quadrupole order is primary, each Np ion is in the eigenstate of the eigenvalue $-q_1$ or $q_1$. For the case of eigenvalue $-q_1$, two eigensates are to be degenerate and give eigenvalues $-t_1$ and $t_1$ to the octupole moment $T_p$, and thereby the net octupole moment becomes zero. For the case of $q_1$, two eigenstates are also to be degenerate and give the eigenvalue 0 to $T_p$. In either case, the quadrupole order carries no octupole order.

### dipole ordering

Although the dipole ordering is ruled out in NpO$_2$, it may be interesting to discuss here what happens in the dipole ordering. Introducing the dipole operators, $$J_{p} = \left\{ \begin{array}{lcl}
\frac{1}{\sqrt{3}} \left( J_{x}+J_{y}+J_{z} \right) 
     & {\textrm for} & p=111, \\
\frac{1}{\sqrt{3}} \left( J_{x}-J_{y}-J_{z} \right) 
     & {\textrm for} & p=\overline{1}11, \\
\frac{1}{\sqrt{3}} \left(-J_{x}+J_{y}-J_{z} \right) 
     & {\textrm for} & p=1\overline{1}1, \\
\frac{1}{\sqrt{3}} \left(-J_{x}-J_{y}+J_{z} \right) 
     & {\textrm for} & p=11\overline{1}, \\
\end{array} \right.  \label{eq.tripleJ}$$ we can construct the dipole ordering state by assigning them to each sublattice in the same way as in the octupole ordering. Note that $J_p$ and $Q_p$ are simultaneously diagonalized, because both operators commute with each other. Within the bases of simultaneous eigenstates of $J_p$ and $Q_p$, the relevant 