Effect of nuclear spin statistics on the IR spectrum of CO2 Specifically, I am talking about $\ce{C^{16}O2}$. With a nuclear spin of zero meaning, it is a boson - it's total wavefunction must be totally symmetric. The total wavefunction is made up of an electronic, vibrational, rotational and a nuclear spin contribution. For this molecule the electronic and vibrational (in ground state) wavefunctions are symmetrical. There is no antisymmetric nuclear spin wavefunction so only the even J rotational levels exist. 

Now consider the degenerate bending vibrational mode. Excitation of this mode leads to excitation into the excited vibrational state of this mode and the IR spectrum has a corresponding P, R, and crucially a Q branch (due to additional angular momentum from the degenerate bending mode). 

I have been asked whether there are similar J restrictions in the excited state as there are in the ground state. I initially jumped into assessing the symmetry of the contributions to the wavefunctions and concluded that the vibrational mode is antisymmetric and thus the corresponding excited state wavefunction is antisymmetric leaving the possibility of only odd J values in the excited state. However, there is a Q branch so this must be incorrect since the Q branch indicates the presence of J=0 in the excited state.

Why is there seemingly no restriction on the excited state?
Because the degenerate bending mode has vibrational angular momentum, it is a vibrational $\Pi_\text{u}$ state and just as in electronic $\Pi$ states there exist levels with $+$ and $-$ parity. As a consequence, the degenerate bending mode has $\Pi_\text{u}^\pm$ symmetry and the vibrational wave functions can be both symmetric and antisymmetric. 

See also Herzberg Vol 2 (p372 and Fig 99 on p373).