 Solution

Take the two magnetic moments along $z$ to be $\pm \mu$, so $H=-\mu B_z{\sigma }_z$ and the two energies are $\mp \mu B_z$. The quantum Gibbs state at inverse temperature $\beta =1/(k_{B}T)$ isIts Bloch vector is $(0,0,\tanh \beta \mu B_z)$. At $B_z=0$, or in the high-temperature limit, it approaches $I/2$; at $T\to 0^{+}$ with nonzero $\mu B_z$, it approaches the lower-energy pure state. Here $\mu$ is the magnitude or signed coefficient multiplying ${\sigma }_z$ in the magnetic Hamiltonian, rather than a moment with an additional unstated factor of two.