Spin-one-half quadratic anisotropy is constant ($S_x^2=S_y^2=S_z^2={\hslash }^{2}I/4$)

For spin one-half, the Pauli representation makes every diagonal quadratic component a multiple of the identity. Thus $A{S}_z^2+B(S_x^2-S_y^2)=A{\hslash }^{2}I/4$ cannot split the doublet or cause spin reversal. This conclusion is special to this quadratic Hamiltonian and does not exclude linear Zeeman dynamics.