ABINIT Discussion Forums

Dear users,
How can I define the electric field direction of the incident light for my dielectric function calculation? In the experiment the crystal is excited perpendicular to the ac-plane, c being the optical axis, and therefore two polarization directions should be distinguished. Some of the relevant keywords for Abinit input are

iscf -3
nline 0
nqpt 1
qpt 0.0 0.0 0.0
rfelfd 2
rfdir1 1 0 0
rfdir2 0 1 0 # For the ac-plane I think only this direction is enough
rfdir3 0 0 1

and for the Optic input

0.002 ! Value of the smearing factor, in Hartree
0.0003 0.74 ! Difference between frequency values (in Hartree), and maximum frequency
0.055 ! Scissor shift if needed, in Hartree
0.002 ! Tolerance on closeness of singularities (in Hartree)
3 ! Number of components of linear optic tensor to be computed
11 22 33 ! Linear coefficients to be computed (x=1, y=2, z=3)
0 ! Number of components of nonlinear optic tensor to be computed

What should I add/change to include anisotropy? Is the incident light considered unpolarized by default?
Thank you

Ok, let me ask this. If I shine light on the ac-plane of my sample such that the electric field is parallel to the c-axis, is the c-axis the direction of perturbation, i.e. rfdir 0 0 1? What is the physical meaning of rfdir in the dielectric function calculation?

you should always use rfdir 1 1 1 which means calculate all perturbations (x y z). The optic output should be tensorial, and hence anisotropic.

What you have added rfdir1 rfdir2 etc... are for different datasets, but this is not necessary.

M.

Thank you for the answer. I now see that it is indeed safer to always use rfdir 1 1 1. Although for the given case (exciting the ac-plane of the crystal with light polarized in the c-direction) using just rfdir 0 0 1 produces the same result so my discussion above should be basically correct.