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?

  • $\begingroup$ Do you mean electronic excited state ? If not look at this answer chemistry.stackexchange.com/questions/70715/… $\endgroup$
    – porphyrin
    Commented Apr 6, 2017 at 21:16
  • $\begingroup$ Thank you but that does not answer my question. I am talking about the excited vibrational state and why nuclear spin statistics appear not to affect this state (i.e they only effect the ground state). $\endgroup$
    – RobChem
    Commented Apr 6, 2017 at 21:19
  • $\begingroup$ @RobChem As the energy scales responsible for either IR spectroscopy and NMR spectroscopy are so remote from each other, and routine ATR-FTIR spectrometers have a resolution of about 0.5 wavenumbers, I am at least surprised other IR spectroscopes were able to discern the (in comparision) tiny differences of energy levels, caused by the coupling with different nuclear spins of the sample. – At least as the original question does not mention an external magnetic field excersing the Zeeman splitting on the later. What did I miss here? $\endgroup$
    – Buttonwood
    Commented Apr 6, 2017 at 21:30
  • $\begingroup$ This has absolutely nothing to do with NMR. $\endgroup$
    – RobChem
    Commented Apr 6, 2017 at 21:31
  • $\begingroup$ @RobChem Thank you, this lowers / relaxes my eyebrow "back to ground state". $\endgroup$
    – Buttonwood
    Commented Apr 6, 2017 at 21:33

1 Answer 1


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).


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