 Solution

Specify the magnetic coupling before fixing its sign: let the magnetic moment be $\mathbf{\mu }=\gamma \mathbf{S}$, where $\gamma$ is a signed gyromagnetic ratio. The Zeeman effect gives $H_Z=-\gamma \mathbf{B}\cdot \mathbf{S}$. For spin one-half, $\mathbf{S}=\hslash \mathbf{\sigma }/2$, so with $B=|\mathbf{B}|$ and $u=\beta \gamma \hslash B/2$,This is the quantum Gibbs state. Its eigenvalues are $(1\pm \tanh u)/2$, so its purity is $(1+{\tanh }^{2}u)/2$. 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.