\nonumber \\\end{aligned}$$ Condition $\tilde A_{xy}(j)=0$ determines $\phi_j$’s, which take values between $\theta_{B}^{(j)}$ and $\theta_{C}^{(j)}$. For example, assuming $q_B=q_C=-(3/4)e$, $q_{Dy}=3e$ (e: proton charge), $d_B=0.01$Å, $d_C=0.02$Å, we have $\phi_1=31.8^{\circ}$, $\phi_2=180^{\circ}-\phi_1$, $\phi_3=90^{\circ}+\phi_1$, $\phi_4=90^{\circ}+\phi_2$. The principal axes thus estimated correspond well with the directions of the ordered magnetic moments in Phase III.[@Yamauchi]

The equivalent operator method allows us to write the crystal field energy $H_{crys}(j)$ at site $j$ within the subspace of angular momentum $J$ as $$H_{\rm crys}(j) = D_{J}[3\tilde J_z^2(j)-J(J+1)]+E_{J}[\tilde J^2_x(j) 
                 - \tilde J^2_y(j)].
\label{eq.crystal}$$

The $5d$ states are forming an energy band with width $\sim 15$ eV through a hybridization with $s$ and $p$ states of neighboring B and C atoms as well as $5d$ states of neighboring Dy atoms. We need the density of states (DOS) for calculating the RXS intensity. We assume a Lorentzian with full width of half maximum 5 eV for the DOS’s projected onto symmetries $xy$, $x^2-y^2$, $yz$, $zx$, and $3z^2-r^2$. The center of each component of the DOS is separate to each other in accordance with the first term of eq. (\[eq.crystal\]), although the first term need not be respected so much because of the large band effect. Explicitly they are assumed to be $(1/\pi)\Delta/((\epsilon-2.5)^2+\Delta^2)$ for 