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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 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.

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 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.

A distinct experiment in An Automated, Highly Selective Reaction Monitoring Instrument Using Molecular Rotational Resonance Spectroscopy by Ann Adele Byars and others (Figure 1 and Figure 3) tracked exchange of hydrogen for deuterium in ethyl acetoacetate mixed with deuterated methanol. A pump circulated chemical solution through a valve; a calibrated loop injected aliquots into a temperature-controlled inlet, held at room temperature for this exchange experiment. Neon carried the vaporized sample to a cavity spectrometer. Separately recorded broadband spectra and fitted rotational Hamiltonians identified transitions for the four isotopic forms before targeted monitoring.^[ref]

The readout was each assigned line's signal over reaction time, normalized by the sum of the four signals. Transitions with the same quantum numbers reduced response-factor differences. A solvent wash addressed observed sample carryover. The signals distinguished the unsubstituted, two singly substituted, and doubly substituted forms. Only one gas-phase structural form was detected; the authors explicitly caution that their readout does not recover the structural chemical equilibrium in chemical solution.^[ref]

This example uses nonlinear molecules, so neither the single-constant linear rigid rotor nor the synthetic comb above is a fit to its spectrum. Its connection to the model is the sensitivity of moments of inertia to isotope placement. Line position helps assign a species; signal size additionally depends on sample transfer, initial quantum state populations and instrument response. Matching quantum numbers is a useful control, not a proof of identical response factors. These distinctions explain why resolving species and determining their concentrations are separate inference steps.

Write the linear rigid rotor energy eigenvalues as $E_J=k_{\mathrm{B}}{\theta }_{r}J(J+1)$, where ${\theta }_r={\hslash }^2/(2I{k}_{\mathrm{B}})=hB/k_{\mathrm{B}}$ is the rotational temperature scale, $I$ the moment of inertia, $B$ the frequency-valued rotational constant, and $k_{\mathrm{B}}$ the Boltzmann constant. Put $t=T/{\theta }_r$ and ${ϵ}_J=J(J+1)$. Define sums that include rotational multiplicity but exclude nuclear-spin multiplicity:The canonical partition function for one equilibrating internal rotor is $q_{\mathrm{eq}}=q_{\mathrm{p}}+3q_{\mathrm{o}}$, so its equilibrium orthohydrogen fraction isFor $t\ll 1$, the lowest allowed levels give $q_{\mathrm{p}}\simeq 1$ and $q_{\mathrm{o}}\simeq 3e^{-2/t}$, hence $x_{\mathrm{o}}^{\mathrm{eq}}\simeq 9e^{-2/t}/(1+9e^{-2/t})\to 0$. For $t\gg 1$, the densely populated even and odd ladders have equal leading sums, $q_{\mathrm{p}}\sim q_{\mathrm{o}}\sim t/2$, hence $x_{\mathrm{o}}^{\mathrm{eq}}\to 3/4$. The factor nine belongs to the low-temperature two-level approximation, and the factor three to the high-temperature spin multiplicities.

A frozen mixture requires two separately normalized probability distributions,Multiply these by the prescribed fractions $1-x$ and $x$, respectively. The nuclear-spin factor three cancels from the conditional orthohydrogen distribution. For fixed counts $N_{\mathrm{o}}$ and $N_{\mathrm{p}}$, the internal part of the many-molecule partition function is proportional to $q_{\mathrm{p}}^{N_{\mathrm{p}}}(3q_{\mathrm{o}}{)}^{N_{\mathrm{o}}}$, with combinatorial factors independent of temperature. Its logarithm yields the two conditional means weighted by the fixed counts. Using $q_{\mathrm{p}}+3q_{\mathrm{o}}$ instead would silently permit conversion and change the physical constraint.

Rotational Raman spectroscopy observes a frequency change of scattered light associated with molecular rotation. The coupling uses the induced electric dipole moment, ${\mathbf{d}}_{\mathrm{ind}}=\mathbf{\alpha }\mathbf{E}$, where $\mathbf{\alpha }$ is the electric polarizability tensor. For a linear molecule, the useful anisotropy is ${\alpha }_{\parallel }-{\alpha }_{\perp }$: the response differs for an electric field parallel or perpendicular to the molecular axis. A permanent electric dipole moment is not required.^[ref]

For the spin-free linear rigid rotor in one internal quantum state, the anisotropic tensor depends quadratically on the molecular direction. Its angular part has even parity and degree two in spherical harmonics. The triangle inequality for coupled angular momenta permits $|\Delta J|\le 2$, and even parity removes odd $\Delta J$, leaving $\Delta J=0,\pm 2$. The $\Delta J=0$ contribution is unshifted light; shifted rotational lines have $\Delta J=\pm 2$. Consequently the rigid-model Stokes shifts, in ordinary frequency units, areSuccessive allowed shifts are separated by $4B$, rather than the $2B$ absorption spacing of pure rotational electric-dipole selection rules. In spectroscopic wavenumber units replace $B$ by $\tilde{B}=B/c$ and $\Delta \nu$ by $\Delta \tilde{\nu }=\Delta \nu /c$, using the speed of light in $\mathrm{cm}{\mathrm{s}}^{-1}$ for $\mathrm{c}{\mathrm{m}}^{-1}$.

A homonuclear diatomic molecule such as molecular hydrogen has no permanent electric dipole moment in its symmetric internal state, so it lacks the ordinary electric-dipole microwave ladder. Its anisotropic electric polarizability can nevertheless give rotational Raman spectroscopy lines. Nuclear-spin symmetry can change the populations of even and odd $J$ levels, so the spin-free population example above must not be reused unchanged for molecular hydrogen. The method of coupling, as well as the energy eigenvalues, determines what the experiment can see.

A rotational spectral line assignment associates a measured spectral line with a molecule, an internal quantum state, and an initial and final set of quantum numbers. For the spin-free linear rigid rotor, absorption is labelled $J\to J+1$. A regularly spaced group suggests a candidate rotational constant, but the lowest line inside a scanned band need not be $0\to 1$. Several molecules can contribute to the same spectrum, and an apparent missing member can lie outside the band or below the detection threshold.

First determine line centers and their uncertainties from the instrument's readout, accounting for resolved Doppler shifts or hyperfine structure rather than treating every peak as a separate rotational transition. Then propose integer assignments, fit a rotational Hamiltonian, and predict additional spectral lines. For a weakly distorted linear molecule within one vibrational quantum state, the centrifugal distortion model gives the useful diagnosticHere $B,D,{\nu }_m$ are all frequencies in $\mathrm{Hz}$. Correct assignments make the second expression a straight line against $m^2$, with intercept $B$ and slope $-2D$. This follows by division of the transition formula; no additional model of molecular geometry is required. A direct weighted least squares fit of ${\nu }_m$ to $2Bm-4D{m}^3$ uses the measured uncertainties. If instead the transformed values are fitted, their uncertainties must also be divided by $2m$.^[ref]

For an original example, stipulate $B=5.000000\mathrm{GHz}$ and $D=2.000\mathrm{kHz}$, and suppose a scan covers $38$–$72\mathrm{GHz}$. The model places four consecutive transitions at $39.999488$, $49.999000$, $59.998272$, and $69.997256\mathrm{GHz}$. Their assignments are $m=4,5,6,7$, hence $J=3,4,5,6$. An extra stipulated line at $53.2\mathrm{GHz}$ is unrelated to this series. All these frequencies are calculated example inputs, not measured OCS data.

At $m=4$ and $m=7$, the transformed values are $4999.936$ and $4999.804\mathrm{MHz}$. Their straight line has slope $(4999.804-4999.936)/(49-16)=-0.004\mathrm{MHz}$, recovering $D=0.002\mathrm{MHz}$ and $B=5000\mathrm{MHz}$. Both intermediate transitions satisfy the same line exactly. Merely relabelling the first visible transition $0\to 1$ fails this diagnostic. The model predicts a withheld $7\to 8$ transition at $79.995904\mathrm{GHz}$: an additional measurement there would test the assignment rather than merely refit the original four lines.

An assignment earns confidence when several independently observed spectral lines, including lines outside the fitted interval, agree within uncertainty and expected strong accessible lines are present. Inspect the residual of an inverse problem for each line against $J$: systematic curvature can signal inadequate centrifugal distortion, whereas isolated discrepancies suggest a blend or wrong assignment. Comparing a known sample, a blank, and an isotope shift of a linear rotor supplies different checks. These are general inference controls; the actual reported controls of the NIST trace-gas FTMW validation experiment are described separately. One isolated match or one fitted rotational constant alone cannot certify molecular identity.

Rotational spectroscopy measures transitions between molecular rotation quantum states, usually in low-pressure gaseous molecules. The measured spectral lines become a molecular fingerprint only after rotational spectral line assignment: identify which transitions belong together, determine their quantum numbers, fit a rotational Hamiltonian, and test additional predictions. Line positions constrain moments of inertia, while rotational state populations and transition coupling determine which lines can be detected.

The simplest ordinary microwave absorption ladder belongs to a polar linear molecule obeying pure rotational electric-dipole selection rules. A homonuclear diatomic molecule has no permanent electric dipole moment in its symmetric internal state, but rotational Raman spectroscopy can probe its rotation through anisotropic electric polarizability. The coupling mechanism must therefore be specified before interpreting a missing spectrum.

The existing linear rigid rotor, rotational constant, isotope shift of a linear rotor, and centrifugal distortion explain the forward model. Molecular structure from rotational constants asks the reverse question: which arrangements of chemical atoms could produce the fitted constants? The NIST trace-gas FTMW validation experiment makes the measurement chain concrete and separates identifying a species from measuring its concentration.

A set of spectral lines associated with transitions between molecular rotational energy levels. A polar linear rigid rotor has electric-dipole transitions with $\Delta J=\pm 1$; in the ideal model adjacent absorption frequencies are separated by twice the rotational constant in frequency units.

For the spin-free linear rigid rotor in thermal equilibrium at temperature $T>0$, assume all integer $J\ge 0$ are allowed and ignore nuclear-spin restrictions. Each $J$ level has $2J+1$ quantum states sharing one energy eigenvalue, an instance of degenerate energy levels, so the Boltzmann distribution givesHere $Q_{\mathrm{rot}}$ is a rotational partition function, $h$ the Planck constant, $k_{\mathrm{B}}$ the Boltzmann constant, and $B$ a frequency in $\mathrm{Hz}$. Thus $hB/(k_{\mathrm{B}}T)$ is dimensionless. The $2J+1$ factor counts magnetic quantum numbers; it does not mean that a whole level is one quantum state. For the stipulated $B=5\mathrm{GHz}$, the most populated levels are $J=2$ at $3\mathrm{K}$ and $J=7$ at $30\mathrm{K}$.

Population alone is insufficient to predict intensity. In weak, unsaturated absorption, the integrated absorption coefficient includes the transition frequency, the squared electric dipole moment matrix element, and the difference of lower and upper substate populations because stimulated emission returns energy to the field. A pulsed FTMW spectroscopy signal additionally depends on excitation, quantum coherence, cavity tuning, and receiver response.^[ref]

For an isotropic sample with polarization of an electromagnetic wave along $z$, write $d_z=d\cos \theta$. Normalized spherical harmonics giveThe last equality follows from ${\sum }_{M=-J}^J{M}^2=J(J+1)(2J+1)/3$. Combining this sum with the per-substate Boltzmann distribution and subtracting stimulated emission gives relative integrated absorption areasThe common factor includes $d^2/3$ and cancels upon normalization. This formula describes integrated weak absorption in a thermal, spin-free linear rigid rotor, not arbitrary peak heights. The example's strongest integrated lines originate at $J=3$ and $J=13$ for $3$ and $30\mathrm{K}$, respectively: they differ from the most populated levels. A missing line can reflect population, coupling, coverage, or sensitivity; it does not prove that the corresponding quantum state is absent.

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.

Spin isomers of hydrogen are two forms of molecular hydrogen, ${}^1{\mathrm{H}}_2$, distinguished by the combined nuclear spin of their two protons. They have the same chemical formula and electronic ground state, but different allowed molecular rotation states. Parahydrogen has total nuclear spin zero and even rotational quantum number $J$; orthohydrogen has total nuclear spin one and odd $J$. The difference makes heat capacity a probe of fermionic exchange symmetry, and makes cooling history matter for a tank of molecular hydrogen.

Assume the electronic ground state $X{}^1{\Sigma }_g^{+}$ and the vibrational ground state of a free molecule. The electronic factor is symmetric under exchanging the two protons, and the vibrational factor depends only on their separation, so it is also symmetric. These assumptions concern the exchange of protons; the electronic wavefunction separately obeys fermionic exchange symmetry under exchange of electrons. Neglect electronic excitation, vibrational excitation, external fields, and small interactions involving nuclear spin when calculating rotational energy eigenvalues.

The two protons are identical particles with spin $1/2$. Their complete wavefunction must change sign under their particle exchange operator. Interchanging the two ends reverses the internuclear direction $\hat{\mathbf{r}}$, and the spherical harmonic rotational factor satisfiesThe antisymmetric spin-one-half singlet state therefore requires symmetric, even-$J$ rotation; the symmetric spin-one-half triplet state requires antisymmetric, odd-$J$ rotation. This is exchange symmetry of two identical spin-one-half fermions applied inside a diatomic molecule, not an extra rotational force. A linear rigid rotor may omit spin-dependent splittings while still requiring this restriction on its allowed quantum states.

The University of Florida's Rotation Vibration Spectrum of the HCl Molecule, University of Florida laboratory protocol, printed pp. 2–3 and 10–11, uses a scanning Michelson interferometer for Fourier transform spectroscopy. An air-filled silicon-window cell supplies the background. Supervised preparation adds concentrated hydrogen chloride liquid; the cleaned cell subsequently receives DCl liquid. Acquisition settings are identical.

The readout is transmission or absorbance against spectroscopic wavenumber, resolving molecular rotation alongside stretching. Background addresses instrument and cell contributions; cleaning addresses carryover. Excessive absorption impairs analysis. Assigning isotopes and rotational branches precedes fitting band parameters to test isotope scaling of diatomic vibration. These are prescribed measurements, not numerical observations supplied here.^[ref]

For a transparent ideal model, add the linear rigid rotor to the harmonic vibrational energy levels and assume the same frequency-valued rotational constant $B$ in both vibrational states:For a spin-free diatomic molecule with no electronic angular momentum, electric-dipole selection rules give $\Delta J=\pm 1$ for this vibrational absorption. Subtracting initial from final energy eigenvalues yieldsThese are two rotational branches about the vibrational band origin, not a single stretch line. For stipulated dimensionless values ${\tilde{\nu }}_{\mathrm{harm}}=100$ and $\tilde{B}=1$, the first three R-branch positions are $102,104,106$, and the first three P-branch positions are $98,96,94$. There is no $J=0\to -1$ transition. The numbers illustrate level subtraction; they are not HCl measurements. Real upper and lower rotational constants differ, and centrifugal distortion shifts lines further.