Linear rigid rotor ($E_J={\hslash }^{2}J(J+1)/(2I)$)

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.

For a linear rigid rotor, rotational temperature expresses the rotational energy constant in kelvins: $E_J=k_B{\theta }_{r}J(J+1)$, where $I$ is the perpendicular moment of inertia. With all integer $J\ge 0$ allowed and degeneracies $2J+1$, its rotational canonical partition function is $Z={\sum }_J(2J+1)e^{-{\theta }_{r}J(J+1)/T}$. It approaches $1+3e^{-2{\theta }_r/T}$ at low temperature and $T/{\theta }_r$ at high temperature. The latter limit gives mean rotational energy $k_{B}T$ and heat capacity $k_B$ per rotor.

In this article the rotational constant is a frequency,Its dimensions are ${\mathrm{s}}^{-1}$, since the Planck constant has dimensions $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ and a moment of inertia has dimensions $\mathrm{kg}{\mathrm{m}}^2$. Some authors instead call $hB={\hslash }^2/(2I)$ the energy-valued rotational constant; this is the convention in the earlier synthetic molecular beam electric resonance diagram. A third convention uses the spectroscopic wavenumber $\tilde{B}=B/c$, with $c$ the speed of light. Always check units before using $\nu =2B(J+1)$.

If $B$ is known and the two masses are specified, the rigid two-atom model gives the bond length $r=\sqrt{h/(8{\pi }^2\mu B)}$. This is why rotational spectroscopy can constrain molecular geometry. A real fitted constant can depend on molecular vibration, so this inference alone does not establish an equilibrium bond length.

For the field-free, spin-free linear rigid rotor in one fixed internal quantum state, the electric-dipole interaction is $-\mathbf{d}\cdot \mathbf{E}(t)$. A nonzero permanent electric dipole moment along the molecular axis allows pure rotational photon absorption with $\Delta J=+1$ and photon emission with $\Delta J=-1$. The matrix element between initial and final quantum states must be nonzero: the existence of an energy gap alone is insufficient. Relative to a chosen $z$ axis, polarization of an electromagnetic wave along $z$ selects $\Delta M=0$; transverse polarization of an electromagnetic wave permits $\Delta M=\pm 1$.^[ref]

Hydrogen chloride and carbon monoxide are examples with a permanent electric dipole moment. By symmetry, molecular hydrogen and carbon dioxide have no permanent electric dipole moment in the ideal symmetric internal state. They still rotate and have rotational energy eigenvalues, but the simple permanent-electric dipole spectrum described here is absent. This restriction concerns this interaction and these assumptions, not every possible way of detecting molecular rotation.

Changing an isotope changes a mass and therefore the moment of inertia. If the bond length is held fixed, $B\propto 1/\mu$ and every corresponding transition shifts by the same fractional amount. This leading inverse-reduced mass scaling is also used in the more complete diatomic formulation.^[ref]

For a deliberately stipulated carbon monoxide example, take $m_{12}=12u$, $m_{13}=13u$, $m_{16}=16u$, $u=1.66053906660\times 10^{-27}\mathrm{kg}$, $r=1.13\times 10^{-10}\mathrm{m}$ for both isotopic forms, and $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$. The integer masses are simplifying inputs, not accurate isotopic atomic masses; the shared bond length is an imposed model assumption. Calculation givesIn particular,The heavier form's calculated $J=0\to 1$ line is $110.363\mathrm{GHz}$ rather than $115.438\mathrm{GHz}$, a decrease of about $4.396\%$. These are model predictions from the stated inputs, not published measured frequencies. An isotope changes the spacing as well as the first line; the shift is not a fixed additive offset for the whole spectrum.