= Solution
Discarding the <photon number>[photon-number] result gives a classical mixture of the two outcome branches. The atomic state, in the basis $(|e\rangle,|g\rangle)$, is
$$
\boxed{\rho_A(t)=\cos^2(gt)|e\rangle\langle e|
+\sin^2(gt)|g\rangle\langle g|
=\begin{pmatrix}\cos^2(gt)&0\\0&\sin^2(gt)\end{pmatrix}.}
$$
It 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 <quantum superposition>[coherent superposition] with these populations.
For an atomic <energy> gap $\hbar\omega$, the thermal population ratio is $p_e/p_g=e^{-\hbar\omega/(k_BT)}$. Matching it gives
$$
\boxed{T_{\rm eff}(t)=\frac{\hbar\omega}{k_B\ln[\tan^2(gt)]}.}
$$
A finite positive <temperature> exists when $0<p_e<p_g$, with $T\to0^+$ at $p_e=0$ and $T\to+\infty$ as $p_e\to1/2$ 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 <quantum Gibbs state>[Gibbs distribution] has $p_e\leq1/2$. The fully excited endpoint is the formal limit $T\to0^-$. Matching one <density matrix> does not establish genuine <thermalization> with a bath.
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