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I am trying to treat a pair of inversion-related S=1/2 spins (Cu(II)) interacting through dipolar interactions, each of which also exhibits a hyperfine interaction.

I have constructed an exact model of the interaction matrix based on T.D. Smith, J. R. Pilbrow, Coord. Chem. Rev., 1974, 13, 173-278, eqn 14 (http://www.sciencedirect.com/science/ar ... 4500802556), and whose code I mention in viewtopic.php?f=3&t=474.

While it gives a very good agreement for angles xi = eta = 0 (collinear g-tensors) with Exp.Ordering = +1 and without strains, as soon as I give a non-zero value to xi, the g|| component loses most of its intensity and the Exp.Ordering = +1 parameter is more or less neglected. The agreement breaks down even for infinitesimal xi angles.

My question is whether my approach of constructing an explicit interaction Hamiltonian combined with Exp.Ordering = +1 and/or hyperfines is valid for xi =/= 0. Is there some aspect of the problem's symmetry I overlooked?

Thanks!

Update: I obtain very good simulations for xi =/= 0 when I switch the Exp.Ordering = -10, even for very small angles (e.g. 1deg).

Would the xi angle change the g-tensor orientation of the dimer, requiring to switch ordering vector from z-axis (of single spin) to xy-plane (of tilted dimer)?

It might be useful to set Opt.Symmetry explicitly to exclude any possible problems with the automatic selection.

Thanks Stefan,

I have tried to look for documentation for the Opt.Symmetry option and only found the 'Ci' label to define it. Is it the standard point group symbols? I suppose this is not the same as theExp.CrystalSymmetry option?

My two systems are inversion-related (each has roughly a C2v symmetry and they crystalise in a P-1 space group). Viewing this as a centrosymmetric dimer, I set Opt.Symmetry = 'Ci' and get exactly the same behaviour. Is that expected?

The Opt.Symmetry determines which region of the unit sphere is used in a powder simulation. 'Ci' indicates the upper hemisphere and is the safest, but slowest. 'C2h' gives two octants, and 'D2h' gives one octant. See sphgrid for more documentation. If you are running single-crystal simulations, Opt.Symmetry is irrelevant.