= Solution
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 gap
$$
\boxed{\Delta=BS,}
$$
as confirmed directly in <2/d>. The factor $S$ is present because the source couples the field as $-BS\sum_jS_j^x$ rather than $-B\sum_jS_j^x$. Gapless zero-field <Goldstone modes> need not survive an explicit symmetry-breaking field.
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