The Two-Orbital Mixing Problem

There is another way to solve the two-orbital mixing problem, which does not involve all the calculus shown above. It is a general approach that results from the variational theorem, and can be applied to ab initio, semi-empirical, or Hiickel theory. 

In the two-orbital mixing problem, we showed that when molecular orbitals are defined as linear combinations of atomic orbitals and are put into the Schrbdinger equation, followed by differentiation to minimize E, a series of simultaneous equations in the c s and E results. When mixing two orbitals, only two energies result along with two molecular orbitals. It is not much of a stretch to realize that there will be as many molecular orbitals with distinct energies as the number of atomic orbitals (or basis functions) we use to create the molecular orbitals. Moreover, there will be the same number of secular equations as the number of starting atomic orbitals or basis functions ( ), and hence the secular determinant will be n by n. 

It is actually easier to write the secular determinant first, and then create the secular equations (although we saw in the "Two-Orbital Mixing Problem" that they are derived in the opposite order). All secular determinants have the following form ... 

Putting this all in the context of only two adjacent carbon p orbitals gives the Huckel secular determinant for ethylene shown below. Compare this to the secular determinant for the "Two-Orbital Mixing Problem" given in Section 14.2.2. 

We can perform all the analysis in a manner similar to how the two-orbital mixing problem was addressed (Section 14.2.2). Alternatively, we can use some of the logic we have developed to this point in the chapter, and some of the rules of HMOT, to derive the answer. We choose the latter method here. 

Because the mapping approach treats electronic and nuclear dynamics on the same dynamical footing, its classical limit can be employed to study the phase-space properties of a nonadiabatic system. With this end in mind, we adopt a onemode two-state spin-boson system (Model IVa), which is mapped on a classical system with two degrees of freedom (DoF). Studying various Poincare surfaces of section, a detailed phase-space analysis of the problem is given, showing that the model exhibits mixed classical dynamics [123]. Furthermore, a number of periodic orbits (i.e., solutions of the classical equation of motion that return to their initial conditions) of the nonadiabatic system are identified and discussed [125]. It is shown that these vibronic periodic orbits can be used to analyze the nonadiabatic quantum dynamics [126]. Finally, a three-mode model of nonadiabatic photoisomerization (Model III) is employed to demonstrate the applicability of the concept of vibronic periodic orbits to multidimensional dynamics [127]. 

The most simple and well known equivalence holds for the special case of two electrons in two orbitals, described by GVB(pp) and CASSCF(2,2) wave functions for covalent bonds [41]. When the bond has a mixed ionic-covalent character, the CASSCF(2,2) description is able to account approximately for this effect due to the presence of an extra ionic configuration. This problem is exactly replicated in a larger scale when... 

The molecular orbital answer to this problem, as you may well know, is that all six p orbitals can combine to form (six) new molecular orbitals, one of which (the one lowest in enery ) consists of a ring of electron density above and below the plane of the molecule. Benzene does not resonate between the two Kekuld structures—the electrons are in molecular orbitals spread equally over all the carbon atoms. However the term resonance is still sometimes used (but not in this book) to describe this mixing of molecular orbitals. 

Now, combining the excited state of one alkene with the ground state of another solves the symmetry problem. Mixing the two 7t orbitals leads to two molecular orbitals and two electrons go down in energy while only one goes up. 

Now let the bands interact. The bands repel each other as they did in the H problem of Exercise 6.3. They mix in such a way that the lower band (valence band) will have dominant H character and the higher band (conduction band) dominant Li character. At k= ir/d, there is no interaction by symmetry. Unlike in the regular H problem these two orbitals are no longer degenerate and their linear combinations are not... 

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