Advanced Spectroscopy. Dr. P. Hunt Rm 167 (Chemistry) web-site:

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1 Advanced Spectroscopy Dr. P. Hunt Rm 167 (Chemistry) web-site:

2 Maths! Coordinate transformations rotations! example 18.1 p501 whole chapter on Matrices and linear transformations The Chemistry Maths Book Erich Steiner Gram-Schmidt orthogonalisation Schmidt_process use Gaussian elimination (also to find the inverse) linear algebra texts Diagonalising a matrix section 19.4 Matrix diagonalisation p543

3 Outline The transition dipole moment The vibrational wavefunction The transition dipole moment components Evaluating Integrals the Direct product Integration of even and odd functions Application to vibrational spectroscopy

4 Selection Rules Rules for determining vibrational activity for Infrared: dipole moment must change for Raman: polarizability must change Selection Rule Δν=± 1 (ν is the vibrational quantum number) Applied this by requiring that active modes must have same IR as T vectors (dipole moment) binary functions (polarizability) where do these rules come from?

5 Application to Vibrational Activity Probability of a molecule absorbing or emitting light: 2 Einstein coefficients Depends on: B if = B fi = µ if 6ε 0! 2 Eqn 1 Transition dipole moment For a photon to be absorbed the transition dipole moment must be non-zero µ = Ψ fi µψ dτ 0 f i µ fi = f µ i 0 Eqn 2 Components are: Ψ =ψ el ψ vib ψ rot ψ trans! # µ = # # "# µ x µ y µ z $ & & & %&! # α = # # "# α xx α xy α xz α yx α yy α yz α zx α zy α zz $ & & & %& inital and final wavefunctions dipole moment or polarizability

6 The Wavefunction total wavefunction: product of components Ψ total =ψ el ψ vib ψ rot ψ trans write a Schrödinger equation for each part Eqn 3 H el ψ el = ε el ψ el H vib ψ vib = ε vib ψ vib H rot ψ rot = ε rot ψ rot Additional OPTIONAL notes on the translational and rotational parts of the nuclear wavefuntion H trans ψ trans = ε trans ψ trans we will focus on the vibrational part today and the electronic component in later lectures

7 Transition Dipole Moment total wavefunction µ = Ψ fi µψ dτ = f µ i 0 f i ψ χ electronic wavefunction nuclear wavefunction µ = µ e + µ n Ψ ψχ assume BO approximation, nuclear and electronic components are seperable µ = ψ fi χ (µ + µ )ψ χ dτ dτ f f e n i i e n Eqn 4

8 Transition Dipole Moment ψ electronic wavefunction χ nuclear wavefunction µ = ψ fi χ f f = ψ χ f f = χ f (µ e + µ n )ψ i χ i dτ e dτ n ψ f ( µ )ψ χ dτ dτ + ψ e i i e n χ f f ( µ )ψ dτ e i e χ i dτ n + ψ f ( µ )ψ χ dτ dτ n i i e n χ f ( µ ) χ dτ n i n ψ i dτ e = χ µ χ dτ + ψ f e i n µ ψ dτ f n i e Eqn 4 grouped electronic components together grouped nuclear components together

9 Transition Dipole Moment ψ electronic wavefunction χ nuclear wavefunction µ = ψ fi χ f f = ψ χ f f = χ f (µ e + µ n )ψ i χ i dτ e dτ n ψ f ( µ )ψ χ dτ dτ + ψ e i i e n χ f f ( µ )ψ dτ e i e χ i dτ n + ψ f ( µ )ψ χ dτ dτ n i i e n χ f ( µ ) χ dτ n i n ψ i dτ e = χ µ χ dτ + ψ f e i n µ ψ dτ f n i e Eqn 4

10 Transition Dipole Moment transition on the same electronic state µ = χ fi µ χ dτ + ψ f e i n µ ψ dτ f n i e χ µ χ dτ = χ f e i n f ψ i ( µ )ψ dτ e i n χ i dτ e = 0 because ψ i ( µ )ψ dτ = ψ e i n eˆrψ dτ = i i e e ˆrψ 2 dτ = 0 i e thus µ = ψ fi µ ψ dτ f n i e an odd function (we will explain this shortly, but you have met this before in both your QM and photochemistry courses!) Eqn 5

11 Transition Dipole Moment transition on the same electronic state µ = ψ fi µ ψ dτ f n i e = ψ i χ f ( µ ) χ dτ n i n ψ i dτ e = ψ ψ dτ χ f i e f ( µ ) χ dτ n i n but we are on the same state so ψ f =ψ i = χ f ( µ ) χ dτ because ψ n i n ψ dτ = 1 i i e µ = χ fi f ( µ ) χ dτ n i n Eqn 6

12 Transition Dipole Moment transition between electronic states µ = χ fi µ χ dτ + ψ f e i n µ ψ dτ f n i e ψ µ ψ dτ = ψ f n i e f χ f ( µ ) χ dτ n i n ψ i dτ e because thus = ψ ψ dτ χ f i e f ψ ψ dτ = 0 if f i f i e ( µ ) χ dτ = 0 n i n Eqn 7 we will consider this equation in a later lecture µ = χ fi µ χ dτ = χ f e i n f ψ f ( µ )ψ dτ e i e χ idτ n Eqn 8

13 Transition Dipole Moment transition on the same electronic state Important! µ fi = χ f ( µ ) χ dτ n i n transition between electronic states µ = χ fi f ψ f ( µ )ψ dτ e i e χ idτ n

14 Transition Dipole Moment transition on the same electronic state µ fi = χ f ( µ ) χ dτ n i n vibrational wavefunction dipole moment Harmonic Oscillator solutions!

15 Transition Dipole Moment transition on the same electronic state µ fi = χ f ( µ ) χ dτ n i n vibrational wavefunction dipole moment Harmonic Oscillator solutions!

16 Vibrational Solutions vibrational wavefunction single coordinate solution: transform to mass weighted coordinates diagonalise the Hessian obtain the normal coordinates Q a Schrödinger equation for each Q gives the following harmonic oscillator solutions: H HO = 2 2m! # " d 2 dx 2 $ &+ 1 % 2 kx2 E i = (v ) ω ω = k m Fig 1 Hermite polynomial ϕ i = N i H i (y i )exp y i 2 2 y i = ω i! 1 2 Qi gaussian Q i are the vibrations derived earlier!

17 Vibrational Solutions Hermite polynomial ϕ i = N i H i (y i )exp y i 2 2 y i = ω i! 1 2 Qi Hermite polynomials give us nodes Fig 2 satisfy a set of relationships: see your notes Eqn 10, p3 images from: hosc7.html Gaussian_functionwiki Hermite_polynomials

18 Vibrational Solutions Hermite polynomial ϕ i = N i H i (y i )exp y i 2 2 y i = ω i! 1 2 Qi gaussian Hermite polynomials give us nodes Fig 2 gaussians give us an envelope images from: hosc7.html Gaussian_functionwiki Hermite_polynomials

19 Vibrational Solutions Hermite polynomial " ψ i = N i H i (y i )exp y i $ # 2 2 % ' & " y i = ω i $ # % ' & 1 2 Qi gaussian Hermite polynomials give us nodes gaussians give us an envelope removed due to copyright images from: hosc7.html Gaussian_functionwiki Hermite_polynomials together they give us the vibrational wavefunctions

20 3N-6 normal coordinates Vibrational Solutions total vibrational wavefunction is a product: ψ vib = ϕ i (Q i ) = ϕ ν1 (Q 1 )ϕ ν 2 (Q 2 )ϕ ν 3 (Q 3 ) Eqn 11 i Notation: Σ Π gamma means sum pi means product product of individual solutions for each vibrational mode Q

21 3N-6 normal coordinates Vibrational Solutions total vibrational wavefunction is a product: ψ vib = ϕ i (Q i ) = ϕ ν1 (Q 1 )ϕ ν 2 (Q 2 )ϕ ν 3 (Q 3 ) Eqn 11 i Notation: Σ Π gamma means sum pi means product product of individual solutions for each vibrational mode Q looking ahead: ground state: excited state: consider only 2 modes: transition dipole moment: [ ] [ ] χ i = ϕ ν1 (Q 1 )ϕ ν 2 (Q 2 ) = 0 0 χ f = ϕ ν1 (Q 1 )ϕ ν 2 (Q 2 ) = 1 0 µ fi = 1 0 µ n quanta of energy in the mode Q 1

22 Transition Dipole Moment transition on the same electronic state µ fi = χ f ( µ ) χ dτ n i n vibrational wavefunction dipole moment Harmonic Oscillator solutions!

23 Dipole Moment Taylor expansion " µ % µ = µ 0 + $ ' Q k + 1 # Q k & 2 k 0 " 2 µ % $ 2 ' Q 2 k + # Q k & k 0 Eqn 12 static dipole moment change in dipole moment during a vibration curvature of dipole moment change

24 Transition Dipole Moment µ fi = ϕ i µ 0 ϕ j + µ Q k ϕ i Q ϕ k j + 1 k µ 2 Q k ϕ i Q ϕ k2 j +! k 0 Eqn 13 ϕ i µ 0 ϕ j = µ 0 ϕ i ϕ j = µ 0 δ ij = 0 initial and final state are orthogonal Eqn 14

25 Transition Dipole Moment µ fi = ϕ i µ 0 ϕ j + µ Q k ϕ i Q ϕ k j + 1 k µ 2 Q k ϕ i Q ϕ k2 j +! k 0 Eqn " 2 µ % $ 2 ' ϕ i Q k2 ϕ j + 0 # Q k & k 0 assumed to be small Eqn 15

26 Transition Dipole Moment " µ % µ = ϕ µ ϕ + α i 0 j $ ' ϕ Q ϕ + 1 # Q i k j k & 2 k 0 Eqn 16 " 2 µ % $ ' ϕ Q ϕ + Q 2 i k2 j # k & k 0 Eqn 13 only one part left! k " µ % $ ' # Q k & the dipole must change or vary as the position of the nuclei changes 0 ϕ i Q k ϕ j 0 Eqn 17 selection rule!

27 Integrals i i χ i = ϕ ν1 (Q 1 )ϕ ν 2 (Q 2 )!ϕ i a (Q a )ϕ i i b (Q b )!ϕ ν 3n 6 (Q 3n 6 ) f f χ f = ϕ ν1 (Q 1 )ϕ ν 2 (Q 2 )!ϕ f a (Q a )ϕ f f b (Q b )!ϕ ν 3n 6 (Q 3n 6 ) i i = ϕ ν1 (Q 1 )ϕ ν 2 (Q 2 )!ϕ f a (Q a )ϕ i i b (Q b )!ϕ ν 3n 6 (Q 3n 6 ) χ f Q k χ i f = (ϕ ν1 i = (ϕ ν1 (Q 1 )!ϕ f i k (Q k )!) Q k (ϕ ν1 (Q 1 )!ϕ i k (Q k )!)dq 1!dQ k! (Q 1 )!ϕ f i k (Q k )!) Q k (ϕ ν1 (Q 1 )!ϕ i k (Q k )!)dq 1!dQ k! = i i ϕ ν1 (Q 1 ) ϕ ν1 (Q 1 )dq 1!### "### $ % ϕ f k (Q k )Q k ϕ i k (Q k )dq k! =1 only one integral will not be 1, that for the Q k component = ϕ f k (Q k )Q k ϕ i k (Q k )dq k ϕ f k (Q k ) Q k ϕ i k (Q k ) Eqn 18 initial vibrational state 1 quanta energy in 1 vibration final vibrational state only differes from the initial state in 1 vibration results in the selection rule that the quantum number Δv=±1 see notes for an argument based on Hermite polynomials

28 Raman follow an analogous procedure expand in terms of the induced dipole moment rather than the permanent dipole moment µ fi = k α Q k 0 χ f Q k χ i ε 0 Eqn 20

29 Integrals we need to determine many different integrals transition dipole moment vibrational selection expectation operators orthogonality are they zero or non-zero? χ f Q k χ i χ f χ i χ f Q k 2 χ i µ e = ψ f µ 0 ψ i µ ψ f Q k 0 ψ i 2 µ ψ f 2 Q k ψ f ψ i 0 ψ i A = ψ x A ψ x

30 Integrals Need two key pieces of knowledge The Direct Product Integration of even and odd functions

31 The Direct Product take the direct product of two Irreducible Representations (IRs) symbol: multiply the characters of the two IRs C 3v E 2C 3 3σ v A A E C 3v E 2C 3 3σ v A A A 1 A = A 2 Fig 3 A 1 A 2 = A 2

32 The Direct Product take the direct product of two Irreducible Representations (IRs) symbol: multiply the characters of the two IRs C 3v E 2C 3 3σ v A A E C 3v E 2C 3 3σ v A A A 1 A = A 2 the sum of the characters of the direct product of different IRs is always zero important implications IRs of a point group are orthogonal A 1 A 2 = A 2 [(1 1)! + (2! 1) + (3 "# $ 1)] %$ = 0 E C 3 σ v Fig 3

33 The Direct Product not as simple as might first seem C 3v E 2C 3 3σ v E E E E Fig 4

34 The Direct Product not as simple as might first seem C 3v E 2C 3 3σ v E E E E E E = { A + A + E} 1 2 use the reduction formula: C 3v E 2C 3 3σ v A n A1 = 1 6 A n A1 = 1 6 ( ) + ( ) + ( ) = 6 6 = 1 ( ) + ( ) + ( ) = 6 6 = 1 ( ) + ( ) + ( 3 0 0) E n A2 = = 6 6 = 1 E E Fig 4

35 The Direct Product Short Cuts! page in your character tables has multiplication properties of symmetry groups you must know how to use these General rules: A 1 A 2 = A 2 A A = A 1 2 = 2 Important! A B = B A A = A B B = A 1 2 = = = 1

36 The Direct Product Short Cuts! use last page of character tables has multiplication properties of symmetry groups you must know how to use these General rules: Point Group specific rules: A 1 A 2 = A 2 A A = A 1 2 = 2 Important! A B = B A A = A B B = A 1 2 = = = 1 E E = { A + A + E} 1 2 C 3v E 1 E 1 = E 2 E 2 = A 1 + A 2 + E

37 In Class Problem form the direct product A2 B1 for the C2v point group by (a) multiplying out the characters (b) using the rules" in your character table handout

38 In Class Problem form the direct product A2 B1 for the C2v point group by (a) multiplying out the characters C 2v E C 2 σ v σ v A A B B C 2v E C 2 σ v σ v A B A 2 B = B 2 (b) using the rules" in your character table handout A 2 B 1 = B 2 A B = B 1 2 = 2

39 Odd and Even Functions touched on this in photochemistry course anti-symmetric functions integrate to zero removed due to copyright

40 Odd and Even Functions Fig 5 f is antisymmetric about x=0 f has zero integral f(-x)=-f(x) f is an odd function g is symmetric about x=0 g has non-zero integral g(-x)=g(x) g is an even function

41 Odd and Even Functions Cs E σ A 1 1 A 1 1 A f dτ integrand Fig 6 Fig 5 A integrand must span the totally symmetric IR to be non-zero Important! anti-symmetric f dτ = 0 symmetric g dτ 0

42 Orthogonality and Symmetry can have a zero integral by accident! Fig 7

43 Orthogonality and Symmetry consider integrals of multiple functions g spans A f f spans A g 2 = g g 2 dτ f 2 = f f spans A A = A spans A A = A totally symmetric! Integral 0 (by symmetry) ( gf ) dτ g spans A and f spans A gf = g f spans A A = A NOT totally symmetric Integral=0 Eqn 23

44 Orthogonality and Symmetry Integrals that have 3 components sequentially form the direct product I = f Γ f Γ f Γ dτ i j k Eqn 24 an example C 4v I = d z d { B A B } xy x 2 y and B 2 A 1 = B 2 then B 2 B 1 = A 2 I = 0 Eqn 25 more complex if the components are degenerate: see OPTIONAL extra notes

45 Application to Vibrational Activity Rules for determining vibrational activity for Infrared: dipole moment must change for Raman: polarizability must change Selection Rule Δν=± 1 (ν is the vibrational quantum number) The transition dipole moment for a photon to be absorbed the transition dipole moment must be non-zero µ fi = f λ i 0 λ = µ,α µ = µ x µ y µ z for Infrared important quantity is the dipole moment for Raman important quantity is the polarizability α = α xx α xy α xz α yx α yy α yz α zx α zy α zz

46 Application to Vibrational Activity λ fi = φ f λ φ i 0 λ = µ,α find the direct product of the components Important! { Γ f Γ i } Γ λ where λ = µ or α Eqn 26

47 Application to Vibrational Activity λ fi = φ f λ φ i 0 λ = µ,α find the direct product of the components Important! { Γ f Γ i } Γ λ where λ = µ or α known information: ground vibrational state is always totally symmetric symmetry of the excited vibrational state is determined by the excited mode symmetry of IR component is based on the dipole which relates to the translation symmetry symmetry of Raman is based on the polarizability which relate to the binary function symmetry Γ µ = Γ(x) Γ(y) Γ(z) Γ α = Γ(x 2 ) Γ(y 2 ) Γ(z 2 ) Γ(xz) Γ(xy) Γ(yz)

48 Application to Vibrational Activity example H2O { Γ f Γ i } Γ λ where λ = µ or α know Γ vib = 2A 1 + B 2 know initial state A 1 test each final state A 1 and B 2 Infrared activity A1 mode: { A 1 A 1 } Γ µ = A 1 B 1 B 2 A 1 = { B 1, B 2, A } 1 for A 1 mode : A 1 { B 1, B 2, A } 1 IR active

49 In-Class Problem example H2O { Γ f Γ i } Γ λ where λ = µ or α know Γ vib = 2A 1 + B 2 Infrared activity: is the B2 mode IR active?

50 In-Class Problem example H2O { Γ f Γ i } Γ λ where λ = µ or α know Γ vib = 2A 1 + B 2 Infrared activity: is the B2 mode IR active? { B 2 A 1 } Γ µ = B 2 B 1 B 2 A 1 = { A 2, A 1, B } 2 for B 2 mode : A 1 { A 2, A 1, B } 2 IR active

51 Application to Vibrational Activity example H2O { Γ f Γ i } Γ λ where λ = µ or α know Γ vib = 2A 1 + B 2 know initial state A 1 test each final state A 1 and B 2 Raman activity A1 mode: { A 1 A 1 } Γ α = A 1 A 1 B 1 A 2 B 2 = { A 1, B 1, A 2, B } 2 for A 1 mode : A 1 { A 1, B 1, A 2, B } 2 Raman active

52 In-Class Problem example H2O { Γ f Γ i } Γ λ where λ = µ or α know Γ vib = 2A 1 + B 2 know initial state A 1 test each final state A 1 and B 2 Raman activity: is the B2 mode Raman active?

53 In-Class Problem example H2O { Γ f Γ i } Γ λ where λ = µ or α know Γ vib = 2A 1 + B 2 Raman activity: is the B2 mode Raman active? { B 2 A 1 } Γ α = B 2 A 1 B 1 A 2 B 2 = { B 2, A 2, B 1, A } 1 for B 2 mode : A 1 { B 2, A 2, B 1, A } 1 Raman active

54 Application to Vibrational Activity but the rules are a mode is infrared active if the IR of the vibration is the same as one of those for the translational vectors which represent the components of the dipole moment a mode is Raman active if the IR of the vibration is the same as one of the binary functions which represent the symmetry of the components of the polarizability how do we arrive at the above from the set of irreducible representations spanned by the integrand must contain the totally symmetric irreducible representation

55 Application to Vibrational Activity { Γ f Γ i } Γ λ where λ = µ or α the initial ground state is ALWAYS totally symmetric { Γ f A } 1 Γ λ where A1 represents the totally symmetric IR BUT A1 is like identity Γ f Γ λ Γ f A 1 = Γ f the integral is only non-zero if the cross product contains A1 Γ f Γ λ = A 1,! { } this is only true when the operator is squared Γm Γ m = A 1,! { } THUS Γ f = Γ λ to be IR active a mode must span the same IR as a component of μ (x, y,z)

56 Key Points be able to explain the simplifications we have employed in deriving the transition dipole moment be able to describe the general equation for the nuclear wavefunction be able to show using a Taylor expansion the origin of the rules for IR and Raman activity be able to demonstrate the direct product of IRs giving examples be able to explain, using a simple example, why only totally symmetric integrals are non-zero be able to determine if an integral is zero or non-zero by symmetry starting from an expression of the form given below be able to explain using direct products the origin of the rules that a mode is IR active only if the IR is the same as one to the translational vectors, and similarly for Raman λ α = χ f λ χ i 0 λ = µ,α using the expression below be able to determine when a mode is IR or Raman active { Γ f Γ i } Γ λ where λ = µ or α

57 Resources Web copies course notes download slides model answers links to good web-sites Reading Recommended Texts Kieran Molloy, Group Theory for Chemists, 2004, Harwood Publishing, Chichester. PW Atkins and RS Friedman, Molecular Quantum Mechanics, Oxford University Press, Oxford Secondary Text Alan Vincent, Molecular Symmetry and Group Theory, Second edition, 2001, John Wiley & Sons Ltd, Chichester. elective reading background material that supports lectures Optional specialist material that explains difficult concepts in more detail if you are interested in a wider perspective