10
$\begingroup$

I was wondering if Kohn-Sham orbitals corresponding to a different Bloch wavevector should be orthogonal? I know that we should have $$\int d \boldsymbol{r}\phi_i(\boldsymbol{r}) \phi_j^*(\boldsymbol{r}) = \delta_{ij}\tag{1}$$ but I was wondering whether, if we also considered the $\boldsymbol{k}$ dependence, we should also have $$\int d \boldsymbol{r}\phi_i(\boldsymbol{r}, \boldsymbol{k}) \phi_j^*(\boldsymbol{r},\boldsymbol{k}' ) = \delta_{ij} \delta(\boldsymbol{k} - \boldsymbol{k}')\tag{2}$$

My feeling is that this should be the case and so the appropriate Lagrange multipliers used in the solution of the KS equations would have to be sought for each $\boldsymbol{k}$ point?

$\endgroup$
8
$\begingroup$

You are correct that orbitals from different $k$-points should be orthogonal. $k$-points are irreps of the translation group and, similar to the irreps of point groups for molecules, integrals of two functions with different irreps will always result in $0$.

More generally, we can actually separately solve the SCF equations for each $k$-point, since the Fock matrix can be split into blocks for each $k$-point: $$F^kC^k=S^kC^k\epsilon^k$$ Some details about efficiently solving for the energy and forces of periodic systems using Gaussian type orbitals and transformations back-and-forth from real space to $k$-space are given in 1.

  1. Konstantin N. Kudin and Gustavo E. Scuseria Phys. Rev. B 61, 16440 DOI: 10.1103/PhysRevB.61.16440
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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