Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 60 / 1

Use the real vector space of Hermitian matrices and its Hilbert-Schmidt inner product $⟨A,B⟩=\mathrm{Tr}(AB)$. A Hermitian $N$-by-$N$ matrix has $N$ real diagonal entries and two real parameters for each off-diagonal pair, hence real dimension $N+2\left(\genfrac{}{}{0ex}{}{N}{2}\right)=N^2$. Imposing zero trace removes one real dimension. Each supplied generalized Pauli matrix is Hermitian and traceless, and there are $N(N-1)$ off-diagonal matrices and $N-1$ diagonal matrices, totaling $N^2-1$.

Their assumed orthonormality makes them linearly independent. Since this equals the dimension of the traceless space, they are an orthonormal basis of the traceless Hermitian matrices. To extend the family, the required additional generator is ${\sigma }_0=I/\sqrt{N}$, which is implicit in the indexing of the question. It has unit Hilbert-Schmidt norm and is orthogonal to every traceless generator, so the extended family is an orthonormal basis of all Hermitian matrices.

For a density operator $\rho$, cyclicity of the trace givesThus $s\in {\mathbb{R}}^{N^2-1}$. Since $\mathrm{Tr}\rho =1$, the resulting Generalized Bloch representation is

By orthonormality in the Generalized Bloch representation,A pure state has density operator $\rho =|\psi ⟩⟨\psi |$ with $⟨\psi |\psi ⟩=1$, so ${\rho }^2=\rho$ and its purity of a density operator is one. ConsequentlyEvery pure state therefore maps to the specified radius-$\sqrt{1-1/N}$ sphere in ${\mathbb{R}}^{N^2-1}$, of dimension $N^2-2$. The expansion is unique, so it distinguishes pure-state density operators, while vectors differing only by global phase describe the same point. This proves that pure states lie on the sphere, without asserting that they fill it.

Take any rank-one orthogonal projection $P$ and its generalized Bloch vector $s$. The antipodal vector $-s$ is on the same sphere, but its corresponding trace-one Hermitian matrix isOn the range of $P$, this matrix has eigenvalue $2/N-1$, which is negative for $N>2$. It is therefore not a positive semidefinite matrix and cannot be a density operator. Hence the pure states do not cover the sphere when $N>2$. This also shows directly why the purity bound for generalized Bloch vectors does not by itself guarantee a physical quantum state.

The eigenvalues $p_j$ of a density operator are nonnegative and sum to one. ThereforeFor a non-pure state at least two eigenvalues are positive, so the last expression is strictly positive. Combining this with the Generalized Bloch representation givesThus every non-pure quantum state lies strictly inside the radius sphere for $N\ge 2$. It need not be an interior point of the entire physical state body: for $N>2$, a rank-deficient mixed density operator can lie on that body's boundary while still lying inside the Euclidean sphere.

Write the unitary conjugation as $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. The identity component is unchanged, and the generalized Bloch vector transforms byThese coefficients are real, by the same trace argument as in part (a). Unitary conjugation preserves the Hilbert-Schmidt inner product, so the conjugated traceless generators form another orthonormal basis. It follows that $O(U{)}^{T}O(U)=I$. The unitary group is connected, $O(I)=I$, and the determinant of an orthogonal matrix is $\pm 1$; hence $\det O(U)=1$. Thus Hamiltonian evolution rotates the generalized Bloch vector by an element of $SO(N^2-1)$.

Equivalently, in units $\hslash =1$, the Von Neumann equation givesCyclicity of the trace shows $A_{k\ell }=-A_{\ell k}$, so $A$ is a real skew-symmetric matrix. This is the generator form of Hamiltonian rotations of Bloch vectors, valid also for a time-dependent Quantum Hamiltonian. For $N>2$, Hamiltonians generally realize only a proper subgroup of all rotations of this real space; they preserve the whole density-operator spectrum, not merely its purity of a density operator.

The individual rates describe population relaxation: ${\gamma }_{12}$ transfers population from level one to level two, and ${\gamma }_{21}$ transfers it back. With $p_1+p_2=1$ and the conventional qubit coordinate $r_z=p_1-p_2$,Thus $T_1^{-1}={\gamma }_{12}+{\gamma }_{21}$ is the longitudinal relaxation rate. The parameter $\Gamma =T_2^{-1}$ is the transverse relaxation rate: it damps the off-diagonal coherence, or the $x,y$ components of the Bloch vector. In the usual completely positive two-level model,so $\Gamma$ includes both the coherence loss caused by population relaxation and additional pure dephasing.

There is a normalization switch in the displayed qubit Affine Bloch equation. Its constant term is appropriate to the conventional Pauli matrices and unit-radius coordinate $r_k=\mathrm{Tr}(\rho {\sigma }_k^{\mathrm{Pauli}})$. With the orthonormal generators used earlier, $s=r/\sqrt{2}$, the same physical rates would instead give a constant term $({\gamma }_{21}-{\gamma }_{12})/\sqrt{2}$. The drift matrix is unchanged. We interpret the supplied qubit equation in its conventional $r$ coordinates.

With zero controls and $g={\gamma }_{12}+{\gamma }_{21}>0$, $\Gamma >0$, the steady state is unique:The earlier normalized vector is $s_{\ast }=r_{\ast }/\sqrt{2}$. Indeed $r_x,r_y$ decay as $e^{-\Gamma t}$ and $r_z-r_{\ast ,z}$ as $e^{-gt}$.

For vanishing rates, the full equilibrium conditions of the Affine Bloch equation are $\Gamma r_x=\Gamma r_y=0$ and $g{r}_z={\gamma }_{21}-{\gamma }_{12}$, together with $\Vert r\Vert \le 1$. In particular, for nonnegative rates, $g=0$ means both population rates vanish. If then $\Gamma >0$, all diagonal density operators are stationary; if also $\Gamma =0$, every density operator is stationary. The formal case $g>0$, $\Gamma =0$ permits extra transverse equilibrium coordinates, but is excluded by the completely positive rate bound above.