Stochastic Protocols — Examples

The partition sums are $Z_0=2$ and $Z_1=1+e^{-\beta ϵ}$, so $\mathrm{\Delta }F=-{\beta }^{-1}\log [(1+e^{-\beta ϵ})\mathrm{/}2]$. Since $e^{-\beta ϵ}<1$, the ratio lies strictly between $e^{-\beta ϵ}$ and $1$, hence $0<\mathrm{\Delta }F<ϵ$.

The identity kernel leaves the state unchanged, so the two work values are $0$ and $ϵ$, each with probability $1\mathrm{/}2$. Therefore ${\mathbb{P}}_F(W<\mathrm{\Delta }F)=1\mathrm{/}2$.

Directly, ${\mathbb{E}}_F{e}^{-\beta W}=(1+e^{-\beta ϵ})\mathrm{/}2=e^{-\beta \mathrm{\Delta }F}$, while the event in step 2.1 has positive probability. Thus the equality is an ensemble statement and does not impose $W\ge \mathrm{\Delta }F$ on every path.

The third probability is constant and $p_1+p_2=1\mathrm{/}2$. The difference $d=p_1-p_2$ satisfies $\dot{d}=-2d$ with $d(0)=1\mathrm{/}4$, so $d(t)=e^{-2t}\mathrm{/}4$, $p_1(t)=1\mathrm{/}4+e^{-2t}\mathrm{/}8$, and $p_2(t)=1\mathrm{/}4-e^{-2t}\mathrm{/}8$.

Hence $p(t)\to (1\mathrm{/}4,1\mathrm{/}4,1\mathrm{/}2)\mathrm{\ne }\pi$. The relative entropy tends to $\frac{1}{2}\log (9\mathrm{/}8)>0$, confirming that detailed balance without connectedness does not force convergence to the full reference law.

Solving ${\dot{p}}_1=1-10p_1$ gives $p_1(t)=0.1+0.4e^{-10t}$, which lies in $(0.1,0.5)$ and strictly decreases for every $t>0$.

For $0<p<1\mathrm{/}2$, $H^{\mathrm{\prime }}(p)=\log ((1-p)\mathrm{/}p)>0$ by [F3]. Therefore $dH(p_1(t))\mathrm{/}dt=H^{\mathrm{\prime }}(p_1(t))p_1^{\mathrm{\prime }}(t)<0$ for every $t>0$. The Shannon entropy decreases while detailed balance holds.

The two-state rate graph is connected, both components of $p(0)$ and $\pi$ are positive, and detailed balance holds. Thus [F2] applies to the relative entropy $D(p(t)\mathrm{\parallel }\pi )$ at the same time that step 2.1 shows the Shannon entropy decreases.

Substituting $V=k_{\mathrm{B}}{T}_b\mathrm{/}k$ and $\gamma =\mu k$ into the exact spectrum supplied by [F2] gives $S(\omega )=\mu k_{\mathrm{B}}{T}_b\mathrm{/}(\pi [(\mu k{)}^2+{\omega }^2])$.

The same supplier bound becomes $\mathrm{Var}(T^{-1}{\int }_0^T{X}_{t}dt)\le 2k_{\mathrm{B}}{T}_b\mathrm{/}(\mu k^{2}T)$, which tends to zero as $T\to \mathrm{\infty }$. Both formulas are conditional model predictions, not measurements.

Since $R=\{r_0\}$, its marginal is $a_{r_0}=1$, and the conditional law is $q(1\mathrm{\mid }{r}_0)=p$, $q(0\mathrm{\mid }{r}_0)=1-p$. The entropy definition therefore gives $H(M\mathrm{\mid }R)=H(M)=-p\log p-(1-p)\log (1-p)$.

Therefore the theorem gives $W\ge k_{\mathrm{B}}{T}_b[-p\log p-(1-p)\log (1-p)]$ for this preparation. It reports no work measurement and makes no claim about an apparatus.

The memory marginal is uniform, so $H(M)=\log 2$. Given either value of $R$, the conditional law of $M$ is a point mass; hence each conditional entropy is zero and $H(M\mathrm{\mid }R)=0$.

Therefore the theorem yields only $W\ge 0$ for this reset under its stated thermodynamic assumptions, while the unconditioned memory entropy is $\log 2$. It does not imply a $k_{\mathrm{B}}{T}_b\log 2$ lower bound for this correlated preparation.