Viktor Sukhorukov is correct, of course. The terms of simple MO theory are normally constructed assuming Russell-Saunders coupling of the angular momenta, not of the j-j coupling that might be taken into consideration for High-Z nuclei. In order to discuss spin-orbit coupling, one usually considers the double symmetry group, ie, the double group of C_{\infinity,v} in the case of heteronuclear diatomics. For singlet and triplet terms, one may consider the transformation properties of the spin components (operators) . In C_{\infinity,v} they transform like the rotations R_x, R_y, R_z, that is, according to representations A_2 and E_1, compare

http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=1000&option=4

The spin-part of Singlet terms transforms according to the totally symmetric representation A_1, while that of Triplet terms transforms according to A_2 + E_1. Since the spatial part of sigma+ terms transforms according to A_1, but that of sigma- according to A_2, any singlet sigma+ transforms according to A_1xA_1 = A_1. Thus S_z that transforms like A_2, cannot couple any singlet sigma+ to any singlet sigma+. Similarly for S_x, S_y that form a basis for E_1. Since in Russell-Saunders coupling involves basically not the operators S but the product L.S, one has also to consider the operators L_z, and L_x,L_y that also transform like the rotations, in order to obtain the complete selection rules for spin-orbit coupling for singlet and triplet terms with orbital parts transforming as A_1, A_2, E_1, .... E_n, .... Here, one may also regard the spin-orbit coupling as an (approximate) one-electron operator, and observe that diagonal elements of L_x,L_y,L_z vanish in the MO basis.

Regarding the question whether singlet sigma+ can couple to singlet sigma- via spin-orbit coupling: The answer should be NO. The former transform like A_1 in the double group, the latter like A_2. Since the total spin-orbit coupling operator is totally symmetric, i.e., transforms like A_1, it cannot couple singlet sigma+ with any singlet sigma-.

The spin-orbit operator is proportional to the product L*S. Therefore if both spins (S_a and S_b) in matrix element (a | L*S | b) equals zero (in other words between singlet states) then the matrix element (a | L*S | b)=0.

The treatment of spin-orbit coupling in molecules is not trivial. We have discussed this problem without giving full details on pages 158-162 in our classical monograph " Many-electron Systems: Properties and Interactions " By Serafin Fraga and Gulzari Malli, W.B.Saunders,Philadelphia,1968. This book is available in some university libraries.

Our treatment is based upon the work of Dr.L.Y.C.Chiu and Dr. Peter Fontana ( the references 8 and 15 given on page 162). All of us were in Prof. Mulliken's Lab at the University of Chicago during 1960-1964. GM was a graduate student and Drs. Chiu, Fontana and Fraga ( now deceased) were PDF Research Associates. It is not easy reading but so is theoretical molecular spectroscopy. Good luck. Gulzari Malli

Viktor Sukhorukov is correct, of course. The terms of simple MO theory are normally constructed assuming Russell-Saunders coupling of the angular momenta, not of the j-j coupling that might be taken into consideration for High-Z nuclei. In order to discuss spin-orbit coupling, one usually considers the double symmetry group, ie, the double group of C_{\infinity,v} in the case of heteronuclear diatomics. For singlet and triplet terms, one may consider the transformation properties of the spin components (operators) . In C_{\infinity,v} they transform like the rotations R_x, R_y, R_z, that is, according to representations A_2 and E_1, compare

http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=1000&option=4

The spin-part of Singlet terms transforms according to the totally symmetric representation A_1, while that of Triplet terms transforms according to A_2 + E_1. Since the spatial part of sigma+ terms transforms according to A_1, but that of sigma- according to A_2, any singlet sigma+ transforms according to A_1xA_1 = A_1. Thus S_z that transforms like A_2, cannot couple any singlet sigma+ to any singlet sigma+. Similarly for S_x, S_y that form a basis for E_1. Since in Russell-Saunders coupling involves basically not the operators S but the product L.S, one has also to consider the operators L_z, and L_x,L_y that also transform like the rotations, in order to obtain the complete selection rules for spin-orbit coupling for singlet and triplet terms with orbital parts transforming as A_1, A_2, E_1, .... E_n, .... Here, one may also regard the spin-orbit coupling as an (approximate) one-electron operator, and observe that diagonal elements of L_x,L_y,L_z vanish in the MO basis.

Regarding the question whether singlet sigma+ can couple to singlet sigma- via spin-orbit coupling: The answer should be NO. The former transform like A_1 in the double group, the latter like A_2. Since the total spin-orbit coupling operator is totally symmetric, i.e., transforms like A_1, it cannot couple singlet sigma+ with any singlet sigma-.

Thank you all for your answers. I have also consulted various references and I agree that the matrix elements is ZERO. Victor: the Hamiltonian A(r) L.S is a phenomenological hamiltonian which is widely used by experimentalists but it has different selection rules than the 'true' microscopic hamiltonian derived from the Breit-Pauli equation. For example, using that form would lead to a zero matrix element for , which is incorrect. So I wouldn't consider conclusive conclusions obtained with this simplified hamiltonian (refs: C.M. Marian, Rev. Comp. Chem., 17, 99 (2001) [see pag.146-147]; H. Levebvre-Brion and R.W. Field, Perturbations in the spectra of diatomic molecules, Academic Press 1986 [see pag.88].

In the book by Levebvre-Brion and Field they use for deriving selection rules a spin-orbit hamiltonian of the form

Sum_i A(i) l_i . s_i (i sums over all electrons)

and by applying the Wigner-Eckart theorem they get a Three J factor ThreeJSymbol[{S, -Sigma},{1,Sigma-Sigma'},{S',Sigma'}]; the triangular condition on [S, S', 1] imposes not only |S-S'| = 1, which excludes matrix elements between singlet states, QED. Too bad, the hamiltonian used for the derivation is not the exact one (for one thing, it does not contain two-electron terms) but Levebvre-Brion/Field state (without discussion) that 'if the more complex form of H_SO is used the selection rules are identical'. Gulzari: I'll try to consult you monograph.

Dear Lorenzo,

I presented the oversimplified explanation but used the correct form of the spin-orbit operator. I started from the atomic basis where singlet-singlet matrix elements are always zero. The transformations of the atomic states via the irreducible representation of any molecule result in molecular terms but will not change their spin momenta, thus, singlet-singlet spin-orbit matrix elements will be zero also in molecules. will not be a priori zero because in atomic basis can be contributed from the 1S_0 -- 3P_0 matrix element which is not zero.

Also I do not agree with you about that the Hso ~ l*s is the phenomenological form of the spin-orbit operator. This form (simplified one but describing the largest contribution) can be obtained within the non-relativistic consideration of the theory of an atom (for sure from Dirac theory also). Precise description of this form you can find in the books devoted to the theory of atom.

I wrote my way of thinking only (about the zero result) but you can consider that I have guessed right answer :-)