Force Constant Matrix Calculations

Chemistry 365: Force Constant Calculations©David RonisMcGill UniversityHere is an example of a force constant matrix calculation. We will consider a diatomicmolecule, where the two atoms interact with a potential of the form:U(r 1 , r 2 ) ≡ 1 22 K⎛ ⎝ |r ⎞1 − r 2 | − R 0 ; (1)⎠i.e., a simple Hookian spring. It is easy to take the various derivatives indicated in the NormalMode Analysis handout; here, however, we will explicitly expand the potential in terms of theatomic displacements, ∆ i . Bywriting, r i = R i +∆ i (where R i is the equilibrium position of thei’th nucleus), Eq., (1) can be rewritten as:U(r 1 , r 2 ) = 1 1/222 K⎡ ⎛⎢⎝ R2 12 + 2R 12 ⋅ (∆ 1 −∆ 2 ) + |∆ 1 −∆ 2 | 2 ⎞ ⎤− R⎠ 0⎥⎦ , (2)⎣where R 12 ≡ R 1 − R 2 . Clearly, the equilibrium will have |R 12 | = R 0 . Moreover, we expect thatthe vibrational amplitude will be small, and thus, the terms in the ∆’s inthe square root in Eq. (2)will be small compared with the first term. The Taylor expansion of the square root implieswe can writewhich can be rewritten as√⎺ ⎺⎺⎺⎺ A + B ≈ √⎺⎺A +B +..., (3)2√⎺⎺AU(r 1 , r 2 ) = 1 2 K⎡ ⎢⎣R 12 + 2R 12 ⋅ (∆ 1 −∆ 2 ) + |∆ 1 −∆ 2 | 22R 12− R 0⎤⎥⎦2, (4)U(r 1 , r 2 ) = 1 2 K[ ˆR 12 ⋅ (∆ 1 −∆ 2 )] 2 , (5)where the ˆ denotes a unit vector and where all terms smaller than quadratic in the nuclear displacementshave been dropped. If the square isexpanded, notice the appearance of cross termsin the displacements of 1 and 2.It is actually quite simple to finish the normal mode calculation in this case. To do so,define the equilibrium bond to point in along the x axis. Equation (5) shows that only x-componentsof the displacements cost energy, and hence, there will no force in thy y or z directions(thereby resulting in 4 zero eigen-frequencies). For the x components, Newton’s equationsbecome:⎛ m 1⎝ 00m 2⎞⎠ ⋅⎛∆ ¨1x ⎞⎝ ⎠ =−⎛ K⎝ −K∆ x 2−KK⎞⎠ ⋅ ⎛ ⎝∆1x∆2x⎞⎠ , (6)where m i is the mass of the i’th nucleus. This in turn leads to the following characteristic equationfor the remaining frequencies:Winter Term, 2015

Chemistry 365 -2- Force Constant Calculations0 = det ⎡ ⎛ m 1⎢⎝ ⎣00 ⎞m 2 ⎠ ω 2 − ⎛ ⎝K−K−KK⎞⎤⎠⎥⎦= (m 1 ω 2 − K)(m 2 ω 2 − K) − K 2and= ω 2 (µω 2 − K),where µ ≡ m 1 m 2 /(m 1 + m 2 )isthe reduced mass. Thus we pick up another zero frequency and anonzero one with ω = √⎺ ⎺⎺ K/µ, which is the usual result. (Note that we don’t count ± rootstwice--WHY?).Winter Term, 2015