Embedded two-level quantum rotation ($e^{-i\theta A}=I-P+\cos \theta P-i\sin \theta A,P=A^2$)

For a Hermitian operator satisfying $A^3=A$, its squared operator $P=A^2$ is the projector onto the active eigenvalue-$\pm 1$ subspace. All remaining states have eigenvalue zero. Separate even and odd powers in the matrix exponential to obtain the displayed rotation on the active subspace and the identity on its spectator subspace. Thus an embedded two-level coupling need not square to the full Hilbert-space identity.