Figure 1

(Color online) Schematic diagram of Szilard’s heat engine. A chamber of volume $V$ contains a one-molecule gas, which can be found in either the right or the left part of the box. (a) Initially, the position of the molecule is unknown. (b) Maxwell’s demon inserts a partition at the center and observes the molecule to determine whether it is in the right- or the left-hand side of the partition. He records this information in his memory. (c) Depending on the outcome of the measurement (which is recorded in his memory), the demon connects a load to the partition. If the molecule is in the right part as shown, he connects the load to the right-hand side of the partition. (d) The isothermal expansion of the gas does work upon the load, whose amount is $kT\ln 2$ which we call $1\text{bit}$. Adapted from Fig. 4 in 66.Reuse & Permissions

Figure 2

Potential energy for a binary memory whose state is represented by the direction of the magnetization. When there is no external transverse magnetic field the memory is stable in one of the potential wells, which correspond to “0” and “1” of recorded information. The transverse field lowers the height of the barrier at the center. At a certain point, the profile of the potential becomes bathtublike in shape with a flat bottom, where the direction of the magnetization is sensitive to the longitudinal field component. Adapted from 7.Reuse & Permissions

Figure 3

(Color online) Thermodynamic process to erase information. Binary information is stored in a vessel as the position of the molecule, either $L$ or $R$. A common procedure for both initial states, i.e., removing the partition and halving the whole volume by an isothermal compression towards the standard state $L$, completes the erasure. Adapted from Fig. 3 in 66.Reuse & Permissions

Figure 4

(Color online) Erasure process for an unbalanced probability distribution. The only difference from the case of balanced distribution (Fig. 3) is the expansion from (a) to ($a^{\prime }$), which gives us $H\left(p\right)$ bits of work.Reuse & Permissions

Figure 5

(Color online) The one-molecule heat engine considered by Gabor to show that the light needs to behave like a wave. When the molecule comes into the illuminated region, a piston is automatically inserted and the “gas” expands isothermally to extract work from the heat bath. This process could in principle be repeated infinitely, converting an infinite amount of heat into mechanical work, if the light had only a particlelike nature. Adapted from Fig. 7 in 22.Reuse & Permissions

Figure 6

(Color online) Possible configuration to confirm the continuity of the extractable work. The left- (right-) hand side of a chamber is filled with an $A_1$ $\left(B_1\right)$ gas. Two membranes that distinguish $A_1$ and $A_2$ are used to extract work. The membrane on the left lets the $A_1$ gas pass through it freely, but reflects the $A_2$ gas completely. The other membrane works in the opposite manner. Since the $B_1$ gas changes its state into $A_1$ with probability $q$ when measured by the membrane, the right membrane does not reach the right end of the chamber by a (quasistatic) isothermal expansion.Reuse & Permissions

Figure 7

(Color online) The work-extracting process with semipermeable membranes. In the initial state (a), the vessel is divided into two parts by an impenetrable opaque partition. The left side of the vessel, whose volume is $p_{1}V$, is occupied by the ∣↑⟩ gas, and the right side is filled with ∣↓⟩ gas. By replacing the partition with two semipermeable membranes $M_{\uparrow }$ and $M_{\downarrow }$ we can extract $H\left(p_1\right)=-p_1\log p_1-p_2\log p_2$ bits of work by isothermal expansion. The membranes reach the end of the vessel in the final state (b).Reuse & Permissions

Figure 8

(Color online) Thermodynamic cycle given by Peres to show that distinguishing nonorthogonal quantum states leads to a violation of the second law. The arrows indicate the directions of spin in the Bloch sphere. The use of hypothetical semipermeable membranes, which distinguish nonorthogonal states ∣↑⟩ and ∣→⟩ perfectly, in the step from (b) to (c) is the key to violate the second law. Adapted from Fig. 9.2 in 60.Reuse & Permissions

Figure 9

(Color online) Bondi’s thought experiment for a perpetuum mobile. Excited atoms (on the right side and at the bottom of the belt) emit a photon at the bottom of the belt to lower its energy level. The emitted photon is reflected by the curved mirrors placed so that it will be absorbed by an atom in the lower energy level at the top of the belt.Reuse & Permissions

Figure 10

(Color online) The erasure of classical information carried by quantum states. Each message state interacts with a heat bath at temperature $T$ and reaches thermal equilibrium. The information originally encoded in a state is lost and all states end up in $\omega$, which is the thermal state at the temperature $T$.Reuse & Permissions

Figure 11

(Color online) The thermodynamic cycle, which we use to discuss the second law. The cycle proceeds from the initial state to the final state $\rho$ via the post-work-extraction state $\sigma$, and returns to the initial state with a reversible process.Reuse & Permissions

Figure 12

(Color online) The thermodynamic process to transform the intermediate state $\sigma$ into the final state $\rho$. First, after attaching an empty vessel of the same volume to that containing the gas $\sigma$, the membranes $M_j$ are used to separate two orthogonal states ${\vec{e}}_1$ and ${\vec{e}}_2$ [(a) to (c)]. As the distance between the movable opaque wall and the membrane $M_2$ is kept constant, this process entails no work consumption or extraction. As $\sigma =\sum c_j{\vec{e}}_j{\vec{e}}_j^{†}$, compressing each ${\vec{e}}_j$ gas into the volume of $c_{j}V$ as in (d) makes the pressures of gases equal and this compression requires $S\left(\sigma \right)=-\sum c_j{\log }_2{c}_j$ bits of work. Second, quantum states of gases are isentropically transformed, thus without consuming work, so that the resulting state (e) will have ${\lambda }_{j}N$ molecules in ${\vec{\varphi }}_j$, where $\rho =\sum {\lambda }_j{\vec{\varphi }}_j{\vec{\varphi }}_j^{†}$ is the eigendecomposition of $\rho$. To reach (f), $S\left(\rho \right)$ bits of work can be extracted by using membranes that distinguish ${\vec{\varphi }}_j$. As a result, the work needed for the transformation $\sigma \to \rho$ is $S\left(\sigma \right)-S\left(\rho \right)$ bits.Reuse & Permissions

Figure 13

(Color online) Schematic of the protocol to extract work from correlated pairs. Two pairs represent an ensemble for which Alice and Bob use $A_{\theta }$ and $B_{{\theta }^{\prime }}$ for their measurement and work extraction. For a half of this ensemble, Alice measures her state with $A_{\theta }$ and Bob extracts work from his side along the direction of ${\theta }^{\prime }$, according to Alice’s measurement results. For the other half, they exchange their roles.Reuse & Permissions