there is a problem in defining total magnetic moment of molecule. Because of the lack of spherical symmetry we can define correctly only spin magnetic moment.
But what I'd like to know is not just about "nuclear magnetic moment", but "molecular magnetic moment" that includes not only nuclear part but also electronic part.
My collaborator and co-author Gennady Gutsev is an expert on this. See J. Chem. Phys. 138, 164303 (2013); http://dx.doi.org/10.1063/1.4799917 for example. Section III. C. should contain most of the information you need. Natural Population Analysis (pop=npa in G09) is a very useful tool for this, in my opinion. I have not tried pop=full for this, as recommended by Arfaoui in his response.
there is a problem in defining total magnetic moment of molecule. Because of the lack of spherical symmetry we can define correctly only spin magnetic moment.
I don't quite get your point. Afaik, the magnetic moment (its component in some direction) is given by the derivative of the total energy with respect to an applied field (pointing along that direction). There is no requirement of spherical symmetry in such a definition. This moment may be of both orbital and/or spin character.
Broken spherical symmetry may quench orbital momentom but need not do so if s.o.c. is strong. Take a Ce atom in f1 configuration. In nearly all symmetries it will keep the largely orbital character of its magnetic moment - and accordingly a g-factor slightly smaller than 1.
So I'd be curious about the exact meaning of your statement above. Thanks
You use definition of total magnetic moment (wich is appropriate only for observation distances much larger then the dimensions of observed system) in classical physics. Quantum mechanics is the other case. Operator of total magnetic moment (it's square and one of the projections) commute with Hamiltonian (i.e. measurable in stationary quantum state) only for spherically symmetric systems.
I am getting a little off topic now, I know. Eugene, I am used to taking another approach here. I would write down the full, field dependent Hamiltonian and again calculate the derivatives of energy with redpect to the field. In spherically symmetric systems (at zero field) you end up with the Landé factor times total angular momentum component along the field direction (times the Bohr magneton). Already the field breaks the spherical symmetry, and cylindrical symmetry is left.
And yes, in non-spherical symmetry the eigenstates with the field applied will no longer be a scalar times the operator of total angular momentum component along the field in general. If you insist on the definition of magnetic moment operator(s) to be proportional to angular momentum operators, then we have to content with expectation values. In non spherical symmetry these may be anisotropic. I have no problem calling these magnetic moments, too, they correspond to the result of magnetisation measurements.
when You write down the full, field dependent Hamiltonian, then You have the part with the scalar product of field and magnetic moment. And what is the magnetic moment then?