Rigid rotor

A rigid rotor models rotational motion with fixed internal separations. Its use in the synthetic Stark effect figure omits vibrational motion and additional interactions in real molecules.

Rigidity fixes the molecular geometry, not the orientation, and does not require a spherical mass distribution. Remove motion of the centre of mass, choose principal axes of the inertia tensor, and write the classical rotational kinetic energy asHere $I_a,I_b,I_c$ are the principal moments of inertia, ${\omega }_a,{\omega }_b,{\omega }_c$ components of angular velocity, and $L_a,L_b,L_c$ components of angular momentum. Quantization replaces these last components by angular momentum operators. Different principal moments of inertia produce different spectra: the single-constant linear rigid rotor is a special case, not a formula for every molecule.^[ref]

For a rigid heteronuclear diatomic rotor, $E_J={\hslash }^{2}J(J+1)/(2I)$ and electric-dipole absorption has $\Delta J=+1$. Therefore the $J^{\prime }\to J^{\prime }-1$ line frequency is $h{J}^{\prime }/(4{\pi }^{2}I)$ and adjacent absorption lines are spaced by $h/(4{\pi }^{2}I)$. With reduced mass $\mu$, the bond-length estimate is $r=\sqrt{I/\mu }$. The conversion assumes frequency rather than angular frequency; centrifugal distortion makes real high-J spacings imperfectly uniform.

A rigid rotor restricted to one angular coordinate $\varphi$. Single-valued spatial states $e^{im\varphi }$ have $m\in \mathbb{Z}$ and energy $E_m={\hslash }^2{m}^2/(2I)$, with moment of inertia $I$. The zero-angular-momentum level has state degeneracy one; every nonzero $|m|$ has two directions. For labelled molecular ends its high-temperature rotational partition function is $\sqrt{\pi T/{\theta }_{\mathrm{rot}}}$, where ${\theta }_{\mathrm{rot}}={\hslash }^2/(2I{k}_{\mathrm{B}})$. Equivalent orientations and nuclear spin statistics require a further molecular counting convention. One quadratic angular momentum gives rotational heat capacity $k_{\mathrm{B}}/2$ per molecule, rather than $k_{\mathrm{B}}$ for an unrestricted linear rigid rotor.

For a linear rigid rotor with moment of inertia $I$, the energy eigenvalue at angular momentum quantum number $J$ is ${\hslash }^{2}J(J+1)/(2I)$. Each level has statistical weight $2J+1$ before additional symmetry or nuclear-spin restrictions.

The angular momentum associated with end-over-end molecular rotation. In an open-shell diatomic molecule it is useful to distinguish $\mathbf{N}$ from the total $\mathbf{J}=\mathbf{N}+\mathbf{S}$, which also contains the electronic spin angular momentum.

The rotational symmetry number corrects a rotational phase-space count for indistinguishable molecular orientations. A linear molecule with identical ends, such as carbon dioxide with identical oxygen isotopes, has $\sigma =2$; making its ends isotopically distinct gives $\sigma =1$. In the high-temperature linear rigid rotor approximation, $q_{\mathrm{rot}}=8{\pi }^{2}IkT/(\sigma h^2)$. Holding the spectrum otherwise fixed, doubling $\sigma$ lowers molar rotational entropy by $R\log 2$.

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.

Real chemical bonds can stretch as molecular rotation increases. The resulting larger moment of inertia lowers the energy eigenvalues relative to a rigid rotor. Within one vibrational quantum state, a leading effective correction isBoth $B_v$ and $D_v$ are in frequency units here; $v$ labels the vibrational quantum state. For positive $D_v$, the adjacent-line spacing becomesso the departures grow with $J$. Molecular vibration also changes the effective rotational constant: even the lowest vibrational quantum state does not have motionless atomic nuclei.^[ref]

The quartic correction is a limited expansion, not a model to extrapolate to arbitrary $J$. It is small relative to the rigid contribution only while $(D_v/B_v)J(J+1)\ll 1$; eventually higher terms, changed internal motion, or chemical bond breaking matter. In the synthetic plot, the largest upper-state label is $J=5$, giving a stipulated correction ratio of $0.03$. No universal maximum valid $J$ follows without a particular molecule and precision target.

The quantum numbers also have limits of meaning. Here $J$ describes spin-free rotation. In a molecule with electronic spin, rotation alone is commonly labelled $N$, while $J$ includes electronic angular momentum; including nuclear spin introduces additional total-angular momentum labels and resolved splittings. An applied electric field can mix different field-free $J$ states through the Stark effect. For identical atomic nuclei, the Pauli exclusion principle can restrict combinations of rotational and nuclear-spin states. The simple $2J+1$ degeneracy and every-integer-$J$ ladder therefore describe the stated idealization, not all observed sublevels.^[ref]