at $J=J_c \equiv |B|/2$.

It is not difficult to find the ground state of this model. When $J \leqslant J_c$, it is simply $$\label{eq:PT}
  {\mbox{$| E^{(0)} \rangle$}} =
    \begin{cases}
      {\mbox{$| - \rangle$}}^{\otimes N}, \text{  for }B \geqslant 0 \, \& \, J \leqslant J_c,& \nonumber \\
      {\mbox{$| + \rangle$}}^{\otimes N}, \text{  for }B \leqslant 0 \, \& \, J \leqslant J_c,& \nonumber
    \end{cases}$$ with the corresponding (non-degenerate) ground state energy given by $E^{(0)} = N(J-2J_c)$.

In the regime $J \geqslant J_c$, the situation is only slightly more involved as the parity of $N$ now becomes important. For an *even* number of spins, the (doubly degenerate) ground state energy is given by $E^{(0)} = -NJ$, the corresponding eigenstates being ${\mbox{$| +- \rangle$}}^{\otimes N/2}$ and ${\mbox{$| -+ \rangle$}}^{\otimes N/2}$. For an *odd* number of spins, the ground state energy is $N$-fold degenerate and reads $E^{(0)} = -NJ + 2(J-J_c)$. The ground state is given by $$\label{eq:PT}
  {\mbox{$| E^{(0)} \rangle$}} =
    \begin{cases}
      {\mbox{$| -+-+- \cdots -+- \rangle$}}, \text{  for }B \geqslant 0 \, \& \, J \geqslant J_c,& \nonumber \\
      {\mbox{$| +-+-+ \cdots +-+ \rangle$}}, \text{  for }B \leqslant 0 \, \& \, J \geqslant J_c,& \nonumber
    \end{cases}$$ and all $N$ translations thereof. 