Table \[table.1\]. Because the isotropic parts of the Coulomb interaction $F^{0}$’s are known to be well screened in solids compared to those of the anisotropic parts, the former quantities are multiplied by a factor 0.25 while the latter’s are by 0.8. Within the HFA, the ground state has the ten-fold degeneracy corresponding to $J=9/2$ multiplet. The choice of the multiplying factors does not alter this conclusion. [@Nagao03] Note that these states of $J=9/2$ are slightly deviated from those of the perfect Russell-Saunders (RS) coupling scheme with $L=6$ and $S=3/2$ due to the presence of the strong SOI. For instance, ${\textbf L}^2$ and ${\textbf S}^2$ take values 39.752 and 3.237 respectively, compared to the RS values 42 and 3.75.

  $F^{k}(3d,3d)$         $F^{k}(3d,5f)$        $F^{k}(5f,5f)$   $G^{k}(3d,5f)$
  ---------------------- --------------------- ---------------- ----------------
  $F^0$ 180.1                                  $F^0$ 19.61      
  $F^2$ 92.04                                  $F^2$ 9.909      
  $F^4$ 59.28                                  $F^4$ 6.491      
                                               $F^6$ 4.769      
  $\zeta_{3d}=$ 76.254   $\zeta_{5f}=$ 0.298                    

  : \[table.1\] Slater integrals and spin-orbit interaction parameters in the $(3d)^{10}(5f)^3$ configuration within the HF approximation (in units of eV).[@Cowan81]

In crystal, the ten-fold degeneracy is lifted by the CEF. Under the cubic symmetry, the CEF Hamiltonian $H_{\textrm CEF}$ may be expressed as $$H_{\textrm CEF} = B_4 ( O_4^0 + 5 O_4^4) 
                         +B_6 (O_6^0 - 21 