= Solution
Let $B=\sum_k g_kZ_k$. In the <Zurek spin-bath model>, $H=-Z_D\otimes B$, so <unitary time evolution> with $\hbar=1$ is $U(t)=e^{itZ_D\otimes B}$. The bath Hamiltonian terms commute, and device states $|0\rangle,|1\rangle$ have $Z_D$ eigenvalues $+1,-1$. Hence
$$
|\Psi(t)\rangle=a|0\rangle|E_0(t)\rangle+b|1\rangle|E_1(t)\rangle,
$$
where the two normalized conditional bath states are
$$
|E_0(t)\rangle=\bigotimes_k\left(\alpha_k e^{ig_kt}|\uparrow_k\rangle+\beta_k e^{-ig_kt}|\downarrow_k\rangle\right),\qquad
|E_1(t)\rangle=\bigotimes_k\left(\alpha_k e^{-ig_kt}|\uparrow_k\rangle+\beta_k e^{ig_kt}|\downarrow_k\rangle\right).
$$
Take each bath factor normalized, $|\alpha_k|^2+|\beta_k|^2=1$, and $|a|^2+|b|^2=1$. This entails no restriction: if only the product is initially normalized, divide each nonzero factor by its norm; the product of these norms is one.
Taking the <partial trace> over the bath gives the <reduced density matrix>
$$
\boxed{\rho_D(t)=\begin{pmatrix}|a|^2&ab^*z(t)\\a^*b z(t)^*&|b|^2\end{pmatrix}},\qquad
z(t)=\langle E_1(t)|E_0(t)\rangle.
$$
The orientation of this <conditional environment overlap> fixes the sign of the phase in the upper-right entry. Factorizing the overlap yields the <decoherence factor>
$$
\boxed{z(t)=\prod_{k=1}^N\left(|\alpha_k|^2e^{2ig_kt}+|\beta_k|^2e^{-2ig_kt}\right)
=\prod_{k=1}^N\left[\cos(2g_kt)+i\left(|\alpha_k|^2-|\beta_k|^2\right)\sin(2g_kt)\right]}.
$$
The populations are conserved because $[H,Z_D]=0$. Only phase coherence can be reduced. In particular,
$$
|z(t)|^2=\prod_k\left[1-4|\alpha_k|^2|\beta_k|^2\sin^2(2g_kt)\right]\leq1.
$$
<Quantum decoherence> here results from distinguishable conditional bath states, although the complete system remains in a <pure state> under <unitary time evolution>.
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