= Spin-one reversal under quadratic anisotropy
{title2=$P_{-x}(t)=\sin^2[(A+B)\hbar t/2]$}
For a spin-one <Hamiltonian> $H=AS_z^2+B(S_x^2-S_y^2)$, Cartesian basis energies are $\hbar^2(A-B)$, $\hbar^2(A+B)$ and zero along the $x,y,z$ basis labels. The $S_x=\hbar$ state combines the latter two basis vectors equally. Their relative phase produces the displayed spin-reversal probability. Complete reversal occurs at odd multiples of $\pi/(|A+B|\hbar)$ if $A+B\ne0$; when it vanishes those two components remain degenerate. Individual matrix signs depend on basis conventions, while the <commutators> and physical probability must remain consistent.
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