Molecular Rotation Eigenfunction

The components may be expressed in either a space-fixed axis system (p) ora molecule-fixed system (q). The early literature used cartesian coordinate systems, but for the past fifty years spherical tensors have become increasingly common. They have many advantages, chief of which is that they make maximum use of molecular symmetry. As we shall see, the rotational eigenfunctions are essentially spherical harmonics we will also find that transformations between space- and molecule-fixed axes systems, which arise when external fields are involved, are very much simpler using rotation matrices rather than direction cosines involving cartesian components. 

Equation (4-68) is the same as the 0, equations seen earlier [(4-36) and (4-39)] with p = lIE/h, and so we already know the eigenvalues and eigenfunctions. We noted earlier that p = l l + ). For molecular rotation it is conventional to symbolize the 0 quantum number as J, rather than /. This leads to... 

The theoretical method, as developed before, concerns a molecule whose nuclei are fixed in a given geometry and whose wavefimctions are the eigenfunctions of the electronic Hamiltonian. Actually, the molecular structure is vibrating and rotating and the electric field is acting on the vibration itself. Thus, in a companion work, we have evaluated the vibronic corrections (5) in order to correct and to compare our results with experimental values. 

Representation theory of molecular point groups tells us how a rotation or a reflection of a molecule can be represented as an orthogonal transformation in 3D coordinate space. We can therefore easily determine the irreducible representation for the spatial part of the wave function. By contrast, a spin eigenfunction is not a function of the spatial coordinates. If we want to study the transformation properties of the spinors... 

Although centrifugal distortion is not a perturbation effect, a derivation of the form of the centrifugal distortion terms in Heff provides an excellent illustration of the Van Vleck transformation. If the vibrational eigenfunctions of the nonrotating molecular potential, V(R) rather than [V(R) + J(J + 1)H2/2/j,R2, are chosen as the vibrational basis set, then the rotational constant becomes an operator,... 

At this point the reader may feel that we have done little in the way of explaining molecular symmetry. All we have done is to state basic results, normally treated in introductory courses on quantum mechanics, coimected with the fact that it is possible to find a complete set of simultaneous eigenfunctions for two or more commuting operators. However, as we shall see in section A 1.4.3,2. the fact that the molecular Hamiltonian //commutes with and is intimately connected to the fact that //commutes with (or, equivalently, is invariant to) any rotation of the molecule about a space-fixed axis passing through the centre of mass of the molecule. As stated above, an operation that leaves the Hamiltonian invariant is a symmetry operation of the Hamiltonian. The infinite set of all possible rotations of the ... 

The ability to assign a group of vibrational/rotational energy levels implies that the complete Hamiltonian for these states is well approximated by a zero-order Hamiltonian which has eigenfunctions /,( i)- The are product functions of a zero-order orthogonal basis for the molecule, or, more precisely, product functions in a natural basis representation of the molecular states, and the quantity m represents the quantum numbers defining tj>,. The wave functions are given by... 

The HMO coefficients can be found by solving the usual set of simultaneous equations, but it is simpler to use molecular symmetry. The Oq symmetry operator commutes with the 7T-electron Hamiltonian, so we can choose each MO to be an eigenfunction of this 60° rotation. Since (Oq) = 1, the eigenvalues of Oq are the six sixth roots of unity (Prob. 7.25) ... 

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