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Unformatted text preview: Lecture 9 The Harmonic Oscillator Shrdinger Equation for The Harmonic Oscillator and its solutions Properties of the Harmonic Oscillator September 30 2009 M&S: pp. 157173. Diatomic Molecules A diatomic moleculehas a single degree of freedom, the bond length. If the molecule is stable, the potential energy will have a minimum at what is called the equilibrium bond length, it will rise to infinity as the atoms are pushed closer and closer together (nuclear fusion!), and it will rise to a plateau as the atoms are pulled so far from one another that they no longer interact. The level of this plateau defines the zero of energy. Vibrational Motion of a Diatomic We expect there to be solutions to the timeindependent Schrdinger equation that will correspond to stationary vibrational states of the system behaving as a quantum mechanical system. That is, there will be a wave function from which the probability of finding any particular range of bond lengths can be computed. To find such stationary states, we need to solve (91) where is some sort of effective mass for the vibration, called the reduced mass It is not simply the mass of atom A + atom B: that mass would be appropriate if we were thinking of the molecule AB as a free particle translating through space, but vibration is a different kind of motion, and well get back to later. " h 2 2 d 2 dr 2 + V r ( ) # $ % & ( ) r ( ) = E ) r ( ) Bond Length and Potential (91) The one dimension we care about is the bond length r (we could as easily use x in our notation, but r is more conventional), and the potential energy V is just the curve. " h 2 2 d 2 dr 2 + V r ( ) # $ % & ( ) r ( ) = E ) r ( ) Potential Later in the class, we will get to how one might use quantum mechanics to predict the bonding potential, but for now lets just accept that it is a nice physical observable that can be measured. However, to plug it into our eq. 91, we need some convenient mathematical representation of that curve. One way to represent any function is from knowledge of that functions value, and the value of its derivatives, at a single point. We may then use a socalled Taylor expansion to compute the functions value at any other point. If we choose as our function our bond potential V, our coordinate as r , and our known point as the equilibrium bond distance, we have (92) V r ( ) = V r eq ( ) + dV dr r = r eq r " r eq ( ) + 1 2! d 2 V dr 2 r = r eq r " r eq ( ) 2 + 1 3! d 3 V dr 3 r = r eq r " r eq ( ) 3 + L Taylor Expansion (92) For the sake of simplicity, we now do several things. First, define the zero of energy to be V ( r eq ). That makes term 1 on the r.h.s. equal to zero. Next, notice that r eq defines the local minimum on the bond potential curve....
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