 Spin-one reversal under quadratic anisotropy

For a spin-one Hamiltonian $H=A{S}_z^2+B(S_x^2-S_y^2)$, Cartesian basis energies are ${\hslash }^2(A-B)$, ${\hslash }^2(A+B)$ and zero along the $x,y,z$ basis labels. The $S_x=\hslash$ state combines the latter two basis vectors equally. Their relative phase produces the displayed spin-reversal probability. Complete reversal occurs at odd multiples of $\pi /(|A+B|\hslash )$ if $A+B\ne 0$; when it vanishes those two components remain degenerate. Individual matrix signs depend on basis conventions, while the commutators and physical probability must remain consistent.