= Solution
\b[The Stokes vibrational band has O, Q and S branches], with $\Delta N=-2,0,+2$, respectively. Here $N$ denotes end-over-end <rotational angular momentum>; treating electronic fine structure as unresolved gives the usual elementary rotor calculation. The absolute incident <wavenumber> is
$$
\widetilde\nu_L=\frac{10^7}{1064}=9398.496\ \mathrm{cm}^{-1},
$$
and energy conservation for <Stokes Raman scattering> gives
$$
\widetilde\nu_s=\widetilde\nu_L-1580-B\{N'(N'+1)-N(N+1)\}.
$$
For the actual $^{16}\mathrm O_2$ ground state, only odd $N$ have nonzero nuclear-spin statistical weight. Thus the lowest initial levels are $N=3$ for O and $N=1$ for Q and S. Ignoring the small change in $B$ upon vibration, the lines are:
|| Branch
|| Lowest transition
|| Rotational energy gained, in $B$
|| Absolute scattered wavenumber, $\mathrm{cm}^{-1}$
| O
| $3\to1$
| $-10B$
| \b[7832.5]
| Q
| $1\to1$
| $0$
| \b[7818.5]
| S
| $1\to3$
| $10B$
| \b[7804.5]
The odd-$N$ restriction follows from the zero nuclear spin and exchange symmetry of $^{16}\mathrm O_2$; the electronic spin splits each such $N$ into several $J$ levels without changing this simplified branch-centre calculation.<O2 A-band line parameters to support atmospheric remote sensing Part II by David A. Long and others>[Section 3, manuscript page 4: only odd lower-state N has nonzero statistical weight for oxygen-16 and oxygen-18]{ref}
If the phrase simplest approximation is instead taken to discard nuclear-spin statistics as well, a fictitious unrestricted rotor gives O $2\to0$ at $7826.9\ \mathrm{cm}^{-1}$ and S $0\to2$ at $7810.1\ \mathrm{cm}^{-1}$, with Q still at $7818.5\ \mathrm{cm}^{-1}$. Those even-$N$ lines are not physical ground-state $^{16}\mathrm O_2$ lines.
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