Past exam of the advanced quantum physics course of the University of Cambridge / 2022 / Section B / 6

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.

The orthogonal projection onto the full spin-$\eta$ subspace isHere the identity acts on all oscillator states. The finite-time scattering operator $S(t)$ in the question is the interaction picture time-evolution operator, so $|{\psi }_{\sigma }^I(t)⟩=S(t)|\sigma ,\{n\}⟩$. Projecting its two orthogonal spin components yieldsExpand the right-hand side in the orthonormal basis $|\eta ,\{n^{\prime }\}⟩$. Taking its squared norm, and using the normalization of $|{\varphi }_{\eta }⟩$, givesThus one sums probabilities over unobserved final oscillator states, not their probability amplitudes. Unitarity of $S(t)$ makes the sum over both $\eta$ equal to one.

The Pauli matrix ${\sigma }_x$ flips the spin with unit matrix element. A nonzero oscillator matrix element changes exactly one occupation number, through a creation operator or an annihilation operator:For a spin flip, the spin energy change divided by $\hslash$ is $-s_{\sigma }\Omega$. Thus creation and annihilation in mode $j$ have total frequency changes ${\omega }_j-s_{\sigma }\Omega$ and $-{\omega }_j-s_{\sigma }\Omega$, respectively. Distinct modes or different occupation changes lead to orthogonal final quantum states, so their probabilities add. Fermi's golden rule consequently givesThe Dirac delta function is even, accounting for the sign in the annihilation term. Since all ${\omega }_j,\Omega >0$, an initially up spin can only emit an oscillator quantum, while an initially down spin can only absorb one. Writing the oscillator density of states per unit angular frequency as $g(\omega )={\sum }_j\delta (\omega -{\omega }_j)$, the continuum limit givesThe $+1$ permits emission even from the oscillator vacuum; absorption requires a quantum already present. The notation $n_{\Omega }$ assumes a common or smoothly specified occupation at the resonant frequency. If several resonant modes have different occupations, it denotes their density of states weighted mean, so that $g(\Omega )n_{\Omega }={\sum }_j{n}_j\delta (\Omega -{\omega }_j)$ after continuum smoothing. This is the golden-rule relaxation of a spin coupled to an oscillator bath.

In first-order time-dependent perturbation theory, the resonant time factor before taking the Fermi's golden rule limit isIts central frequency width is of order $1/t$, and its total area is $2\pi t$. To replace it by $2\pi t\delta (\nu )$, the oscillator density of states and relevant occupations must be approximately constant over that width. Thus $t\Delta \omega \gg 1$. At the same time, the perturbative flip probability must remain much smaller than one.

Define ${\Gamma }_0=2\pi {\lambda }^{2}g(\Omega )/{\hslash }^2$, so the two flip rates are ${\Gamma }_0(n_{\Omega }+1)$ and ${\Gamma }_0{n}_{\Omega }$. A sufficient common time window for both initial spins isFor a specified initial spin, the upper condition is simply its own flip rate times $t\ll 1$. Such a window requires weak enough coupling. The continuum approximation also presumes times shorter than recurrences from a resolved finite oscillator spectrum, and long enough to resolve the positive-frequency resonance from its nonresonant counterpart.

The thermal oscillator occupation is the Bose-Einstein distribution. With the rate notation ${\Gamma }_{\eta \sigma }$ for a transition from $\sigma$ to $\eta$, the master equation for the up population isAt thermal equilibrium, assuming ${\Gamma }_0>0$, vanishing time derivative is detailed balance between the two spin states. HenceorTogether with normalization, this gives $p_{\uparrow }=1/(1+e^{\beta \hslash \Omega })$ and $p_{\downarrow }=e^{\beta \hslash \Omega }/(1+e^{\beta \hslash \Omega })$, the Boltzmann distribution for the energy eigenvalues $\pm \hslash \Omega /2$. At zero temperature, absorption vanishes and emission relaxes the spin into its down ground state.

The nonzero-rate condition is necessary: if $\lambda =0$ or $g(\Omega )=0$, both rates vanish, so every initial pair of populations is stationary within this approximation. In that case stationarity alone cannot imply a thermal distribution. The equilibrium conclusion therefore presumes coupling to resonant oscillator modes.