Does the triplet sigma state of a diatomic molecule experience 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.