Rigid Rotor

The rigid rotor is the quantum model of a system whose distance from a fixed center is frozen but whose orientation can change. It is the angular part of a free particle constrained to a sphere and the first model behind rotational spectra of simple molecules. The one-angle predecessor is Particle on a Ring, and the geometry-first sphere model is Particle on a Sphere.

For a diatomic molecule in the rigid-rotor approximation, the bond length is treated as fixed at $R_e$. After separating center-of-mass motion, the relative coordinate has fixed length and only its direction changes. The moment of inertia is

$$I=\mu R_e^2,$$

where $\mu$ is the reduced mass. The approximation ignores vibration, centrifugal stretching, electronic structure, spin couplings, and nuclear exchange effects. Those refinements matter in molecular spectroscopy, but the rigid rotor isolates the angular kinetic energy cleanly.

Helpful background. Orbital Angular Momentum to use orbital angular-momentum eigenvalues.

Configuration Space and Hilbert Space

The orientation of a linear rotor is a point on the unit sphere:

$$\Omega=(\theta,\phi), \qquad 0\le\theta\le\pi, \qquad 0\le\phi\lt2\pi.$$

The Hilbert space is $L^2(S^2,d\Omega)$, with

$$d\Omega = \sin\theta\,d\theta\,d\phi.$$

A rotor wavefunction $\Psi(\theta,\phi)$ is normalized by

$$\int_0^{2\pi} \int_0^\pi \lvert\Psi(\theta,\phi)\rvert^2 \sin\theta\,d\theta\,d\phi =1.$$

This measure is not optional. It is the surface-area measure on the sphere and is the reason spherical harmonics are the natural basis.

Hamiltonian

Classically, a linear rigid rotor has rotational kinetic energy

$$T = \frac{L^2}{2I}.$$

Quantization replaces $L^2$ by the orbital angular-momentum operator on the sphere:

$$\hat H = \frac{\hat L^2}{2I}.$$

Using the angular Laplacian $\Delta_{S^2}$,

$$\hat L^2 = -\hbar^2\Delta_{S^2},$$

so in spherical coordinates

$$\hat H = - \frac{\hbar^2}{2I} \left[ \frac{1}{\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta \frac{\partial}{\partial\theta} \right) + \frac{1}{\sin^2\theta} \frac{\partial^2}{\partial\phi^2} \right].$$

There is no radial equation because the radius has been frozen. The problem is a pure angular eigenvalue problem.

Eigenstates

The normalized eigenfunctions are spherical harmonics:

$$\Psi_{JM}(\theta,\phi) = Y_J^M(\theta,\phi),$$

where this page uses the molecular-rotor notation

$$J=0,1,2,\ldots, \qquad M=-J,-J+1,\ldots,J .$$

The same functions are often written $Y_\ell^m$ in orbital angular momentum and central-potential problems. The notation changes because molecular spectroscopy conventionally uses $J$ for rotational angular momentum.

The eigenvalue equations are

$$\hat L^2Y_J^M = \hbar^2J(J+1)Y_J^M,$$

and

$$\hat L_zY_J^M = \hbar M Y_J^M.$$

Thus the rigid rotor is an exactly solvable angular-momentum system: the Hamiltonian is proportional to the angular-momentum Casimir operator.

Energy Levels and Degeneracy

The energy levels are

$$E_J = \frac{\hbar^2}{2I}J(J+1), \qquad J=0,1,2,\ldots .$$

It is common to define the rotational constant in energy units by

$$B = \frac{\hbar^2}{2I},$$

so that

$$E_J=BJ(J+1).$$

For fixed $J$, the quantum number $M$ has $2J+1$ possible values. In the absence of external fields or anisotropic interactions, all of these states are degenerate:

$$g_J=2J+1.$$

This degeneracy follows from rotational invariance. The energy depends on the total angular momentum magnitude, not on the orientation of that angular momentum with respect to an arbitrary chosen $z$ axis.

The ground state has $J=0$ and energy $E_0=0$ in this ideal model. This is not a contradiction with zero-point motion in the harmonic oscillator. The free rotor has no confining angular potential; its lowest state is the constant wavefunction on the sphere, with no angular variation and hence no rotational kinetic energy.

Physical Interpretation

The symbol $B$ is used for three dimensionally different rotational constants. Keeping a subscript until units are fixed prevents many numerical errors:

Convention	Definition	Spectrum
Energy	$B_E=\hbar^2/(2I)$	$E_J=B_EJ(J+1)$
Frequency	$B_f=B_E/h=h/(8\pi^2I)$	$E_J/h=B_fJ(J+1)$
Wavenumber	$\widetilde B=B_E/(hc)=h/(8\pi^2Ic)$	$E_J/(hc)=\widetilde BJ(J+1)$

A tabulated value in $\mathrm{cm}^{-1}$ is $\widetilde B$, not an energy. The rotational temperature

$$\theta_{\mathrm{rot}}=\frac{B_E}{k_B}$$

compares the level scale with thermal energy.

The rotor levels grow as $J(J+1)$ rather than linearly in $J$. Neighboring level spacings are

$$E_{J+1}-E_J = B[(J+1)(J+2)-J(J+1)] = 2B(J+1).$$

Thus adjacent spacings increase with $J$ in energy units. In rotational spectroscopy, one often measures transition frequencies or wavenumbers rather than absolute energies. For a polar linear molecule with electric-dipole rotational transitions, the dominant selection rule is

$$\Delta J=\pm1,$$

with the allowed $\Delta M$ values determined by the polarization and quantization axis. This produces a ladder of spectral lines whose ideal spacings are controlled by $I$.

For absorption from $J$ to $J+1$,

$$\nu_{J\to J+1}=2B_f(J+1), \qquad \widetilde\nu_{J\to J+1}=2\widetilde B(J+1).$$

The lines are equally spaced by $2B_f$ or $2\widetilde B$, although the energy levels are not equally spaced. Homonuclear diatomic molecules have no permanent electric dipole in their equilibrium electronic state, so the ideal electric-dipole rule does not by itself predict observable pure rotational lines for them.

The moment of inertia carries the physical scale. Larger $I$ means slower rotational motion and smaller level spacings. For a diatomic molecule,

$$I=\mu R_e^2,$$

so heavier reduced mass or longer bond length compresses the rotational spectrum.

At fixed bond length, two isotopologues approximately obey

$$\frac{\widetilde B'}{\widetilde B}\simeq\frac{\mu}{\mu'},$$

up to isotope-dependent vibrational averaging and Born–Oppenheimer corrections.

Thermal Populations and the Nonrigid Correction

For a directed linear rotor (for example, a heteronuclear diatomic) with no additional spin factors, the rotational partition sum is

$$Z_{\mathrm{rot}} =\sum_{J=0}^{\infty}(2J+1) e^{-B_EJ(J+1)/(k_BT)}.$$

The degeneracy factor is essential. In the high-temperature classical regime,

$$Z_{\mathrm{rot}}\simeq\frac{T}{\theta_{\mathrm{rot}}}.$$

For indistinguishable orientations, the classical count is divided by the molecular symmetry number $\sigma$. This is a different count from the unrestricted quantum sum above. In a homonuclear molecule, nuclear-spin statistics select and weight even/odd $J$ sectors and must be included explicitly.

A general three-dimensional rigid body has configuration space $SO(3)$ and an additional orientation angle. Its symmetric- and asymmetric-top spectra are developed in Rotations of Molecules; the $S^2$ model here is the linear rotor.

The leading centrifugal-distortion correction is commonly written

$$E_J\simeq B_EJ(J+1)-D_EJ^2(J+1)^2, \qquad D_E\gt0.$$

It is a controlled warning about the rigid approximation, not part of the ideal rotor Hamiltonian.

Relation to Other Angular Problems

The Particle on a Ring: First Encounter has one angular coordinate and eigenfunctions $e^{in\phi}$. The rigid rotor has two angular coordinates and eigenfunctions $Y_J^M(\theta,\phi)$. In both cases, periodicity and geometry quantize angular motion.

The hydrogen atom uses the same spherical harmonics, but there they multiply a nontrivial radial wavefunction. In the rigid rotor, the radius is frozen, so the spherical harmonic is the whole configuration-space wavefunction.

This is why the rigid rotor is a clean bridge between spherical harmonics as special functions and orbital angular momentum as a quantum symmetry.

Common Mistakes

• Forgetting the $\sin\theta$ factor in normalization integrals.
• Using half-integer $J$ values for an ordinary scalar rotor. Half-integer labels belong to spinor representations, not single-valued scalar wavefunctions on the sphere.
• Confusing the rotational constant $B$ with a magnetic field.
• Treating $M$ as changing the energy of a free isotropic rotor.
• Assuming all molecules show the ideal pure rotational dipole spectrum. Homonuclear diatomic molecules lack a permanent electric dipole, and real spectra include additional selection rules and corrections.
• Interpreting a stationary rotor eigenfunction as a classical spinning rod with a definite orientation.

Exercises

1. List the allowed $M$ values and degeneracy for the $J=3$ level.

Solution

For fixed $J$, the allowed values are

$$M=-J,-J+1,\ldots,J.$$

For $J=3$, this gives

$$M=-3,-2,-1,0,1,2,3.$$

There are $2J+1=7$ states, so the degeneracy is $7$ in the free isotropic rotor.

2. Show that the energy difference between adjacent rotor levels is $2B(J+1)$.

Solution

Using $E_J=BJ(J+1)$,

$$\begin{aligned} E_{J+1}-E_J &= B(J+1)(J+2)-BJ(J+1) \\ &= B(J+1)[(J+2)-J] \\ &= 2B(J+1). \end{aligned}$$

3. A diatomic molecule has moment of inertia $I=\mu R_e^2$. If $R_e$ is doubled while $\mu$ is unchanged, how do the rotational energies change?

Solution

Doubling $R_e$ changes the moment of inertia to

$$I' = \mu(2R_e)^2 = 4I.$$

Since

$$E_J = \frac{\hbar^2}{2I}J(J+1),$$

all rotational energies and adjacent transition energies are divided by $4$.

References

• B. H. Bransden and C. J. Joachain, Physics of Atoms and Molecules, 2nd ed., Pearson, 2003.
• J. M. Brown and A. Carrington, Rotational Spectroscopy of Diatomic Molecules, Cambridge University Press, 2003.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• G. Herzberg, Molecular Spectra and Molecular Structure I: Spectra of Diatomic Molecules, 2nd ed., Van Nostrand, 1950.
• R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.