Stationary states of boundary driven quantum systems: some exact results

• URL: http://arxiv.org/abs/2408.06887v2
• Date: Sun, 22 Sep 2024 17:53:31 GMT
• Title: Stationary states of boundary driven quantum systems: some exact results
• Authors: Eric A. Carlen, David a. Huse, Joel L. Lebowitz,
• Abstract summary: We study finite-dimensional open quantum systems whose density matrix evolves via a Lindbladian, $dotrho=-i[H,rho]+mathcal Drho$. We show that any stationary density matrix $barrho$ on the full system which commutes with $H$ must be of the product form $barrho=hatrho_Aotimesrho_B$.
• Score: 0.40964539027092906
• License: http://arxiv.org/licenses/nonexclusive-distrib/1.0/
• Abstract: We study finite-dimensional open quantum systems whose density matrix evolves via a Lindbladian, $\dot{\rho}=-i[H,\rho]+{\mathcal D}\rho$. Here $H$ is the Hamiltonian of the isolated system and ${\mathcal D}$ is the dissipator. We consider the case where the system consists of two parts, the "boundary'' $A$ and the ``bulk'' $B$, and ${\mathcal D}$ acts only on $A$, so ${\mathcal D}={\mathcal D}_A\otimes{\mathcal I}_B$, where ${\mathcal D}_A$ acts only on part $A$, while ${\mathcal I}_B$ is the identity superoperator on part $B$. Let ${\mathcal D}_A$ be ergodic, so ${\mathcal D}_A\hat{\rho}_A=0$ only for one unique density matrix $\hat{\rho}_A$. We show that any stationary density matrix $\bar{\rho}$ on the full system which commutes with $H$ must be of the product form $\bar{\rho}=\hat{\rho}_A\otimes\rho_B$ for some $\rho_B$. This rules out finding any ${\mathcal D}_A$ that has the Gibbs measure $\rho_\beta\sim e^{-\beta H}$ as a stationary state with $\beta\neq 0$, unless there is no interaction between parts $A$ and $B$. We give criteria for the uniqueness of the stationary state $\bar{\rho}$ for systems with interactions between $A$ and $B$. Related results for non-ergodic cases are also discussed.

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