the quadratic terms in creation and annihilation operators to Eqs. (\[eq:3.14\]), (\[eq:3.15\]) is examined in Refs. [@Tolmachev; @JPhysB] for atoms with zero spin.

Ferromagnetic ground state and excitations
==========================================

Now we use the obtained equations to study the ground state properties and corresponding single-particle excitations of a weakly interacting Bose gas of spin-1 atoms. In order to introduce the condensate density $n_{0}$, consider the normalized state vector $\zeta_{\alpha}$, $$\label{eq:4.1}
\Psi_{\alpha}=\sqrt{n_{0}}\zeta_{\alpha}, \quad \zeta^{*}_{\alpha}\zeta_{\alpha}\equiv(\zeta^{*}\zeta)=1.$$ In the Cartesian basis, the ferromagnetic ordering is specified by the following vector [@Ohmi]: $$\label{eq:4.2}
\zeta={1\over\sqrt{2}}(1,\,i,\,0).$$ As we have shown, the description of a many-body system of spin-1 constituents requires the introduction of additional parameters along with the ordinary magnetization vector. These parameters are determined by Eqs. (\[eq:2.9\]), (\[eq:2.10\]). In the problem under consideration, the above ferromagnetic state vector generates the ordinary magnetization along $z$–direction, $$\label{eq:4.3}
\langle S^{i}\rangle=\Psi^{*}S^{i}\Psi=n_{0}\delta_{iz}, \quad S^{i}= (S^{x}\equiv\lambda^{7},\,S^{y}\equiv-\lambda^{5},\,S^{z}\equiv\lambda^{2})$$ and one more parameter associated with the quadrupole degrees of freedom, $$\label{eq:4.4}
\langle Q^{b}\rangle=\Psi^{*}q^{b}\Psi={n_{0}\over\sqrt{3}}\delta_{b8}.$$ The quadrupole tensor for the ferromagnetic state becomes $$\label{eq:4.5}
\langle Q^{ik}\rangle=\Psi^{*}Q^{ik}\Psi=n_{0}\left(
  \begin{array}{ccc}
    -1/3 & 0 & 0 \\ [-1.5ex]
    0 & -1/3 & 0 \\ [-1.5ex]
    0 & 0 & 2/3 \\
  \end{array}
\right).$$ Since $\langle Q^{xx}\rangle=\langle Q^{yy}\rangle$, the order parameter $\zeta$ is 