= Solution
Specify the magnetic coupling before fixing its sign: let the <magnetic moment> be $\boldsymbol\mu=\gamma\mathbf S$, where $\gamma$ is a signed gyromagnetic ratio. The <Zeeman effect> gives $H_Z=-\gamma\mathbf B\cdot\mathbf S$. For <spin one-half>, $\mathbf S=\hbar\boldsymbol\sigma/2$, so with $B=|\mathbf B|$ and $u=\beta\gamma\hbar B/2$,
$$
\boxed{\rho=\frac{e^{-\beta H_Z}}{\operatorname{Tr}e^{-\beta H_Z}}
=\frac12\left[I+\tanh u\,\hat{\mathbf B}\cdot\boldsymbol\sigma\right],\qquad
Z=2\cosh u.}
$$
This is the <quantum Gibbs state>. Its <eigenvalues> are $(1\pm\tanh u)/2$, so its <purity> is $(1+\tanh^2u)/2$. For nonzero level splitting it approaches a <pure state> as $|u|\to\infty$: positive <inverse temperature> selects the lower <energy>, while negative <inverse temperature> selects the upper one. It is a <maximally mixed state> at $u=0$, including infinite <temperature> or zero magnetic splitting. At zero field the Gibbs state remains $I/2$; lowering <temperature> alone does not select a unique <pure state> from that degeneracy.
<Negative temperature> is meaningful for an internally equilibrated spin sector with a bounded <energy> spectrum and <population inversion>. It must be sufficiently isolated from unbounded motional degrees of freedom: the oscillator's unbounded spectrum, for instance, makes its partition sum diverge at negative <inverse temperature>. The two-level spin partition sum itself stays finite for either sign.
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