Abstract

The Lindblad form guarantees complete positivity of a Markovian quantum master equation(QME). However, its microscopic derivation for a quantum system weakly interacting with a thermal bath requires several approximations, which may result in inaccuracies in the QME. Recently, various Lindbladian QMEs were derived without resorting to the secular approximation from the Redfield equation which does not guarantee the complete positivity. Here we explicitly calculate, in a perturbative manner, the equilibrium steady states of these Lindbladian QMEs. We compare the results with the steady state of the Redfield equation obtained from an analytic continuation method, which coincides with the so-called mean force Gibbs (MFG) state. The MFG state is obtained by integrating out the bath degrees of freedom for the Gibbs state of the total Hamiltonian. We explicitly show that the steady states of the Lindbladian QMEs are different from the MFG state. Our results indicate that manipulations of the Redfield equationneeded to enforce complete positivity of a QME drives its steady state away from the MFG state. We also find that, in the high-temperature regime, both the steady states of the Lindbladian QMEs and MFG state reduce to the same Gibbs state of a system Hamiltonian under certain conditions.

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