= Solution
For real $\alpha,\beta$, the equilibrium equations are $-\beta^2x-\alpha z=0$, $y=0$ 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, gives
$$
\boxed{s_*=(x_*,y_*,z_*)^T=\frac1D(\alpha\beta,0,-\beta^3)^T}.
$$
Substitution 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 $\beta^2(\alpha^2+\beta^4)/D^2$, and
$$
D^2-4\beta^2(\alpha^2+\beta^4)=(\alpha^2+\beta^4-\beta^2)^2\ge0.
$$
Thus $\|s_*\|\le1/2$, which is inside the qubit physical <Bloch vector> body; in the Hilbert-Schmidt normalization of part (b), that body has radius $1/\sqrt2$.
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 are
$$
\boxed{\lambda_\pm=-\beta^2-\frac12\pm\sqrt{\frac14-\alpha^2}}.
$$
If $|\alpha|>1/2$, they have real part $-\beta^2-1/2<0$. If $|\alpha|\le1/2$, 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 $\boxed{s_*\text{ is globally attractive}}$.
Back to article page
