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

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.