Degenerate vibrations vibrational angular momentum

Asymmetric tops have no degenerate vibrational modes (this will be shown in Chapter 9) hence we need not worry about vibrational angular momentum for them. 

There are several important effects associated with degenerate vibrational levels. We begin by considering the doubly degenerate v2a and v2b vibrational modes of a linear triatomic molecule. In these modes, the three atoms each vibrate in a plane perpendicular to the molecular axis, the vibrations being either in the xz or the yz plane, where the z axis is the molecular axis (Fig. 6.2). Classically, if both these modes are excited, the vibrations may give rise to vibrational angular momentum about the internuclear axis for example, if v2a and v2b are of equal amplitude and differ in phase by 90°, then the resultant motion of each nucleus is a circle about the molecular axis, as shown in Fig. 6.8 (see Problem 6.18). 

Quantum mechanically, there is also the possibility of nuclear vibrational angular momentum for these degenerate modes. We denote the normal coordinates for the doubly degenerate vibrations by Qx and Qy. From Fig. 6.2, we have... 

The functions (6.85) each correspond to a specific pair of the normal-mode vibrational quantum numbers v2a and v2b, whereas the correct zeroth-order functions (6.87) do not. Hence the quantum numbers v2a and v2b are not particularly physically significant. The quantum numbers of physical significance are v2 = v2a -I- v2b, which (together with c, and t>3) specifies which degenerate vibrational level we are dealing with, and the vibrational angular-momentum quantum number. 

A linear molecule in a vibrational level with vibrational angular momentum quantum number / can be viewed as a symmetric top with Ih replacing K h as the absolute value of the component of nuclear angular momentum about the symmetry axis. Replacing A with / in (5.67), we have as the rotational energy of a linear molecule for a degenerate vibrational level... 

S. A. Rice Have you studied any excited ion states with degenerate vibrations to ascertain the coupling of the vibrational angular momentum to the electron angular momentum ... 

This term describes the rotational Zeeman effect, that is, the coupling between the external field and the magnetic moment of the rotating nuclei. We note that there is no corresponding vibrational contribution since R a k is zero. The physical reason for this lack is that it is not possible to generate vibrational angular momentum in a diatomic molecule because it possesses only one, non-degenerate, vibrational mode. 

The normal vibrations of H3 are shown in Fig. 1. The totally symmetric Vj mode is infrared inactive, and the doubly degenerate mode is infrared active. The doubly degenerate mode has unit vibrational angular momentum (fj = — 1) as initially shown by Teller. The vibrational states of H3 are specified by two vibrational quantum numbers and and the vibrational angular momentum quantum number 1. The vibrational energy structure of H3 relevant for astronomical observations is shown in Fig. 2. The transitions observed in the laboratory are shown by upward-pointing arrows, while the emissions observed in astronomical objects are shown by bold downward-pointing arrows. The numbers in parentheses are 111. 

A discussion of molecular magnetic moments due to vibrational angular momentum in molecules with degenerate vibrational states has been given by Moss, R. E., Perry, A. J. Mol. Phys. 25, 1121 (1973). 

CO2 Laser A section of the level diagram is illustrated in Fig. 5.31. The vibrational levels (i>i, I s) are characterized by the number of quanta in the three normal vibrational modes. The upper index of the degenerate vibration V2 gives the quantum number of the corresponding vibrational angular momentum / [5.45]. Laser oscillation is achieved on many rotational lines within two vibrational transitions (vi, =... 

For a symmetric rotor, in the present approximation, only the z component of in, the vibrational angular momentum, needs to be considered. The problem may be treated as a perturbation employing zero-order wave functions which are products of rigid rotor and harmonic oscillator functions. When the molecule is in a state such that vka + Vkb — 1, where Qka and Qw> are degenerate, it is necessary to solve the secular determinant... 

For example, molecules that belong to low-symmetry point groups without any degenerate representations cannot have electronic or vibrational angular momentum, and therefore the photon angular momentum must be absorbed by the rotations, so A/ = 1 in that case. 

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