1. Symmetry and Spectroscopy

2. P. R. Bunker and Per Jensen: Molecular Symmetry and Spectroscopy, 2 nd Edition, 2 nd Printing, NRC Research Press, Ottawa, 2006 (ISBN 0-660-19628-X). $49.95 for 747 pages. paperback. BJ1 P. R. Bunker and Per Jensen: Fundamentals of Molecular Symmetry, IOP Publishing, Bristol, 2004 (ISBN 0-7503-0941-5). $57.95 paperback. BJ2

3. Examples of point group symmetry H2OH2O CH 3 F C 60 C3H4C3H4 C 2v C 3v D 2d IhIh

4. Examples of point group symmetry H2OH2O CH 3 F C 60 C3H4C3H4 C 2v C 3v D 2d IhIh

5. Point group symmetry of H 2 O z y (-x) The point group C 2v consists of the four operations E, C 2y,  yz, and  xy The word ´group´ is loaded. To see how we do two operations in succession

6. Point groups: Number of rotation axes and reflection planes.

7. z y (-x) 1 2 σ yz C2C2 z y (-x) 1 2 z y

8. z y σ xy = C 2 σ yz 1 2 σ yz σ xy C2C2 z y (-x) 1 2 z y

9. Multiplication Table for H 2 O Point group z y (-x) C 2v = {E, C 2,  yz,  xy } Multiplication table (=R row R column, in succession) Use multiplication table to prove that it is a “group.” σ xy = C 2 σ yz

10. {E, C 2,  yz,  xy } forms a “group“ if it obeys the following GROUP AXIOMS : All possible products RS = T belong to the group Group contains identity E (which does nothing) The inverse of each operation R  1 (R  1 R = RR  1 = E ) is in the group Associative law (AB )C = A(BC ) holds C 2v

11. Fermi: {E, C 2,  yz,  xy } forms a “group“ if it obeys the following GROUP AXIOMS : All possible products RS = T belong to the group Group contains identity E (which does nothing) The inverse of each operation R  1 (R  1 R = RR  1 = E ) is in the group Associative law (AB )C = A(BC ) holds C 2v ‘‘Group theory is just a bunch of definitions‘‘

12. All possible products RS = T belong to the group Group contains identity E (which does nothing) The inverse of each operation R  1 (R  1 R = RR  1 = E ) is in the group Associative law (AB )C = A(BC ) holds Not a GROUP

13. All possible products RS = T belong to the group Group contains identity E (which does nothing) The inverse of each operation R  1 (R  1 R = RR  1 = E ) is in the group Associative law (AB )C = A(BC ) holds Is a GROUP (subgroup of C 2v ) Rotational subgroup

14. PH 3 at equilibrium Symmetry operations: C 3v = { E, C 3, C 3 2,  1,  2,  3 } Symmetry elements: C 3,  1,  2,  3 C 3 Rotation axis  k Reflection plane

15. C 3 2 =  2  1 Multiplying C 3v symmetry operations Reflection Rotation

16. Multiplication table for C 3v C 3 2 = σ 2 σ 1

17. Multiplication table for C 3v C 3 2 = σ 2 σ 1 Note that C 3 = σ 1 σ 2

18. Multiplication table for C 3v Rotational subgroup

19. Multiplication table for C 3v 3 classes

20. A matrix group ´ M 1 = M 4 = ´ M 5 = ´ M 6 = ´ M 2 = ´ M 3 =

21. Multiplication table for the matrix group Products are M row M column

22. E C 3 C 3 2 σ 1 σ 2 σ 3 This matrix group forms a “representation” of the C 3v group These two groups are isomorphic. Multiplication tables have the ‘same shape’

23. Irreducible Representations The matrix group we have just introduced is an irreducible representation of the C 3v point group. The sum of the diagonal elements (character) of each matrix in an irreducible representation is tabulated in the character table of the point group.

24. The characters of this irreducible rep. ´ M 1 = E M 4 =  1 ´ M 5 =  2 ´ M 6 =  3 ´ M 2 = C 3 ´ M 3 = C 3 2 C 3v 2 0 0 0

25. The characters of this irreducible rep. The 2D representation M = {M 1, M 2, M 3,....., M 6 } of C 3v is the irreducible representation E. In this table we give the characters of the matrices. E C 3 σ 1 C 3 2 σ 2 σ 3 Elements in the same class have the same characters 3 classes

26. Character Table for the point group C 3v The 2D representation M = {M 1, M 2, M 3,....., M 6 } of C 3v is the irreducible representation E. In this table we give the characters of the matrices. E C 3 σ 1 C 3 2 σ 2 σ 3 Elements in the same class have the same characters Two 1D irreducible representations of the C 3v group

27. The matrices of the E irreducible rep. ´ M 1 = E M 4 =  1 ´ M 5 =  2 ´ M 6 =  3 ´ M 2 = C 3 ´ M 3 = C 3 2 C 3v

28. The matrices of the A 1 + E reducible rep. ´ M 1 = E M 4 =  1 ´ M 5 =  2 ´ M 6 =  3 ´ M 2 = C 3 ´ M 3 = C 3 2 C 3v 1 0 0 0 1 0 0 0 1 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 ‘ ‘ ‘ ‘ ‘ ‘

29. ‘ The matrices of the A 2 + E reducible rep. ´ M 1 = E M 4 =  1 ´ M 5 =  2 ´ M 6 =  3 ´ M 2 = C 3 ´ M 3 = C 3 2 C 3v 1 0 0 0 1 0 0 0 1 -1 0 0 0 1 0 0 0 1 0 0 0 -1 0 0 0 -1 0 0 0 ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘

30. Character table for the point group C 2v E C 2 σ yz σ xy Irreducible representations are “symmetry labels”

31. Some of Fermi’s definitions Group Subgroup Multiplication table of group operations Classes Representations Irreducible and reducible representations Character table See, for example, pp 14-15 and Chapter 5 of BJ1

32. Some of Fermi’s definitions Group Subgroup Multiplication table of group operations Classes Representations Irreducible and reducible representations Character table See, for example, pp 14-15 and Chapter 5 of BJ1

33. Irreducible representations The elements of irrep matrices satisfy the „Grand Orthogonality Theorem“ (GOT). We do not discuss the GOT here, but we list three consequences of it: Number of irreps = Number of classes in the group. Dimensions of the irreps, l 1, l 2, l 3 … satisfy l 1 2 + l 2 2 + l 3 2 + … = h, where h is the number of elements in the group. Orthogonality relation

34. Irreducible and reducible representations These are used as ‘‘symmetry labels‘‘ on energy levels. Which energy levels can ‘‘interact‘‘ and which transitions can occur. Can also determine whether certain terms are in the Hamiltonian.

35. IN SOME CIRCUMSTANCES THERE ARE PROBLEMS IF WE TRY TO USE POINT GROUP SYMMETRY TO DO THESE THINGS BUT

36. How do we use point group symmetry if the molecule rotates and distorts? H3+H3+ D 3h C 2v

37. 2 3 1 1 3 2 Or if tunnels? NH 3 C 3v D 3h

38. What are the symmetries of B(CH 3 ) 3, CH 3.CC.CH 3, (CO) 2, (NH 3 ) 2,…? Nonrigid molecules (i.e. molecules that tunnel) are a problem if we try to use a point group.

39. What should we do if we study transitions (or interactions) between electronic states that have different point group symmetries at equilibrium? Also

40. Point groups used for classifying: The electronic states for any molecule at a fixed nuclear geometry (see BJ2 Chapter 10), and The vibrational states for molecules, called “rigid” molecules, undergoing infinitesimal vibrations about a unique equilibrium structure (see BJ2 Pages 230-238).

41. Rotations and reflections Permutations and the inversion J.T.Hougen, JCP 37, 1422 (1962); ibid, 39, 358 (1963) H.C.Longuet-Higgins, Mol. Phys., 6, 445 (1963) P.R.B. and Per Jensen, JMS 228, 640 (2004) [historical introduction] To understand how we use symmetry labels and where the point group goes wrong we must study what we mean by “symmetry” See also BJ1 and BJ2

42. Symmetry not from geometry since molecules are dynamic Centrifugal distortion eg. H 3 + or CH 4 dipole moment Nonrigid molecules: eg. ethane, ammonia, (H 2 O) 2, (CO) 2,… Breakdown of BOA: eg. HCCH* - H 2 CC Also symmetry applies to atoms, nuclei and fundamental particles. Geometrical point group symmetry is not possible for them. We need a more general definition of symmetry

43. Symmetry Based on Energy Invariance Symmetry operations are operations that leave the energy of the system (a molecule in our case) unchanged. Using quantum mechanics we define a symmetry operation as follows: A symmetry operation is an operation that commutes with the Hamiltonian: (RH – HR)  n = [R,H]  n = 0

44. Uniform Space ----------Translation Isotropic Space----------Rotation Identical electrons------Permute electrons Identical nuclei-----------Permute identical nuclei Parity conservation-----Inversion (p,q,s) (-p,-q,s) P(E*) Reversal symmetry-----Time reversal (p,q,s) (-p,q,-s) T Ch. conj. Symmetry-----Particle antiparticle C Symmetry Operations (energy invariance)

45. Uniform Space ----------Translation Symmetry Operations (energy invariance) Separate translation… Translational momentum Ψ tot = Ψ trans Ψ int int = rot-vib-elec.orb-elec.spin-nuc.spin

46. Uniform Space ----------Translation Isotropic Space----------Rotation Identical electrons-------Permute electrons Identical nuclei-----------Permute identical nuclei Parity conservation-----Inversion (p,q) (-p,-q) P(E*) Reversal symmetry-----Time reversal (p,s) (-p,-s) T Ch. conj. Symmetry-----Particle antiparticle C Symmetry Operations (energy invariance) K(spatial) group, J, m J or F,m F labels

47. Uniform Space ----------Translation Isotropic Space----------Rotation Identical electrons-------Permute electrons Identical nuclei-----------Permute identical nuclei Parity conservation-----Inversion (p,q) (-p,-q) P(E*) Reversal symmetry-----Time reversal (p,s) (-p,-s) T Ch. conj. Symmetry-----Particle antiparticle C Symmetry Operations (energy invariance) Symmetric group S n

48. For the BeH molecule (5 electrons) Ψ orb-spin transforms as D(0) of S 5 PEP Slater determinant ensures antisymmetry so do not need S 5

49. Uniform Space ----------Translation Isotropic Space----------Rotation Identical electrons-------Permute electrons Identical nuclei-----------Permute identical nuclei Parity conservation-----Inversion (p,q) (-p,-q) P(E*) Reversal symmetry-----Time reversal (p,s) (-p,-s) T Ch. conj. Symmetry-----Particle antiparticle C Symmetry Operations (energy invariance)

50. Uniform Space ----------Translation Isotropic Space----------Rotation Identical electrons-------Permute electrons Identical nuclei-----------Permute identical nuclei Parity conservation-----Inversion (p,q) (-p,-q) P(E*) Reversal symmetry-----Time reversal (p,s) (-p,-s) T Ch. conj. Symmetry-----Particle antiparticle C Symmetry Operations (energy invariance) CNPI group = Complete Nuclear Permutation Inversion Group

51. Uniform Space ----------Translation Isotropic Space----------Rotation Identical electrons-------Permute electrons Identical nuclei-----------Permute identical nuclei Parity conservation-----Inversion (p,q) (-p,-q) P(E*) Reversal symmetry-----Time reversal (p,s) (-p,-s) T Ch. conj. Symmetry-----Particle antiparticle C Symmetry Operations (energy invariance) CNPI group = Complete Nuclear Permutation Inversion Group EXAMPLE: The CNPI group for H 2 O is C 2v (M) = {E, (12), E*, (12)*}

52. The CNPI Group for the Water Molecule The Complete Nuclear Permutation Inversion (CNPI) group for the water molecule is C 2v (M) = {E, (12), E*, (12)*} H1H1 H2H2 O e + H2H2 H1H1 O e + H1H1 H2H2 O e - (12)E* (12) *

53. We compare C 2v and this CNPI group Multiplication table (R row R column ) C 2v and CNPI are isomorphic! C 2v CNPI

54. We compare C 2v and this CNPI group Rotational subgroup Permutation subgroup

55. CNPI group of water: Character table This group is called C 2v (M)

56. Why RH = HR used to Define Symmetry? RH  = RE  HR  = ER  H  = E  Thus R  = c  since E is nondegenerate. However R 2 = E, so R(RΨ) = Ψ, but R(RΨ) = c 2 Ψ. Thus c 2 = 1, c = ±1 and R  = ±  For the water molecule ( nondegenerate and R 2 = E for all R) : Symmetry of H restricts symmetry of eigenfunctions Ψ 

57. + Parity - Parity x Ψ 1 + (x) x Ψ 3 + (x) x Ψ 2 - (x) Ψ + (-x) = Ψ + (x) Ψ - (-x) = -Ψ - (x) Eigenfunctions of H must satisfy E*Ψ = ±Ψ R=E*

58. ”Why RH = HR used to Define Symmetry?” continued…… RH  = RE  HR  = ER  H  = E  Thus R  = c . However R 2 = E, so c = ±1 and R  = ±  Allows us to SYMMETRY LABEL the energy levels using the irreps of the symmetry group For the water molecule (with nondegenerate states): Symmetry of H restricts symmetry of eigenfunctions Ψ RH=HR implies

59. There are four symmetry types of H 2 O wavefunction (12) E* 1 1 1 -1 -1 -1 -1 1 E1111E1111 (12)* 1 1 A1A2B1B2A1A2B1B2 R  = ±  A 2 x B 1 = B 2, B 1 x B 2 = A 2, B 1 x A 2 x B 2 = A 1 ∫ Ψ a HΨ b dτ = 0 if symmetries of Ψ a and Ψ b are different. ∫ Ψ a μΨ b dτ = 0 if symmetry of product is not A 1 Possible labels would be (1,1), (1,-1), (-1,-1), and (-1,1). However. More generally systematic are the irreducible representation labels (or symmetry labels) from the symmetry group.

60. The Symmetry Labels of the C 2v (M) Group of H 2 O (12) E* 1 1 1 -1 -1 -1 -1 1 E1111E1111 (12)* 1 1 A1A2B1B2A1A2B1B2 R  = ±  A 2 x B 1 = B 2, B 1 x B 2 = A 2, B 1 x A 2 x B 2 = A 1 ∫ Ψ a HΨ b dτ = 0 if symmetries of Ψ a and Ψ b are different. ∫ Ψ a μΨ b dτ = 0 if symmetry of product is not A 1

61. The Symmetry Labels of the C 2v (M) Group of H 2 O (12) E* 1 1 1 -1 -1 -1 -1 1 E1111E1111 (12)* 1 1 A1A2B1B2A1A2B1B2 R  = ±  because of: The vanishing integral theorem BJ1 pp114-117, BJ2 pp 136-139 Labeling is not just bureaucracy…it is useful

62. The Symmetry Labels of the C 2v (M) Group of H 2 O (12) E* 1 1 1 -1 -1 -1 -1 1 E1111E1111 (12)* 1 1 A1A2B1B2A1A2B1B2 R  = ±  because of: The vanishing integral theorem BJ1 pp114-117, BJ2 pp 136-139 Labeling is not just bureaucracy…it is useful But first let’s look at three things we overlooked: R n =E with n>2, degenerate states, symmetry of a product

63. Suppose R n = E where n > 2.  = e i2  /3 We still have RΨ = cΨ for nondegenerate Ψ, but now R n Ψ = Ψ. Thus c n = 1 and c = n √ 1 If n = 3 we introduce and c = 1, ε or ε 2 (= ε *) e iπ = -1 C 3 C 3 2 e i2π = 1

64. Suppose R n = E where n > 2.  = e i2  /3 We still have RΨ = cΨ for nondegenerate Ψ, but now R n Ψ = Ψ. Thus c n = 1 and c = n √ 1 If n = 3 we introduce and c = 1, ε or ε 2 (= ε *) C 3 C 3 2 A pair of separably degenerate irreps. Degenerate because of T

65. For nondegenerate states we had this as the effect of a symmetry operation on an eigenfunction: RH  = RE  HR  = ER  H  = E  Thus R  = c  since E is nondegenerate. For the water molecule ( nondegenerate) :  What about degenerate states?

66. R Ψ nk = D[R ] k1 Ψ n1 + D[R ] k2 Ψ n2 + D[R ] k3 Ψ n3 +…+ D[R ] kℓ Ψ nℓ For each relevant symmetry operation R, the constants D[R ] kp form the elements of an ℓ  ℓ matrix D[R ]. ForT = RS it is straightforward to show that D[T ] = D[R ] D[S ] The matrices D[T ], D[R ], D[S ] ….. form an ℓ-dimensional representation that is generated by the ℓ functions Ψ nk The ℓ functions Ψ nk transform according to this representation degenerate energy level with energy E n ℓ - fold

67. Symmetry of a product:C 2v (M) example B 1 x B 2, A 1 x A 2, B 1 x A 2, B 2 x A 2, B 1 x B 1,… A 2 A 2 B 2 B 1 A 1 The symmetry of the product of two nondegenerate states is easy:

68. Symmetry of a product. Example: C 3v E  E: 4 1 0 A 1  A 1 = A 1 A 1  A 2 = A 2 A 2  A 2 = A 1 A 1  E = E A 2  E = E E  E = A 1  A 2  E Characters of the product representation are the products of the characters of the representations being multiplied. See pp 109-114 in BJ1 Reducible representation

69. Symmetry of a product. Example: C 3v E  E: 4 1 0 A 1  A 1 = A 1 A 1  A 2 = A 2 A 2  A 2 = A 1 A 1  E = E A 2  E = E E  E = A 1  A 2  E Characters of the product representation are the products of the characters of the representations being multiplied. See pp 109-114 in BJ1 Reducible representation We say that E x E A 1

70. Back to the vanishing integral theorem

71. + Parity - Parity x Ψ + (x) x x Ψ - (x) Ψ + (-x) = Ψ + (x) Ψ - (-x) = -Ψ - (x) ∫ Ψ + Ψ - Ψ + d x = 0 - parity

72. + Parity - Parity x Ψ + (x) x x Ψ - (x) Ψ + (-x) = Ψ + (x) Ψ - (-x) = -Ψ - (x) ∫ Ψ + Ψ - Ψ + d x = 0 - parity ∫f(τ)dτ = 0 if symmetry of f(τ) does not contain A 1 The vanishing integral theorem

73. Diagonalizing the molecular Hamiltonian Schrödinger equation To apply the vanishing integral rule we look at symmetry of Eigenvalues and eigenfunctions are found by diagonalization of a matrix with elements

74. Diagonalizing the molecular Hamiltonian H mn vanishes if Γ( ) and Γ( ) are different The Hamiltonian matrix factorizes, for example for H 2 O if Γ(integrand) does not contain Γ (s) = 0

75. F(E i ) = [ g i e -E i /kT ] / g j e -E j /kT Boltzmann factor ∑ j S(f ← i) = | ∫ Φ f * μ A Φ i d τ | 2 Line strength ∑ A=X,Y,Z R stim (f→i) = 1 – exp (-h ν if /kT ) Stimulated emission I(f ← i) = ∫ 8π 3 N a ______ (4π ε 0 ) 3hc 2 F(E i ) S(f ← i) R stim (f→i) Integrated absorption intensity for a line is: = ν if ~ line Frequency factor ε(ν)dν ~

76. Selection rules for transitions The intensity of a transition is proportional to the square of For the integral to be non-vanishing, the integrand must have a totally symmetric component. μ Z = Σ q i Z i Z i Product of symmetries of Φs must contain that of μ Z

77. Symmetry of  Z  Z has symmetry  *  * ??? What is

78. Symmetry of  Z  Z has symmetry  *  * has character +1 under all permutations P  1 under all permutation-inversions P*

79. Symmetry of  Z for H 2 O  * = A 2

80. The Symmetry Labels of the C 2v (M) Group of H 2 O (12) E* 1 1 1 -1 -1 -1 -1 1 E1111E1111 (12)* 1 1 A1A2B1B2A1A2B1B2 Symmetry of μ Z R  = ±  ∫ Ψ a * HΨ b dτ = 0 if symmetries of Ψ a and Ψ b are different. As a result the Hamiltonian matrix is block diagonal. ∫ Ψ a * μΨ b dτ = 0 if symmetry of product Ψ a Ψ b is not Γ * Symmetry of H Using symmetry labels and the vanishing integral theorem we deduce that: Γ(μ Z ) = Γ * 0 00 0 Γ(H) = Γ (s)

81. Example of using the symmetry operation (12): H1H1 H2H2 r1´r1´ r2´r2´ ´´  (12) We have (12) (r 1, r 2,  ) = (r 1 ´, r 2 ´,  ´) We see that (r 1 ´, r 2 ´,  ´) = (r 2, r 1,  ) Determining symmetry and reducing a representation

82. 2 3 1 1 3 2 3 1 2 1 3 2 3 1 2 1 3 2 1 3 2 1 3 2 r2´r2´ r1´r1´ ´´ r1r1 r2r2  r1r1 r2r2  r1r1 r2r2  r1r1 r2r2  r1´r1´ r2´r2´ ´´ r1´r1´ r2´r2´ ´´ r2´r2´ r1´r1´ ´´ (12) E E* (12)*

83. (12) E E* (12)* R a = a´ = D[R] a  = 3  = 1  = 3  = 1

84. a A1 = ( 1  3 + 1  1 + 1  3 + 1  1) = 2a A2 = ( 1  3 + 1  1  1  3  1  1) = 0a B1 = ( 1  3  1  1  1  3 + 1  1) = 0a B2 = ( 1  3  1  1 + 1  3  1  1) = 1  = 2 A 1  B 2 A reducible representation Γ = Σ a i Γ i i

85. a A1 = ( 1  3 + 1  1 + 1  3 + 1  1) = 2a A2 = ( 1  3 + 1  1  1  3  1  1) = 0a B1 = ( 1  3  1  1  1  3 + 1  1) = 0a B2 = ( 1  3  1  1 + 1  3  1  1) = 1  = 2 A 1  B 2 A reducible representation Γ = Σ a i Γ i i i

86. a A1 = ( 1  3 + 1  1 + 1  3 + 1  1) = 2a A2 = ( 1  3 + 1  1  1  3  1  1) = 0a B1 = ( 1  3  1  1  1  3 + 1  1) = 0a B2 = ( 1  3  1  1 + 1  3  1  1) = 1  = 2 A 1  B 2 A reducible representation Γ = Σ a i Γ i i i

87. a A1 = ( 1  3 + 1  1 + 1  3 + 1  1) = 2a A2 = ( 1  3 + 1  1  1  3  1  1) = 0a B1 = ( 1  3  1  1  1  3 + 1  1) = 0a B2 = ( 1  3  1  1 + 1  3  1  1) = 1  = 2 A 1  B 2 A reducible representation Γ = Σ a i Γ i i i

88. We know now that r 1, r 2, and  generate the representation 2 A 1  B 2 Consequently, we can generate from r 1, r 2, and  three „symmetrized“ coordinates: S 1 with A 1 symmetry S 2 with A 1 symmetry S 3 with B 2 symmetry For this, we need projection operators Pages 102-109 of BJ1

89. Projection operators: General for l i -dimensional irrep  i Diagonal element of representation matrix Symmetry operation Simpler for 1 -dimensional irrep  i Character 1

90. Projection operators: General for l i -dimensional irrep  i Diagonal element of representation matrix Symmetry operation Simpler for 1 -dimensional irrep  i Character 1

91. Projection operators: General for l i -dimensional irrep  i Diagonal element of representation matrix Symmetry operation Simpler for 1 -dimensional irrep  i Character 1 E (12) E* (12)* A 1 1 1 1 1 P A 1 = (1/4) [ E + (12) + E* + (12)* ]

92. Projection operators: General for l i -dimensional irrep  i Diagonal element of representation matrix Symmetry operation Simpler for 1 -dimensional irrep  i Character 1 E (12) E* (12)* A 1 1 1 1 1 B 2 1 -1 1 -1 P A 1 = (1/4) [ E + (12) + E* + (12)* ] P B 2 = (1/4) [ E – (12) + E* – (12)* ]

93. Projection operator for A 1 acting on r 1 = [ r 1 + r 2 + r 1 + r 2 ] = [ r 1 + r 2 ] S 1 = P 11 A1 r 1 = [ E + (12) + E* + (12)* ]r 1 = [  +  +  +  ] =  S 2 = P 11 A1  = [ E + (12) + E* + (12)* ]  = [ r 1  r 2 + r 1  r 2 ] = [ r 1  r 2 ] S 3 = P 11 B2 r 1 = [ E  (12) + E*  (12)*] r 1 = [    +    ] = 0 P 11 B2  = [ E  (12) + E*  (12)* ]   Is „annihilated“ by P 11 B2 PA1PA1 PA1PA1 PB2PB2 PB2PB2 PB2PB2

94. Projection operators for A 1 and B 2 = [ r 1 + r 2 + r 1 + r 2 ] = [ r 1 + r 2 ] S 1 = P 11 A1 r 1 = [ E + (12) + E* + (12)* ]r 1 = [  +  +  +  ] =  S 2 = P 11 A1  = [ E + (12) + E* + (12)* ]  = [ r 1  r 2 + r 1  r 2 ] = [ r 1  r 2 ] S 3 = P 11 B2 r 1 = [ E  (12) + E*  (12)*] r 1 = [    +    ] = 0 P 11 B2  = [ E  (12) + E*  (12)* ]   Is „annihilated“ by P 11 B2 PA1PA1 PA1PA1 PB2PB2 PB2PB2 PB2PB2

95. Projection operators for A 1 and B 2 = [ r 1 + r 2 + r 1 + r 2 ] = [ r 1 + r 2 ] S 1 = P 11 A1 r 1 = [ E + (12) + E* + (12)* ]r 1 = [  +  +  +  ] =  S 2 = P 11 A1  = [ E + (12) + E* + (12)* ]  = [ r 1  r 2 + r 1  r 2 ] = [ r 1  r 2 ] S 3 = P 11 B2 r 1 = [ E  (12) + E*  (12)*] r 1 PA1PA1 PA1PA1 PB2PB2 Aside: S 1, S 2 and S 3 have the symmetry and form of the normal coordinates. See pp 269-277 in BJ1, and 232-233 in BJ2

96. [H,R] Symmetry and conservation laws (see chapter 14 of BJ2) i ħ ∂Ψ/∂t = HΨ where Ψ is a function of q and t Does symmetry change with time? ∂ /∂t = + = + = [ - ] iħiħ __ = (H is Hermitian) = 0 iħiħ __

97. So Far: Point group (geometrical) symmetry H 2 O and PH 3 point groups used as examples Group theory definitions: Irreducible reps and Ch. Tables Reducible representations and projection operators Problems using point groups: Rotation, tunneling,… Use [H,R]=0 to define R as a symmetry operation Introduce the CNPI group Explain why [H,R]=0 used to define symmetry Can label energy levels (the Ψ generate a representation.) Vanishing integral theorem Forbidden interactions and forbidden transitions Conservation of symmetry

98. Where are we going? HΨ n = E n Ψ n H = H 0 + H’ where H 0 Ψ n 0 = E n 0 Ψ n 0 E n 0 is -fold degenerate: Eigenfunctions are Ψ n1 0,Ψ n2 0,…,Ψ n 0 We want to symmetry label the energy levels using the irreducible representations of a symmetry group. We do this because it helps us to do many things: Which E n 0 can be mixed by H’: Block diagonalize H-matrix. Selection rules for transitions: Only if connected by Γ*. Nuclear spin statistics and intensity ratios Tunneling splittings, Stark effect, Zeeman effect, Breakdown of Born-Oppenheimer approximation… ℓ ℓ

99. The ℓ-fold degenerate eigenfunctions Ψ n1 0,Ψ n2 0,…,Ψ nℓ 0 GENERATE an ℓ-fold irreducible representation of the symmetry group (this labels the energy level E n 0 ). The basis of what we do using symmetry is that: Ψ n1 0 Ψ n2 0. Ψ n ℓ 0 Ψ n1 0 Ψ n2 0. Ψ n ℓ 0 R = D(R) The matrices D(R) form an irreducible representation To obtain the matrices D(R), and hence the irreducible rep. label, we need to know the Ψ ni 0 and to know how the symmetry ops transform the coordinates in the Ψ ni 0. The above follows from the fact that [H,R] = 0.

100. But first of all we must decide on the symmetry group that we are going to use. It could be the CNPI group BUT…

101. There are problems with the CNPI Group Number of elements in the CNPI groups of various molecules C 6 H 6, for example, has a 1036800-element CNPI group, but a 24-element point group at equilibrium, D 6h Huge groups and (as we shall see) multiple symmetry labels