Lec9 - Chem 3502/5502 Physical Chemistry II (Quantum...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2010 Laura Gagliardi Lecture 9, September 29, 2010 The Harmonic Oscillator Consider a diatomic molecule. Such a molecule has 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 (see figure below).
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9-2 If we consider the vibrational motion of the molecule, 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 2 2 μ d 2 dr 2 + V r ( ) Ψ r ( ) = E Ψ r ( ) (9-1) where μ is some sort of effective mass for the vibration (called the “reduced mass”; it’s 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), 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 in the above figure. 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. We could somehow develop a spline fit or other elegant form, but that might make the solution of the differential equation quite difficult. For the moment, let’s keep things really simple. 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 V r ( ) = V r eq ( ) + dV dr r = r eq r r ( ) + 1 2! d 2 V dr 2 r = r eq r r ( ) 2 + 1 3! d 3 V dr 3 r = r eq r r ( ) 3 + (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. So, now the second term on the r.h.s. is zero. The next term, the quadratic term, is not zero. So, let’s throw away all
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This note was uploaded on 04/06/2011 for the course CHEM 3502 taught by Professor Staff during the Fall '08 term at Minnesota.

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Lec9 - Chem 3502/5502 Physical Chemistry II (Quantum...

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