I'm aware that the Q branch ($\Delta j = 0$) is usually not observed in rotational-vibrational spectra of diatomic molecules in the gas state, however, I've heard different things about when exactly this transition is forbidden.

I've sometimes been told that the transition is not allowed in general, for any diatomic molecule, but I've sometimes heard that this is only true if the electronic state has a $\Lambda \neq0$, so which is the correct selection rule here?

I'd love to have this clarified for me. If it is possible I would also like to know when is a Q branch observed in vibronic spectra of heteronuclear and homonuclear diatomics.

  • $\begingroup$ You are correct in that the electronic angular momentum $\Lambda \ne 0$ for a Q branch to be observed in a diatomic molecule, i.e. the molecule is not in a $\Sigma$ state. The rule is then $\Delta J =0; \Delta v = \pm1$. (Homo-nuclear diatomic molecules do not have an ir spectrum as $d\mu/dx = 0$, as the dipole is zero) $\endgroup$
    – porphyrin
    Sep 22, 2017 at 14:32
  • 1
    $\begingroup$ ahhh, I know homo nuclear diatomics don't have an IR spectrum, there was a typo in the question, thanks. Are you sure about the $ \Lambda$ selection rule? is a Q branch always permitted if $\Lambda \neq 0$? Is there some easy derivation of this or somewhere where I could find a derivation? I can't see where this comes from from a simple transition moment integral. $\endgroup$
    – Ignacio
    Sep 22, 2017 at 15:47
  • $\begingroup$ A photon carries one unit of angular momentum which is conserved. When the photon is destroyed on absorption, this ang mom must be transferred to the molecule in one way or another, in your case $\Lambda_i +1=\Lambda_f $ where i and f are initial and final states. This is the basis of the transition moment integral and so selection rules. $\endgroup$
    – porphyrin
    Sep 24, 2017 at 7:21
  • $\begingroup$ I understand that argument, however, $L$ is not really a good quantum number in diatomic molecules and $\Lambda$ is not the angular momentum, but the projection of the angular momentum in the direction of the internuclear axis, so I don't think that the fact that the total anguar momentum is conserved helps a lot. $\endgroup$
    – Ignacio
    Sep 25, 2017 at 15:12
  • $\begingroup$ In Hydrogen, for instance, although $\Delta l$ is constrained to be $\pm 1$, $\Delta m_l $, the z projection, can be zero, and this doesn't break angular momentum conservation. $\endgroup$
    – Ignacio
    Sep 25, 2017 at 15:26

1 Answer 1


To determine the intensity of a rotation-vibration transition, you need to consider the transition momentum integral $\int \psi_f \hat{\mu} \psi_i d\tau$, where $\psi_f$ is the final state, $\psi_i$ is the initial state, $\hat{\mu}$ is the transition dipole moment operator, and $d\tau$ indicates integration over all space. Assuming separation of vibrational and rotational motion, this factors into the product of two terms: one that specifies the intensity of the vibrational transition and another that specifies the intensity of the rotational transition. The vibrational integral leads to a selection rule $\Delta v = \pm 1$ (in the harmonic approximation) and the rotational integral leads to a selection rule $\Delta J = \pm 1$. So no Q branch occurs. However, if there is some additional source of angular momentum (e.g., from vibrations or electronic motion) then it is possible to have a Q branch because now that extra motion can contribute to the total angular momentum in a way that allows the Q branch to occur.

So here is the rule that you ask for: a Q branch will not occur in transitions between $\Sigma$ electronic states of a diatomic molecule, but can occur in transitions between $\Sigma$ and $\Pi$ states. Since vibration-rotation transitions by definition occur within a single electronic state, Q branches do not occur in vibration-rotation spectra of diatomic molecules in a $\Sigma$ state. In a polyatomic molecule, however, vibrational modes can carry angular momentum so even within a single electronic state that carries no electronic angular momentum it may be possible to observe Q branches in a vibration-rotation spectrum so long as the vibration being excited is one which carries nuclear orbital angular momentum (e.g., from a vibrationless ground state to a vibrationally excited bending mode).

  • $\begingroup$ I think this is a solid answer. It might be nice if you can expand a bit on the vibrational and rotational integrals in the first paragraph. While I know Q branches can exist, I'd love to know if there are example molecules / spectra in which they have been observed. $\endgroup$ Dec 21, 2022 at 6:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.