Schmidt orthogonalization. χ R 6

17 i Summary of main Group integration formulas i * j G αµ βν = δδ ij αβ δµν mi GOT : D R D R R G ( J) * ( j) 8π dω D ( αβγ ) D ( αβγ ) = δ δ δ j + ab a' b' jj' aa' bb' * : i j LOT ( R) ( R) NG R G Number of times irrep i is present in basis: i * n i = χ R χ R NG R Projector : = N () i () i * G χ ( C) χ ( C') = δcc' nc N χ χ = δ () i i * P χ R R R ij Burnside theorem mi = NG i 7 7

18 Simpler Uses of Group Theory N C I C σ g = 6 v v E ( xy) 0, 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 = σa = 0 0 σb = 0 0 σc =

20 a b alias s s s c E = 0 0 C = 0 0 C = σa = 0 0 σb = 0 0 σc = Traces : I, C 0, σ i G R * χ ni = χ R R N C I C σ N = 6 v z E 0 ( xy, ) Γ( H ) 0 v G R z *+ *0+ * n = χ R = = 6 R 6 n E = (* **0 + *0*) = 6 n = (* + **0 + **( )) = 0 6 Γ ( 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 σa = 0 0 σb = 0 0 σc = projectors on irreps: = () i i * P χ R R R P ( ) = R R P ( ) = R = R cting with P ( ) on any Hydrogen one finds the obvious ψ = s + s + s C I C σ g = 6 v v E. ( xy) 0, z R z

22 Projection operators on irrep E 0 0 s 0 s s E = 0 0 C = 0 0 C = 0 0 σa = 0 0 σb = 0 0 σc = projectors on irreps: = χ () i i * P R R R C I C σ g = 6 v v E ( xy) 0, z R z P ( E) = =

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

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

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

26 However now the geometry is: a b D ( C) =, DC ( ) = remain 0 σa( xy, ) = ( x, y), that is, σa = 0 σ σ σ c = σ C a = b = C a = c New matrices for irrep E D( C ) =, DC ( ) = σ a 0 = 0 σ b = σ 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 = σa = 0 0 σb = 0 0 σc = Recall: Γ ( H ) = E 7 7

28 R Operators R on Hydrogen basis E = 0 0 C = 0 0 C = 0 0 σa = 0 0 σb = 0 0 σc = D ( C) =, DC ( ) = P E xx Input data: E E * P xx Dxx R R 6 R 0 σ a = 0 = σ b = σ c = E P xx = [ E ( C + C) + σa ( σb + σc )] = E P yy =

29 original basis: 7 functions s,p x,p y,p z,s,s,s 7 7 Determinant is broken: basis: s, p z,ψ E bases: (p x,p y ),(ψ,ψ ) H matrices: (p x, ψ),(py,ψ ) By Schur lemma, the two E determinants are identical (elements cannot depend on component, so there can be no difference between x and y). 9 9

30 Molecular Orbitals of CH CH 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 = operations. There are C and C rotations around CH bonds,c and S 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 6σ 6S N = d d G 0 0 (, ) E z r x y T 0 ( R, R, R ) T 0 ( xyz,, ) r x y z 0 0

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

32 Tetrahedron: T d 8C Class moved unmoved χ E 0 8C C 6σ 6S d E 8C

33 C Class moved unmoved χ E 0 8C C 0 0 6σ 6S d E 8C C

34 6σ d one for each pair of H Class moved unmoved χ E 0 8C C 0 0 6σ 6S d E 8C C 6σ d

35 6S Class moved unmoved χ E 0 8C C 0 0 6σ d 6S 0 0 E 8C C 6σ 6S d 5 5

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

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

38 How to build Symmetry adapted orbitals from H states using Projectors We can build the actual states very simply. Class moved unmoved destinations atom T E 8C C 6σ 6S N = E 0 8C,,,,,,, C 0,, 6σ,,,,, d 6S 0,,,,, d d G 0 0 (, ) E z r x y T 0 ( R, R, R ) T 0 ( xyz,, ) Projector : = P ( ) r x y z () i i * P χ R R = R R R ψ = (obviously) 8 8

39 T E 8C C 6σ 6S N = d d G 0 0 (, ) E z r x y T 0 ( R, R, R ) T 0 ( xyz,, ) r x y z Class moved unmoved destinations atom E 0 8C,,,,,,, C 0,, 6σ,,,,, d 6S 0,,,,, Symmetry adapted orbitals Projector : = () i i * P χ R R ( T ) P = C + σ S d C σ d S Start for instance from atom R [ { + + } ( + ) ( T ) P = ( T T ) = [ hus, P + + ] to be normalized ] 9 9

40 T ( xyz,, ) : How can we find the components? ( T ) T i P D R R xx G R T d ( T ) T i P D R R yy = = m N m N G R T 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 z has eigenvalue under C rotation, like >+ >. ( T ) Start from the action of = + σ on, P ( T ) To obtain ψ z symmetrize on P C S C d σ S d = [ + + ] + [ ] [ + + ] 0 0

41 ( T ) Start from the action of P C S on, P ( T ) = + σ C d σ S d = [ + + ] + [ ] [ + + ] Simplify and do the same on P = [ + + ] P = [ + + ] ( T ) ( T ) ψ P + = z P P ( T ) ( T ) = [ + + ] = [ + + ] ( T ) [ ]

42 ψ P + = z ( T ) ( T ) [ ] T + simplify and norm ψ = z T + T + ψ = ψ x y = In a similar way, P P ( T ) = [ + + ] = [ + + ] Without symmetry, 8*8 determinant, With group thy determinants * T T T C, ψ C, ψ C, ψ C, ψ 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.

43 Broken representations! C v : (x,y) basis for E has subgroup Z C I C σ g = 6 v v E ( xy) 0, z R z Z I C C * ε ε E * ε ε ε z ( xy, ) However Z is abelian and degeneracy is broken. Basis fo conjugate representations iϕ ϕ m = = x + iy = e = cosϕ+ i sinϕ iϕ ϕ m = = x iy = e = cosϕ i sinϕ are not mixed by Z π π π i i i iϕ iϕ iϕ iϕ iϕ : : : C e e e C e e e e e π π i i iϕ iϕ iϕ iϕ : : C e e e C e e e Rule: H G subgroup irreps of G are still representations of H, but some representations become reducible. (x,y) is an irreducible basis for E in C but is broken in Z. v

44 Important example: angular momentum eigenstates { } j Characters and crystal field splitting jm, irrep of O(), with j m 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 L jm, = m jm, z j j j ilzα im jα R( α ) jm, = exp jm, = e jm, z j j j ( α ) ( j) im jα matrix of j representation is D ( Rz ) diag( e ) = j ( j) im jα TrD ( Rz ( α) ) e. m = j χα = = j

45 Trick to evaluate the sum: α i i j α α im jα i i ijα i( j ) α ijα ijα i( j ) α ijα sin e = e ( e + e + + e ) e ( e + e + + e ) m j = j j ( j) im j TrD ( Rz ( α) ) e α. m = j χα = = j ( ) ( α im ) jα i j+ α i j+ α sin e = ( e e ) = sin[ j+ α] m = j i j ll terms cancel except two: for instance for J= α e = e e + e + + e + e e e + e + + e + e i i α α im jα i i iα iα iα iα iα iα iα iα sin ( ) ( ) m = j 5 iα/ iα/ iα/ iα/ iα/ iα/ iα/ iα/ iα/ 5 iα/ = ( e + e + e + e + e ) ( e + e + e + e + e ) = i i 5 iα/ 5 iα/ 5α = ( e e ) = sin. i j 5 5

46 χα j sin[ j + α] im j = e α = α mj = j sin The { jm, } j basis is reducible in point Groups Number of times irrep i is present in basis: i ni = χ R χ R N G R * In this way the orbitals of atoms that are non displaced by symmetry operations can be analyzed in Irreps of the point Group. 6 6

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 8 operations (twice as many as in T d ) D 7 7

48 The Cubic group O h The O h Group is the point Group of many interesting solids, including complexes like CuSO 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 5H O where the central Cu ++ is bound to H O molecules and SO ions, and some superconducting cuprates.. CuSO 5H O Iron(III) chloride. 8 By contrast, Iron(II) choride is green 8

49 Cube: O h Group 6C coordinate axes, clockwise or counterclockwise E 6C 9 9

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

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

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

53 Cube: O h Group i:( xyz,, ) ( xyz,, ) i = σ * R π E C 6C 6 C ' 8C i 5 5

54 6S E C 6C 6 C ' 8C i 6S 5 5

55 σ h E C 6C 6 C ' 8C i 6S σ h 55 55

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

57 8S 6 π 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 σ E C 6C 6 C ' 8C i 6S σ h 6σ 8S d