Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 60 / 2

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 $⟨X,Y⟩=-\mathrm{Tr}(XY)$ 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.

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 $\phi$ with $e^{iN\phi }=\det V$; then $e^{-i\phi }V\in SU(N)$, proving that every gate is available up to global phase. But all generated Quantum Hamiltonians are traceless, andso 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.

The stated relation places every generator, and therefore the whole Dynamical Lie algebra, inside the standard compact symplectic Lie algebra $\mathfrak{sp}(\ell )\subset \mathfrak{su}(2\ell )$. Indeed the relation is closed under linear combinations and commutators, and differentiating evolution givesThus it supplies a symplectic dynamical symmetry in quantum control, rather than a claim that $J$ commutes with every Quantum Hamiltonian.

For $\ell \ge 2$, b1 is false: even the full compact symplectic group does not act transitively on every mixed-state spectral orbit. Its algebra has dimension $\ell (2\ell +1)<(2\ell {)}^2-1$. An explicit obstruction is the symplectic invariant of a density operatorwhich is unchanged by a symplectic conjugation. For $N=4$ and the printed $J$, let $a,b>0$, $a\ne b$, $2a+2b=1$. The isospectral operators $\mathrm{diag}(a,a,b,b)$ and $\mathrm{diag}(a,b,a,b)$ have invariants $4ab$ and $2(a^2+b^2)$, respectively, and cannot be interconverted. For the exceptional case $\ell =1$, $\mathfrak{sp}(1)=\mathfrak{su}(2)$, so b1 cannot be decided from inclusion alone; equality gives density-state controllability.

b2 is not implied by the displayed relation: more information is needed. It permits the full symplectic algebra but also a single commuting generator, or even zero generators. The full $Sp(\ell )$ is pure-state controllable: identify ${\mathbb{C}}^{2\ell }$ with ${\mathbb{H}}^{\ell }$, extend any unit vector to a quaternionic orthonormal basis, and map one such basis to another. A one-axis diagonal subgroup, in contrast, cannot change level populations.

The required extra information is the generated algebra, obtained by computing its real span of iterated commutators. Within this specified symplectic inclusion, equality $\mathfrak{g}=\mathfrak{sp}(\ell )$ gives pure-state controllability; its dimension is $\ell (2\ell +1)$. The mere inclusion does not establish equality. For $N=2$, equality also answers b1 affirmatively. None of these strictly symplectic generators supplies arbitrary global phase.

c1 is true for density-state controllability. Every qubit density operator can be written $\rho ={\lambda }_{2}I+({\lambda }_1-{\lambda }_2)|\psi ⟩⟨\psi |$ using its eigenvalues. A control taking one pure orthogonal projector to any other therefore also connects every pair of qubit density operators with the same spectrum. In Lie-algebra language, $\mathfrak{sp}(1)=\mathfrak{su}(2)$; there is no larger-dimensional symplectic exception. This does not supply arbitrary global phase: $SU(2)$ alone is already a counterexample to an implication involving strict operator controllability.

c2 is false. For every even $N=2\ell \ge 4$, the defining $Sp(\ell )$ action is transitive on unit vectors, so a system generating this algebra is pure-state controllable. It still preserves additional mixed-state invariants and lacks density operator controllability, as the preceding explicit four-dimensional example demonstrates. pure states have fewer orbit requirements than general mixed states.

True for the finite bound-level Morse model with nonzero adjacent control couplings. The Morse oscillator energies have the formOver the physical bound-level range these adjacent gaps are positive, and $B\ne 0$ makes them pairwise distinct. Label the retained levels consecutively. With $H_1={\sum }_j{d}_j(E_{j,j+1}+E_{j+1,j})$, all $d_j\ne 0$, the squared adjoint action of $i{H}_0$ has eigenvalue $-(E_{j+1}-E_j{)}^2$ on each adjacent skew-Hermitian coupling. A polynomial chosen by Lagrange interpolation can therefore isolate any one adjacent coupling from $i{H}_1$.

Commuting an isolated coupling once with the drift produces its independent antisymmetric quadrature. The bracket between the two quadratures yields a traceless diagonal generator; brackets along the connected chain produce every nonadjacent off-diagonal generator. Thus connected nondegenerate transition chain generates special unitary control proves $\mathfrak{su}(N)\subseteq \mathfrak{g}$, which gives density-state controllability. If the physical drift has nonzero trace, subtracting its traceless part also produces the identity direction, giving $\mathfrak{u}(N)$ and full operator controllability. Shifting the energy origin may remove that extra phase direction but does not change state controllability. The assertion concerns the finite-level model in the question, not the entire bound-plus-dissociation Hilbert space of the untruncated Morse potential.