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

In units $\hslash =1$, a bilinear Hamiltonian control system haswhere the Hermitian matrices $H_0,H_k$ are fixed, and the real functions $u_k$ are admissible control inputs. The corresponding vector or density operator dynamics are $\dot{\psi }=-iH(u)\psi$ or $\dot{\rho }=-i[H(u),\rho ]$. The dynamics are linear in the state at fixed input and affine in the inputs at fixed state; this is the bilinear terminology.

The reachable set from an initial state $x_0$ at time $T$ is ${\mathcal{R}}_T(x_0)=\{x(T;x_0,u):u\text{ admissible}\}$. With freely chosen duration, take $\mathcal{R}(x_0)={\bigcup }_{T\ge 0}{\mathcal{R}}_T(x_0)$. Controllability means every allowed target is reachable from every allowed initial state. The allowed space and phase convention must be specified. For closed quantum control, density operators remain on their initial unitary orbit of a density operator, rather than becoming arbitrary density operators with different eigenvalues.

A real Lie algebra is a real vector space with a bilinear antisymmetric bracket satisfying the Jacobi identity. For matrices, the bracket is the commutator $[X,Y]=XY-YX$. The Dynamical Lie algebra isIt is found by repeatedly adjoining commutators and taking real linear spans until no new independent generators appear. Let $G$ be its connected dynamical Lie group. Under the standard assumptions of freely timed controls with unrestricted real amplitudes, the following are the relevant Lie criteria.

For pure-state controllability, $G$ must act transitively on normalized state vectors, or on their rays when global phase is disregarded. The standard finite-dimensional classification is, up to a unitary change of basis,Here $\mathfrak{sp}(n)$ is the compact symplectic Lie algebra in its defining complex $2n$-dimensional representation. Equivalently, the infinitesimal action spans the tangent directions to the whole pure-state orbit. The symplectic cases are important because pure-state reachability alone need not give complete mixed-state reachability.

For density operator controllability, the action must be transitive on every unitary orbit of a density operator, meaning on every fixed-spectrum class. The criterion is $\mathfrak{g}=\mathfrak{su}(N)$ or $\mathfrak{u}(N)$. This concerns all density operators, not merely one unusually degenerate spectrum.

For exact unitary operator controllability, the criterion is $\mathfrak{g}=\mathfrak{u}(N)$. If only the physical gate up to global phase matters, it is enough that the image after removing the scalar direction is $\mathfrak{su}(N)$, equivalently that $\mathfrak{g}+\mathbb{R}iI=\mathfrak{u}(N)$. A traceless system can implement all special-unitary gates while lacking independent control of the overall phase.

Set $X_j=i{H}_j$. The symplectic Lie condition is $X_j^{T}J+J{X}_j=0$, equivalently $H_j^{T}J+J{H}_j=0$. For the supplied diagonal $H_0$, the nonzero entries of $J$ pair levels $1,4$ and $2,3$; the corresponding energy sums are $-\alpha +\alpha =0$ and $-\beta +\beta =0$. Thus $H_{0}J+J{H}_0=0$ for all real $\alpha ,\beta$.

For the supplied real symmetric control matrix, direct multiplication givesBoth generators therefore satisfy the same alternating-form condition. That condition is closed under real linear combinations, and for two such generatorsas follows by substituting $X^{T}J=-JX$ and $Y^{T}J=-JY$ into $(XY{)}^{T}J-(YX{)}^{T}J$. Hence the whole Dynamical Lie algebra obeys it.

If $\dot{U}=X(t)U$ and $U(0)=I$, differentiating givesThus every reachable propagator satisfies $U^{T}JU=J$. Also $J^T=-J$, $J^{†}J=I$, so this is a symplectic dynamical symmetry in quantum control: the generated group lies in the corresponding compact symplectic group, unitarily equivalent to $Sp(2)$. It is not necessary to claim equality with $Sp(2)$, and special choices of $\alpha ,\beta$ can give a smaller group.

The two density operators have the same eigenvalues $a,a,b,b$, so a permutation of basis vectors makes them unitarily equivalent. A dynamical equivalence must additionally respect the symplectic dynamical symmetry in quantum control.

Define the symplectic dual $\tilde{\rho }=J{\rho }^T{J}^{†}$. For a unitary $U$ preserving $J$, both $U^{T}JU=J$ and $UJ{U}^T=J$ hold. In particular $J{U}^{\ast }=UJ$, and thereforeIt follows by cyclicity of the trace that the symplectic invariant of a density operatoris unchanged by every reachable unitary conjugation. The given $J$ reverses the diagonal order, soConsequentlyHence the states are not dynamically equivalent, despite their identical ordinary spectra. This works for all the allowed values, including $a=0$, and for any subgroup satisfying the symmetry in part (c).

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 $Sp(1)=SU(2)$.