= Embedded two-level quantum rotation
{title2=$e^{-i\theta A}=I-P+\cos\theta\,P-i\sin\theta\,A,\quad P=A^2$}
For a <Hermitian operator> satisfying $A^3=A$, its squared operator $P=A^2$ is the projector onto the active eigenvalue-$\pm1$ 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.
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