Rotational Symmetry of the Rigid Rotor

The rigid rotor is the pure angular-motion problem. For an oriented linear rotor, such as an ideal heteronuclear diatomic molecule, the radius or bond length is fixed, so the only remaining degrees of freedom are the orientation angles on a sphere. The Hamiltonian is proportional to the angular-momentum Casimir:

$$H_{\rm rot} = \frac{L^2}{2I}.$$

This makes the rotor a clean model of how rotational symmetry organizes quantum states. The full wave-mechanics model, normalization, molecular interpretation, and spectroscopy details live in Rigid Rotor in the Wave Mechanics volume. This page focuses on the symmetry content.

We assume a fixed moment of inertia $I$ and no external field. Symmetric and asymmetric tops have additional orientation variables and different Hamiltonians; the two-angle model here describes a linear rotor.

Configuration Space

An oriented linear rotor has a distinguishable head and tail, so its orientation is a point on the sphere:

$$\Omega=(\theta,\phi)\in S^2.$$

The Hilbert space is

$$L^2(S^2,d\Omega),$$

with

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

The wavefunction $\Psi(\theta,\phi)$ is a scalar function on the sphere. For this ideal scalar rotor, ordinary single-valuedness on $S^2$ implies integer angular-momentum labels. For identical nuclei, reversing the molecular axis exchanges the nuclei. The allowed rotational states must then be combined with nuclear spin and other internal states to satisfy exchange symmetry; one cannot assume that every $J$ occurs with the same statistical weight.

Hamiltonian as a Casimir

The rotational kinetic energy of an isotropic linear rotor is

$$H_{\rm rot} = \frac{L^2}{2I},$$

where $I$ is the moment of inertia. Since

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

the Hamiltonian is the angular Laplacian on the sphere multiplied by $-\hbar^2/(2I)$.

The important symmetry point is that $L^2$ is the Casimir operator of the rotation algebra. It commutes with every component:

$$[L^2,L_i]=0.$$

Therefore

$$[H_{\rm rot},L_i]=0, \qquad i=x,y,z.$$

The ideal rotor is fully rotationally invariant. Its energy cannot depend on the orientation label $M$ relative to an arbitrary chosen $z$ axis.

Eigenstates and Labels

The simultaneous eigenstates of $L^2$ and $L_z$ are spherical harmonics:

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

They satisfy

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

and

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

The allowed labels for an ordinary scalar rotor are

$$J=0,1,2,\ldots,$$

and

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

This page uses the molecular-rotor convention $J,M$. The same mathematical labels are often written $\ell,m$ in central-potential wave mechanics.

Energy Levels

Acting with $H_{\rm rot}$ on $Y_J^M$ gives

$$H_{\rm rot}Y_J^M = \frac{\hbar^2}{2I}J(J+1)Y_J^M.$$

Thus

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

It is common to define the rotational constant

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

so that

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

Here $B$ has energy units. Spectroscopic tables may instead quote $B/h$ in frequency units or $B/(hc)$ in inverse-length units; check the units before inserting a tabulated value.

The spacing between adjacent levels is

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

The levels are not evenly spaced in energy. The transition lines of the ideal dipole rotor can still form a regular ladder because allowed transitions connect adjacent $J$ values.

Degeneracy

For fixed $J$, there are

$$2J+1$$

allowed $M$ values. Since $H_{\rm rot}$ depends only on $L^2$, all $M$ states in the same $J$ multiplet have the same energy:

$$g_J=2J+1.$$

This degeneracy is the representation-theoretic signature of rotational invariance. A weak external electric or magnetic field selects a physical axis and can split the $M$ levels. Anisotropic molecular environments can also split it. By contrast, isotropic centrifugal distortion changes the $J$ dependence of the energies but retains the $M$ degeneracy.

Relation to Central Potentials

The angular part of a central-potential wavefunction is also a spherical harmonic:

$$\psi(r,\theta,\phi) = R_{\alpha\ell}(r)Y_\ell^m(\theta,\phi).$$

The difference is that a central-potential particle still has radial motion. The rigid rotor freezes the radial coordinate, so the angular factor is the whole configuration-space wavefunction.

This makes the rotor useful pedagogically:

• it isolates the $L^2$ eigenvalue problem;
• it shows the $2J+1$ degeneracy without a radial equation;
• it gives a physical model where spherical harmonics are literal stationary states;
• it prepares the selection-rule machinery used in rotational spectroscopy.

Selection-Rule Preview

For a polar linear rotor, electric-dipole rotational transitions are controlled by how the dipole operator transforms under rotations. The dipole is a vector operator, so it carries angular momentum rank one.

Rank one permits $\Delta J=0,\pm1$ subject to the triangle rule. The odd parity of the electric dipole, together with rotor parity $(-1)^J$, removes $\Delta J=0$. The resulting rule for a polar linear rotor in one fixed electronic and vibrational state is

$$\Delta J=\pm1,$$

with $\Delta M=0,\pm1$ depending on polarization and the chosen quantization axis. The detailed tensor-operator derivation belongs to Applications to Molecular Rotations and the line-position application belongs to Rotational Spectra.

A homonuclear molecule has no permanent electric dipole, so this pure rotational electric-dipole spectrum is absent even though rotational levels exist. Other processes, such as Raman scattering, have different rules.

This rule is not a statement that the rotor can only have neighboring energy levels. It is a statement about which matrix elements of the dipole operator are allowed by angular momentum.

Common Mistakes

• Using half-integer $J$ labels for an ordinary scalar rotor on $S^2$.
• Forgetting that the $M$ degeneracy follows from rotational invariance and can be split by external fields.
• Confusing the rotational constant $B$ with a magnetic field.
• Treating the spherical harmonic shape as a classical orientation of a rigid rod.
• Assuming the ideal rotor includes vibration, centrifugal distortion, spin, nuclear exchange symmetry, or electronic structure.
• Applying $\Delta J=\pm1$ as a universal rule for every perturbation; it is the electric-dipole rule for the ideal linear polar rotor.

Exercises

1. Find the energy and degeneracy of the $J=2$ level.

Solution

The energy formula is

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

For $J=2$,

$$E_2=B(2)(3)=6B.$$

The degeneracy is

$$g_J=2J+1,$$

so

$$g_2=5.$$

The allowed $M$ values are

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

2. Explain why the ideal rotor energy does not depend on $M$.

Solution

The ideal rotor Hamiltonian is

$$H_{\rm rot}=\frac{L^2}{2I}.$$

It depends on the Casimir $L^2$, not on any chosen component such as $L_z$. Since

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

all states with the same $J$ and different $M$ have the same energy. Physically, the free rotor has no preferred spatial axis, so different orientations of the angular momentum multiplet are degenerate.

3. Compare the rigid rotor with a central-potential bound state.

Solution

Both use spherical harmonics as angular eigenfunctions. For a central potential,

$$\psi_{\alpha\ell m}(r,\theta,\phi) = R_{\alpha\ell}(r)Y_\ell^m(\theta,\phi),$$

so the radial function carries additional dynamics. For the rigid rotor, the radius is fixed and the wavefunction is just

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

Thus the rotor isolates the angular part of the central-potential separation problem.

4. The $J=0\to1$ absorption line is at frequency $\nu_0$. Where is the $J=1\to2$ line in the ideal model, and what moment of inertia follows from $\nu_0$?

Solution

Since $h\nu_{J\to J+1}=2B(J+1)$, the second line is at $2\nu_0$. Also $B=h\nu_0/2$ and $B=\hbar^2/(2I)$ give

$$I=\frac{\hbar^2}{h\nu_0}=\frac{\hbar}{2\pi\nu_0}.$$

The units are mass times length squared. Equally spaced lines do not mean equally spaced energy levels: each line measures a different adjacent-level gap.

References

• J. M. Brown and A. Carrington, Rotational Spectroscopy of Diatomic Molecules, Cambridge University Press, 2003.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.