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The negative sign of the exchange term and $J>0$ make this a Heisenberg ferromagnet. Each classical bond minimizes its energy when its two fixed-length spins are parallel. On the connected cubic lattice all ground states are therefore uniform:using length $S$ for the classical large-spin convention. Any two orientations are related by a common spatial rotation of every spin. The global spin-rotation symmetry is $SO(3)$, and a chosen direction retains its $SO(2)$ subgroup of rotations about that axis.

Spontaneous symmetry breaking means selecting one member of this continuous set in the thermodynamic limit, for example by taking volume to infinity before sending an aligning field to zero. The zero-field Hamiltonian operator remains rotationally invariant even though the selected state is not. There are gapless transverse spin waves. Although two rotation generators are broken, an isotropic ferromagnet has one quadratic magnon branch, whose two transverse spin components form a conjugate dynamical pair. It is not necessary to invent two independent linear-dispersion branches from the number of broken generators. The long-wavelength dispersion is derived in d.

The field term explicitly selects the positive $x$ direction. For $B>0$ the uniform minimizing state is ${\mathbf{S}}_j=S\hat{\mathbf{x}}$, with no remaining continuous degeneracy of its orientation. Rotations about the $x$ axis still leave the Hamiltonian operator invariant, and the selected state respects that residual symmetry.

Tilting the magnetization now costs a field energy. Accordingly the transverse spin wave acquires a gapas confirmed directly in d. The factor $S$ is present because the source couples the field as $-BS{\sum }_j{S}_j^x$ rather than $-B{\sum }_j{S}_j^x$. Gapless zero-field Goldstone modes need not survive an explicit symmetry-breaking field.