#
A Cost/Speed/Reliability Tradeoff to Erasing^{ †}

^{†}

## Abstract

**:**

## 1. Introduction

## 2. The Erasing Problem

#### Kullback–Leibler Cost

## 3. Solution to the Erasing Problem

## 4. Interpreting the KL Cost

#### 4.1. Path Space Szilard–Landauer Correspondence

#### 4.2. Thermodynamic Interpretation

#### 4.2.1. Thermodynamics on a Two-State Markov Chain

- Consider again the two-state continuous-time Markov chain with passive dynamics given by transition rates ${k}_{01}$ and ${k}_{10}$.Let ${E}_{0}$ and ${E}_{1}$ denote the internal energy of states “0” and “1”, respectively. Then, the equilibrium distribution is given by ${\pi}_{0}\propto {e}^{-{E}_{0}/{k}_{B}T}$ and ${\pi}_{1}\propto {e}^{-{E}_{1}/{k}_{B}T}$. We also have ${k}_{01}{\pi}_{0}={k}_{10}{\pi}_{1}$ from detailed balance. Together this yields$$\begin{array}{c}\hfill {E}_{0}-{E}_{1}={k}_{B}Tlog\frac{{k}_{01}}{{k}_{10}}.\end{array}$$
- Now consider the same two-state system with a control applied to it by means of a field of potential $\varphi \left(t\right)=({\varphi}_{0}\left(t\right),{\varphi}_{1}\left(t\right))$ so that the potential energy in state i becomes ${E}_{i}+{\varphi}_{i}\left(t\right)$. The transition rates due to the control become ${u}_{01}\left(t\right)$ and ${u}_{10}\left(t\right)$. By a reasoning similar to how we derived Equation (9), we get$${E}_{0}+{\varphi}_{0}-{E}_{1}-{\varphi}_{1}={k}_{B}Tlog\frac{{u}_{01}}{{u}_{10}}.$$Combining with Equation (9), this yields$$\begin{array}{c}\hfill {\varphi}_{0}-{\varphi}_{1}={k}_{B}Tlog\frac{{u}_{01}{k}_{10}}{{u}_{10}{k}_{01}}.\end{array}$$
- Given a distribution $p=({p}_{0},{p}_{1})$ on the states, we can define the following thermodynamic quantities:
- Expected internal energy $E\left(p\right)={p}_{0}{E}_{0}+{p}_{1}{E}_{1}$.
- Entropy $S\left(p\right)=-{p}_{0}log{p}_{0}-{p}_{1}log{p}_{1}$.
- Nonequilibrium free energy $F\left(p\right)=E\left(p\right)-{k}_{B}TS\left(p\right)$.

- Given a transition from state i to state j in the presence of the control field, we can define the following thermodynamic quantities:
- Heat dissipated ${Q}_{ij}\left(t\right)={E}_{i}+{\varphi}_{i}\left(t\right)-{E}_{j}-{\varphi}_{j}\left(t\right)$.
- Entropy increase of the system ${S}_{ij}\left(t\right)=log\frac{{p}_{i}\left(t\right)}{{p}_{j}\left(t\right)}$.

- Suppose the system is described at time t by a distribution $p\left(t\right)=({p}_{0}\left(t\right),{p}_{1}\left(t\right))$. Define the Current ${J}_{ij}\left(t\right)={p}_{i}\left(t\right){u}_{ij}\left(t\right)-{p}_{j}\left(t\right){u}_{ji}\left(t\right)$ so that ${\dot{p}}_{0}\left(t\right)=-{J}_{01}\left(t\right)$.
- We can further compute$$\begin{array}{cc}\hfill \frac{dE}{dt}& ={J}_{01}\left(t\right)({E}_{1}\left(t\right)-{E}_{0}\left(t\right))={k}_{B}T{J}_{01}\left(t\right)log\frac{{k}_{10}}{{k}_{01}},\phantom{\rule{4pt}{0ex}}\hfill \\ \hfill \frac{dW}{dt}& ={J}_{01}\left(t\right){W}_{01}\left(t\right)={J}_{01}\left(t\right)({\varphi}_{0}\left(t\right)-{\varphi}_{1}\left(t\right))={k}_{B}T{J}_{01}\left(t\right)log\frac{{u}_{01}\left(t\right){k}_{10}}{{u}_{10}\left(t\right){k}_{01}},\phantom{\rule{4pt}{0ex}}\hfill \\ \hfill \frac{dQ}{dt}& ={J}_{01}\left(t\right){Q}_{01}\left(t\right)={J}_{01}\left(t\right)({E}_{0}+{\varphi}_{0}\left(t\right)-{E}_{1}-{\varphi}_{1}\left(t\right))={k}_{B}T{J}_{01}\left(t\right)log\frac{{u}_{01}\left(t\right)}{{u}_{10}\left(t\right)},\hfill \\ \hfill \frac{dS}{dt}& ={J}_{01}\left(t\right){S}_{01}\left(t\right)={J}_{01}\left(t\right)log\frac{{p}_{0}\left(t\right)}{{p}_{1}\left(t\right)},\phantom{\rule{4pt}{0ex}}\hfill \\ \hfill \frac{dF}{dt}& ={k}_{B}T{J}_{01}\left(t\right)log\frac{{k}_{10}{p}_{1}\left(t\right)}{{k}_{01}{p}_{0}\left(t\right)}.\hfill \end{array}$$
- Define Total Entropy Production ${S}_{\text{tot}}\left(t\right)$ to be the total entropy produced from time 0 to time t. In other words, ${S}_{\text{tot}}\left(0\right)=0$ and$$\frac{d{S}_{\text{tot}}\left(t\right)}{dt}=\frac{1}{{k}_{B}T}\frac{dQ}{dt}+\frac{dS}{dt}.$$After simplification,$$\begin{array}{c}\hfill \frac{d{S}_{\text{tot}}\left(t\right)}{dt}=({p}_{0}\left(t\right){u}_{01}\left(t\right)-{p}_{1}\left(t\right){u}_{10}\left(t\right))log\frac{{p}_{0}\left(t\right){u}_{01}\left(t\right)}{{p}_{1}\left(t\right){u}_{10}\left(t\right)}\ge 0,\end{array}$$
- The following identity is immediate$$\begin{array}{c}\hfill \frac{dW}{dt}=\frac{dF}{dt}+\frac{d{S}_{\text{tot}}}{dt}\end{array}$$

#### 4.2.2. Thermodynamic Cost for Rapid Erasing of a Reliable Bit

#### 4.2.3. Link between KL-Cost and Thermodynamic Work

**Definition 1**

**Theorem 1.**

**Proof.**

**Theorem 2.**

**Proof.**

#### 4.3. Large Deviations Interpretation

#### 4.4. Gibbs Measure

## 5. Conclusions

## Acknowledgments

## Conflicts of Interest

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Gopalkrishnan, M.
A Cost/Speed/Reliability Tradeoff to Erasing. *Entropy* **2016**, *18*, 165.
https://doi.org/10.3390/e18050165

**AMA Style**

Gopalkrishnan M.
A Cost/Speed/Reliability Tradeoff to Erasing. *Entropy*. 2016; 18(5):165.
https://doi.org/10.3390/e18050165

**Chicago/Turabian Style**

Gopalkrishnan, Manoj.
2016. "A Cost/Speed/Reliability Tradeoff to Erasing" *Entropy* 18, no. 5: 165.
https://doi.org/10.3390/e18050165