= Rotational state populations
{title2=$P_J=(2J+1)e^{-hBJ(J+1)/(k_{\rm B}T)}/Q_{\rm rot}$}
For the spin-free <linear rigid rotor> in <thermal equilibrium> at <temperature> $T>0$, assume all integer $J\geq0$ 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> gives
$$
Q_{\rm rot}=\sum_{J=0}^{\infty}(2J+1)e^{-hBJ(J+1)/(k_{\rm B}T)},\qquad
P_J=\frac{(2J+1)e^{-hBJ(J+1)/(k_{\rm B}T)}}{Q_{\rm rot}}.
$$
Here $Q_{\rm rot}$ is a rotational <partition function>, $h$ the <Planck constant>, $k_{\rm B}$ the <Boltzmann constant>, and $B$ a <frequency> in $\mathrm{Hz}$. Thus $hB/(k_{\rm B}T)$ 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.<Absorption lineshapes by Andrei Tokmakoff>[Andrei Tokmakoff, §§6.1–6.2, printed pp. 6-2–6-6, equations (6.8)–(6.10), (6.27)–(6.29): population difference and rotational dipole matrix elements]{ref}
For an isotropic sample with <polarization of an electromagnetic wave> along $z$, write $d_z=d\cos\theta$. Normalized <spherical harmonics> give
$$
|\langle J+1,M|\cos\theta|J,M\rangle|^2
=\frac{(J+1)^2-M^2}{(2J+1)(2J+3)},\qquad
\sum_{M=-J}^{J}|\langle J+1,M|\cos\theta|J,M\rangle|^2=\frac{J+1}{3}.
$$
The 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 areas
$$
A_J\propto\frac{\nu_J(J+1)}{Q_{\rm rot}}e^{-hBJ(J+1)/(k_{\rm B}T)}\left(1-e^{-h\nu_J/(k_{\rm B}T)}\right),\qquad \nu_J=2B(J+1).
$$
The 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.
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