# Does the triplet sigma state of a diatomic molecule experience spin-orbit coupling?

I know that states with spin S=0 in a diatomic molecule have no spin orbit coupling, independent on the value of the projection of the total electronic angular momentum.

I expect the same is true if the absolute value of $\Lambda$ is equal to zero independent on the spin of the diatomic molecule.

Is it correct that a diatomic molecule with a $^3\Sigma$ state has no spin orbit coupling?

Yes I would think so because the projection of orbital angular momentum is zero in forming $\Omega=\Lambda + \Sigma$. This would seem to correspond to Hund's case (b) where $\Lambda =0$ but $S \ne 0$, so no spin orbit coupling (unlike Hund's case (a)(c)(d)). The spin angular momentum remains fixed in space and the molecule rotates under it. If $R$ is the whole body (molecular) angular momentum this couples with the spin so that each rotational level splits into $2S+1$ components and the total angular momentum $J$ takes values from $R+S$ to $R-S$. The splitting is scaled with a spin rotation constant rather than spin-orbit constant but in $^3\Sigma$ there is also a $s_1\cdot s_2$ spin-spin coupling term.
First, I think one needs to specify what you mean when you say that a certain electronic state, e.g. one with $$S=0$$, "has no spin orbit coupling".
In a less strict interpretation, we might merely require a weak dependence on the spin-orbit operator. In practice this usually means that, in a perturbation-theory treatment, the first-order effect of the spin-orbit operator is zero. In this sense you a right: in diatomic molecules all singlet states (any $$\Lambda$$) and all $$\Sigma$$ states (any multiplicity) have a zero first-order spin-orbit energy shift.
In the specific case of $$^3\Sigma^\pm$$ states, as I said above the effect of spin-orbit is zero at first order of perturbation, so it can be said that they "have no spin orbit coupling". At higher order, they have non-zero matrix elements with the following terms: $$^{3,5}\Sigma^\mp$$, $$^{1,3,5}\Pi$$. If a state with one of these symmetries is energetically close it'll have some effect on the $$^3\Sigma^\pm$$.