5.80 Small-Molecule Spectroscopy and Dynamics

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1 MIT OpenCourseWare Small-Molecule Spectroscopy and Dynamics Fall 2008 For information about citing these materials or our Terms of Use, visit:

2 5.76 Lecture #28 Page 1 of 8 pages Lecture #28: Polyatomic Vibrations IV: Symmetry What is a normal mode? all atoms undergoing oscillation at the same frequency and phase and with amplitudes determined by the eigenvectors of the GF matrix. Q = L 1 S s tα = s tα (t) ê stα we built in specific relative lengths s tα (t) cos(λt + δ) of displacements -1 S i eigenvectors are e.g. Q j = L ji rows of L 1 i If we represent S i by a set of s iα vectors of prescribed lengths and directions, then L 1 tells us how to weight and add the vectors at each atom associated with the various internal displacements e.g. H 2 O (page 4 of H 2 O example in 4/24/96 notes) Q 1 = (S 2 S 1 ) Q 2 = (S 1 + S 2 ) S 3 Q 3 = 0.655S 3 S 1 = S 2 = S 3 =

3 5.76 Lecture #28 Page 2 of 8 pages Q 1 = = Q 2 = = Q 3 = names of normal modes from pictures or from equations? Note that each of the pictures involves some change of bond angle. So which mode is the bend? Q 1 involves dominantly a compression of one bond and an expansion of the other equivalent one; Q 2 involves dominantly two equivalent bonds expanding and contracting in phase; Q 3 pure internal bend. equations: S 1 ± S 2 symmetric antisymmetric mixed character no S 3 in Q 1 some S 3 character in Q 2 Why no (S 1 + S 2 ) character in Q 3? Actually, there is some, but very small.

4 5.76 Lecture #28 Page 3 of 8 pages 1 0 0? L 1 L NON-LECTURE = actually small, not 0 by symmetry = Why are there mixtures of internal coordinates in the normal coordinates? This perturbation theory kind of argument is OK for symmetric matrix. Some are due solely to symmetry. Others are due to the structure of the F and G matrices. The detailed character of the modes depends on ratios of off diagonal matrix elements of GF to differences between diagonal values, just as for H (even though GF is not symmetric). Best to see the cause of mixed character normal modes by going to symmetrized internal coordinates. Let S = U S 2 1/2 2 1/2 0 U = 2 1/2 2 1/ S F S = S U UFU U S = S F S

5 5.76 Lecture #28 Page 4 of 8 pages similarly F = UFU G = UGU md / Å md / Å md F = md / Å md / Å md md md md Å (md = millidyne, dyne is the cgs unit of force, now illegal.) Note that each term in expansion of determinant F has same units: md 3 /Å even though individual terms in F do not. An easy way for humans, not computers, to compute new F matrix is term by term rather than multiplying out. F 11 = 2 1/2 (S 1 + S 2 ) F (S 1 + S 2 ) 2 1/2 1 = [F 11 + F F 12 ] = 8.454md / Å 0.200md / Å 2 = 8.254md / Å 1 = 2 2 F 12 = 1 (S 1 S 2 ) F (S 1 + S 2 ) [F 11 F 22 + F 12 F 21 ] = 0 etc md / Å md F = md / Å md md Å

6 5.76 Lecture #28 Page 5 of 8 pages rearrange 8.254md / Å 0.317md 0 1 F = sym 0.697md Å md / Å 2 Similarly for amu amu amu 1 Å 1 G = amu amu amu 1 Å amu amu 1 Å amu 1 Å 2 and, in re-arranged form amu amu 1 Å G = sym amu 1 Å amu 1 2 Notice that both F and G are block diagonalized. This is a symmetry effect Group Theory. (torsions in would have been a separate block in symmetry coordinates) What would cause the coupling between S 1 and S 3 to get larger or smaller? IVR Look at G 13 and F 13.

7 5.76 Lecture #28 Page 6 of 8 pages G 13 = 2 1/2 ( e ) 1 sinθ e 1 r 32 mass & m 3 geometry center atom small for heavy end atoms effects 0 at linear max at 90 small for very heavy center atom F 13 with respect to F 11 F 33 bend and stretch have very different F ii but two stretches or two bends have more similar ones Physical basis for bend vs. stretch as dominant character in a normal mode, even though there is no symmetry reason that stretches and bends should not mix strongly. Alternate approach to vibrational analysis. See Bernath pages with respect to G 11 G 33 Work in mass weighted Cartesian displacement representation rather than internal coordinates. Convenient for electronic structure calculations. No insight. No transferability. 3N 3N f matrix f is symmetric f + = Λ eigenvalues of f matrix (6 are zero) + = 1 unitary Q = q Once we obtain {λ i } and can get to S and F representations if desired. Now for quantum mechanics and treatment beyond harmonic level. We know what individual atom motions are involved in each Q. Set up matrix representation of H(P,Q).

8 5.76 Lecture #28 Page 7 of 8 pages 3N 6 H = h n 0 (Q n,p n ) + V (Q 1, Q 3N 6 ) n=1 1 / 2 0 find 3 / 2 h 0 = hω 0 n n v n +1 / 2 out about this! P = i Q T = P 2 2 because? V = F ijk Q i Q j Q k + F ijkl Q i Q j Q k Q l ijk ijkl matrix elements Hv 1 v 3N 6 ; v 1 v 3N 6 infinite matrix how to truncate? how to organize? * in order of increasing energy * polyads What is a polyad? 2 mode frequencies are near integer multiples of each other, e.g. 2 : 1. etc. 2 : 1 polyad ω 1 2ω 2

9 5.76 Lecture #28 Page 8 of 8 pages hexad (0,10) (1,8) (2,6) (3,4) (4,2) (5,0) triad (0,4) (1,2) (2,0) (0,3) (1,1) dyad (0,2) (1,0) (0,1) (v 1,v 2 ) = (0,0) # of levels in polyad increase monotonically, but all matrix elements are related to 1 st example. H 0,2;1,0 F 122 (2 1) 1/2 ( 1) = 2 1/2 F 122 H nm;n+1 m 2 F 122 (m m 1) 1/2 (n +1) 1/2 m(m 1)(n +1) = H 02;10 2 superpolyad two interlocking polyads, as in acetylene ω 1 : ω 2 : ω 3 : ω 4 : ω 5 = 5 : 3 : 5 : 1 : 1 1/2 Darling-Dennison Q 2 1 Q Q 2 Q 3 Q 4 Q 5 Resonance Vectors Basis Vectors (v 1, v 2, v 3, v 4, 4, v 5, 5 ) 7 dimensional vector each harmonic oscillator product state represented by a 7 dimensional vector each coupling term represented by a vector that describes its selection rules. Q 1 2 Q 2 3 ( ) Q Q 3 ψ 1 ψ 2 Find conserved quantum numbers by listing all relevant resonance vectors, then find directions to all of those. In HCCH n res n s are the conserved quantities: polyad quantum numbers. Tells you which block of H to diagonalize.