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:

$$S(\rho \| \sigma) \equiv \operatorname{tr}\{\rho \ln \rho\}-\operatorname{tr}\{\rho \ln \sigma\}$$

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

$$S(\rho \| \rho^{(1)} \otimes \rho^{(2)}) = S(\rho^{(1)}) +S(\rho^{(2)}) - S(\rho)$$

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 and and , there is .

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

$$S\left(\sum_{i} \lambda_{i} \rho_{i}\right) \geq \sum_{i} \lambda_{i} S\left(\rho_{i}\right)$$

1.2 Dynamical semigroup

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

$$V(t) \rho_{S}=\sum_{\alpha, \beta} W_{\alpha \beta}(t) \rho_{S} W_{\alpha \beta}^{\dagger}(t)$$ $$\sum_{\alpha , \beta} W_{\alpha \beta}^\dagger(t) W_{\alpha \beta}(t) = I_S$$ $$Tr_S\{V(t)\rho_S\} = Tr_S\rho_S = 1$$

Semigroup property:

1.3 Irreversibility and entropy production rate (EPR)

For a dynamical map , we have

$$S(V(t) \rho \| V(t) \rho_0) = S(\rho \| \rho_0).$$

And dynamical map do not increase the relative entropy,

$$S(V(t) \rho \| \rho_0) \leq S(\rho \| \rho_0).$$

The equality holds for stationary state .

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

$$\sigma(\rho) \equiv -k_{\mathrm{B}}\frac{d}{dt}S(\rho(t) \| \rho_0 ) =-k_{\mathrm{B}} \operatorname{tr}\{\mathcal{L}(\rho) \ln \rho\}+k_{\mathrm{B}} \operatorname{tr}\left\{\mathcal{L}(\rho) \ln \rho_{0}\right\} \geq 0$$

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

$$\sigma = \frac{dS}{dt} +J$$

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 and .

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 . Transition rate from i to j is , and . So the probability distribution is: .

Suppose the system has an unique stationary state, . It satisfies . The EPR at is:

$$\sigma = k_B \sum_{i<j} (\pi_i W_{ij} - \pi_j W_{ji}) \log (\frac{\pi_i W_{ij}}{\pi_j W_{ji}})$$

• For isothermal system with a bath, the free energy dissipation rate required is .
• For euqilibrium system when the detailed balance is satisfied, .

2.2 The waiting time distribution

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