= Solution
Here “controllable” means <density operator controllability>, so all transformations on a fixed spectral orbit are available. Strict <unitary operator controllability>, including an independently specified <global phase>, is a stronger condition. Assume the usual finite-dimensional real controls and freely chosen duration.
The dimension hypothesis actually identifies the algebra, not just its size. For completeness, put the invariant inner product $\langle X,Y\rangle=-\operatorname{Tr}(XY)$ on $\mathfrak u(N)$ and write $\mathfrak g^\perp=\mathbb R Z$. Since $\mathfrak g$ is closed under <commutators>, $[X,Z]$ lies in $\mathfrak g^\perp$ for every $X\in\mathfrak g$. The adjoint action is skew-adjoint for this inner product, so $[X,Z]$ is also orthogonal to $Z$ and must vanish. If $Z$ were nonscalar, its orthogonal eigenspaces would give a centralizer of dimension at most $(N-1)^2+1<N^2-1$, too small to contain $\mathfrak g$. Therefore $Z$ is scalar and $\mathfrak g=\mathfrak{su}(N)$. This proves <codimension-one quantum dynamical algebra is special unitary>.
\b[a1 is true for density-state controllability; a2 is true; a3 is false.] The algebra $\mathfrak{su}(N)$ generates every determinant-one unitary. For any $V\in U(N)$, choose $\varphi$ with $e^{iN\varphi}=\det V$; then $e^{-i\varphi}V\in SU(N)$, proving that every gate is available up to <global phase>. But all generated <Quantum Hamiltonians> are traceless, and
$$
\frac d{dt}\det U(t)=-i\operatorname{Tr}H(t)\det U(t)=0,
$$
so evolution starting at the identity always has determinant one. Arbitrary determinant, requiring the additional central direction $iI$, cannot be generated. If a1 is instead read as phase-sensitive unitary-operator controllability, it is false for precisely the same reason as a3; the physical distinction is explicit here.
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