Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 52 / 3

A steady state is a time-independent physical state: its derivative vanishes. With $n=N^2-1$, the Affine Bloch equation therefore requiresAs a real linear equation, it has a solution exactly when $\mathrm{rank}A=\mathrm{rank}[A\mid -c]$. If $s_p$ is one solution, all algebraic solutions are $s_p+\ker A$, an affine set of dimension $n-\mathrm{rank}A$. Physical steady states are its intersection with the set of Bloch vectors corresponding to positive trace-one density operators. An arbitrary affine equation need not have a steady state: $A=0$, $c\ne 0$ is a counterexample.

The equation here comes from a finite-dimensional time-independent Lindblad equation, however, so it always has at least one physical steady state. To prove this, start from any density operator and average its trajectory:These averages remain positive with trace one. That set is a compact set, so some sequence has a limit ${\rho }_{\ast }$. If $\mathcal{L}$ is the time-independent generator, thenbecause density operators are bounded. Continuity gives $\mathcal{L}({\rho }_{\ast })=0$. This proves directly that Finite-dimensional Lindbladians have stationary states.

If $A$ has full rank, its algebraic solution $s_{\ast }=-A^{-1}c$ is unique and existence makes that solution physical. For these Lindblad dynamics the converse also holds, even for a boundary steady state. If $A$ is singular, choose a nonzero traceless Hermitian stationary direction $X$. The two parts of its positive-negative decomposition have the same trace $\tau >0$, so $X=\tau ({\rho }_{+}-{\rho }_{-})$ for two density operators. Time-average their trajectories along a common convergent subsequence. Their limits are stationary by the preceding argument, and their difference remains $X/\tau$, since $X$ itself is stationary. The limits are therefore distinct physical stationary states.

Thus the finite-dimensional physical Bloch equation has a unique steady state exactly when $\mathrm{rank}A=n$. These steady states of an affine Bloch equation must be distinguished from convergence to that steady state, which requires the additional spectral condition in (b). The possibly nonorthonormal coordinate basis does not affect existence or uniqueness, since it still gives an invertible parametrization of the traceless Hermitian space.

Subtract a steady state to obtain $e(t)=s(t)-s_{\ast }$ with $\dot{e}=Ae$, hence $e(t)=e^{At}e(0)$. The steady state attracts every initial condition exactly whenIn other words, $A$ must be a Hurwitz stable matrix. To see sufficiency without assuming diagonalizability, each block in its Jordan normal form contributes terms of the form $t^k{e}^{\mu t}$ with finite $k$. Negative real part makes every such term tend to zero, so the whole matrix exponential tends to zero.

For necessity, an eigenmode with positive real part grows, and a mode with zero real part persists or oscillates instead of decaying. For a complex eigenvalue of a real matrix, take the real and imaginary parts of its eigenvector to obtain real nondecaying solutions. Thus any eigenvalue in the closed right half-plane prevents convergence of all deviations. Physical density operators span the trace-one affine space, so convergence from all physical initial states also forces decay on every traceless direction.

This gives global asymptotic stability for the affine linear system and uniqueness of its steady state. A merely negative-semidefinite symmetric part proves only nonincrease of a particular norm, not this necessary-and-sufficient condition.

Take the driving and feedback strengths to be real, so the feedback Quantum Hamiltonian is Hermitian. The lowering operator satisfies ${\sigma }^2=({\sigma }^{†}{)}^2=0$. Using the supplied convention for ${\sigma }_y$,The additional Quantum Hamiltonian term is consequentlyWith no drift Quantum Hamiltonian, the total Quantum Hamiltonian is just $\alpha {\sigma }_y$. The feedback-modified Lindblad operator isSubstitution into the measurement-feedback equation givesThus the feedback affects the Lindblad dissipator even though its extra coherent Quantum Hamiltonian contribution cancels. This is a concrete mechanism of Markovian reservoir engineering, rather than merely a classical control force added to an unchanged dissipative equation.

First fix a source convention needed for physical coordinates. The PDF's displayed definition of ${\sigma }_z$ omits a factor of $i$: half the commutator of the two Hermitian transverse matrices is anti-Hermitian and cannot supply a real Bloch coordinate. The Hermitian choice consistent with the supplied dynamical matrix isWith this convention the three Pauli expectation values use the unit-radius normalization, unlike the orthonormal coordinates of Question 2. For example, writing $K=\left(\begin{array}{cc}0& a\\ b& 0\end{array}\right)$ with $a=1+\lambda$, $b=-\lambda$, the dissipator gives rates $-(a-b{)}^2/2$ for $x$, $-(a+b{)}^2/2$ for $y$, and population equation $\dot{z}=b^2-a^2-(a^2+b^2)z$. The Quantum Hamiltonian adds $2\alpha z$ to $\dot{x}$ and $-2\alpha x$ to $\dot{z}$. This independently reproduces the supplied real Bloch dynamics.

Put $q=2\lambda +1$. Since $y$ decays independently, every steady state has $y_{\ast }=0$. The reduced steady-state equations areTheir coefficient determinant isIt is strictly positive unless $q=0$ and $\alpha =0$. Inverting this two-by-two system givesThus the reduced system has a unique steady state except at $\alpha =0$, $\lambda =-1/2$. At that exceptional point the equations reduce to $\dot{x}=0$, $\dot{z}=-z/2$, so every $(x,0)$ is stationary; physical coordinates restrict $-1\le x\le 1$.

As a consistency check on positivity in the nonsingular case,The calculated state is therefore in the unit Bloch disk. The original PDF correctly labels the second steady-state expression as $z_{\ast }$; the TeX conversion's repeated $x_{\ast }$ is another transcription defect, not a second formula for $x_{\ast }$.

Write $S=\sin {\theta }_d$, $C=\cos {\theta }_d$. The chosen feedback parameters give $q=-C$ and $\alpha =SC/4$. Substitute the desired state directly into the dynamics, rather than dividing by a determinant that can vanish:Taking $y=0$ makes its derivative zero as well. Hence the prescribed state is stationary for every target angle, including the equator. It is pure because $S^2+C^2=1$ in this Pauli normalization.

For $C\ne 0$, $D=C^2(C^2+1)+S^2{C}^2=2C^2$, and the expressions in (d) reduce to $x_{\ast }=S$, $z_{\ast }=C$, confirming uniqueness. For $C=0$, both $q$ and $\alpha$ vanish and $D=0$. The desired equatorial state is still stationary by direct substitution, but belongs to the whole stationary segment $(x,0)$; the singular formula cannot be used to claim unique stabilization.

With $q=2\lambda +1$, the reduced coefficient matrix and its symmetric part areIts diagonal entries are both strictly negative exactly when $q\ne 0$, equivalently $\boxed{\lambda \ne -1/2}$. The coherent driving cancels from the symmetric part because it is a rotation generator.

For a steady state $s_{\ast }$ put $e=s-s_{\ast }$. Differentiating the squared Euclidean distance givesFor $q\ne 0$ the derivative is strictly negative whenever $e\ne 0$. Integrating the inequality proves the symmetric-part contraction criterion here explicitly:Thus every nonstationary initial state approaches the unique steady state, with strictly decreasing distance; the steady state itself has constant zero distance. If the decoupled $y$ coordinate is retained, it contributes $-e_y^2$ to the squared-distance derivative and also decays, so the conclusion holds for the full qubit state.

For the target-dependent settings in (e), $q=-\cos {\theta }_d$. Every nonequatorial target is therefore globally attractive. At an equatorial target, $q=\alpha =0$: $x(t)=x(0)$, $z(t)=z(0)e^{-t/2}$, and $y(t)=y(0)e^{-t/2}$. A general state approaches $(x(0),0,0)$ rather than the specified pure target $x=\pm 1$. This proves the equatorial exception, not merely failure of a sufficient estimate.

The parameter dependence matters: $q=0$ alone does not exclude attractivity for every possible driving strength. If $\alpha \ne 0$, the reduced matrix has negative trace and positive determinant $4{\alpha }^2$, so both eigenvalues have negative real parts even though its symmetric part is only semidefinite. The nonattractive case asserted for the equatorial target also uses its prescribed $\alpha =0$.