# When are the off-diagonal terms of magnetic susceptibility tensor non-zero?

The susceptibility tensor is defined as

$$\bf{M}_i = \chi_{ij} \bf{H}_j.$$

From looking at a material's geometry and crystalographic properties, is it possible to say whether the off-diagonal terms of the suceptability tensor will be non-zero? Framed slightly differently, if it was calculated that $$\chi_{i, j} \neq 0$$ for some $$i \neq j$$, what would this tell us about the structure of the material? Is this ever a property of ideal crystaline materials or does it always suggest impurities?

• This ties closely to symmetry of the crystal, indeed. Compare with the existence of off-diagonal mechanical property elements. – Jon Custer Sep 11 '19 at 16:31

## 1 Answer

Indeed, the terms of the susceptibility tensor are related to the symmetry of the material (see Neumann's principle). Some basic facts about the diagonal elements of this tensor are:

• Cubic crystal has $$\chi_{11} =\chi_{22} = \chi_{33}$$
• Uniaxial crystals (tetragonal, hexagonal, trigonal symmetry) have a preferred magnetization direction (easy axis), let's say along the z-axis, such that $$\chi_{11} =\chi_{22} < \chi_{33}$$
• Biaxial crystals (orthorhombic, monoclinic, triclinic symmetry) have two preferred magnetization directions, so all diagonal components are different

The off-diagonal elements are more complicated. Of course, these off-diagonal terms can be non-zero for complicated symmetries. Nevertheless, the crystal symmetry might not tell us everything, for example, there might be magnetic interactions (e.g. crystal field) that result in nonzero off-diagonal terms.

NOTE: The relation you stated assumes that the magnetic response of the material is linear, which is the case when the magnetic field is sufficiently small. At high magnetic fields, non-linear magnetic effects might occur.