Molecules with Degenerate Vibrations

If a molecule has a doubly degenerate vibrations they have the same frequencies $\omega_1=\omega_2=\omega_i$ and the formula for the term values can be written as

$$G(v_1, v_2) = \omega_1(v_1 + \frac{1}{2}) + \omega_2(v_2 + \frac{1}{2})=\omega_i(v_i +1),$$ (15)

where $v_i=v_1+v_2$ can be treated as a new vibrational quantum number.

The corresponding total vibrational eigenfunction can be written as (see eq. (14))

$$\Phi_i = N_{vi}e^{-\frac{\alpha_i}{2}(Q_1^2+Q_2^2)} H_{v1}(\sqrt{\alpha_i}Q_1)H_{v2}(\sqrt{\alpha_i}Q_2),$$ (16)

where $\alpha_i = \omega_i/h$.

If $v_1=v_2=v_i=0$, than $H_0(\sqrt{\alpha_i}Q) = constant$ and there is only one function $\Phi_i$ in eq. (16). Thus, the zero-point vibrational $v_i=0$ does not introduce a degeneracy. In this case, the same relations apply as in the previous section.

If the degenerate vibration if excited by only one quantum, we have either $v_1=1, v_2=0$, or $v_1=0, v_2=1$ for which the wavefunctions $\Phi_i$ in eq. (16) are not the same. That is, there are two eigenfunctions for the state $v_i=v_1+v_2=1$ with the energy $G_{vi}= 2\omega_i$, see eq. (15). Therefore, the state $v_i$ is doubly degenerate. Note, that any linear combination of the two wavefunctions in eq. (16) is also an eigenfunction of the same energy level.

If two quanta are excited ($v_i=2$), we may have either $v_1=2, v_2=0$, or $v_1=1, v_2=1$, or $v_1=0, v_2=2$, that is there is a triple degeneracy. In general, the degree of degeneracy if $v_i$ quanta of the double degenerate vibration are excited, is equal to $v_i+1$.

Important result: Total vibrational eigenfunctions, corresponding to a degenerate vibration are neither symmetric, nor antisymmetric, but can in general be transformed under a symmetry operation as a linear combination of each other. However, there is only one zero-point $v=v_1=v_2=0$ vibration wavefunction $\Phi_0$ which must be either symmetric, or antisymmetric under a symmetry operation.

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