Knowlpedia

Quantum Gibbs state

For a finite quantum system, the Gibbs state is exp(-beta H) normalized by its trace.

Let HH be a on a finite-dimensional Hilbert space and let β>0\beta>0. The quantum Gibbs state at inverse temperature β\beta is the

ρβ=e−βHZ(β),Z(β)=Tr⁡(e−βH).\rho_\beta=\frac{e^{-\beta H}}{Z(\beta)}, \qquad Z(\beta)=\operatorname{Tr}(e^{-\beta H}).

Its expectation functional is ωβ(A)=Tr⁡(ρβA)\omega_\beta(A)=\operatorname{Tr}(\rho_\beta A).

Spectral form

If H=∑nEnPnH=\sum_nE_nP_n, then

ρβ=∑ne−βEnPn∑ne−βEnTr⁡(Pn).\rho_\beta=\frac{\sum_ne^{-\beta E_n}P_n}{\sum_ne^{-\beta E_n}\operatorname{Tr}(P_n)}.

Thus the Gibbs weight of an energy eigenspace is proportional to e−βEne^{-\beta E_n} times its degeneracy.

Variational characterization

For density operators ρ\rho, define

Fβ(ρ)=Tr⁡(ρH)−β−1S(ρ),S(ρ)=−Tr⁡(ρlog⁡ρ).\mathcal F_\beta(\rho)=\operatorname{Tr}(\rho H)-\beta^{-1}S(\rho), \qquad S(\rho)=-\operatorname{Tr}(\rho\log\rho).

Then ρβ\rho_\beta uniquely minimizes Fβ\mathcal F_\beta, because

Fβ(ρ)−Fβ(ρβ)=β−1D(ρ∥ρβ)≥0,\mathcal F_\beta(\rho)-\mathcal F_\beta(\rho_\beta) =\beta^{-1}D(\rho\|\rho_\beta)\ge0,

where DD is .

Remarks

The state ρβ\rho_\beta commutes with HH, hence is stationary. On the full matrix algebra it is also the unique state satisfying the for the dynamics generated by HH.