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12          The Quantized Harmonic



                        Oscillator: Vibrational

                        Spectroscopy






            INTRODUCTION
            Now that we have learned some of the principles which apply in quantum mechanics, we move on
            to the next more difficult problem of a quantized harmonic oscillator. The goal here is to provide a
            rigorous application of the polynomial method of solving differential equations on a relatively
            simple case and to provide some insight into how the Schrödinger equation was first solved [1].
            Then we proceed to application in the form of worked examples. We have tried to give sufficient
            details of this solution to allow a student to follow the derivation with pencil and paper but do not
            forget to ponder over the spectroscopic applications! This author would agree that it is more
            important to absorb the main conclusions of this material than to master the derivations. In fact,
            the highest recommendation of this author is to always ask ‘‘What does this mean?’’ and absorb the
            conclusions for a future activity called ‘‘thinking’’ rather than just memorizing facts.
              The previous particle-in-a-box (PIB) and particle-on-a-ring (POR) problems both had V ¼ 0 and
            only dealt with the kinetic energy operator. The essence of the harmonic oscillator is a parabolic
                             kx 2
                                where k is the ‘‘spring constant’’ or ‘‘restoring force constant.’’ According
            potential energy V ¼
                              2
            to  the  idea  that  a  force  is  the  negative  derivative  of  a  potential,  we  have
                   dV     2kx
                              ¼ kx so k is the proportionality factor of the force of a spring stretched
            f (x) ¼   ¼
                    dx     2
            or compressed away from x ¼ 0. The classical case can be solved easily by equating two
                                                                                 "     ! #
                                                                                    r ffiffiffiffi
                             2
                            d x                                                       k
            forces f ¼ ma ¼ m  ¼ kx. The solution from sophomore physics is x ¼ A sin    t
                            dt 2                                                      m
                                                                            "     ! #
                                                                     r ffiffiffiffi    r ffiffiffiffi
                                                                dx      k        k
            and  this  can  be  shown   by  direct  substitution.  ¼ A   cos        t   and
                                                                dt     m         m
                           "     ! #
                              r ffiffiffiffi
                                                    2
             2

            d x       k         k         k        d x
               ¼ A       sin       t ¼       x.So m    ¼ kx, Q.E.D.
            dt 2     m          m         m        dt 2
              This could have been solved to obtain the same result in more complicated ways but we want to
            build on your experience from problems in physics and we note that ‘‘A ¼ amplitude.’’
              Here we are faced with the quantum mechanical treatment of the harmonic oscillator. This
            potential does not take into account the fact that a chemical bond will eventually break when
            stretched more than about 10 Å but it does provide a very good approximation to the small
            vibrations of a bond in low energy states. An interesting fact plays a role here in that on planet
            Earth at average temperatures around 258C, almost all molecules are in very low vibrational states.
            This is fortuitous for the use of a potential which is most accurate at low energies. We see in
            Figure 12.1 a sketch of a parabolic potential superimposed on a model of a dissociative bond
            potential. Here the value of the potential is a minimum at r ¼ r 0 and the energy is negative for a
            ‘‘bound’’ state but we will measure the energy as positive above the minimum of the parabolic
            potential. We see that the parabolic potential makes a good fit to a real potential for a considerable
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