In quantum mechanics, a rotational transition is an abrupt change in angular momentum. Like all other properties of a quantum particle, angular momentum is quantized, meaning it can only equal certain discrete values, which correspond to different rotational energy states. When a particle loses angular momentum, it is said to have transitioned to a lower rotational energy state. Likewise, when a particle gains angular momentum, a positive rotational transition is said to have occurred. Rotational transitions are important in physics due to the unique spectral lines that result. Because there is a net gain or loss of energy during a transition, electromagnetic radiation of a particular frequency must be absorbed or emitted. This forms spectral lines at that frequency which can be detected with a spectrometer, as in rotational spectroscopy or Raman spectroscopy.
Diatomic molecules Molecules have rotational energy owing to rotational motion of the nuclei about their center of mass. Due to quantization, these energies can take only certain discrete values. Rotational transition thus corresponds to transition of the molecule from one rotational energy level to the other through gain or loss of a photon. Analysis is simple in the case of diatomic molecules.
Nuclear wave function Quantum theoretical analysis of a molecule is simplified by use of Born–Oppenheimer approximation. Typically, rotational energies of molecules are smaller than electronic transition energies by a factor of m/M ≈ 10−3–10−5, where m is electronic mass and M is typical nuclear mass. From uncertainty principle, period of motion is of the order of the Planck constant h divided by its energy. Hence nuclear rotational periods are much longer than the electronic periods. So electronic and nuclear motions can be treated separately. In the simple case of a diatomic molecule, the radial part of the Schrödinger Equation for a nuclear wave function Fs(R), in an electronic state s, is written as (neglecting spin interactions)
[ − ℏ 2 2 μ R 2 ∂ ∂ R ( R 2 ∂ ∂ R ) + ⟨ Φ s | N 2 | Φ s ⟩ 2 μ R 2 + E s ( R ) − E ] F s ( R ) = 0 {\displaystyle \left[-{\frac {\hbar ^{2}}{2\mu R^{2}}}{\frac {\partial }{\partial R}}\left(R^{2}{\frac {\partial }{\partial R}}\right)+{\frac {\langle \Phi _{s}|N^{2}|\Phi _{s}\rangle }{2\mu R^{2}}}+E_{s}(R)-E\right]F_{s}(\mathbf {R} )=0}
where μ is reduced mass of two nuclei, R is vector joining the two nuclei, Es(R) is energy eigenvalue of electronic wave function Φs representing electronic state s and N is orbital momentum operator for the relative motion of the two nuclei given by
N 2 = − ℏ 2 [ 1 sin Θ ∂ ∂ Θ ( sin Θ ∂ ∂ Θ ) + 1 sin 2 Θ ∂ 2 ∂ Φ 2 ] {\displaystyle N^{2}=-\hbar ^{2}\left[{\frac {1}{\sin \Theta }}{\frac {\partial }{\partial \Theta }}\left(\sin \Theta {\frac {\partial }{\partial \Theta }}\right)+{\frac {1}{\sin ^{2}\Theta }}{\frac {\partial ^{2}}{\partial \Phi ^{2}}}\right]}
The total wave function for the molecule is
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