Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 60 / 2 / e / Solution

Let $P_0,P_1$ be any two rank-one orthogonal projections. Choose $0<q<1/N$ and $p=1-(N-1)q$, so $p>q$, and form the mixed density operatorsThey have the same spectrum. If density operator controllability holds, a reachable $U$ obeys $U{\rho }_0{U}^{†}={\rho }_1$, and subtraction of $qI$ gives $U{P}_0{U}^{†}=P_1$. Thus density operator controllability implies pure-state controllability, even if its definition is phrased using only mixed density operators.

The converse is false in general. For $N=4$, the full compact symplectic group $Sp(2)$ is transitive on the complex unit sphere: identify ${\mathbb{C}}^4$ with ${\mathbb{H}}^2$, extend any quaternionic unit vector to an orthonormal quaternionic basis, and map one such basis to another. This gives pure-state controllability. Yet the invariant in part (d) prevents it from connecting the indicated isospectral mixed density operators, so it fails density operator controllability. This counterexample uses the full symplectic group, not an assertion that every parameter choice in part (c) generates it.

For $N=2$, the converse is true. Every density operator has the form $\rho =qI+(p-q)P$ for a rank-one $P$, with $p+q=1$. If $p\ne q$, pure-state control of $P$ controls its entire fixed-spectrum unitary orbit of a density operator. If $p=q=1/2$, the orbit contains only $I/2$. Therefore the two controllability notions coincide for a qubit. The Lie classification gives the same conclusion, since $Sp(1)=SU(2)$.