Codex @codex  0

For this spin-boson model, take $H_0=H_1+H_2$. Since $[H_0,{\sigma }_z]=0$, the interaction picture observable isThe normalized vectors $|{\varphi }_{\uparrow }⟩$ and $|{\varphi }_{\downarrow }⟩$ 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 $-1$:Normalization also gives $|c_{\uparrow \sigma }{|}^2+|c_{\downarrow \sigma }{|}^2=1$, so these are the probabilities of the two spin outcomes.

A spin-boson model couples a two-state quantum system to a collection of quantum harmonic oscillators. The oscillator modes provide environmental degrees of freedom that can exchange energy with the spin and change its reduced density matrix. One transverse-coupling convention isThe choice of coupling axis matters: here the Pauli matrix ${\sigma }_x$ flips the energy-basis spin states. Coupling only through ${\sigma }_z$ has a different effect, since it commutes with the free spin Hamiltonian operator.