as happens for the charged ones, the magnetic field intensity $B$ enters in the energy coupled with the perpendicular momentum (see the last term in the previous equation). This coupling reflects the axial symmetry imposed to the system by the magnetic field. The ground state energy of the neutral spin one boson ($s=1$ and $p_3=p_{\perp}=0$) is $$\varepsilon(0, B)=\sqrt{m^2-2\kappa B m}=m\sqrt{1-b}.\label{massrest}$$ being $b=\frac{B}{B_c}$ and $B_c=\frac{m}{2\kappa}$.

From Eq. (\[massrest\]) follows that the rest energy of the system decreases with the magnetic field and becomes zero for $B=B_c$. For the values or $m$ and $\kappa$ we are considering, $m=2m_e$ and $\kappa=2\mu_B$, $B_c=m_e^2/e=4.41 \times 10^{13}~G$ which is the Schwinger critical field. Let us note that the ground state of the charged vector boson has a similar instability (see [@ROJAS1996148]). Eq.(\[massrest\]) allows us to define and effective magnetic moment as

$$d=-\frac{\partial \varepsilon(0,B)}{\partial B}=\frac{\kappa m}{\sqrt{m^2-2m\kappa B}}=\frac{\kappa}{\sqrt{1-b}}\label{magmoment}.$$

The system has a paramagnetic behavior because $d>0$. It will be also important for the discussion below the fact that $d$ grows with the increasing of the magnetic field and diverges for $b \rightarrow 1$ $(B \rightarrow B_c)$.

Thermodynamical properties {#thermo}
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The general expression for the thermodynamical potential of the NVBG has the form

$$\label{Grand-Potential-4}
\Omega(B,\mu,T)= \sum_{s=-1,0,1}\frac{1}{\beta}\left[\sum_{p_4}\int\limits_{-\infty}^{\infty}\frac{p_{\perp}dp_{\perp}dp_3}{(2\pi)^2} \ln \det 