 Solution

Discarding the photon-number result gives a classical mixture of the two outcome branches. The atomic state, in the basis $(|e⟩,|g⟩)$, isIt is also the reduced state before measuring the field; a nonselective operation on the field does not change the atomic reduced density matrix. The atom is not in a coherent superposition with these populations.

For an atomic energy gap $\hslash \omega$, the thermal population ratio is $p_e/p_g=e^{-\hslash \omega /(k_{B}T)}$. Matching it givesA finite positive temperature exists when $0<p_e<p_g$, with $T\to 0^{+}$ at $p_e=0$ and $T\to +\infty$ as $p_e\to 1/2$ from below. Equal populations correspond to zero inverse temperature. If $p_e>1/2$, the expression is a negative effective temperature for a bounded two-level system, not an equilibrium state at any ordinary positive temperature. For a concrete counterexample to an unrestricted positive-temperature interpretation, at $t=0$ the atom is certainly excited, whereas every positive-temperature Gibbs distribution has $p_e\le 1/2$. The fully excited endpoint is the formal limit $T\to 0^{-}$. Matching one density matrix does not establish genuine thermalization with a bath.