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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.

The two protons are identical spin-one-half fermions, so the total state is antisymmetric under exchange. In the electronic and vibrational ground state, an even rotational $J$ has a symmetric spatial wavefunction and therefore uses the antisymmetric nuclear-spin singlet state; odd $J$ uses the symmetric triplet state. Thus parahydrogen has one nuclear-spin state, and orthohydrogen has three. Each rotational level also has spatial state degeneracy $2J+1$.

The translational partition functions and all common electronic/vibrational factors cancel in the ratio of equilibrium populations. Using the same zero of energy for the two spin species givesSubtracting the $J=1$ energy separately from all odd levels without restoring that offset in the equilibrium ratio would incorrectly remove the low-temperature preference for parahydrogen. The nuclear spin statistics account for the factor three, independently of the spatial factor $2J+1$.

The spin isomers of hydrogen illustrate how nuclear spin statistics constrain the product of electronic, vibrational, rotational and nuclear-spin exchange symmetries. In the ground electronic and vibrational states of these homonuclear molecules, the first two factors are symmetric. Exchanging the nuclei reverses the internuclear direction, so a rotational state has parity $(-1{)}^J$.

For ${}^1{\mathrm{H}}_2$, the protons are spin-one-half fermions and the total nuclear-exchange wavefunction must be antisymmetric. The antisymmetric spin singlet pairs with even $J$ (para-hydrogen, one spin state); the symmetric spin triplet pairs with odd $J$ (ortho-hydrogen, three spin states). For ${}^2{\mathrm{H}}_2$, spin-one deuterons are bosons, so the total must be symmetric. Nuclear spins $I=0,2$ supply six symmetric states paired with even $J$ (ortho-deuterium); $I=1$ supplies three antisymmetric states paired with odd $J$ (para-deuterium). The high-temperature ortho/para statistical ratios are $3:1$ for hydrogen and $2:1$ for deuterium, while low-temperature equilibrium populations also depend on rotational energies and whether spin conversion can occur.

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.