Geometric fluctuation theorem for a spin-boson system

We derive an extended fluctuation theorem for geometric pumping of a spin-boson system under periodic control of environmental temperatures by using a Markovian quantum master equation. We obtain the current distribution, the average current, and the fluctuation in terms of the Monte Carlo simulation. To explain the results of our simulation we derive an extended fluctuation theorem. This fluctuation theorem leads to the fluctuation dissipation relations but the absence of the conventional reciprocal relation.

Figure 1

The plots of (a) the average $\langle J\rangle$ and (b) the fluctuation ${\langle J^2\rangle }_c/\left(2\pi \right|\varepsilon \left|\right)$ obtained from the Monte Carlo simulation of Eqs. (33) and (34), where the number of samples is 20 000 000 for each $\varepsilon$. The parameters are modulated by ${\left({\beta }_{\text{L}}\left(t\right)\hslash {\omega }_0\right)}^{-1}=0.7+0.35cos(\mathrm{\Omega }t+\pi /4), {\left({\beta }_{\text{R}}\left(t\right)\hslash {\omega }_0\right)}^{-1}=0.7+0.35sin(\mathrm{\Omega }t+\pi /4)$, and $\mathrm{\Gamma }=0.01$. We have eliminated the effects of initial conditions by omitting the data in the transient processes to reach the steady states. The best fitting of the behavior for the fluctuation is ${\langle J^2\rangle }_c\propto {\left|\varepsilon \right|}^{0.85}$.

Figure 2

The plots of the current distributions for (a) $\varepsilon =0.001$, (b) $\varepsilon =0.01$, and (c) $\varepsilon =0.1$, where the set of the other parameters is equivalent to that used in Fig. 1. The average $\langle J\rangle$ is small at $O\left(\right|\varepsilon \left|\right)$ but it is the finite value from Fig. 1 and Eq. (36).

Figure 3

The plot of the deviation from the Gaussian $P_G{\left(J\right)\sim exp[-a\left(0\right)(J-\langle J\rangle )}^2/\left(2\right|\varepsilon \left|\right)]$ for $\mathrm{\Omega }/\mathrm{\Gamma }=0.1$. The line represents the fourth-order fit curve. From this figure we can estimate the coefficients of $J^4$ in Eq. (35) as $b\left(0\right)\simeq 5.4\times {10}^6$.

Figure 4

(a) The plot of Eq. (37) versus $J$ in $\varepsilon =0.1$ corresponding to Fig. 2. The set of the other parameters is equivalent to that used in Fig. 1. Dots in (b) are calculated by the arithmetic average of $\left|\varepsilon \right|ln[P_{\varepsilon }\left(J\right)/P_{\varepsilon }(-J)]/J$ defined as ${\sum }_{\left|J\right|\le J_{max},J\ne 0}\left\{\left|\varepsilon \right|ln[P_{\varepsilon }\left(J\right)/P_{\varepsilon }(-J)]/J\right\}/N$, where $N$ is the number of $J\in [-J_{max},J_{max}]$ and $J_{max}$ is the threshold excluding large deviations of $P_{\varepsilon }(\pm J)$, which are $J_{max}=3.9\times {10}^{-4}, N=106$ for $\varepsilon =0.005, J_{max}=1.3\times {10}^{-3}, N=82$ for $\varepsilon =0.02, J_{max}=2.1\times {10}^{-3}, N=52$ for $\varepsilon =0.05, J_{max}=2.3\times {10}^{-3}, N=40$ for $\varepsilon =0.07, J_{max}=2.9\times {10}^{-3}, N=36$ for $\varepsilon =0.1, J_{max}=3.7\times {10}^{-3}, N=32$ for $\varepsilon =0.14$, and $J_{max}=3.9\times {10}^{-3}, N=24$ for $\varepsilon =0.2$. From this plot, we obtain $A\simeq 4\times {10}^{-2}$.