invariant with respect to rotations about $z$-axis, as it should be in the ferromagnetic state. Next, multiplying Eq. (\[eq:3.15\]) by $\Psi_{\alpha}^{*}$ and performing the summation over $\alpha$, one can obtain the relation between the chemical potential and condensate density for the above state vector $\zeta$: $$\label{eq:4.6}
\mu=n_{0}\left(U(0)+{4\over 3}J(0)\right)-h.$$ In a similar manner, the thermodynamic potential density determined by Eq. (\[eq:3.14\]) is written as $$\omega^{(0)}={n_{0}^{2}\over 2}\left(U(0)+{4\over 3}J(0)\right)-n_{0}(h+\mu),$$ or eliminating the condensate density by using Eq. (\[eq:4.6\]), one finds $$\label{eq:4.7}
\omega^{(0)}=-{1\over 2}{(\mu+h)^{2}\over{U(0)+(4/3)J(0)}}.$$ In order for the equilibrium state to be stable, the thermodynamic potential density $\omega^{(0)}$ must be negative (the pressure is positive) that implies $U(0)+(4/3)J(0)>0$. Since the necessary stability condition for a condensed gas of spinless particles requires the predominance of repulsive forces over attractive forces, we also consider $U(0)>0$. Therefore, from Eq. (\[eq:4.7\]), it becomes evident that the condition $J(0)<0$ must be satisfied for the stability of ferromagnetic state under consideration.

Having defined the ferromagnetic ground state, we now address the issue of single-particle excitations. To obtain the corresponding spectra, let us return to the Hamiltonian given by Eqs. (\[eq:3.6’\])-(\[eq:3.9\]), (\[eq:3.12\]). Eliminating the chemical potential by using Eq. (\[eq:4.6\]) and taking into account the explicit form for the ground state 