= Solution
For this <spin-boson model>, take $H_0=H_1+H_2$. Since $[H_0,\sigma_z]=0$, the <interaction picture> observable is
$$
\sigma_z^I(t)=e^{iH_0t/\hbar}\sigma_ze^{-iH_0t/\hbar}=\sigma_z.
$$
The normalized vectors $|\phi_\uparrow\rangle$ and $|\phi_\downarrow\rangle$ belong to different <eigenspaces> of the <Hermitian operator> $\sigma_z$, and are therefore orthogonal. This remains true even though each vector includes the oscillator degrees of freedom. Hence all cross terms in the <expectation value> vanish, while the diagonal terms have <eigenvalues> $+1$ and $-1$:
$$
\boxed{\langle\psi_\sigma^I(t)|\sigma_z^I(t)|\psi_\sigma^I(t)\rangle
=|c_{\uparrow\sigma}(t)|^2-|c_{\downarrow\sigma}(t)|^2.}
$$
Normalization also gives $|c_{\uparrow\sigma}|^2+|c_{\downarrow\sigma}|^2=1$, so these are the <probabilities> of the two <spin> outcomes.
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