# NLO question (second hyperpolarizability tensor)

1. Jun 22, 2007

### pascal

I'm working on a calculation of the second hyperpolarizability "\beta" of a compound with D(3) symmetry, and I am trying to figure out how the tensor \beta transforms under a coordinate rotation about the z axis (the three-fold symmetry axis). I know that if we have two bases x and x' with a linear transformation tensor T such that x' = xT, that the "ordinary" polarizability tensors "\alpha" and "\alpha^{\prime}" in the two coordinate systems are related by \alpha^{\prime} = T^{-1} \alpha T, but I'm not sure what the transformation rule is for a rank-3 tensor when the coordinate system is rotated. Thanks for your help.

2. Jul 16, 2007

### olgranpappy

ah, yes. there are three indices on $$\beta$$ so you can't treat it as a matrix which transforms by being sandwiched between two other matrices. We have to generalize a bit. Consider the transformation you do know:
$$\alpha'=T^{-1}\alpha T$$

Now, the inverse of T is the same as the transpose so I can rewrite this using indices as
$$\alpha'_{ij}=\sum_{mn}T^{-1}_{im}\alpha_{mn}T_{nj}=\sum_{mn}\alpha_{mn}T_{mi}T_{nj}$$

So, apparently the correct transformation for a quantity with three indices is
$$\beta'_{ijk}=\sum_{mnp}\beta_{mnp}T_{mi}T_{nj}T_{pk}$$

this is easily generalized to high numbers of indices, or lower, e.g. you could reporduce the transformation of a vector. Cheers.

3. Jul 16, 2007

### pascal

Thanks, Adam. That does make sense. The only question I have now is which of the indices are contravariant, and which are covariant? I've looked through MANY NLO books and cannot find any resource that addresses this question.

4. Jul 16, 2007

### olgranpappy

No probelm.

Yes, you need to know whether they are contra- or co-variant.

Can you write down the equation for me so that I can see how it should go? I.e., is it something like:
$$P_i=\alpha_{ij}E_j+\beta_{ijk}E_jE_k\;?$$
Because you can figure out how the indices transform since you know that E and P are vectors.