Global spin-rotation symmetry about one axis ($[H,S_{\mathrm{tot}}^z]=0$)

If the Hamiltonian commutes with total $S^z$, simultaneous spin rotations $e^{-i\varphi S_{\mathrm{tot}}^z}$ form a circle group symmetry. In a state of definite total $S^z$, the transverse one-point averages vanish because $S^x\pm i{S}^y$ change that eigenvalue. A chosen classical orientation or symmetry-breaking limit requires separate care when constructing a linear spin-wave approximation.