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?