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论文:Estimating entropy production from waiting time distributions

Skinner, D. J., & Dunkel, J. (2021). Estimating entropy production from waiting time distributions. Physical Review Letters, 127(19), 198101. https://doi.org/10.1103/PhysRevLett.127.198101
虽然介绍的都是经典模型,但是对于量子介观输运有一定借鉴意义。

首先是对于一些基本概念的review,参考H. P. Breuer的开放系统理论那本书3.2章节。

1. Entropy Production Rate in Open Systems

1.1 Relative Entropy

For a given pair of density matrices and the relative entropy is defined by:

The physical meaning of the relative entropy is, for a composite system,

Where is the von Neumann entropy of . Recall that for von Neumann entropy, there’s a subaddition condition: . The equality holds iff .

Important properties of relative entropy:

  • . The equality holds iff .
  • Invariant under unitary: .
  • Jointly convex: for $\rho = \lambda \rho1 +(1-\lambda)\rho_2\sigma = \lambda \sigma_1 +(1-\lambda) \sigma_20\leq \lambda \leq 1S(\rho | \sigma) \leq \lambda S\left(\rho{1} | \sigma{1}\right)+(1-\lambda) S\left(\rho{2} | \sigma_{2}\right)$.

Note: Recall that the von Neumann entropy is a concave functional for : for and ,

1.2 Dynamical semigroup

Dynamical map , is a trace-preserving map of density matrix.

Semigroup property:

1.3 Irreversibility and entropy production rate (EPR)

For a dynamical map , we have

And dynamical map do not increase the relative entropy,

The equality holds for stationary state .

The entropy production rate (EPR) can be defined as the non-negative time derivative of the relative entropy:

The EPR is non-negative and vanishes at stationary state. In non-equilibrium thermodynamics, the EPR satisfies:

Here is the von Neumann entropy and is the entropy flux, with denotes that entropy flows from the system to the environment. We can see that $\frac{dS}{dt} = -k{\mathrm{B}} \operatorname{tr}{\mathcal{L}(\rho) \ln \rho}J = k{\mathrm{B}} \operatorname{tr}\left{\mathcal{L}(\rho) \ln \rho_{0}\right}$.

2. Estimating EPR from waiting time distributions

2.1 The Model

Assume the system can be described by a Markovian stochastic dynamics on a finite set of discrete states ${1,\dots,NT}W{ij}W{ii} = -\sum{j \neq i} W{ij}\frac{d}{dt} p_i = \sum_i p_j W{ji}$.

Suppose the system has an unique stationary state, $\pi = (\pii)\frac{d}{dt} \pi_i = \sum_j \pi_j W{ji} = 0\pi$ is:

  • For isothermal system with a bath, the free energy dissipation rate required is .
  • For euqilibrium system when the detailed balance $\pii W{ij} = \pij W{ji}\sigma = 0$.

2.2 The waiting time distribution

Let be the distribution of time spent in A and B, with some set of states.