In vibrational spectroscopy, transitions are observed between different vibrational states. In a fundamental vibration, the molecule is excited from its ground state (v = 0) to the first excited state (v = 1). The symmetry of the ground-state wave function is the same as that of the molecule. It is, therefore, a basis for the totally symmetric representation in the point group of the molecule. It follows that, for a vibrational transition to be allowed, the symmetry of the excited state wave function must be the same as the symmetry of the transition moment operator.
In infrared spectroscopy, the transition moment operator transforms as either x and/or y and/or z. The excited state wave function must also transform as at least one of these vectors. In Raman spectroscopy, the operator transforms as one of the second-order terms in the right-most column of the character table, below.
The molecule methane, CH, may be used as an example to illustrate the application of these principles.  The molecule is tetrahedral and has T symmetry. The vibrations of methane span the representations A + E +  2T. Examination of the character table shows that all four vibrations are Raman-active, but only the T vibrations can be seen in the infrared spectrum.
In the harmonic approximation, it can be shown that overtones are forbidden in both infrared and Raman spectra. However, when anharmonicity is taken into account, the transitions are weakly allowed.
In Raman and infrared spectroscopy, the selection rules predict certain vibrational modes to have zero intensities in the Raman and/or the IR. Displacements from the ideal structure can result in relaxation of the selection rules and appearance of these unexpected phonon modes in the spectra. Therefore, the appearance of new modes in the spectra can be a useful indicator of symmetry breakdown.