operators, $$\langle k|S^{i}|l\rangle\equiv (S^{i})_{kl}=-i\varepsilon_{ikl},$$ whence $$\label{eq:2.7}
S^{x}=\left(
  \begin{array}{ccc}
    0 & 0 & 0 \\ [-2ex]
    0 & 0 & -i \\ [-2ex]
    0 & i & 0 \\
  \end{array}
\right), \quad
S^{y}=\left(
  \begin{array}{ccc}
    0 & 0 & i \\[-2ex]
    0 & 0 & 0 \\[-2ex]
    -i & 0 & 0 \\
  \end{array}
\right), \quad
S^{z}=\left(
  \begin{array}{ccc}
    0 & -i & 0 \\ [-2ex]
    i & 0 & 0 \\ [-2ex]
    0 & 0 & 0 \\
  \end{array}
\right).$$ One can easily seen that $S^{x}=\lambda^{7}$, $S^{y}=-\lambda^{5}$, and $S^{z}=\lambda^{2}$ (see Eqs. (\[eq:A1\])), so that subalgebra of these matrices generates an SU(2) subgroup of SU(3) group. The remaining five Gell-Mann matrices, due to their properties given by Eq. (\[eq:A6\]), can be expressed in terms of the quadratic combinations of spin operators: $$\begin{gathered}
\lambda^{1}=-\{S^{x},S^{y}\}, \quad \lambda^{3}=(S^{y})^{2}-(S^{x})^{2}, \quad \lambda^{4}=-\{S^{x},S^{z}\}, \nonumber \\
\lambda^{6}=-\{S^{y},S^{z}\}, \quad \lambda^{8}=\sqrt{3}(S^{z})^{2}-{2\over\sqrt{3}}\,I, \label{eq:2.8}\end{gathered}$$ where $\{\ldots,\ldots\}$ denotes an anticommutator and $I$ is the unit $3\times3$ matrix. Since the traceless quadrupole matrix $Q^{ik}\equiv S^{i}S^{k}+S^{k}S^{i}-(4/3)\,\delta_{ik}$ is determined by the above five independent components, $$Q^{ik}={1\over\sqrt{3}}\left(
  \begin{array}{ccc}
    -\lambda^{8}-\sqrt{3}\lambda^{3} & -\lambda^{1} & -\lambda^{4} \\ [-1.5ex]
    -\lambda^{1} & -\lambda^{8}+\sqrt{3}\lambda^{3} & -\lambda^{6} \\ [-1.5ex]
    -\lambda^{4} & -\lambda^{6} & 2\lambda^{8} \\
  \end{array}
\right),$$ we call them the quadrupole operators. These operators can 