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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?

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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.

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