1
$\begingroup$

For instance, I know that $\chi^{2}$ materials need to lack inversion symmetry, but what about $\chi^{3}$ materials?

$\endgroup$

1 Answer 1

3
$\begingroup$

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: $M\left(a\right): \mathbb{R}^3\to \mathbb{R}^3$, where $a\in G$

Then the induced representation for $\chi^{(3)}$ is $K\left(a\right)=M\left(a\right)\otimes M\left(a\right) \otimes M\left(a\right) \otimes M\left(a\right)$. If you want to know the independent components of $\chi^{(3)}$, simply define the projection operator, to project into trivial irreps:

$P=\frac{1}{\#G}\sum_{a\in G} K\left(a\right)$

$P:\mathbb{R}^3 \otimes \mathbb{R}^3 \otimes \mathbb{R}^3 \otimes \mathbb{R}^3\to\mathbb{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:

$n_{allowed}=Tr\left(P\right)=\frac{1}{\#G}\sum_{a\in G} Tr\left(K\left(a\right)\right)=\frac{1}{\#G}\sum_{a\in G} Tr\left(M\left(a\right)\right)^4$


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).

$\endgroup$
1
  • $\begingroup$ One more thing. Representation of the identity will always be the identity matrix, so its trace will be nonzero. Representation of group over the real-valued vectors will be real-valued (all entries in $M(a)$ will be real numbers). In case of $\chi^{(3)}$ one therefore will end up having a sum of real-valued things to 4-th power, with at least one of them being non-zero, so you will always end up with $n_{allowed}>0$ $\endgroup$
    – Cryo
    Commented Oct 31, 2022 at 16:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.