Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 50 / 2

The Hermitian operators form a real vector space of dimension $N^2$. To see that the same basis spans all complex operators, decompose an arbitrary $A$ as $A=B+iC$, where $B=(A+A^{†})/2$ and $C=(A-A^{†})/(2i)$ are Hermitian. Expand $B,C$ in the real orthonormal basis and combine their coefficients. The resulting complex expansion is unique because $\mathrm{Tr}({\sigma }_m{\sigma }_k)={\delta }_{mk}$. Taking the Hilbert-Schmidt inner product with ${\sigma }_m$ givesIf $A$ is Hermitian, then $a_k^{\ast }=\mathrm{Tr}(({\sigma }_{k}A{)}^{†})=\mathrm{Tr}(A{\sigma }_k)=a_k$ by cyclicity. Hence $\boxed{\vec{a}\in {\mathbb{R}}^{N^2}\text{ for Hermitian }A}$. The basis is a real basis of Hermitian operators and its complexification is a complex basis of all operators.

Write $\rho ={\sum }_n{r}_n{\sigma }_n$ in the Generalized Bloch representation. Linearity of the Lindbladian gives ${\dot{r}}_m={\sum }_n\mathrm{Tr}({\sigma }_m\mathcal{L}({\sigma }_n))r_n$. For the Hamiltonian term, cyclicity yieldsFor one Lindblad operator $V_d$, the jump term becomes $\mathrm{Tr}({\sigma }_m{V}_d{\sigma }_n{V}_d^{†})=\mathrm{Tr}(V_d^{†}{\sigma }_m{V}_d{\sigma }_n)$, and the two remaining terms giveThusEvery coefficient is real because $\mathcal{L}({\sigma }_n)$ is Hermitian. In addition, $L_{mn}=-L_{nm}$, although a dissipative $D^{(d)}$ need not be symmetric.

The Hamiltonian commutator has zero trace, and each dissipator obeys $\mathrm{Tr}(V\rho V^{†})=\mathrm{Tr}(V^{†}V\rho )$, so $\mathrm{Tr}\mathcal{L}(\rho )=0$. Consequently $\boxed{{\dot{r}}_{N^2}=0,r_{N^2}=1/\sqrt{N}}$ for normalized density operators. Orthogonality to $I/\sqrt{N}$ makes every other ${\sigma }_k$ traceless. Define $s=(r_1,\dots ,r_{N^2-1}{)}^T$. With $M=L+{\sum }_d{D}^{(d)}$, trace preservation gives the block formHere the indices of $A$ range from $1$ to $N^2-1$. Since $\mathcal{L}(I)={\sum }_d(V_d{V}_d^{†}-V_d^{†}{V}_d)$, the offset can also be written $c_m=N^{-1}\mathrm{Tr}({\sigma }_m{\sum }_d[V_d,V_d^{†}])$. This is the Affine Bloch equation; its offset is zero for unital dynamics, which preserve the identity.

A steady state is a time-independent density operator satisfying $\mathcal{L}({\rho }_{\ast })=0$, equivalently a constant physical Bloch vector $s_{\ast }$ satisfying $\boxed{A{s}_{\ast }=-c}$. Let $K=N^2-1$. By the rank criterion for a linear system, this equation has a solution exactly whenWhen it is consistent, every algebraic solution has the form $s_0+v$ with $v\in \ker A$, so the solution affine space has dimension $K-\mathrm{rank}A$ by the rank-nullity theorem. Physical steady states are its intersection with the positive density operator body. The algebraic equilibrium is unique exactly when $\mathrm{rank}A=K$, in which case $\boxed{s_{\ast }=-A^{-1}c}$.

An arbitrary affine differential equation need not have an equilibrium: $A=0$, $c\ne 0$ is inconsistent. For the finite-dimensional, time-independent Lindblad equation under discussion, however, a physical steady state always exists. Indeed, the averages ${\bar{\rho }}_T=T^{-1}{\int }_0^T\rho (t)dt$ remain positive with trace one. Compactness supplies a convergent subsequence as $T\to \infty$, andThe limit is a physical steady state, proving that Finite-dimensional Lindbladians have stationary states and that the rank consistency condition is automatically satisfied for this physical generator.

In this setting, full rank also characterizes uniqueness among physical steady states. To justify the converse when $A$ is singular, a nonzero $v\in \ker A$ gives a fixed traceless Hermitian operator $X={\sum }_k{v}_k{\sigma }_k$. Write its positive-negative decomposition $X=X_{+}-X_{-}$. Time-average the evolutions of the two positive operators along a common convergent subsequence; the limits $Y_{\pm }$ are positive stationary operators and satisfy $X=Y_{+}-Y_{-}$. Their traces are equal because $\mathrm{Tr}X=0$. If there were only one stationary density operator, both $Y_{\pm }$ would be this matrix times the same trace, forcing $X=0$, a contradiction. Hence $\boxed{\text{unique physical steady state}⟺\mathrm{rank}A=N^2-1}$.

A steady state is attractive here if every physical initial density operator approaches it. Subtracting its Affine Bloch equation from the trajectory equation gives $\delta \dot{s}=A\delta s$, so $\delta s(t)=e^{At}\delta s(0)$. The necessary and sufficient condition isSuch an $A$ is a Hurwitz stable matrix. To prove sufficiency without assuming diagonalizability, put it in Jordan normal form. Each Jordan block contributes $e^{\lambda t}{\sum }_{j=0}^{q-1}{t}^j{N}^j/j!$, and the negative real part makes the exponential dominate all polynomial factors. For necessity, an eigenmode with positive real part grows; an eigenmode with zero real part remains constant or oscillates, so it does not tend to zero. A nontrivial Jordan block on the imaginary axis also prevents decay.

The physical Bloch vector body has nonempty interior, so differences of physical initial vectors from $s_{\ast }$ span the full real coordinate space. Thus convergence for every physical initial state also forces $e^{At}\to 0$ on that space. This rules out hiding a nondecaying eigenmode outside the physical set. The strict spectral condition entails uniqueness and global asymptotic stability of the steady state.

For real $\alpha ,\beta$, the equilibrium equations are $-{\beta }^{2}x-\alpha z=0$, $y=0$ and $\alpha x-({\beta }^2+1)z=\beta$. The determinant of the coefficient matrix is $-D$, where $D={\alpha }^2+{\beta }^2({\beta }^2+1)>0$ whenever $(\alpha ,\beta )\ne (0,0)$. Solving the two-by-two system, including cases with one parameter zero, givesSubstitution verifies the first equation by cancellation of $-\alpha {\beta }^3$ and $+\alpha {\beta }^3$, and the third because ${\alpha }^2\beta +{\beta }^3({\beta }^2+1)=\beta D$. The nonzero determinant proves uniqueness. Its squared length is ${\beta }^2({\alpha }^2+{\beta }^4)/D^2$, andThus $\Vert s_{\ast }\Vert \le 1/2$, which is inside the qubit physical Bloch vector body; in the Hilbert-Schmidt normalization of part (b), that body has radius $1/\sqrt{2}$.

The $y$ mode has eigenvalue $-1$. The $xz$ block has characteristic polynomial $(\lambda +{\beta }^2)(\lambda +{\beta }^2+1)+{\alpha }^2$, so the other eigenvalues areIf $|\alpha |>1/2$, they have real part $-{\beta }^2-1/2<0$. If $|\alpha |\le 1/2$, the square root is at most $1/2$, with equality only when $\alpha =0$; the larger eigenvalue is still negative unless $\alpha =\beta =0$. At $|\alpha |=1/2$ an eventual Jordan factor does not change decay because its eigenvalue is negative. Therefore for every allowed parameter pair all eigenvalues have negative real part, and $\boxed{s_{\ast }\text{ is globally attractive}}$.