A Statistical View of Entropy — Question 10

Question 10

Resetting biased memory has a probability-dependent thermodynamic bound

A memory bit is initially 0 with probability .900 and 1 with probability .100. It is reset to 0 while coupled to a 300 K bath. Use $S_{\mathrm{bit}}=-k_B\sum p\ln p$ and $k_B=1.381\times10^{-23}$ J/K. Assume energetically degenerate logical states, zero net internal-energy change, no initial correlations, and a controller returned to its initial state. Positive $Q_b$ is heat delivered to the bath.

1. Use the total entropy balance and energy conservation to find the lower bounds on bath heat and work input per reset.

2. Evaluate the bounds for $10^9$ independent bits, and compare the single-bit result with a fair bit. State the total entropy generation in the ideal reversible limit.

Question 10 - Solution

Resetting biased memory has a probability-dependent thermodynamic bound

(a) Method and calculation

$$h=-.9\ln.9-.1\ln.1\approx.325083,\quad\Delta S_{\mathrm{bit}}=-k_Bh.$$$$S_{\mathrm{gen}}=Q_b/T-k_Bh\ge0\Rightarrow Q_b\ge k_BTh,\quad W_{\mathrm{in}}=Q_b.$$

(b) Interpretation and further calculation

$$W_{\min}=Q_{b,\min}\approx1.34682\times10^{-21}\,\mathrm J\ \text{per bit}.$$$$W_{\min}(10^9\text{ bits})\approx1.34682\times10^{-12}\,\mathrm J,\quad W_{\min,\mathrm{fair}}=k_BT\ln2\approx2.87171\times10^{-21}\,\mathrm J.$$$$S_{\mathrm{gen,min}}=0.$$In the ideal reversible limit, the bath entropy increase exactly balances the memory entropy decrease. Positive heat rejection does not imply strictly positive total entropy generation.

Physical check

This bound concerns the stated reset operation and assumptions, not an energy cost for every computational step.