The Rigid Rotor and Rotational Energy Levels

Point a microwave beam at a sample of gaseous HCl and something remarkable happens. The molecule absorbs energy only at a specific comb of frequencies, spaced with almost metronomic regularity. That regularity is not an accident of HCl; it is the fingerprint of a rotating body governed by quantum mechanics. The rigid rotor model explains this pattern from first principles. It takes a diatomic molecule, freezes its bond length, and asks how a rigid stick can store rotational energy according to quantum rules. The answer is a strikingly simple formula, and that formula is the entire basis of rotational spectroscopy. This article derives the rotational term $F(J) = BJ(J+1)$ from the rigid rotor Hamiltonian. You will see where the rotational constant $B$ comes from. Then you will see why absorption lines appear as an evenly spaced ladder. Finally, you will work two full numeric examples using real molecular data.

Key takeaways

The rigid rotor models a diatomic molecule as two point masses fixed at distance $r$ , rotating about their center of mass.
Quantization of angular momentum gives discrete energy levels $E_J = BhcJ(J+1)$ , labeled by the rotational quantum number $J = 0, 1, 2, \ldots$
The rotational constant $B$ depends only on the moment of inertia $I$ , and thus only on masses and bond length.
The selection rule $\Delta J = \pm1$ (for molecules with a permanent dipole) turns this level ladder into an absorption spectrum.
Adjacent absorption lines are separated by exactly $2B$ , giving a spectrum of evenly spaced lines rather than a continuum.
Key idea: Quantizing a rotating rigid body produces energy levels that grow as $J(J+1)$ , and this simple quadratic spacing is directly what a microwave spectrometer measures as a uniform comb of lines.

The core insight

Because the rigid rotor's energy depends on $J(J+1)$ rather than on $J$ alone, the gaps between successive levels grow linearly with $J$ , but the gaps between successive allowed transitions (governed by $\Delta J = +1$ ) come out perfectly uniform at $2B$ , which is why rotational spectra look like an evenly spaced picket fence rather than a crowded, unevenly spaced ladder.

The physical picture

Picture a diatomic molecule such as CO or HCl as a lightweight, rigid dumbbell: two point masses $m_1$ and $m_2$ held at a fixed separation $r$ by a bond that we pretend cannot stretch. This is the rigid rotor. Real bonds do stretch slightly as the molecule spins faster, but that correction is small and is treated separately (as centrifugal distortion); the rigid rotor captures the dominant physics. Classically, such a body can spin with any angular momentum and any energy—rotation is "unquantized" in classical mechanics. But a molecule is a quantum object. When the Schrödinger equation is solved for a rigid rotor (the same mathematical problem as a particle constrained to the surface of a sphere), the allowed energies turn out to be quantized, indexed by a single non-negative integer $J$ , the rotational quantum number. The key structural fact is that all the molecule-specific information—both masses and the bond length—collapses into one number, the moment of inertia $I$ . Everything about the rotational energy-level pattern then follows from $I$ alone.

Figure

The rigid rotor: two masses on a rigid stick, rotating about the center of mass

The core relations

I = \mu r^2, \qquad \mu = \frac{m_1 m_2}{m_1 + m_2}

where: $I$ is the moment of inertia, $\mu$ the reduced mass, $r$ the bond length.

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

where: $E_J$ is the rotational energy of level $J$ , $\hbar = h/2\pi$ .

F(J) = \frac{E_J}{hc} = BJ(J+1)

where: $F(J)$ is the rotational term (energy expressed as a wavenumber), $B$ the rotational constant.

B = \frac{h}{8\pi^2 c I}

where: $B$ is in units of wavenumber (cm⁻¹) once $h$ , $c$ , $I$ are in consistent CGS/SI units.

\tilde{\nu}_{J \to J+1} = F(J+1) - F(J) = 2B(J+1)

where: $\tilde{\nu}$ is the transition wavenumber, valid under the selection rule $\Delta J = +1$ .

Derivation AFrom the rigid-rotor Hamiltonian to F(J) = BJ(J+1)

Goal. Show that quantizing the angular momentum of a rigid diatomic rotor forces its energy to take the form $E_J = (\hbar^2/2I)J(J+1)$ , and recast this as the spectroscopic term $F(J) = BJ(J+1)$ .

Step 1: Reduce two bodies to one. A diatomic molecule rotating about its center of mass is mechanically equivalent to a single particle of reduced mass $\mu = m_1m_2/(m_1+m_2)$ moving on the surface of a sphere of radius $r$ , the bond length. This equivalence is exact for two-body problems and lets us write the classical rotational energy as

E = \frac{L^2}{2I}

where $L$ is the angular momentum and $I = \mu r^2$ the moment of inertia. No potential energy term appears because the bond length is fixed—there is no restoring force to store potential energy in as the molecule rotates.

Step 2: Quantize the angular momentum. The Hamiltonian for a particle constrained to a spherical surface is

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

This is exactly the angular part of the Schrödinger equation solved for the hydrogen atom and for any central-force problem. Its eigenfunctions are the spherical harmonics $Y_J^{m_J}(\theta,\phi)$ , and the eigenvalues of $\hat{L}^2$ are known from the general theory of angular momentum to be

\hat{L}^2 Y_J^{m_J} = \hbar^2 J(J+1) Y_J^{m_J}, \qquad J = 0, 1, 2, \ldots

with $m_J$ running over $2J+1$ integer values from $-J$ to $+J$ .

Step 3: Read off the energy levels. Substituting the eigenvalue into $\hat H = \hat L^2/2I$ gives immediately

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

Each level $J$ is $(2J+1)$ -fold degenerate, since energy does not depend on $m_J$ —physically, a given rotational speed can point in $2J+1$ distinguishable spatial orientations.

Step 4: Convert to a spectroscopic term. Spectroscopists conventionally express energies as wavenumbers via $E = hc\tilde{\nu}$ , so divide through by $hc$ :

F(J) = \frac{E_J}{hc} = \frac{\hbar^2}{2Ihc}J(J+1)

Writing $\hbar = h/2\pi$ and collecting constants defines the rotational constant,

B \equiv \frac{h}{8\pi^2cI}

which finally gives the compact spectroscopic result

\boxed{F(J) = BJ(J+1)}

Everything about the molecule's rotational spectrum—the entire ladder of energy levels—is packed into the single constant $B$ , itself packed entirely into the moment of inertia $I$ .

Worked examples

Example 1: Computing B and the first few rotational terms for ¹²C¹⁶O. Carbon monoxide has bond length $r = 112.8$ pm $= 1.128 \times 10^{-10}$ m. Atomic masses: $m_C = 12.000$ u, $m_O = 15.995$ u, with $1\ \text{u} = 1.6605 \times 10^{-27}$ kg. Step 1 — reduced mass:

\mu = \frac{(12.000)(15.995)}{12.000+15.995}\ \text{u} = \frac{191.94}{27.995}\ \text{u} = 6.856\ \text{u}

\mu = 6.856 \times 1.6605\times10^{-27}\ \text{kg} = 1.1385\times10^{-26}\ \text{kg}

Step 2 — moment of inertia:

I = \mu r^2 = (1.1385\times10^{-26}\ \text{kg})(1.128\times10^{-10}\ \text{m})^2

I = (1.1385\times10^{-26})(1.2724\times10^{-20})\ \text{kg m}^2 = 1.4485\times10^{-46}\ \text{kg m}^2

Step 3 — rotational constant:

B = \frac{h}{8\pi^2 c I} = \frac{6.6261\times10^{-34}\ \text{J s}}{8\pi^2 (2.9979\times10^{10}\ \text{cm s}^{-1})(1.4485\times10^{-46}\ \text{kg m}^2)}

Carrying the units carefully (working in cm gives $B$ directly in cm⁻¹):

8\pi^2 = 78.96, \qquad 78.96 \times 2.9979\times10^{10} \times 1.4485\times10^{-46} = 3.430\times10^{-34}

B = \frac{6.6261\times10^{-34}}{3.430\times10^{-34}}\ \text{cm}^{-1} = 1.932\ \text{cm}^{-1}

This matches the experimental literature value for CO, $B \approx 1.9313$ cm⁻¹, confirming the calculation. Step 4 — first rotational terms:

F(0) = 0, \quad F(1) = 2B = 3.864\ \text{cm}^{-1}, \quad F(2) = 6B = 11.59\ \text{cm}^{-1}, \quad F(3) = 12B = 23.18\ \text{cm}^{-1}

Step 5 — predicted absorption lines ( $\Delta J = +1$ ):

\tilde\nu_{0\to1} = 2B(1) = 3.864\ \text{cm}^{-1}, \quad \tilde\nu_{1\to2} = 2B(2) = 7.728\ \text{cm}^{-1}, \quad \tilde\nu_{2\to3} = 2B(3) = 11.59\ \text{cm}^{-1}

Each successive line is exactly $2B = 3.864$ cm⁻¹ higher than the last—the hallmark equal spacing.

Example 2: Which J level is most populated at 298 K for HCl, and how does this shape the observed spectrum? For H³⁵Cl, $B = 10.59$ cm⁻¹ (bond length 127.5 pm, $\mu = 0.9799$ u). Rotational level populations follow the Boltzmann distribution, weighted by degeneracy $(2J+1)$ :

N_J \propto (2J+1)\exp\left(-\frac{BhcJ(J+1)}{k_BT}\right)

To find the most populated level, differentiate the exponent-and-degeneracy product with respect to $J$ and set it to zero. Treating $J$ as continuous gives the standard result:

J_{\max} = \sqrt{\frac{k_BT}{2Bhc}} - \frac{1}{2}

Compute $k_BT$ at 298 K: $k_BT = (1.3807\times10^{-23}\ \text{J/K})(298\ \text{K}) = 4.115\times10^{-21}$ J. Convert to cm⁻¹ units matching $Bhc$ : divide by $hc = 1.9864\times10^{-23}$ J·cm, giving $k_BT = 207.2$ cm⁻¹.

J_{\max} = \sqrt{\frac{207.2}{2(10.59)}} - 0.5 = \sqrt{9.783} - 0.5 = 3.128 - 0.5 = 2.63

Rounding, the most heavily populated rotational level is $J = 3$ (occasionally reported as $J=2$ or $3$ depending on rounding convention, since the true maximum of a discrete distribution needn't fall exactly on the continuous formula's output). This means the strongest line in the HCl microwave/far-IR rotational spectrum is not the $J=0\to1$ transition but roughly the $J=3\to4$ transition, at

\tilde\nu_{3\to4} = 2B(4) = 8(10.59) = 84.7\ \text{cm}^{-1}

Sanity check: this is comfortably within the far-infrared/microwave range where HCl rotational lines are experimentally observed (tens to low hundreds of cm⁻¹), and the intensity envelope rising then falling with $J$ matches textbook HCl rotational spectra exactly.

Limiting cases & common traps

Situation	What happens	Why
$J=0$	$F(0) = 0$ , a true rotational ground state	No angular momentum, no energy, consistent with quantum zero-point behavior of angular motion (unlike vibration, there is no rotational zero-point energy)
Large $J$	Level spacing $F(J{+}1)-F(J) = 2B(J{+}1)$ grows linearly	Energy itself grows as $J^2$ , so gaps between adjacent levels widen even though transition spacings stay at $2B$
Heavier atoms / longer bonds	$B$ shrinks	$B \propto 1/I$ and $I$ grows with $\mu$ and $r^2$
Homonuclear diatomics (N₂, O₂, H₂)	No pure rotational microwave spectrum	No permanent dipole moment, so $\Delta J = \pm1$ electric-dipole transitions are forbidden

The chemist behind it

The rigid rotor's spectroscopic payoff—turning a line spectrum into bond lengths—was made experimentally decisive by Charles Townes and collaborators, who pioneered precision microwave spectroscopy in the 1940s. Their measurements of rotational transition frequencies, matched against $F(J) = BJ(J+1)$ , gave some of the first sub-picometer-precision bond lengths for simple molecules, and the microwave techniques Townes developed for spectroscopy directly seeded his later invention of the maser.

Origins · 1 minCharles Hard Townes and how this was discovered

Charles Hard Townes

1915–2015 · pioneer of microwave spectroscopy who used rotational transitions to extract precise molecular structures, later inventing the maser

Ask the tutor: Once you measure $2B$ from a spectrum, how exactly do you turn that single number into a bond length accurate to a hundredth of a picometer?

Why does F(J) depend on J(J+1) instead of just J squared or J?

This form comes directly from quantum mechanics, not from an approximation. The operator $\hat L^2$ for angular momentum has eigenvalues $\hbar^2 J(J+1)$ , a mathematical fact proven from the commutation relations of angular momentum operators, the same relations that govern orbital angular momentum in atoms. The $J(J+1)$ dependence is universal to any three-dimensional rigid rotor, appearing identically in atomic physics, molecular rotation, and nuclear spin systems.

Does every molecule show a pure rotational absorption spectrum?

No. Absorption requires a permanent electric dipole moment that can interact with the oscillating electric field of microwave radiation, plus the selection rule $\Delta J = \pm1$ . Heteronuclear diatomics like CO and HCl qualify. Homonuclear diatomics like N₂ and O₂, and any molecule with zero permanent dipole (like CO₂ in its linear ground state), show no pure rotational microwave absorption spectrum, though they can still show rotational structure in Raman spectroscopy.

How is the rotational constant B actually measured, and what is it used for?

$B$ is extracted directly from experiment by measuring the spacing between adjacent rotational absorption lines, which equals $2B$ . Once $B$ is known, it can be inverted through $B = h/(8\pi^2cI)$ to obtain the moment of inertia $I$ , and then through $I = \mu r^2$ to obtain the bond length $r$ to extraordinary precision, often better than 0.1 pm for simple diatomics. This is the basis of the companion topic on determining bond lengths from rotational spectra.

Is the rigid rotor model exact?

No, it is an idealization. Real bonds stretch slightly as the molecule spins faster, an effect called centrifugal distortion, which adds a small correction term $-D_J J^2(J+1)^2$ to $F(J)$ that grows more important at high $J$ . For most rotational spectroscopy at moderate $J$ , though, the rigid rotor term $BJ(J+1)$ dominates and centrifugal distortion is a small, well-characterized correction rather than a qualitative failure.

Why is there no rotational zero-point energy, unlike vibration?

Vibrational zero-point energy exists because the harmonic oscillator potential curves upward on both sides of the minimum, forcing the ground state wavefunction to have nonzero curvature and kinetic energy even at $J=0$ -equivalent conditions. Rotation has no potential energy term at all in the rigid rotor Hamiltonian—only kinetic energy from angular momentum—so the $J=0$ state, having zero angular momentum, genuinely has zero energy.