vector and for matrices $\lambda^{a}$ (see Eqs. (\[eq:4.1\]), (\[eq:4.2\]), (\[eq:A1\])), the quadratic Hamiltonian is reduced to $$\label{eq:4.8}
{\cal H}^{(2)}(n_{0})={\cal H}^{(2)}_{1}(n_{0})+{\cal H}^{(2)}_{2}(n_{0}),$$ where $$\label{eq:4.9}
{\cal H}^{(2)}_{1}(n_{0})=\sum_{{\bf p}\neq 0}\left[\varepsilon_{\bf p}+h+2n_{0}J({\bf p})-2n_{0}J(0)\right]a^{\dagger}_{{\bf p}z}a_{{\bf p}z},$$ and $$\label{eq:4.10}
{\cal H}^{(2)}_{2}(n_{0})=\sum_{{\bf p}\neq 0}a^{\dagger}_{{\bf p}\alpha}A_{\alpha\beta}({\bf p})a_{{\bf p}\beta}+{1\over 2}\sum_{{\bf p}\neq 0}a^{\dagger}_{{\bf p}\alpha}B_{\alpha\beta}({\bf p})a^{\dagger}_{-{\bf p}\beta}+{1\over 2}\sum_{{\bf p}\neq 0}a_{{\bf p}\alpha}B^{*}_{\alpha\beta}({\bf p})a^{\dagger}_{-{\bf p}\beta}, \quad \alpha,\,\beta=x,\,y.$$ Here $A=A^{\dagger}$ and $B=B^{T}$ are the Hermitian and symmetric matrices, respectively, $$\label{eq:4.11}
A_{\alpha\beta}({\bf p})=\left(
  \begin{array}{ccc}
    A({\bf p}) & i{\cal A}({\bf p}) \\ [-1.0ex]
    -i{\cal A}({\bf p}) & A({\bf p}) \\
      \end{array}
\right), \quad
B_{\alpha\beta}({\bf p})=\left(
  \begin{array}{ccc}
    B({\bf p}) & iB({\bf p}) \\ [-1.0ex]
    iB({\bf p}) & -B({\bf p}) \\
      \end{array}
\right)$$ with the following matrix elements: $$\begin{gathered}
A({\bf p})=\varepsilon_{\bf p}+h+{5\over 3} n_{0}J({\bf p})-n_{0}J(0)+{1\over 2}{n_{0}U({\bf p})}, \nonumber \\
{\cal A}({\bf p})={1\over 3}n_{0}J({\bf p})-{1\over 2}n_{0}U({\bf p})-n_{0}J(0)+h, \nonumber \\
B({\bf p})={2\over 3}n_{0}J({\bf p})+{1\over 2}n_{0}U({\bf p}). \label{eq:4.12}\end{gathered}$$ The first part ${\cal H}_{1}^{(2)}(n_{0})$ of the total quadratic Hamiltonian has already diagonal form with the following spectrum of single-particle excitations: $$\label{eq:4.13}
\omega^{({\rm I})}_{{\bf p}z}=\varepsilon_{\bf p}+h+2n_{0}\left[J({\bf p})-J(0)\right],$$ while ${\cal H}_{2}^{(2)}(n_{0})$ should be diagonalized in creation and annihilation operators. Note that since ${\cal H}_{2}^{(2)}(n_{0})$ commutes with ${\cal H}_{1}^{(2)}(n_{0})$, it can be diagonalized independently. To this end, we apply the Bogoliubov canonical transformation method which 