 Solution

Use spin one-half states $|\uparrow ⟩,|\downarrow ⟩$ with $S^z=\pm 1/2$, so ${\sigma }^i=2S^i$ and $H=-J{\sum }_n({\sigma }_n^x{\sigma }_{n+1}^x+g{\sigma }_n^z)$. At $g=0$, every ferromagnetic bond has its lowest energy $-J$ when adjacent $x$ spins agree. On the connected periodic chain the normalized ground states arewhere $|\pm x⟩=(|\uparrow ⟩\pm |\downarrow ⟩)/\sqrt{2}$. Their span is the two-dimensional ground-state subspace; any normalized superposition is another ground state. They are exchanged by the global spin-flip discrete symmetry generated by ${\prod }_n{\sigma }_n^z$.

As $g\to +\infty$, the field dominates and selectsThe limiting polarized state is unique. At large finite $g$, exchange admixes virtual spin flips, so the product state is the limiting wavefunction rather than an exact finite-field eigenstate. The bulk expansion below gives $E_0/N=-J[g+1/(4g)+O(g^{-3})]$. The exact twofold degeneracy stated at $g=0$ should be distinguished from thermodynamic spontaneous symmetry breaking in the ordered phase: a finite chain at nonzero field can have a split pair of symmetry eigenstates.