Physics Asked by Connor Skehan on July 8, 2021
For instance, I know that $chi^{2}$ materials need to lack inversion symmetry, but what about $chi^{3}$ materials?
Long answer:
Basically, $chi^{(3)}$ is a rank four tensor with 81 components. You then apply the Neuman principle saying that symmetries of the crystal that supports $chi^{(3)}$ should also be the symmetries of the tensor. Then you look for independent components of this tensor, which boils down to looking for trivial irreducible sub-representations in the representation of the symmetry group of the crystal over that rank four tensor.
So lets say the representation of your crystal symmetry group ($G$) over vectors is given by: $Mleft(aright): mathbb{R}^3to mathbb{R}^3$, where $ain G$
Then the induced representation for $chi^{(3)}$ is $Kleft(aright)=Mleft(aright)otimes Mleft(aright) otimes Mleft(aright) otimes Mleft(aright)$. If you want to know the independent components of $chi^{(3)}$, simply define the projection operator, to project into trivial irreps:
$P=frac{1}{#G}sum_{ain G} Kleft(aright)$
$P:mathbb{R}^3 otimes mathbb{R}^3 otimes mathbb{R}^3 otimes mathbb{R}^3tomathbb{R}^3 otimes mathbb{R}^3 otimes mathbb{R}^3 otimes mathbb{R}^3 $
The eigenvectors, with eigenvalues=1, of this projection operator will be the allowed components of $chi^{(3)}$, if you only want to know the number of allowed components $n_{allowed}$, it is given by the trace of the projection operator:
$n_{allowed}=Trleft(Pright)=frac{1}{#G}sum_{ain G} Trleft(Kleft(aright)right)=frac{1}{#G}sum_{ain G} Trleft(Mleft(aright)right)^4$
Short answer:
Have a look in a suitable nonlinear optics book, e.g. Popov, Svirko, Zheludev "Susceptibility Tensors for Nonlinear Optics", Appendix F. They list all the allowed components of $chi^{(3)}$ for all crystallographic classes. There is no class where $chi^{(3)}$ is not allowed, but in some cases the number of allowed components can be as low as 2 (Cubic system, class 432) or even 1 (isotropic medium).
Correct answer by Cryo on July 8, 2021
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