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A heteronuclear diatomic molecule consisting of carbon and sulfur. Its permanent electric dipole moment permits an electric-dipole rotational spectrum. The linear rigid rotor model relates its rotational constant to its reduced mass and bond length.

A diatomic molecule contains two chemical atoms; hydrogen chloride and carbon monoxide are examples. Their fixed-separation moment of inertia is $I=\mu r^2$ in the linear rigid rotor model.

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.

A linear molecule has collinear atomic nuclei in its idealized molecular geometry. Both a diatomic molecule and carbon dioxide can supply linear rigid rotor examples, although their electric-dipole selection rules need not give an observable permanent-electric dipole spectrum.

For a diatomic molecule, let $m_1,m_2$ be the two masses and $r$ their fixed separation. The distances from the centre of mass are $r_1=m_{2}r/(m_1+m_2)$ and $r_2=m_{1}r/(m_1+m_2)$. A perpendicular axis through the centre of mass therefore has moment of inertiaThe reduced mass $\mu$ here is not the electric dipole moment, for which we use $\mathbf{d}$. For a longer linear molecule, replace $\mu r^2$ by ${\sum }_i{m}_i{\rho }_i^2$, where ${\rho }_i$ is each distance along the axis from the centre of mass. In the point-mass idealization, the axial moment of inertia is zero; turning about that axis changes no nuclear positions, so no independent axial rotational degree of freedom is included.

For the field-free linear rigid rotor, assume a fixed internal quantum state with no electronic angular momentum or electronic spin, and omit couplings to nuclear spin. The rotational Hamiltonian is ${\hat{H}}_{\mathrm{rot}}={\hat{\mathbf{L}}}^2/(2I)$. Its orientation wavefunctions are spherical harmonics $Y_{JM}$, and its angular momentum eigenstates satisfywhere $J=0,1,2,\dots$ and $M=-J,-J+1,\dots ,J$. Each $E_J$ belongs to $2J+1$ angular momentum eigenstates distinguished by the magnetic quantum number $M$, an instance of degenerate energy levels. The Planck constant is $h=2\pi \hslash$; $B$ is the frequency-valued rotational constant. Notice that the energy eigenvalues scale as $J(J+1)$, not $J^2$.

Subtracting consecutive energy eigenvalues givesThus increasing level gaps produce a regularly spaced series of lines at $2B,4B,6B,\dots$, with spacing $2B$. Allowed positions do not guarantee that every line is visible: electric-dipole selection rules, occupation of the initial quantum state, and instrument response also matter.

The rotational state populations of a spin-free polar linear rigid rotor are proportional to $(2J+1)e^{-{\theta }_{\mathrm{rot}}J(J+1)/T}$. Differentiating their logarithm with respect to continuous $J$ gives the displayed maximum when it is nonnegative. The actual maximum must be selected among nonnegative integers by comparing adjacent populations. This population maximum generally differs from a maximum in measured spectral line intensity.

For hydrogen approximated as a linear rigid rotor, the $\ell =1$ to $\ell =0$ energy decrease is ${\hslash }^2/I$, with moment of inertia $I\simeq m_p{r}^2/2$. A sample of mass $M$ initially containing an orthohydrogen fraction $x_o$ therefore releases $Q\simeq x_{o}M{\hslash }^2/(m_p^2{r}^2)$ on complete conversion to the lowest parahydrogen level. This estimate neglects the electron contribution to molecular mass and changes of the rotational constant with environment. It describes an internal heat load that external insulation cannot remove.

Parahydrogen uses the antisymmetric spin-one-half singlet state,Here $I_{\mathrm{n}}$ labels total nuclear spin; it is not the moment of inertia $I$ in the linear rigid rotor formula. There is one nuclear-spin state for every rotational quantum state. The lowest allowed rotational level is $J=0$, and the next is $J=2$. The description “opposite spins” alone is insufficient: the relative minus sign distinguishes this singlet state from the zero-projection member of the triplet state.

For the linear rigid rotor, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ and the electric electric dipole moment interaction is $V=-E{\hat{d}}_z$. The isotropic $\ell =0$ state has no first-order shift. Its only coupled rotational state is $\ell =1,m_{\ell }=0$, at gap ${\Delta }_{\mathrm{rot}}={\hslash }^2/I$, giving the quadratic Stark shift of a polar linear rotorUsing $1\mathrm{kV}\mathrm{c}{\mathrm{m}}^{-1}=10^5\mathrm{V}{\mathrm{m}}^{-1}$ and the supplied I and d givesThe negative sign is the expected second-order Stark effect of a nondegenerate ground state. The mixing parameter $dE/(\sqrt{3}{\Delta }_{\mathrm{rot}})\simeq 0.074$ is small, supporting the weak-field estimate. The constants h and $\hslash =h/(2\pi )$ use their defined SI values; no molecular input beyond the supplied data is needed.

For an axis through the center of mass perpendicular to the bond, the point-mass moment of inertia isThe prefactor is the reduced mass $\mu$; this is the standard reduction of a two-body linear rigid rotor to $I=\mu r^2$.

Place the carbon and sulfur nuclei on a line separated by $r$, with masses $m_{\mathrm{C}}$ and $m_{\mathrm{S}}$. The center of mass lies between them and closer to the heavier sulfur nucleus. In the linear rigid rotor approximation, $r$ remains fixed while the molecule rotates.

The rotational state populations start at $N_0/N_0=1$. Their $2J+1$ statistical weight factor initially favors higher $J$, while the Boltzmann factor eventually suppresses them, producing one maximum followed by a tail tending to zero. The physically allowed values occur at integer $J$; a continuous curve only guides the eye.

A planar rigid rotor has angular eigenfunctions $e^{im\varphi }$ and single-valuedness imposes $m=0,\pm 1,\pm 2,\dots$. Its energy depends on $m^2$, so if the level is labelled by $J=|m|$, thenThe two states at $J>0$ represent opposite directions of angular momentum. If the signed quantum number itself is called $J$, each signed state is counted once instead; one must not both sum over both signs and multiply by two again.

For an unrestricted three-dimensional linear rigid rotor, $J=0,1,2,\dots$ has $M=-J,\dots ,J$ and state degeneracy $2J+1$, with $E_J={\hslash }^{2}J(J+1)/(2I)$. Planar confinement removes one rotational coordinate and changes both the spectrum and degeneracy. These are the spatial counts before nuclear spin statistics. For homonuclear hydrogen, the allowed even/odd spatial states must also be paired with the appropriate parahydrogen or orthohydrogen nuclear states, as considered explicitly in A8.

Define ${\theta }_{\mathrm{rot}}={\hslash }^2/(2I{k}_{\mathrm{B}})$. Summing the spatial states counted in part (c),When $T\gg {\theta }_{\mathrm{rot}}$ the closely spaced terms can be replaced by a Gaussian integral:Here labelled means that the two ends of the molecular axis were distinguished in the angular count. In the usual high-temperature molecular convention that removes equivalent orientations and factors out nuclear-spin multiplicity, divide by the rotational symmetry number $s=2$ for homonuclear hydrogen:The full exact spin isomers of hydrogen count is instead the even-$m$ spatial sum plus three times the odd-$m$ sum; its high-temperature limit is $2\sqrt{\pi T/{\theta }_{\mathrm{rot}}}$, including the four nuclear-spin states. These conventions differ only by temperature-independent factors in this limit. All give $U_{\mathrm{rot},\mathrm{m}}=RT/2$ and $C_{V,m}^{\mathrm{rot}}=R/2$, consistent with one quadratic rotational degree of freedom. By comparison, the free linear rigid rotor has $q_{\mathrm{rot}}\propto T$ and $C_{V,m}^{\mathrm{rot}}=R$.

For a high-temperature linear rigid rotor, $q_{\mathrm{rot}}=T/{\theta }_{\mathrm{rot}}$ up to temperature-independent symmetry/spin factors. Therefore $q^{\prime }/q=1/T$ and $q^{\prime \prime }=0$, givingThis agrees with the equipartition theorem for two quadratic rotational coordinates. It is the rotational contribution; adding free three-dimensional translation gives $C_{V,m}=5R/2$ when vibrations remain unexcited. A constant rotational symmetry number does not affect this derivative.

A polar diatomic molecule can absorb microwaves through its permanent electric dipole moment. In the linear rigid rotor approximation,The electric-dipole selection rule is $\Delta J=+1$ in absorption. It givesso adjacent lines are separated by $2B_0$. ThereforeAssign a line near $m\times 21.999\mathrm{GHz}$ to $J=m-1\to m$ for positive integer $m$. The first member is $0\to 1$, the next $1\to 2$, and so on; if the recorded window misses the lowest lines, their absolute frequencies fix the integer offset. At high temperature many rotational states are populated, so the strongest line need not be $0\to 1$.

This is the leading rotational pattern, not a claim that real ZnF lacks spin or hyperfine structure. Fine splittings and centrifugal distortion can perturb the ideal lines; they do not explain the three systematically displaced progressions as economically as vibrational excitation.

For a linear rigid rotor, dividing the angular momentum squared eigenvalue by twice the moment of inertia givesThe quantum degeneracy counts the values $M=-J,\dots ,J$ with the same energy. It excludes nuclear-spin multiplicity and any fine or hyperfine splitting absent from this model.

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.

The rotational kinetic energy is $K=L^2/(2I)$. For a three-dimensional linear rigid rotor, angular momentum operator replaces the squared magnitude by ${\hslash }^{2}l(l+1)$, where $l=0,1,2,\dots$. HenceThe three lowest distinct energy levels are thereforeTheir magnetic quantum degeneracy values are $1,3,5$. A single moment of inertia specifies this rigid rotor model; an arbitrary asymmetric body would require its other principal moments.

For the supplied linear rigid rotor spectrum, sum over every nonnegative integer $J$ and include all $2J+1$ magnetic substates:It is convenient to introduce the rotational temperatureso that the first terms are $Z_{\mathrm{rot}}=1+3e^{-2{\theta }_r/T}+5e^{-6{\theta }_r/T}+\cdots$. Here $I$ is the moment of inertia and $\hslash$ is the reduced Planck constant, as confirmed in the original PDF. The stipulated degeneracies are used as given, without adding nuclear-spin weights or omitting selected $J$ values.