 Solution

The fixed-particle-number sector is spanned by $|1⟩=c_1^{†}|0⟩$ and $|2⟩=c_2^{†}|0⟩$. In this basis,The relevant quantum Gibbs state is normalized within this two-dimensional sector, not over all particle numbers. With $\beta =(k_{B}T{)}^{-1}$ and $x=\beta t$, the exponential of a Pauli vector givesThereforeIts eigenstates are $|\pm ⟩=(|1⟩\pm |2⟩)/\sqrt{2}$, with energies $\mp t$ and probabilities $p_{\pm }=e^{\pm x}/(2\cosh x)$. These are nonnegative and sum to one, so the matrix is Hermitian and has trace one. Its purity of a density operator isFor $t\ne 0$, the zero-temperature limit is the pure ground-state projector,with purity one. As $T\to \infty$, $x\to 0$ and $\rho \to I/2$, with purity $1/2$. At every strictly positive finite temperature and finite $t$, both eigenvalues are nonzero, so the state is mixed. If $t=0$, the two levels are degenerate and the Gibbs state is $I/2$ at all temperatures, including its zero-temperature limit.

Using dimensionless Von Neumann entropy, $S=-\mathrm{Tr}(\rho \ln \rho )$, givesFor nonzero $t$, $S\to 0$ as $T\to 0$, while $S\to \ln 2$ as $T\to \infty$. More precisely, the high-temperature expansion is $S=\ln 2-x^2/2+O(x^4)$, and the low-temperature form is $S\sim (1+2|x|)e^{-2|x|}$. The thermodynamic entropy includes an overall factor $k_B$.