Equivalence between the second order steady state for the spin-Boson model and its quantum mean force Gibbs state

TL;DR

This paper demonstrates that for a spin-boson model, the second-order steady state matches the quantum mean force Gibbs state at any temperature, providing a precise analytical link between dynamics and thermodynamics in open quantum systems.

Contribution

It analytically proves the equivalence between the second-order steady state and the quantum mean force Gibbs state for the spin-boson model, extending understanding of open quantum system equilibrium.

Findings

Steady state deviations are calculated up to second order in coupling.

The steady state exactly matches the quantum mean force Gibbs state.

Results are applicable to a wide range of physical systems.

Abstract

When the coupling of a quantum system to its environment is non-negligible, its steady state is known to deviate from the textbook Gibbs state. The Bloch-Redfield quantum master equation, one of the most widely adopted equations to solve the open quantum dynamics, cannot predict all the deviations of the steady state of a quantum system from the Gibbs state. In this paper, for a generic spin-boson model, we use a higher-order quantum master equation (in system environment coupling strength) to analytically calculate all the deviations of the steady state of the quantum system up to second order in the coupling strength. We also show that this steady state is exactly identical to the corresponding generalized Gibbs state, the so-called quantum mean force Gibbs state, at arbitrary temperature. All these calculations are highly general, making them immediately applicable to a wide class of systems well modeled by the spin-Boson model, ranging from various condensed phase processes to quantum thermodynamics. As an example, we use our results to study the dynamics and the steady state of a double quantum dot system under physically relevant choices of parameters.