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Both the drift and local $Z_k$ control commute with $S$, so $[H_S+f(t)Z_k,S]=0$ at every time. The propagator consequently obeys $U(t{)}^{†}SU(t)=S$, as follows by differentiating this expression. Thus every reachable unitary operator preserves the magnetization sectors of a spin chain. If $P_q$ is a spectral projector of $S$, then $U$ commutes with $P_q$ and $\mathrm{Tr}(P_q\rho (t))=\mathrm{Tr}(P_q\rho (0))$.

The eigenspaces are nontrivial proper subspaces: $|0\cdots 0⟩$ has eigenvalue $N$, while $|10\cdots 0⟩$ has eigenvalue $N-2$. One cannot transfer the first pure state to the second, although their density operators have the same spectrum. Nor can one implement a unitary that mixes these two sectors. Hence $\boxed{\text{the full system has neither operator, density, nor pure-state controllability}}$. Its Dynamical Lie algebra lies in the block-diagonal commutant of $S$. Controllability within a fixed sector, such as the single-excitation subspace, is still possible and must be tested separately.

Use the computational basis ordered as $|000⟩,|001⟩,|010⟩,|011⟩,|100⟩,|101⟩,|110⟩,|111⟩$. Since $Z|0⟩=|0⟩$ and $Z|1⟩=-|1⟩$,For arbitrary $N$, a computational basis vector with $q$ ones has $N-q$ contributions $+1$ and $q$ contributions $-1$, giving eigenvalue $N-2q$. All $q=0,\dots ,N$ occur, soThe binomial coefficient counts which $q$ sites are excited. Thus there are exactly $N+1$ distinct magnetization sectors of a spin chain, and their dimensions sum to ${\sum }_q\left(\genfrac{}{}{0ex}{}{N}{q}\right)=2^N$, the full Hilbert-space dimension.