Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 52 / 2

A basis of Hermitian matrices is a real basis for the Hermitian subspace, but it also supplies a complex basis for all matrices. Indeed every operator has the decompositionwith $B,C$ Hermitian. Expand each in the given real basis, then combine their coefficients to expand $A$. Orthogonality for the Hilbert-Schmidt inner product determines those complex coefficients uniquely:ConsequentlyIf $A$ is Hermitian, cyclicity of the trace givesThus its coordinate vector is real. The product ${\sigma }_{j}A$ need not itself be Hermitian; trace cyclicity is what justifies the conclusion.

A density operator is Hermitian, positive semidefinite and has trace one. Part (a) therefore makes all its coordinates real, and the identity basis vector fixes the final coordinate:The remaining basis vectors are traceless by orthogonality to $I/\sqrt{N}$. The Generalized Bloch representation is consequentlyUsing orthonormality to expand the squared Hilbert-Schmidt norm gives the purity bound for generalized Bloch vectors:If $p_1,\dots ,p_N$ are its nonnegative eigenvalues, then ${\sum }_j{p}_j=1$ andIt follows thatEquality requires every pairwise product $p_j{p}_k$ to vanish, so exactly one eigenvalue equals one and all others are zero. Conversely, a rank-one projector has precisely that spectrum and attains equality. Thus equality holds exactly for a pure state. This normalization is different from the commonly used unit-radius qubit Bloch vector.

For $N=2$, take the normalized Pauli operators as the traceless basis. The matrix associated with a vector $s$ is $\rho =I/2+s\cdot \sigma$, where these basis matrices are the usual Pauli matrices divided by $\sqrt{2}$. Their multiplication relations giveAt radius $1/\sqrt{2}$, these eigenvalues are one and zero. Every sphere point therefore gives a positive rank-one projector. Any other orthonormal Hermitian basis merely changes the real coordinates by an orthogonal transformation, so the result is basis independent.

For every $N>2$, choose a rank-one projector $P$ and consider the Hermitian trace-one matrixIts eigenvalues are $2/N-1$ once and $2/N$ with multiplicity $N-1$. They sum to one and satisfyBy orthonormality its traceless coordinate vector has squared length $1-1/N$, so it is on the specified sphere. But $2/N-1<0$, so $A$ is not a density operator. Thus for $N>2$, the pure states occupy a proper subset of the sphere; positivity excludes some of its points. In fact, this example is the antipodal Bloch vector to that of $P$.

The usual assertion presumes $N\ge 2$. If the trivial dimension $N=1$ is admitted, the radius-zero sphere consists of the single coordinate point, also corresponding to the single pure state, so that dimension is an additional trivial exception.

Let $\mathcal{G}=-i[H,\cdot ]+{\sum }_d\mathcal{D}[V_d]$ denote the Lindbladian. Substituting $\rho ={\sum }_n{r}_n{\sigma }_n$ and projecting with the Hilbert-Schmidt inner product givesFor the Quantum Hamiltonian term, cyclicity of the trace yieldsFor each Lindblad dissipator, the same cyclic rearrangement givesThese are the required coefficient matrices, so $\dot{r}=(L+{\sum }_d{D}^{(d)})r$. They are real because both generators map Hermitian matrices to Hermitian matrices, whose coordinates are real by (a).

The trace of a commutator vanishes. AlsoThus the last row of $L+{\sum }_d{D}^{(d)}$ is zero and $\boxed{{\dot{r}}_{N^2}=0}$. It is trace preservation, not preservation of the purity of a density operator, that produces this constant coordinate.

Write $G=L+{\sum }_d{D}^{(d)}$ and $n=N^2-1$. Since $r_{N^2}=1/\sqrt{N}$, the remaining coordinates obey the Affine Bloch equationBecause $[H,I]=0$ and $\mathcal{D}[V](I)=[V,V^{†}]$, its offset can also be writtenIt vanishes for unital dynamics, but need not vanish for a general dissipative evolution.

For a Hermitian Quantum Hamiltonian, $[{\sigma }_m,{\sigma }_n{]}^{†}=-[{\sigma }_m,{\sigma }_n]$. ConsequentlySwapping $m,n$ changes the sign of the commutator, so $L_{nm}=-L_{mn}$. The generator is therefore real and antisymmetric. Its last column vanishes because the identity commutes with $H$, so $c=0$ and the traceless block satisfies $A^T=-A$.

For constant $H$, the Bloch vector evolves by $s(t)=R(t)s(0)$ with $R(t)=e^{At}$. Direct differentiation givesHence $R^{T}R=I$. Its determinant is continuously connected to one and cannot change between the two possible values $\pm 1$, so $\det R=1$. This proves that the evolution is a rotation in the special orthogonal group. A time-dependent Hermitian Quantum Hamiltonian has the same conclusion from $\dot{R}=A(t)R$.

These Hamiltonian rotations of Bloch vectors preserve the Euclidean norm and purity of a density operator. For $N>2$, not every rotation of the ambient sphere is a physical unitary conjugation: the higher-dimensional positivity constraints and full spectrum of the density operator impose further restrictions.

Density operator controllability cannot defeat unitary invariants: it grants access to the unitary orbit of a density operator, whose members have the same eigenvalues, rather than to every density operator. In the Hilbert-Schmidt inner product orthonormal normalization of this question,Quantum Hamiltonian evolution preserves this norm, so the target is not reachable from the stated initial state. Equivalently, their values of purity of a density operator are $1/2+1/4=3/4$ and $1/2+1/2=1$: the initial state is a mixed state and the target a pure state. Dissipation, an appropriate measurement protocol or an auxiliary system with additional resources would be needed to change the spectrum; freely controlling the closed-system Quantum Hamiltonian alone does not suffice.