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A bond-dissociation enthalpy is the molar enthalpy change for the specified homolysis at temperature $T$, with reactant and products in specified standard states. Its usual units are $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The superscript $\circ$ specifies standard states; it does not fix $T$ to room temperature.^[ref]

Write $\delta h_i(T)=H_{m,i}^{\circ }(T)-H_{m,i}^{\circ }(0)$ for each species' molar thermal enthalpy increment. With $D_0$ an energy per parent molecule,Here $N_{\mathrm{A}}$ is the Avogadro constant. The thermal increments include translational kinetic energy, molecular rotation, thermally excited molecular vibration and electronic states when populated, and the $pV$ contribution to enthalpy. They are additional to the zero-point energies already included in $D_0$. If values of standard enthalpy of formation at the same $T$ are available, Hess's law gives the same quantity as ${\Delta }_f{H}^{\circ }(A)+{\Delta }_f{H}^{\circ }(B)-{\Delta }_f{H}^{\circ }(AB)$.

Units alone do not identify the convention. Multiplying either $D_e$ or $D_0$ by the Avogadro constant gives molar energy, but does not add thermal enthalpy. One electronvolt per dissociation corresponds to $96.485332\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. A spectroscopic wavenumber $\tilde{\nu }$ represents energy $hc\tilde{\nu }$, using the Planck constant $h$ and speed of light $c$; $1\mathrm{c}{\mathrm{m}}^{-1}$ corresponds to $0.011962657\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref]

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]

Let $r$ be the instantaneous bond length, $r_e>0$ its equilibrium value, and $q=r-r_e$ the stretching displacement. Assume one fixed electronic quantum state, a smooth potential energy $V(r)$ with a nondegenerate local minimum at $r_e$, and omit molecular rotation initially. The Taylor series has no linear term because $V^{\prime }(r_e)=0$:Keeping only the quadratic term is the harmonic approximation near a potential minimum. The bond force constant $k$ has units $\mathrm{N}{\mathrm{m}}^{-1}$. It measures local stiffness; it is not the bond-dissociation energy, which concerns the depth and distant shape of the entire well.

The two-particle relative coordinate explains the reduced mass. For stretching along a fixed axis in a stationary centre of mass frame, let $u_1,u_2$ be the nuclear displacements from equilibrium. Then $u_2-u_1=q$ and $m_1{u}_1+m_2{u}_2=0$, soHere $T$ is the internal kinetic energy, and $m_1,m_2>0$ are the two nuclear masses. The lighter partner moves farther; the reduced mass is smaller than either individual mass. The restoring force $-dV/dq\simeq -kq$ gives simple harmonic motion:The angular frequency is $\omega$, the ordinary frequency is $\nu$, and ${\tilde{\nu }}_{\mathrm{harm}}$ is the corresponding spectroscopic wavenumber; $c$ is the speed of light. Use $c$ in $\mathrm{cm}{\mathrm{s}}^{-1}$ to obtain ${\tilde{\nu }}_{\mathrm{harm}}$ in $\mathrm{c}{\mathrm{m}}^{-1}$. The symbol $k$ here is a bond force constant, not an angular wavenumber.

For an original dimensionless example, choose $m_1=m_2=2$ and $k=4$ in consistent model units. Then $\mu =1$ and $\omega =2$. Releasing at $q(0)=0.1$ with $\dot{q}(0)=0$ gives $q(t)=0.1\cos (2t)$ and $u_1=-q/2$, $u_2=q/2$. Each atomic nucleus moves only half the stretching amplitude, while the conserved internal mechanical energy above $V_0$ is $\frac{1}{2}k(0.1{)}^2=0.02$. Fixing one partner instead would give $\omega =\sqrt{2}$, demonstrating why a one-body calculation changes the model.

Electric polarizability relates an induced electric dipole moment to an applied electric field. In a weak-field model with linearity, ${\mathbf{d}}_{\mathrm{ind}}=\mathbf{\alpha }\mathbf{E}$. A tensor response allows different electric field directions to induce electric dipole moments in different directions. A response that changes with molecular rotation enables rotational Raman spectroscopy.

Quantizing the stretching coordinate gives a quantum harmonic oscillator with reduced mass $\mu$:Here $v$ is the vibrational quantum number and $\hslash =h/(2\pi )$ the reduced Planck constant. Consecutive energy eigenvalues have gap $\hslash \omega$; the ground state has zero-point energy $\hslash \omega /2$ above the well minimum. The offset $V_0$ cancels from every transition energy.

A short proof uses creation and annihilation operators: with $\ell =\sqrt{\hslash /(\mu \omega )}$, write $q=\ell (a+a^{†})/\sqrt{2}$. The Quantum Hamiltonian becomes $\hat{H}-V_0=\hslash \omega (a^{†}a+1/2)$. Nonnegativity of $a^{†}a$ gives the lowest level; repeated application of $a^{†}$ raises its energy by $\hslash \omega$ each time. This also gives $⟨q{⟩}_v=0$ andThe ground state therefore has a nonzero spread of bond lengths, even though its mean stretching displacement vanishes. A stationary energy eigenstate has a time-independent position probability density; the classical oscillation curve is not a trajectory of an atomic nucleus in that state.

The parabolic model formally extends $q$ to the whole real line, whereas an actual bond length satisfies $r>0$. It is useful for low levels concentrated near $r_e$, with typical stretching displacements much smaller than $r_e$. It fails near breakup of the chemical bond and when neglected anharmonicity becomes substantial.

Keep the same bond force constant $k$ and approximate the nuclear masses of ${}^1\mathrm{H}$, ${}^2\mathrm{H}$, and ${}^{35}\mathrm{Cl}$ by $1u$, $2u$, and $35u$, respectively. Here $u$ is the unified atomic mass unit, and $\mathrm{D}$ denotes deuterium. These are chosen mass number approximations, not measured atomic masses or natural-abundance averages. With the same chlorine isotope in both molecules,This is the isotope scaling of diatomic vibration prediction. Doubling the light partner's mass nearly doubles the reduced mass, lowering the frequency by about $28.3\%$. The factor differs from $1/\sqrt{2}$ because the chlorine partner also moves.

For the same small initial stretch $A$ and zero velocity, both models start with the same internal mechanical energy $k{A}^2/2$, but the DCl oscillation period is longer by $\sqrt{72/37}\approx 1.394972$. In the centre of mass frame, $u_{\mathrm{H}}=-35q/36$ and $u_{\mathrm{Cl}}=q/36$ for HCl; $u_{\mathrm{D}}=-35q/37$ and $u_{\mathrm{Cl}}=2q/37$ for DCl. These ratios explain which partner moves and are separate from the change in frequency.

The common-$k$ assumption belongs to the leading Born-Oppenheimer approximation: fixed nuclear positions and electric charges determine the electronic potential energy, while nuclear masses govern motion on that curve. Quantitative comparison requires actual atomic masses, departures from that approximation, and separation of anharmonicity and molecular rotation from the measured band. Even with common $k$, the quantum harmonic oscillator predicts a different zero-point energy for each isotope.^[ref]

An actual infrared spectroscopy route is to compare gaseous hydrogen chloride containing ordinary hydrogen with its deuterium counterpart. A documented teaching experiment uses a stopcock-equipped gas cell with potassium bromide windows and a Nicolet Avatar instrument for Fourier transform spectroscopy. A nitrogen-purged flask and drying tube supply the mixture; the sample also contains sulfur dioxide. An empty sample-chamber background precedes the filled-cell scan. The absorbance trace and peak-picked spectroscopic wavenumbers are the readout. The analysis assigns and indexes ${}^1{\mathrm{H}}^{35}\mathrm{Cl}$ and ${}^2{\mathrm{H}}^{35}\mathrm{Cl}$ transitions separately and propagates transition uncertainties. The instructions distinguish resolvable chlorine isotope peaks from adequate precision for the intended parameter analysis. ^[ref]

A separate published teaching dataset reports a Nicolet Nexus 670 instrument, a $10\mathrm{cm}$ gas cell at $20\mathrm{Torr}$, and chlorine isotope peaks resolved to baseline. These apparatus details describe that dataset, not the preceding experiment. ^[ref]

The inference connects several models. Fit correctly assigned transitions to separate molecular rotation from molecular vibration, then compare the inferred vibrational scales using isotope scaling of diatomic vibration. Agreement supports the approximate separation between a nearly unchanged chemical bond stiffness and a changed nuclear mass. It does not make a real molecule an exact classical spring: anharmonicity shifts the fundamental transition away from the local harmonic oscillator scale, and coupling to molecular rotation shifts individual lines.

For an independent test, calculate the reduced masses from the specified isotopes and compare the inferred values $k=\mu (2\pi c{\tilde{\nu }}_{\mathrm{harm}}{)}^2$. Agreement within uncertainty tests the common-curvature assumption; choosing separate $k$ values just to fit each frequency would not test it. The documented background addresses instrument and ambient contributions, but does not by itself subtract window absorption, identify contaminating species, or calibrate the spectroscopic wavenumber scale. Those require separate controls. Finite spectral resolution, blended peaks, incorrect isotope assignments, and departure from the harmonic approximation near a potential minimum limit the comparison. No numerical measured line positions are assumed here.

Let $u_s=⟨ϵ{⟩}_s$ and $v_s={\mathrm{Var}}_s(ϵ)$ for the conditional distributions of parahydrogen and orthohydrogen, with $s=\mathrm{p},\mathrm{o}$. Differentiating the normalized Boltzmann factors gives $d{u}_s/dt=v_s/t^2$. Thus the rotational heat capacity at constant volume per molecule of a composition frozen at $x$ isAt conversion equilibrium, the law of total variance adds a nonnegative contribution from the changing species populations:The last term is the contribution from conversion. Compare the first two terms with a frozen sample at the same instantaneous composition; a frozen $3:1$ sample generally has different weights too. This formula follows directly from the variance of a mixture: subtract the overall mean inside each species, expand the square, and the cross terms vanish because each conditional deviation has zero mean. It also proves nonnegative rotational heat capacity in either model.

For an original worked example at $t=1/2$, summing the allowed ladders gives $q_{\mathrm{p}}\approx 1.00003072$, $q_{\mathrm{o}}\approx 0.05494692$, and $x_{\mathrm{o}}^{\mathrm{eq}}\approx 0.14150982$. The calculated rotational heat capacities are $c_{\mathrm{eq}}/k_{\mathrm{B}}\approx 1.94719633$ and $c_{\mathrm{fr}}/k_{\mathrm{B}}\approx 0.00110733$ for frozen normal hydrogen. The contrast is large because changing species accesses the $J=0$ to $J=1$ separation, whereas fixed species must use the larger $J=0$ to $2$ or $J=1$ to $3$ separations. These are model numbers at a chosen dimensionless temperature, not measured hydrogen data.

Both rotational heat capacities approach zero as $t\to 0$, although frozen normal hydrogen retains nonzero rotational internal energy. Both approach $k_{\mathrm{B}}$ as $t\to \infty$ within the rotor model. The high-temperature limit is a formal limit with all other excitations suppressed; actual molecular hydrogen eventually needs molecular vibration, nonrigidity, and dissociation. Multiplying the per-molecule result by the Avogadro constant gives rotational molar heat capacity in units of the molar gas constant $R$. A dilute classical ideal gas additionally contributes $3R/2$ from translation to its heat capacity at constant volume.

Near a stable equilibrium bond length $r_e$, a Taylor expansion of a diatomic molecule's potential energy gives $U(r)\simeq U(r_e)+k(r-r_e{)}^2/2$, with $k=U^{\prime \prime }(r_e)>0$. Treat the internal stretch as a harmonic oscillator of reduced mass, initially omitting molecular rotation. If replacing an isotope leaves this local potential energy curvature unchanged, thenHere $c$ is the speed of light and ${\tilde{\nu }}_{\mathrm{harm}}$ a spectroscopic wavenumber; use consistent length units for $c$ and ${\tilde{\nu }}_{\mathrm{harm}}$. The quantum harmonic oscillator gives adjacent energy eigenvalues separated by $h\nu$, with $h$ the Planck constant. Consequently a vibrational photon absorption can probe the same frequency scale, subject to electric-dipole selection rules.

For an explicitly approximate isotope example, assign the atomic masses of ${}^1\mathrm{H}$, ${}^2\mathrm{H}$, ${}^{35}\mathrm{Cl}$, and ${}^{37}\mathrm{Cl}$ the mass-number values $1u,2u,35u,37u$, respectively, with $u$ the unified atomic mass unit. These are stipulated approximations to individual atomic masses, not measured masses of molecules or natural-abundance averages. They giveThe symbol $\mathrm{D}$ means deuterium, ${}^2\mathrm{H}$. Doubling the light partner's approximate atomic mass therefore produces a large predicted shift, whereas changing the already heavy chlorine partner by two mass units produces a much smaller one. A quantitative comparison with infrared spectroscopy requires actual isotope-specific atomic masses, fitted band parameters, and uncertainties, rather than identifying any arbitrary peak with ${\tilde{\nu }}_{\mathrm{harm}}$.

Molecular rotation changes a molecule's orientation. Its rotational kinetic energy depends on the arrangement and masses of its chemical atoms. A rigid rotor freezes internal distances to isolate this motion; rotational spectroscopy tests the resulting energy eigenvalues through transitions between quantum states. The molecular beam electric resonance experiment provides a complementary probe of the coupling to an electric field.

Molecular vibration changes internal separations and bond angles, unlike the fixed internal molecular geometry of a rigid rotor. Its coupling to molecular rotation makes measured rotational constants depend on the vibrational quantum state.

A diatomic molecule has one internal stretching coordinate: its two atomic nuclei approach and separate about an equilibrium bond length. The diatomic bond stretch turns this motion into a particularly useful model. The local potential energy curvature supplies the restoring force, while the reduced mass accounts for both moving partners. Thus the same chemical bond can vibrate more slowly after an isotope substitution even when its electronic structure is nearly unchanged.

This separation makes infrared spectroscopy informative: transition frequencies constrain chemical bond stiffness, and isotope scaling of diatomic vibration tests the mass dependence. The University of Florida HCl and DCl FTIR experiment illustrates the measurement. Detectability also requires vibrational infrared activity; having vibrational energy eigenvalues alone does not guarantee an infrared absorption line.

A molecule, using the International Union of Pure and Applied Chemistry (IUPAC) convention, is an electrically neutral entity containing two or more chemical atoms. Its constituents form a bound unit through chemical bonds; merely drawing a boundary around neighboring particles does not make a molecule. The neutrality requirement concerns total electric charge, not the absence of internal charge separation: water is a polar molecule with zero total electric charge.^[ref]

One water molecule, ${\mathrm{H}}_2\mathrm{O}$, contains two hydrogen and one oxygen chemical atom. One molecular oxygen molecule, ${\mathrm{O}}_2$, contains two oxygen chemical atoms; a molecule need not contain different chemical elements. One carbon dioxide molecule, $\mathrm{C}{\mathrm{O}}_2$, contains one carbon and two oxygen chemical atoms. The molecular formula records these inventories, while molecular geometry describes their arrangement. Water and carbon dioxide are chemical compounds; molecular oxygen is a form of one chemical element.

Ordinary crystalline sodium chloride, $\mathrm{NaCl}$, contains an extended arrangement of sodium cations $\mathrm{N}{\mathrm{a}}^{+}$ and chlorine anions $\mathrm{C}{\mathrm{l}}^{-}$. Its formula unit specifies one of each for compositional accounting, without selecting a unique isolated pair. Bulk neutrality therefore does not imply a material consists of neutral molecules. This statement concerns the ionic crystal; the physical state and conditions matter when deciding what particles a chemical formula represents.^[ref]

A polyatomic ion is a charged bound group of chemical atoms, such as hydronium, ${\mathrm{H}}_3{\mathrm{O}}^{+}$. A molecular ion formed by removing an electron from intact water, ${\mathrm{H}}_2{\mathrm{O}}^{\bullet +}$, retains the same atom inventory but has net electric charge $+e$. Neither satisfies the neutral-molecule definition. Distinguishing these entities makes molecular count and amount of substance and mass spectrometry interpretable: say whether the count is of chemical atoms, neutral molecules, ions, or formula units.

The useful surprise is that a molecule can be treated as one particle for motion and counting while retaining internal structure. Its molar mass connects a particle inventory to bulk mass; molecular rotation and molecular vibration reveal internal organization. The continuum approximation uses bulk fields in place of individual molecular positions and motions.

This is a rovibrational spectrum: the vibrational excitation is accompanied by quantized molecular rotation. The high-wavenumber R branch has $\Delta J=+1$, and the low-wavenumber P branch has $\Delta J=-1$. In the simplest common-$B$ approximation,Consecutive lines within either branch are therefore separated by roughly $2B$, while the gap between P(1) and R(0) is roughly $4B$. There is no Q branch: for this diatomic parallel $\Sigma \to \Sigma$ transition, the electric-dipole rotational selection rule allows $\Delta J=\pm 1$, and the $\Delta J=0$ line strength vanishes.

The intensity envelope comes from rotational populations, approximately $(2J+1)e^{-hcBJ(J+1)/(k_{B}T)}$, together with the rotational line strengths. The paired peaks arise from the two chlorine isotopologues, ${\mathrm{H}}^{35}Cl$ and ${\mathrm{H}}^{37}Cl$: their different reduced masses shift both rotational and vibrational spacings. The more abundant ${}^{35}\mathrm{Cl}$ gives the stronger member. The changing spacing away from the band centre reflects $B_1\ne B_0$ and, at higher $J$, centrifugal distortion; it is not evidence for additional vibrational fundamentals.

Absorption depends on an allowed transition with a nonzero transition dipole moment and appreciable line strength at wavelengths carrying terrestrial thermal radiation. Molecular vibration and molecular rotation, level populations and pressure/temperature-dependent line shapes determine the spectrum. A band in an atmospheric window can be more effective incrementally than one already strongly absorbed by water or carbon dioxide. Symmetry can make a vibration infrared inactive even when the molecule has many normal modes. Atmospheric lifetime affects the persistence of a released gas, not the intrinsic absorption strength of one molecule.

Exchange constraint. Exchange of two identical particles cannot change physical predictions. For the usual three-dimensional bosonic or fermionic state space, the exchange operator obeys $P_{12}^2=1$, so its allowed eigenvalues are $\pm 1$. bosons use the symmetric overall wavefunction, $P_{12}\Psi =+\Psi$; fermions use the antisymmetric overall wavefunction, $P_{12}\Psi =-\Psi$. Exchange includes the spatial coordinates and all internal labels. The identity $P_{12}^2=1$ alone does not prove which particle species has which statistics; that assignment is the Spin-statistics theorem.

Factored fermion states. If $\Psi ={\psi }_{\mathrm{space}}{\chi }_{\mathrm{spin}}$ and both factors have definite exchange symmetry, their exchange signs must multiply to $-1$. Thus a symmetric spatial factor requires an antisymmetric spin factor, and an antisymmetric spatial factor requires a symmetric spin factor. For two spin-one-half particles, the former spin factor is the singlet $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$; the latter factors are the three triplet states. A more general entangled spin-space state need not factor, but must still be antisymmetric overall.

hydrogen spin isomers. The two protons are identical spin-one-half fermions. parahydrogen has total nuclear spin $I_{\mathrm{n}}=0$, the antisymmetric singlet; orthohydrogen has $I_{\mathrm{n}}=1$, the symmetric triplet. Here $I_{\mathrm{n}}$ is a spin quantum number, distinct from the moment of inertia used below.

Rotational restriction. In the electronic and vibrational ground state, these two factors are symmetric under nuclear exchange. Exchanging the protons reverses the internuclear axis, $(\theta ,\varphi )\mapsto (\pi -\theta ,\varphi +\pi )$. The spherical harmonic rotational state contributes the exchange sign $(-1{)}^{\ell }$. Overall antisymmetry therefore requiresHence parahydrogen has even $\ell$; orthohydrogen has odd $\ell$. The $\ell$ here labels molecular rotation, rather than the angular momentum obtained by also coupling the nuclear spins.

Conversion heat. The ortho-para conversion heat follows from the linear rigid rotor approximation, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ with $I=\mu r^2\simeq m_p{r}^2/2$. Thus conversion of one molecule from $\ell =1$ to $\ell =0$ releasesNeglecting the small electron contribution to the molecular mass, a sample of mass $M$ contains $M/(2m_p)$ molecules. Only the initial orthohydrogen fraction releases this energy, givingThe estimate uses $r=74$ pm, $m_p\simeq 1.67262\times 10^{-27}$ kg and $\hslash \simeq 1.05457\times 10^{-34}$ J s. Handbook-level rounding and the rigid rotor approximation justify about three significant figures, not a high-precision thermochemical prediction.^[ref]

Storage consequence. The internal conversion is a delayed heat source even in a well insulated container, causing warming or additional evaporation of the liquid. Insulation reduces external heat transfer but does not remove this source. Converting to predominantly parahydrogen during liquefaction, while removing the released heat, reduces the subsequent storage heat load. No evaporation fraction can be calculated without a latent heat and a model for how the heat is removed.

For an ideal gas of fixed molecular composition, translational energy is supplemented by rotational, vibrational and eventually electronic excitations as their level spacings become thermally accessible. Activating more internal degrees of freedom generally increases the heat capacity. In a harmonic vibrational mode the contribution rises from nearly zero when $k_{B}T\ll \hslash \omega$ to its classical limit when $k_{B}T\gg \hslash \omega$; summing modes with different frequencies gives a smooth increase over an intermediate temperature range.

A monatomic gas has no molecular rotations or vibrations, so its mass-specific $c_p$ is nearly constant until electronic excitation or ionization becomes important: $B$ is then small. A simple diatomic gas can also have a small slope when its rotations are already active but its vibrational modes remain frozen. Large flexible polyatomic molecules have many internal modes, often with relatively low excitation energies, and can give a larger slope over their activation range.

The linear law is a local approximation, not an asymptotic result. A finite set of modes approaches finite classical heat-capacity contributions, whereas $A+BT$ grows without bound. At still higher temperatures dissociation and ionization change the composition and gas constant, invalidating the fixed-molecule ideal-gas model as well. The constants have units $[A]=\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $[B]=\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-2}$.

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.

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.

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.

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.