= Solution
The fixed-particle-number sector is spanned by $|1\rangle=c_1^\dagger|0\rangle$ and $|2\rangle=c_2^\dagger|0\rangle$. In this basis,
$$
H=-t\begin{pmatrix}0&1\\1&0\end{pmatrix}=-t\sigma_x.
$$
The relevant <quantum Gibbs state> is normalized within this two-dimensional sector, not over all particle numbers. With $\beta=(k_BT)^{-1}$ and $x=\beta t$, the <exponential of a Pauli vector> gives
$$
e^{-\beta H}=e^{x\sigma_x}=I\cosh x+\sigma_x\sinh x,\qquad
Z=2\cosh x.
$$
Therefore
$$
\boxed{\rho_\beta=\frac12(I+\tanh x\,\sigma_x)
=\frac12\begin{pmatrix}1&\tanh x\\\tanh x&1\end{pmatrix}}.
$$
Its <eigenstates> are $|\pm\rangle=(|1\rangle\pm|2\rangle)/\sqrt2$, 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> is
$$
\boxed{\operatorname{Tr}\rho_\beta^2=\frac12(1+\tanh^2x)}.
$$
For $t\ne0$, the zero-temperature limit is the pure ground-state projector,
$$
\rho_{T\to0}=\frac12[I+\operatorname{sgn}(t)\sigma_x],
$$
with purity one. As $T\to\infty$, $x\to0$ 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>, $S=-\operatorname{Tr}(\rho\ln\rho)$, gives
$$
\boxed{S(T)=-p_+\ln p_+-p_-\ln p_-
=\ln(2\cosh x)-x\tanh x}.
$$
For nonzero $t$, $S\to0$ as $T\to0$, while $S\to\ln2$ as $T\to\infty$. More precisely, the high-temperature expansion is $S=\ln2-x^2/2+O(x^4)$, and the low-temperature form is $S\sim(1+2|x|)e^{-2|x|}$. The thermodynamic entropy includes an overall factor $k_B$.
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