MIT OpenCourseWare http://ocw.mit.edu 5.80 Small-Molecule Spectroscopy and Dynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.5.80 Lecture #27 Fall, 2008 Page 1 of 8 pages Lecture #27: Polyatomic Vibrations III: s-Vectors and H2O Last time: F-matrix: too many Fij’s even at quadratic-only level Internal coordinates: types 3N–6 independent ones constraints * translation * rotation s-vectors ∇αSt ≡ stα * direction of fastest increase    * magnitude resulting from unit displacementin optimum direction ρ ρstα α α N( )= St { } ∑ • α=1  rigid translation ρα = εfor all α constraint ∑ stα = 0 no center of mass translation α Rα× drigid rotation by dΩ ρα  )= (dΩ Ω  stα× Rα constraint ∑ stα× Rαe= 0 ECKART α (minimizes vibrational angular momentum)If the normal displacements are built from stα vectors that satisfy these constraints, then, forinfinitesimal displacements from equilibrium, there is no rotation. For large displacements, or for smalldisplacements away from a non-equilibrium configuration, there is a small vibrational angularmomentum. This definition of vibrations embeds a specific partitioning between rotation and vibration. TODAY: G from stα ’s  Examples of stα ’s 1. valence bond stretch ∆r  2. valence angle bend ∆φ G matrix using diagrams and tables from WDC pages 304 and 305H2O FG handout G ≡ DD† recall |S〉 = B|ξ〉 = D|q〉 = DΜ1/2|ξ〉B = DM1/ 2 BM−1/ 2 = D G = DD† = BM−1/ 2 (M−1/ 2 )† B† = BM−1B† St (dΩ)= d∑ Ω • α5.80 Lecture #27 Fall, 2008 Page 2 of 8 pages Gtt′ = ∑ 1 BtiB*t i ′i=13Nmi definition of |S〉 = B|ξ〉 →⎝⎜⎛∂∂ξSit ⎠⎟⎞ 0 ⎜⎝⎛∂∂ξSti ′ ⎟⎠⎞ 0 = ∑ 1 (∇αSt )0·(∇αSt′ )0 α=1Nmα N1 st ′α stα · Gtt′ = ∑ α=1 mα This way to derive G is convenient * locally defined stα . Easy to compute stα · st′α .   * Each St involves small number of stα’s (only the involved atoms). * Small number of topological cases for internal displacements. All analyzed in WDC, pages303-306. s -Vector Method. WDC pages 54-63. * start with all atoms at equilibrium positions;* direction of stα is direction of α’th atom must move to yield maximum increase in St; * magnitude of stα is increase in St that results from unit displacement of atom α in optimaldirection;* must verify or impose the 6 constraints (3 Cartesian components for the two vector constraintequations). e∑stα = 0 ∑Rα× stα = 0  αα5.80 Lecture #27 Fall, 2008 Page 3 of 8 pages Several possible types of internal displacements. 1. Bond Stretch 2. Valence angle bend3. angle between a bond and a plane (non-planar A1A2 state of H2CO) defined by 2 bonds4. torsion → trans-bent excited A1Au state of HCCH 1. Bond Stretch St ≡ ∆r e12 e12 st1 st2 r12 21 only 2 nonzero s vectors (even in a long linear chain)!Atom 1 unit displacementst1 st 2 (displacements of all other atoms have no effect on ∆r12) ρ − ρ1 2 These are the vector representations of S .∆rAre the constraints satisfied? ∑stα = st1 + st 2 = −eˆ12 + ˆ    e21 = −e21 e12 1 = − st1 = = Atom 2 st 2 = 1 =  ({ )= eˆ21} St ( ) ξα = 0! e12 α = R1e × −eˆ12( (eˆ12∑ R2e ×)+ ) eRα × stαα = R2 e − ( center of mass R2 e R1 e ) × eˆ12 eR1    e e e– = R2 R1 R12e∴ R12 × 12  e12 = 0!5.80 Lecture #27 Fall, 2008 Page 4 of 8 pages 2. Valence Angle Bend St ≡ ∆φ ∆φ φ 3 r32r31 st1 st2 12 Exactly 3 atoms are involved. 3 nonzero stα ’s.  How to move each atom to increase φ by maximum amount? tan∆ φ ≈ ∆ φ = st1 r31 How to define a UNIT VECTOR pointing in correct direction? eˆst1 = eˆ31 × eˆ31sin ×φ eˆ32 Recall eˆ31 × eˆ32 = sin φ right hand rule ⊥ to plane, up out of board Rules for vector triple product eˆ=(eˆ31·eˆcos φ 32 )eˆ31 − (eˆ31·eˆ31 1 )eˆ32 st1 sin φ eˆ= cos φeˆ31 − eˆ32 st1 sin φ  Now, how much does unit displacement of atom 1 in eˆ direction increase St?st1 tan ∆ St ≈ ∆ St = unity 1 = = st1 r31 r31 ∴ st1 = st1 eˆst1 = cos φeˆ31 − eˆ31 r31sin φ this is a vector of specifiedlength and direction1 5.80 Lecture #27 Fall, 2008 Page 5 of 8 pages similarly for atom 2 cos φeˆ32 − eˆ31=st 2 r32 sin φ now for the hard one: atom 3! Easy way: impose constraint ∑ stα = 0 α ∴st 3 = −(st1 + st 2 )=(r31 − r32 cos φ)eˆ31 +(r32 − r31cos φ)eˆ32   r31r32 sin φ Hard way: move atom 3 1 unit in optimal direction, then translate deformed molecule rigidly to putatom 3 back at its original position. This evidently leaves atoms 1 and 2 displaced by st1 and st 2 respectively.  3 2 move atom 3  st 2  st1 −  st 3  st 3 = −  translate distorted structure back to put atom 3 in original location st1 + st 2 ( ) This obviously satisfies ∑ stα = 0 α eIt is harder to show that it also satisfies 0 = ∑Rα× stα . Grind out the algebra! (see Non-Lecture on next page)5.80 Lecture #27 Fall, 2008 Page 6 of 8 pages Alternative definition of S∆θ as a linear displacement rather than an angular displacement is possible. e.g. r31∆φ, r32∆φ, or (r31r32)1/2∆φ. Then S∆φ would have dimension of length and all bending force constants would have same units asstretching ones. The derivation of S∆φ would follow same path, but each stα gets multiplied by therelevant length factor, r31 or r32 or (r31r32)1/2. NON-LECTURE Proof that stα ’s satisfy Eckart Condition cos φeˆ31 − eˆ32=st1 st 2 st 3   r31sin φ cos φeˆ32 − eˆ31= r32 sin φ (r31 − r32 cos φ)eˆ31 +(r32 − r31cos φ)eˆ32   R2 + R23 R1 + R13 R3 × st 3 st 2 = r31r32 sin φ 0 ? R1e × st1 + R2e × st 2 + Re3 ×   st 3 R3 R1 R2 R3 R3 = =    R3 − R13 R3 − R23 st 2 + s +t 2R3 × (st1 + st 3 ) R13 × st1 + R23 ×  0=ˆˆRRes××13 3123 32 = 0 = 0 0?R13 ×⎛⎜ −eˆ32 ⎞⎟− R23 ×⎛⎜ −eˆ31 ⎞ ⎝r31sin φ⎠ ⎝r32 sin φ⎠⎟ R13 × eˆ32 = r31eˆ13 × eˆ32 = −r31eˆ31 × eˆ32 R23 × eˆ31 = r32eˆ23 × eˆ31 QED 0?( ) ×