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

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 $x_{\ast }$.