 Linear rigid rotor

For a diatomic molecule, let $m_1,m_2$ be the two masses and $r$ their fixed separation. The distances from the centre of mass are $r_1=m_{2}r/(m_1+m_2)$ and $r_2=m_{1}r/(m_1+m_2)$. A perpendicular axis through the centre of mass therefore has moment of inertiaThe reduced mass $\mu$ here is not the electric dipole moment, for which we use $\mathbf{d}$. For a longer linear molecule, replace $\mu r^2$ by ${\sum }_i{m}_i{\rho }_i^2$, where ${\rho }_i$ is each distance along the axis from the centre of mass. In the point-mass idealization, the axial moment of inertia is zero; turning about that axis changes no nuclear positions, so no independent axial rotational degree of freedom is included.

For the field-free linear rigid rotor, assume a fixed internal quantum state with no electronic angular momentum or electronic spin, and omit couplings to nuclear spin. The rotational Hamiltonian is ${\hat{H}}_{\mathrm{rot}}={\hat{\mathbf{L}}}^2/(2I)$. Its orientation wavefunctions are spherical harmonics $Y_{JM}$, and its angular momentum eigenstates satisfywhere $J=0,1,2,\dots$ and $M=-J,-J+1,\dots ,J$. Each $E_J$ belongs to $2J+1$ angular momentum eigenstates distinguished by the magnetic quantum number $M$, an instance of degenerate energy levels. The Planck constant is $h=2\pi \hslash$; $B$ is the frequency-valued rotational constant. Notice that the energy eigenvalues scale as $J(J+1)$, not $J^2$.

Subtracting consecutive energy eigenvalues givesThus increasing level gaps produce a regularly spaced series of lines at $2B,4B,6B,\dots$, with spacing $2B$. Allowed positions do not guarantee that every line is visible: electric-dipole selection rules, occupation of the initial quantum state, and instrument response also matter.