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Unformatted text preview: Quantum Mechanics: Vibration and Rotation of Molecules 5th April 2010 I. 1Dimensional Classical Harmonic Oscillator The classical picture for motion under a harmonic potential (mass attached to spring attached to surface; two massess connected by spring) is deter mined by solutions to Newton’s equations of motion: F = ma = m d 2 x dx 2 = dV ( x ) dx = kx where k is a force constant for the spring connecting the masses, and V ( x ) = 1 2 kx 2 is the harmonic potential (Hooke’s Law),and vector notation has been dropped for this onedimensional case. General solutions to this equation are of the form: x = x M cos ( ωt φ ) The behavior of the amplitude, or deviation from equilibrium separation, is sinusoidally varying with amplitude x M and angular frequency ω . ω is related to the mass and force constant through the relation ω = q k m . The kinetic energy of the mass is T = 1 2 m dx dt 2 = p 2 2 m with p = m dx dt being the particle momentum. The total energy is then (recalling that k = mω 2 ): E = T + V ( x ) = p 2 2 m + 1 2 kx 2 = p 2 2 m + 1 2 mω 2 x 2 1 Substituting the general solution equation discussed above, we see that the total energy is independent of time (as required for conservative sys tems ): E = 1 2 mω 2 x 2 M Thus, the energy is nonzero, continuous, and constant. The maximum dis placement is related to the energy as x M = q 2 E mω 2 . Aside: Though we will not consider the following in terms of reduced mass for an oscillator (or rigid rotor when considering angular momentum), we present here the reduced mass for reference. Reduced mass is the effective inertial mass appearing in the twobody problem of Newtonian mechanics. This is a quantity with the units of mass, which allows the twobody problem to be solved as if it were a onebody problem. Note however that the mass determining the gravitational force is not reduced. In the computation one mass can be replaced by the reduced mass, if this is compensated by replacing the other mass by the sum of both masses. Given two bodies, one with mass m 1 , and the other with mass m 2 , they will orbit the barycenter of the two bodies. The equivalent onebody problem, with the position of one body with respect to the other as the unknown, is that of a single body of mass m red = μ = 1 1 m 1 + 1 m 2 = m 1 m 2 m 1 + m 2 where the force on this mass is given by the gravitational force between the two bodies. This can be proven easily. Use Newton’s second law, the force exerted by body 2 on body 1 is F 12 = m 1 a 1 The force exerted by body 1 on body 2 is F 21 = m 2 a 2 2 According to Newton’s third law, for every action there is an equal and opposite reaction: F 12 = F 21 Therefore, m 1 a 1 = m 2 a 2 and a 2 = m 1 m 2 a 1 The relative acceleration between the two bodies is given by a...
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This note was uploaded on 02/02/2012 for the course CHEM 444 taught by Professor Dybowski,c during the Fall '08 term at University of Delaware.
 Fall '08
 Dybowski,C
 Physical chemistry, Mole, pH

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