Also interested only in internal energies Uel (R) only internal forces, has symmetry of molecule--that is source of potential.

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1 IV. Molecular Vibrations IV-1 As discussed solutions, ψ, of the amiltonian, (Schrödinger Equation) must be representations of the group of the molecule i.e. energy cannot change due to a symmetry operation, so [,R] = 0. Earlier we showed that B-O Approximation Ψ = φ e (r,r)χ υ (R) vib χ υ = (T N + U el (R)) χ υ = E υ χ υ can approximately separate nuclear and electronic coordinates. Also interested only in internal energies Uel (R) only internal forces, has symmetry of molecule--that is source of potential. But still left with 3N-6 coordinates could consider as a (3N-6) dimension space to be spaced by χ υ functions. Actual function must have 3N-6 coordinates, span with many kinds of functions. Ideal would be product (3N-6) 1-D function, χ υ(q n ) n = 1,2, χ υ must be representation of group since vib is totally symmetric [ vib, R] = 0 A. Questions: Which representations correspond to χ υ must span space What are the coordinates? χ υ transform as 3N-6 coordinate. 1st go: Representing nuclear motion in Cartesian space. Solution 3N coordinates are x,y,z motion on each atom. Suggest analysis as direct product representation (Γ atoms ) (Γ xyz ) [Note this is 3N-dimensional, will correct later to get 3N-6] Γ xyz Γ x Γ y Γ z from character point of view: χ xyz = χ x + χ y + χ z Γ atoms reprsentation relate to interchange matrix, only has diagonal if atom not move character: χ atom(r) = number atoms not moving under R Put it together Γ 3N = Γ atom Γ xyz But more simply χ 3N = χ atom χ xyz Thus we have characters of a reducible representation for all vibrations in a molecule plus six extra coordinates: 3 trans + 3 rotational To get just vibration: χ 3N-6 = χ 3N - χ Tr - χ rot χ Tr = χ xyz translation is just motion in Cartesian space χ rot = χ Rx,Ry,Rz = χ Rx + χ Ry + χ Rz Rotation about x,y,z also in Character Table

2 IV-2 Aside: Vibrational Spectra Review Solution to amiltonian of internal nuclear motion (3N-6 dimension) vib χ υ = (T N + U el (R)) χ υ = E υ χ υ but X υ 3N - 6 dimension x,y,z on each atom span space [ vib, R] = 0 R operation in group χ υ must correspond to representation of group xyz not representation will need new coordinates symmetry coordinates from projection of vibrational motion: q = c a i rai i xyz Classify Group Theory Γ xyz Γ atoms = Γ 3N count atoms NOT momentum vibrations Γ 3N-6 = Γ 3N - Γ trans - Γ rot as character: χ 3N = χ xyz χ atom then reduce to get linear combination irreducible reps: a i Γ i i Can categorize subspaces stretches χ str = bonds that do not move -- same as exchange idea bends, etc etc. again reduce to sum irred. Reps -- idea these pretty different energies Reducing Γ 3N always works, but starting in sub spaces may or may not span the space must pick carefully include all motion Group Theory on Γ 3N provides test do you get all representations? To determine the irreducible representation that corresponds to 3N-6 vibrations: just reduce using: a i = 1 h χ 3N-6 (R) χ * i (R) R Example 1: 2 O, 3N = 9, 3N-6 = 3: 3 internal motions, 2 O- stretches, 1 O bend z r 1 O α r 2 y Character Table: Reducible Representations: A z χ atom = A R z χ xyz = B x R y χ 3N = B y R x χ Tr + χ r = χ 3N-6 = Reduce by formula (or just look at reps and see) : χ 3N-6 = 2a 1 + b 2

3 Note: Could just reduce (r 1 + r 2 + α) representation of exchange: χ vib = IV-3 Therefore 3 vibrational functions will span space of vibrational degrees freedom Vibs: 2 (a 1 ) total symmetry (can interact via vib ) (this means motions can be mixed) 1 (b 2 ) symmetry separate must transform like y Could now use Projection operator to get form of vibs, or Guess kinds of motion approximation to real vibs, do not know level of mixing α angle open a 1 (will be mixed with sym. Str.) r 1 + r 2 symmetric str a 1 (will be mixed with bend) r 1 - r 2 asymmetric str b 2 (unique, transform like y, has a direction) Example 2: N N = 4x3 = 12 3N-6 = 6, expect 3 N- stretches and some mix of bends r N r 3 1 α 3 α 1 E 2C 3 3σ υ xyz rot A z χ atom = A R 2 χ xyz = E (x,y) (R y,r x ) χ 3N = X xyz χ Tr + χ r = X rot χ 3N-6 = Reduce: χ 3N-6 = 2a 1 + 2e Guess: all coupled motions a 1 : r 1 + r 2 + r 3 (symmetry str) α 1 + α 2 + α 3 (umbrella) 6 modes, 2x1 + 2x2, but only 4 energy levels, e modes degenerate To get e modes use P E operator e str : (r 1 - r 2, 2r 3 - r 1 - r 2 ) and e bend : (α 1 -α 2, 2α 3- α 1- α 2) B. Sub space representation Above -- representation of entire 3N-6 space of vibrational degrees of freedom. Can imagine subspaces and determine representation Example: r i -- N- Stretches α i -- N 2 angle bends from group frequency considerations know that these motions correspond to quite different energies (typically factor 2, N- str ~3300 cm -1, N bend ~ 1500 cm -1 )

4 Also know that if these are independent and if they span 3N-6 space they provide a set of basis functions for expanding IV-4 ex: χ υ = 3 (c i χ Ni + b i χ i N2 ) i=1 2 O -- Simple example can do by inspection inspection 2 O α bend total symmetry A 1 r 1, r 2 two identical coordinates, always take (r 1 + r 2 ) (r 1 r 2 ) A 1 B 2 recall χ 3N-6 = 2A 1 + B 2 so α, (r 1 + r 2 ), (r 1 r 2 ) basis function/coordinates/span space C. Projection Operator -- Alternate approach to guessing p Γ (r) = c i r i will give linear comb. of equiv. r, [could be bonds, angles, torsions] Example 3: go on to C 4 T d -- 3N = 15 3N-6 = 9 internal motions (vibrations) 1 C T d E 8C 3 3C 2 6S 4 6σ d Reducible rep A χ xyz A χ atom E χ 3N T R x R y R z -(χ trans + χ rot ) T x y z χ 3N Again χ 3N = χ xyz χ atoms Reduce χ 3N-6 a A 1 = 1 [ (1) ] = a A 2 = 1 [ ] = 0 a E = 1 [ ] = = 1 a T 1 = 1 [ ] = 0 a T 2 = 1 [ ] = 48 = 2 = 1 thus Γ 3N-6 = A 1 + E + 2T 2 see how = 9 = 3N-6 coordinates but yields 4 separate levels (IR or Raman bands)

5 IV-5 Recall: from above for N 3 we got: χ N- = χ N = (Note reflection in place bisect reduce: A 1 + E reduce: A 1 + E angle gives χ(σ) = +1) recall χ 3N-6 = 2A 1 + 2E 2-1D + 2-2D 6 dimensions or 3N - 6 = 12-6 = 6 so these 3N + 3N span the space C 4 is a bit harder see attached andout χ 3N-6 = A 1 + E + 2T 2 1D + 2D + 2 3D 9 dimensions 3N - 6 = 15-6 = 9 Now could choose C- str 4 -C- bend 6 10 problem extra dimension since more internal coordinate then 3N-6 these cannot be all independent χ C- = implies 2A 1 + E + 2T 2 get one too many A 1 coordinates reduces to A 1 + T 2 implies a) one is not independent or b) one is zero, such as: χ C = χ C = α 12 + α 13 + α 21 + α 23 + α + α 34 = 0 reduces to A 1 + E + T 2 can t all open at once