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lecture09_umn - Lecture 9 The Harmonic Oscillator Shrdinger...

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Lecture 9 The Harmonic Oscillator Shrödinger Equation for The Harmonic Oscillator and its solutions Properties of the Harmonic Oscillator September 30 2009 M&S: pp. 157-173.
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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.
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Vibrational Motion of a Diatomic We expect there to be solutions to the time-independent Schrödinger 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 (9-1) where μ 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 we’ll get back to μ later. " h 2 2 μ d 2 dr 2 + V r ( ) # $ % & ( ) r ( ) = E ) r ( )
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Bond Length and Potential (9-1) 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 μ d 2 dr 2 + V r ( ) # $ % & ( ) r ( ) = E ) r ( )
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Potential Later in the class, we will get to how one might use quantum mechanics to predict the bonding potential, but for now let’s just accept that it is a nice physical observable that can be measured. However, to plug it into our eq. 9-1, we need some convenient mathematical representation of that curve. One way to represent any function is from knowledge of that function’s value, and the value of its derivatives, at a single point. We may then use a so-called Taylor expansion to compute the function’s 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 (9-2) 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
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Taylor Expansion (9-2) 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. Since it is a critical point on the curve, we know that the first derivative of V must be zero at that point.
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