Thermodynamic and logical reversibilities revisited

We consider a memory that has several logical states, in which information is stored. A computational process is performed by changing an input logical state into an output one with a certain algorithm. Let $\mathcal M$ and $\mathcal M'$ be the sets of input and output logical states, respectively. We assume that these are finite sets. For example, if the input consists of n bits, we set $\mathcal M = \{ 0,1 \}^n$. We note that the logical states do not have one-to-one correspondence to the physical phase space, as discussed in section 4 in detail.

We next formulate (deterministic) computational processes. In this paper, a computational process $\hat C$ is defined as a map

$$\begin{equation} \hat C: \mathcal M \to \mathcal M'. \end{equation} \tag{ 13 }$$

We also call $\hat C$ as a gate. We note that $\hat C$ is not necessarily a surjection.

In order to discuss the logical reversibility and its relationship to thermodynamics, we do not need the detailed characterization of computable functions in terms of computability theory [81]; the following argument is applicable to any map $\hat C : \mathcal M \to \mathcal M'$.

We now define the logical reversibility [8]–[11]:

A computational process $\hat C$ is logically reversible if and only if it is an injection. In other words, $\hat C$ is logically reversible if and only if, for any output logical state, there is a unique input logical state. Otherwise, $\hat C$ is logically irreversible.

We note that, if $\mathcal M' = \mathcal M$, a computational process is reversible if and only if it is a bijection (i.e., an injection and a surjection). In general, a computational process is reversible if and only if we can precisely estimate the input state from the output state. In fact, if a computational process is reversible and $\hat C$ is an injection, we can define the reversed computational process

$$\begin{equation} \hat C^{-1}: \mathcal M' \to \mathcal M, \end{equation} \tag{ 14 }$$

where the domain of $\hat C^{-1}$ is given by the image of $\hat C$ denoted as $\hat C (\mathcal M) \subset \mathcal M'$.

We next discuss several simple examples of computation. We first consider the case that both the input and output are one bit so that $\mathcal M = \mathcal M' = \{ 0,1 \}$. A simple example of reversible gate is NOT, which is defined as

$$\begin{equation}0 \mapsto 1, \quad\qquad 1 \mapsto 0. \end{equation} \tag{ 15 }$$

This is clearly a bijection, and the reversal of NOT is also NOT. A simple example of irreversible gate is the information erasure that is referred to as ERASE:

$$\begin{equation}0 \mapsto 0, \quad\qquad 1 \mapsto 0. \end{equation} \tag{ 16 }$$

This is not a bijection, as the logical state is always 0 after the computation; we cannot estimate the input state from the output state.

We next consider the case that both the input and output are two bits so that $\mathcal M = \mathcal M' = \{ 0,1 \}^2 = \{ 00, 01, 10, 11 \}$. A simple example of reversible gate is CNOT (controlled-NOT), which is defined as

$$\begin{equation}00 \mapsto 00, \quad\qquad 01 \mapsto 01, \quad\qquad 10 \mapsto 11, \quad\qquad 11 \mapsto 10, \label {CNOT} \end{equation} \tag{ 17 }$$

where the first (left) bit is the control bit and the second (right) bit is the target bit. If the control gate is 1, CNOT behaves as NOT on the target bit. Otherwise, CNOT is just identity. CNOT is a bijection and its reversal is also CNOT.

CNOT can be used for a measurement (i.e., the copy of information); if the input of the target bit is 0, the input of the control bit is copied to the output of the target bit by CNOT:

$$\begin{equation}00 \mapsto 00, \quad\qquad 10 \mapsto 11. \label {CNOT_measurement} \end{equation} \tag{ 18 }$$

CNOT can also be used for feedback control, where the control bit is the feedback controller and the target bit is to be controlled. In the case of feedback control, we exchange the roles of the bits from the case of measurement (18); we regard the first (left) bit as the target bit and the second (right) bit as the control bit for feedback control. Before the feedback, the input states of the two bits are assumed to be the same, which is realized after the measurement. The target bit is then flipped if the control bit is 1:

$$\begin{equation}00 \mapsto 00, \quad\qquad 11 \mapsto 01, \label {CNOT_feedback} \end{equation} \tag{ 19 }$$

where the output of the target bit is 0 irrespective of its input.

We next consider the case that the input is two bit and the output is one bit, where $\mathcal M = \{ 0,1 \}^2$ and $\mathcal M' = \{ 0, 1 \}$. In this case, any computational process is irreversible. In fact, the numbers of the elements in $\mathcal M$ and $\mathcal M'$ are 4 and 2, respectively, and therefore any map from $\mathcal M$ to $\mathcal M'$ cannot be an injection. An example of such an irreversible gate is XOR, which is defined as

$$\begin{equation}00 \mapsto 0, \quad\qquad 01 \mapsto 1, \quad\qquad 10 \mapsto 1, \quad\qquad 11 \mapsto 0. \end{equation} \tag{ 20 }$$

We next show that any irreversible computation can be extended to a reversible computation, which has been discussed by Bennett in detail [9]. In fact, we can show the following theorem:

For any $\hat C : \mathcal M \to \mathcal M'$, there exist a finite set $\mathcal M''$ and a map $\hat C_{\rm ex}: \mathcal M \to \mathcal M' \times \mathcal M''$, such that $\hat C_{\rm ex}$ is logically reversible and the restriction of $\hat C_{\rm ex}$ on $\mathcal M \to \mathcal M'$ is equivalent to $\hat C$.

Here, $\mathcal M''$ can be regarded as an ancilla or an environment. We call $\hat C_{\rm ex}$ the reversible extension of $\hat C$. We note that the reversible extension is not unique.

Before the proof of the theorem, we illustrate the reversible extension of ERASE. Let $\mathcal M'' = \{ 0,1 \}$. We then define $\hat C_{\rm ex} : \{ 0,1 \} \to \{ 0,1 \}^2$ by

$$\begin{equation}0 \mapsto 00, \quad\qquad 1 \mapsto 01, \label {ERASE_ex} \end{equation} \tag{ 21 }$$

where the first (left) bit of the output is in $\mathcal M$. $\hat C_{\rm ex}$ is clearly an injection and therefore reversible. Moreover, its restriction on $\mathcal M \to \mathcal M'$ is equivalent to ERASE. Intuitively, this extension describes that the erased information can be kept in ancilla $\mathcal M''$ (i.e., can remain in the environment).

We now show a simple constructive proof of the theorem: we set $\mathcal M'' = \mathcal M$ and define $\hat C_{\rm ex}$ as

$$\begin{equation} \hat C_{\rm ex} (m) = (\hat C(m), m ) \in \mathcal M' \times \mathcal M'', \label {reversible_ex} \end{equation} \tag{ 22 }$$

which satisfies the condition of the theorem.

The extension of ERASE (21) is a special case of extension (22). We note that extension (22) may be redundant in general. For example, in the case of XOR, the extension (22) is given by

$$\begin{equation}00 \mapsto 000, \quad\qquad 01 \mapsto 101, \quad\qquad 10 \mapsto 110, \quad\qquad 11 \mapsto 011, \end{equation} \tag{ 23 }$$

where $\mathcal M'' = \{ 0, 1 \}^2$. However, there is a simpler reversible extension of XOR:

$$\begin{equation}00 \mapsto 00, \quad\qquad 01 \mapsto 10, \quad\qquad 10 \mapsto 11, \quad\qquad 11 \mapsto 01, \end{equation} \tag{ 24 }$$

where $\mathcal M'' = \{ 0 , 1 \}$.

We next consider the concept of logical entropy, which is defined by the Shannon entropy of the logical states.

We consider the probability distribution on $\mathcal M$ (i.e., P[m] for $m \in \mathcal M$) with $\sum _{m \in \mathcal M} P[m] = 1$. After computation $\hat C$, the probability distribution on $\mathcal M'$ (i.e., P^'[m^'] for $m' \in \mathcal M'$) is given by

$$\begin{equation}P' [m'] = \sum _{m: \ \hat C (m) = m'} P[m], \label {prob_com1} \end{equation} \tag{ 25 }$$

where the sum in the right-hand side is taken over m satisfying $\hat C (m ) = m'$. If the computation is reversible, equation (25) reduces to

$$\begin{equation}P' [m'] = P[\hat C^{-1} (m') ] \quad\qquad (\mbox {if } m' \in \hat C (\mathcal M)), \quad\qquad P' [m'] = 0\ {\rm (otherwise)}, \label {prob_com2} \end{equation} \tag{ 26 }$$

which describes the conservation of probability.

We now define the initial and final logical entropies by [79, 80]

$$\begin{eqnarray}&&H (\mathcal M) := - \sum _{m \in \mathcal M} P[m] \ln P[m],\end{eqnarray} \tag{ 27 }$$

$$\begin{eqnarray}&&H' (\mathcal M') := - \sum _{m' \in \mathcal M'} P'[m'] \ln P'[m']. \end{eqnarray} \tag{ 28 }$$

If the computation is logically reversible, the logical entropy does not change. In fact, by using equation (26), we obtain

$$\begin{eqnarray}H' (\mathcal M') &=& - \sum _{m' \in \hat C (\mathcal M)} P[\hat C^{-1} (m') ] \ln P[\hat C^{-1} (m') ] \nonumber \\ &=& - \sum _{m \in \mathcal M} P[m] \ln P[m] = H (\mathcal M). \end{eqnarray} \tag{ 29 }$$

In contrast, if the computation is logically irreversible, the logical entropy states is decreased:

$$\begin{equation}H(\mathcal M) \geq H'(\mathcal M'). \label {Shannon_decrease} \end{equation} \tag{ 30 }$$

In fact, by noting that $P'[m'] = \sum _{m : \ \hat C (m) = m'} P[m]$ (i.e., P[m]/P^'[m^'] is a probability distribution over the m that satisfy $\hat C (m) = m'$), we obtain

$$\begin{equation}H(\mathcal M) - H'(\mathcal M') = - \sum _{m'} P' [m'] \sum _{m : \ \hat C (m ) = m'} \frac {P[m]}{P'[m']} \ln \frac {P[m]}{P'[m']} \geq 0. \end{equation} \tag{ 31 }$$

We note that inequality (30) is a special case of the data processing inequality in information theory [80].

In the case of the reversible extension (22), the logical entropy of the extended logical states does not change: $H(\mathcal M ) = H' (\mathcal M' \times \mathcal M'')$. From the subadditivity of the Shannon entropy, we have

$$\begin{equation}H'(\mathcal M' \times \mathcal M'') \leq H'(\mathcal M') + H'(\mathcal M''), \label {subadditivity} \end{equation} \tag{ 32 }$$

and therefore, together with inequality (30),

$$\begin{equation}H'(\mathcal M') \leq H(\mathcal M ) \leq H'(\mathcal M') + H'(\mathcal M''). \end{equation} \tag{ 33 }$$

We discuss simple examples with $\mathcal M = \mathcal M' = \{ 0,1 \}$. Let p : = P[0] and p^' : = P^'[0] be the probabilities of the input 0 and output 0, respectively. In the case of NOT, p^' = 1 − p and therefore $H(\mathcal M) = - p \ln p - (1-p) \ln (1-p) = H'(\mathcal M')$. In the case of ERASE, p^' = 1 for any p. Therefore, $H(\mathcal M) = - p \ln p - (1-p) \ln (1-p)$ and $H' (\mathcal M') = 0$, which implies that the logical entropy is decreased by $H(\mathcal M)$ by information erasure. In the case of the reversible extension of erasure (21), the equality in (32) is achieved with $H'(\mathcal M') = 0$ and $H'(\mathcal M'') = - p \ln p - (1-p) \ln (1-p)$. Therefore, the decrease in the entropy of the memory is compensated for by the increase in the entropy of the ancilla or environment, such that the total entropy is conserved.

We first consider the physical structure of the memory. In general, the logical states do not have one-to-one correspondence to the physical states of the memory; there may be a lot of microscopic physical states that correspond to a single logical state.

Let $\mathcal Y$ be the phase space of the memory, where each phase-space point $y \in \mathcal Y$ describes a microscopic physical state of the memory, and let $\mathcal M$ be the set of the possible input logical states. To relate the physical states to logical ones, we decompose $\mathcal Y$ into subspaces $\mathcal Y_m$ s ($m \in \mathcal M$), where $\mathcal Y_m$ and $\mathcal Y_n$ do not overlap with each other for m≠n, i.e., $\cup _{m \in \mathcal M} \mathcal Y_m = \mathcal Y$ and $\mathcal Y_m \cap \mathcal Y_n = \phi$ (m≠n) with ϕ the empty set. If the phase-space point of the memory is in $\mathcal Y_m$ before the computation, we define that the input logical state is m. We call $\mathcal Y_m$ an input logical subspace associated with m.

Figure 3 shows simple examples of phase-space separations with binary potentials of the memory. Figure 3(a) shows a symmetric potential, which is the same as the memory illustrated in figure 1(a). In this case, 0 and 1 correspond to the left and right well, respectively, and the phase-space volume of $\mathcal Y_0$ is the same as that of $\mathcal Y_1$. Figure 3(b) shows an asymmetric potential, where 0 and 1 correspond to the left and right well, respectively. In this case, the phase-space volume of $\mathcal Y_0$ is different from that of $\mathcal Y_1$.

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We consider probability distributions of the physical states in $\mathcal Y$. Let $y \in \mathcal Y$ be the initial phase-space point before the computation, P[y] be its probability, $m \in \mathcal M$ be the initial logical state such that $y \in \mathcal Y_m$, and P[m] be its probability that satisfies

$$\begin{equation}P[m] = \int _{y \in \mathcal Y_m} {\rm d} y\, P[y]. \end{equation} \tag{ 40 }$$

The probability distribution of y under the condition of m is written as P[y|m], which takes nonzero value only if $y \in \mathcal Y_m$. We then have

$$\begin{equation}P [y] = \sum _{m \in \mathcal M} P [y|m] P [m]. \end{equation} \tag{ 41 }$$

We also consider output logical states and phase-space points after the computation. Let $\mathcal M'$ be the set of the output logical states, and $\mathcal Y'_{m'}$ be the logical subspace associated with $m' \in \mathcal M'$, where $\cup _{m' \in \mathcal M'} \mathcal Y'_{m'} = \mathcal Y$ and $\mathcal Y'_{m'} \cap \mathcal Y'_{n'} = \phi$ (m^'≠n^'). Let $y' \in \mathcal Y$ be the final phase-space point after the computation, P^'[y^'] be its probability, $m' \in \mathcal M'$ be the final logical state such that $y' \in \mathcal Y'_{m'}$, and P^'[m^'] be its probability. The probability of y^' under the condition of m^' is written as P^'[y^'|m^'], which satisfies

$$\begin{equation}P'[y'] = \sum _{m' \in \mathcal M'} P' [y'|m'] P'[m']. \end{equation} \tag{ 42 }$$

We next consider the changes in entropies during the computation. Before the computation, the Shannon entropy of the physical states is given by

$$\begin{equation}S (\mathcal Y ) = -\int _{y \in \mathcal Y} {\rm d} y\, P [y ] \ln P [y ], \end{equation} \tag{ 43 }$$

and the logical entropy of the input is given by

$$\begin{equation}H (\mathcal M) = -\sum _{m \in \mathcal M} P [m] \ln P [m]. \end{equation} \tag{ 44 }$$

The conditional Shannon entropy inside $\mathcal Y_m$ is given by

$$\begin{equation}S (\mathcal Y | m) = -\int _{y \in \mathcal Y_m} {\rm d} y\, P [y | m] \ln P [y| m], \end{equation} \tag{ 45 }$$

whose ensemble average over $m \in \mathcal M$ is

$$\begin{equation}S (\mathcal Y | \mathcal M) := \sum _m P [m] S (\mathcal Y | m). \end{equation} \tag{ 46 }$$

Due to a general formula in probability theory [80], these entropies satisfy

$$\begin{equation}S (\mathcal Y) = H (\mathcal M) + S (\mathcal Y | \mathcal M), \label {entropy_balance1} \end{equation} \tag{ 47 }$$

which implies that the total entropy is given by the sum of the logical entropy of $\mathcal M$ and the average of the conditional entropies of the $\mathcal Y_m$. Intuitively, the fluctuation over the whole phase space can be decomposed into that over the logical states and that over the internal physical states in the individual logical subspaces.

We also consider the entropies after the computation in the same manner:

$$\begin{eqnarray}&&H' (\mathcal M') = -\sum _{m' \in \mathcal M'} P' [m'] \ln P' [m'],\end{eqnarray} \tag{ 48 }$$

$$\begin{eqnarray}&&S' (\mathcal Y ) = -\int _{y' \in \mathcal Y} {\rm d} y'\, P'[y' ] \ln P' [y' ].\end{eqnarray} \tag{ 49 }$$

$$\begin{eqnarray}&&S' (\mathcal Y | m') = -\int _{y' \in \mathcal Y'_{m'}} {\rm d} y'\, P' [y' | m'] \ln P' [y'| m'], \end{eqnarray} \tag{ 50 }$$

and

$$\begin{equation}S' (\mathcal Y | \mathcal M') := \sum _{m'} P' [m'] S' (\mathcal Y | m'). \end{equation} \tag{ 51 }$$

They also satisfy the decomposition formula:

$$\begin{equation}S' (\mathcal Y) = H' (\mathcal M') + S' (\mathcal Y | \mathcal M'). \label {entropy_balance2} \end{equation} \tag{ 52 }$$

Therefore, the total entropy change during the computation is decomposed as

$$\begin{equation} \Delta S = \Delta H + \Delta S_{\rm in}, \label {total_entropy_change} \end{equation} \tag{ 53 }$$

where

$$\begin{eqnarray} &&\Delta S := S' (\mathcal Y) - S(\mathcal Y), \label {entropy_change1}\end{eqnarray} \tag{ 54 }$$

$$\begin{eqnarray} &&\Delta H := H'(\mathcal M') - H(\mathcal M),\end{eqnarray} \tag{ 55 }$$

$$\begin{eqnarray} &&\Delta S_{\rm in} := S' (\mathcal Y | \mathcal M') - S(\mathcal Y | \mathcal M). \end{eqnarray} \tag{ 56 }$$

Here, ΔS is the change in the total entropy, ΔH is the change in the logical entropy, and ΔS_in is the average of the change in the physical entropy in the individual logical subspaces. We note that ΔH ≤ 0 holds for logically irreversible computations, while ΔH = 0 holds for logically reversible computations.

As a special case, we consider the information erasure. We refer to one of the logical states in $\mathcal M'$ as the 'standard state', which we denote by $0 \in \mathcal M'$. The information erasure is defined as the process in which the output logical state is in the standard state with unit probability (i.e., P^'[0] = 1 and P^'[m^'] = 0 if m^'≠0) for any probability distribution P[m] of the input logical states. In this case, $H' (\mathcal M') = 0$ holds by definition, and therefore $\Delta H = - H(\mathcal M')$. The change in the physical entropy is given by

$$\begin{equation} \Delta S_{\rm in} = S' (\mathcal Y' | 0) - S(\mathcal Y | \mathcal M). \end{equation} \tag{ 57 }$$

We note that $\Delta S \neq - H(\mathcal M')$ if ΔS_in≠0.

To clarify the role of ΔS_in, we consider a simple case with $\mathcal M = \mathcal M'$ and $\mathcal Y_m = \mathcal Y'_m$. We assume that the initial and final distributions inside the individual logical subspaces are the same (i.e., P[y|m] = P^'[y|m] for any y and m), and therefore $S(\mathcal Y| m) = S'(\mathcal Y|m)$ holds for any m. On the other hand, the probability distribution over $\mathcal M$ changes from P[m] to P^'[m] during the computation. In this case, we have

$$\begin{equation} \Delta S_{\rm in} = \sum _{m \in \mathcal M} (P'[m] - P[m]) S(\mathcal Y | m). \end{equation} \tag{ 58 }$$

If $S(\mathcal Y | m)$ does not depend on m, we have ΔS_in = 0 for any P[m] and P^'[m]. For example, in the case of the symmetric memory in figure 3(a), $S(\mathcal Y | 0) = S(\mathcal Y | 1)$ holds if the conditional probability distributions in the individual wells are in thermal equilibrium. If ΔS_in = 0, the internal fluctuations inside the individual logical subspaces do not contribute to the change in the total entropy (i.e., ΔS = ΔH).

In contrast, if $S(\mathcal Y | m)$ depends on m, ΔS_in≠0 in general. For example, in the case of the asymmetric memory in figure 3(b), $S(\mathcal Y | 0)$ and $S(\mathcal Y | 1)$ are different to each other. If ΔS_in≠0, the internal fluctuations inside the individual wells contribute to the change in the total entropy, and therefore ΔS≠ΔH. The role of the asymmetry of the potential has been discussed in [31]–[35].

We now consider the second law of thermodynamics for computation. We assume that the memory is attached to a single heat bath at inverse temperature β during the computation. The total entropy production is given by equation (2), which leads to

$$\begin{equation} \Delta S_{\rm tot} = \Delta H + \Delta S_{\rm in} - \beta Q, \label {total_entropy1} \end{equation} \tag{ 59 }$$

where Q is the average heat that is absorbed by the memory from the bath. Therefore, the second law (4) is given by

$$\begin{equation} \Delta H \geq \beta Q - \Delta S_{\rm in}, \label {g_Landauer1} \end{equation} \tag{ 60 }$$

or equivalently,

$$\begin{equation} -\beta Q \geq - \Delta H - \Delta S_{\rm in}, \label {g_Landauer2} \end{equation} \tag{ 61 }$$

which gives the fundamental lower bound of the heat emission into the bath during the computation. We refer to inequalities (60) and (61) as the generalized Landauer principle, and the right-hand side of (61) as the generalized Landauer bound. Several inequalities that are similar to or essentially equivalent to (61) have been obtained in [33]–[35].

In the special case of ΔS_in = 0, inequality (61) reduces to

$$\begin{equation} -\beta Q \geq - \Delta H. \end{equation} \tag{ 62 }$$

In this case, if the computation is logically irreversible, the heat emission − Q is non-negative because of ΔH ≥ 0.

In the case of the information erasure, the heat emission is bounded as

$$\begin{equation} -\beta Q \geq H (\mathcal M) - \Delta S_{\rm in}, \label {g_Landauer3} \end{equation} \tag{ 63 }$$

where ΔS_in is the modification term to the original Landauer principle due to the change in the fluctuations inside the individual logical subspaces. If ΔS_in = 0, we obtain

$$\begin{equation} -\beta Q \geq H (\mathcal M), \end{equation} \tag{ 64 }$$

which is the conventional Landauer principle [7], [16]–[18].

In the conventional setup of the Landauer principle with a symmetric memory such as in figures 1(a) and 3(a), the decrease in the entropy of the logical states should be compensated for by the increase in the entropy of the bath, which is accompanied by at least $\beta ^{-1} H (\mathcal M)$ of heat emission into the bath. In contrast, in the case of the asymmetric memories such as figure 3(b), the decrease in the entropy of the logical states can be compensated for not only by the increase in the entropy of the bath, but also by that inside the individual logical subspaces. Here, the heat emission is determined only by the change in the entropy of the bath, but not by that inside the logical subspaces. Therefore, the lower bound of the heat emission can be different from $\beta ^{-1} H (\mathcal M)$ as shown in inequality (63); in particular, the heat emission can be smaller than $\beta ^{-1} H (\mathcal M)$ if ΔS_in > 0.

To illustrate the above situation, we consider the situation with $\mathcal M = \mathcal M' = \{ 0, 1 \}$ and $\mathcal Y_m = \mathcal Y'_m$ for m = 0,1. We assume that $S (\mathcal Y | 0) = S' (\mathcal Y | 0 )$ holds, which implies that the probability distribution inside the standard state is the same before and after the erasure. We then have

$$\begin{eqnarray} \Delta S_{\rm in} &=& S (\mathcal Y | 0) - \left (P[0] S (\mathcal Y | 0) + P[1] S (\mathcal Y | 1) \right )\nonumber \\ &=& P[1] \left (S (\mathcal Y | 0) - S (\mathcal Y | 1) \right ). \end{eqnarray} \tag{ 65 }$$

Therefore, if P[1]≠0 and $S (\mathcal Y | 0) > S (\mathcal Y | 1)$, we have ΔS_in > 0. It is natural to assume that $S (\mathcal Y | 0) \neq S (\mathcal Y | 1)$ in asymmetric memories.

We now summarize the relationship between the thermodynamic and logical reversibilities in the general setup. On the basis of the argument in section 5.2, the conventional Landauer principle discussed in section 4 needs to be modified in general, and the generalized Landauer principle is stated as follows:

The decrease in the entropy of the logical states during a logically irreversible computation can be compensated for not only by the increase in the entropy of the heat bath, but also by that of the physical states inside the individual logical subspaces, where only the former determines the amount of the heat emission.

Since the generalized Landauer principle (60) is equivalent to the second law of thermodynamics (3), the equality in (60) can be achieved in the quasi-static limit (i.e., in the case of the quasi-static computation) where the total entropy production is zero. In this case, the computational process is thermodynamically reversible. Therefore, we conclude that:

Any logically irreversible computation can be performed in a thermodynamically reversible manner in the quasi-static limit, and therefore, the thermodynamic and logical reversibilities are not equivalent with each other.

These are generalizations of the arguments in section 4. We again stress that our observations here do not contradict the conventional Landauer principle; the logical reversibility is related to the change in the entropy of the logical states, while the thermodynamic reversibility is related to the change in the entropy of the whole universe that consists of the logical and physical states of the memory and the physical states of the heat bath.

We next consider the thermodynamic work needed for computation. Let E[y] with $y \in \mathcal Y$ be the initial Hamiltonian of the memory. We define the conditional free energy in logical subspace $\mathcal Y_m$ as

$$\begin{equation}F_m := -\beta ^{-1} \int _{y \in \mathcal Y_m} {\rm d} y\, {\rm e} ^{-\beta E[y]}. \end{equation} \tag{ 66 }$$

We assume that the memory is initially in the conditional canonical distribution in the individual logical subspaces, which is given by

$$\begin{equation}P [y | m ] = {\rm e} ^{\beta (F_m - E [y])} \end{equation} \tag{ 67 }$$

for $y \in \mathcal Y_m$, and otherwise P[y|m] = 0. The total probability distribution is then given by

$$\begin{equation}P [y] = \sum _{m \in \mathcal M} P[m] \chi (y, m) {\rm e} ^{\beta (F_m - E [y])}, \end{equation} \tag{ 68 }$$

where χ(y, m) is the characteristic function which takes one if $y \in \mathcal Y_m$ and zero otherwise. The conditional entropy is then given by

$$\begin{equation}S(\mathcal Y | m) = \beta (E_m - F_m), \end{equation} \tag{ 69 }$$

where

$$\begin{equation}E_m := \int _{y \in \mathcal Y_m} {\rm d} y\, P[y|m] E[y]. \end{equation} \tag{ 70 }$$

Therefore, we obtain

$$\begin{equation}S(\mathcal Y | \mathcal M) = \beta (E - F ), \end{equation} \tag{ 71 }$$

where

$$\begin{equation}E := \sum _{m \in \mathcal M} P[m] E_m = \int _{y \in \mathcal Y} {\rm d} y\, P[y] E[y] \end{equation} \tag{ 72 }$$

is the average energy, and

$$\begin{equation}F := \sum _{m \in \mathcal M} P[m] F_m \end{equation} \tag{ 73 }$$

is the average free energy.

We also consider the final Hamiltonian E^'[y^'] after the computation. Correspondingly, every argument about the final state is parallel to that about the initial one in the previous paragraph. To show that a quantity is about the final states after the computation, we use the notation of a prime.

We then have

$$\begin{equation} \Delta S_{\rm in} = \beta (\Delta E - \Delta F ), \label {Sint_F} \end{equation} \tag{ 74 }$$

where ΔE : = E^'− E is the change in the average energy, and ΔF : = F^'− F is the change in the average free energy. We note that, if the final distribution is different from the conditional canonical distribution, we can show an inequality

$$\begin{equation} \Delta S_{\rm in} \leq \beta (\Delta E - \Delta F ), \label {entropy} \end{equation} \tag{ 75 }$$

where the equality in (75) is achieved if and only if the output state is in the conditional canonical distribution. Therefore, the total entropy production (59) satisfies

$$\begin{equation} \Delta S_{\rm tot} \leq \beta (W - \Delta F) + \Delta H, \label {total_entropy2} \end{equation} \tag{ 76 }$$

where W is the work performed on the memory, and we have used the first law of thermodynamics

$$\begin{equation} \Delta E = Q + W. \end{equation} \tag{ 77 }$$

Therefore, by applying the second law (3), we obtain the generalized Landauer principle in terms of the work:

$$\begin{equation} \beta W \geq - \Delta H + \beta \Delta F, \label {g_LandauerW1} \end{equation} \tag{ 78 }$$

which gives the minimal work needed for the computation. In the case of information erasure, inequality (78) reduces to

$$\begin{equation} \beta W_{\rm eras} \geq H (\mathcal M) + \beta \Delta F, \label {g_Landauer_work} \end{equation} \tag{ 79 }$$

where ΔF is the modification term to the conventional Landauer principle (36). As shown in equation (74), there are two contributions to ΔF: the changes in the entropy and the energy inside the individual logical subspaces, where the former is ΔS_in and the latter is ΔE. We can also regard W − ΔF as the energy cost needed for the information erasure, whose lower bound is given by the conventional Landauer bound $\beta ^{-1} H(\mathcal M)$. Several inequalities that are similar to or essentially equivalent to the generalized Landauer principle (79) have been obtained in [33]–[35].

In general, we call the memory as symmetric if F_m does not depend on m. The memory shown in figure 3(a) is symmetric in this sense because of F₀ = F₁. In contrast, the memory shown in figure 3(b) is asymmetric because of F₀≠F₁. When $\mathcal M = \mathcal M'$, $\mathcal Y_m = \mathcal Y'_{m'}$ with m = m^', $F_m = F'_{m'}$ with m = m^', and F_m does not depend on m, then we have ΔF = 0 for any P[m] and P^'[m].

As a simple example of information erasure with an asymmetric memory, we consider a model of the memory shown in figure 4(a) that is in contact with a heat bath at inverse temperature β. If the barrier is much higher than the thermal fluctuation, this model is idealized by a memory with two boxes with different volumes as shown in figure 4(b). Let t : 1 − t (0 < t < 1) be the ratio of the volumes of the boxes, which characterizes the ratio of the phase-space volumes inside the individual logical states. If the memory is symmetric, t = 1/2. We assume that $\mathcal M = \mathcal M' = \{ 0,1 \}$ and $\mathcal Y_m = \mathcal Y'_m$ with m = 0,1, and write p : = P[0]. We note that t is in general different from p, because the initial state is not necessarily in global thermal equilibrium over the whole phase space. The protocol of the information erasure is as follows (see also figure 4(c)). We first move the wall quasi-statically to the position where the ratio of the two volumes is given by p: 1 − p. During this process, we perform the work of β⁻¹[pln(t/p) + (1 − p)ln(1 − t/1 − p)] on average. We then remove the wall without any work. We next compress the box from the right quasi-statically, and the ratio of the two volumes is given by t : 1 − t in the final stage, where the logical state is '0' with unit probability. During this compression, we perform the work of − β⁻¹ lnt. The total work performed on the memory is then given by

$$\begin{equation} \beta W_{\rm eras} = - p \ln p - (1-p) \ln (1-p) + \left (1-p \ln \frac {1-t}{t}\right ). \label {erasure_asymmetric} \end{equation} \tag{ 80 }$$

We note that the initial logical entropy is given by $H(\mathcal M) = - p \ln p - (1-p ) \ln (1-p)$. The free-energy difference between two logical states are given by F₀ − F₁ = β⁻¹ ln[(1 − t)/t], and therefore ΔF = β⁻¹(1 − p)ln[(1 − t)/t]. Therefore, equation (80) achieves the generalized Landauer bound in (79), which implies that this protocol of the information erasure is thermodynamically reversible.

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We next consider a simple example of the information erasure with $\mathcal M \neq \mathcal M'$. We assume that $\mathcal M = \{ 0,1 \}$ and that the memory is initially binary symmetric as shown in figure 1(a), which is idealized by the two-box memory with equal volumes as shown in figure 1(b). We then assume that $\mathcal M' = \{ 0 \}$ in the final stage, i.e., the output logical state is only the standard state; the corresponding potential model is shown in figure 5(a). This is idealized by a memory with a single box as shown in figure 5(b). For simplicity, we set P[0] = 1/2. In this case, the information erasure is just the removal of the wall as shown in figure 5(c), and therefore W_eras = 0. This erasure protocol achieves the generalized Landauer bound in (79), since $H(\mathcal M) = \ln 2$, $F_0 - F'_0 = F_1 - F'_0 = \beta ^{-1} \ln 2$, and ΔF = −β⁻¹ ln2. Therefore, this erasure protocol is thermodynamically reversible. In fact, if we insert a wall at the center of the single box after the erasure, the probability distribution of the logical states returns to the initial one, where a particle is in one of the two boxes with equal probability 1/2.

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If p : = P[0]≠1/2 in the above model, we quasi-statically move the wall to the position where the ratio of the volumes p: 1 − p, and then remove the wall (see all figure 5(d)). In this case, the work performed on the memory is given by βW_eras = −ln2 − plnp − (1 − p)ln(1 − p). Since $H(\mathcal M) = - p \ln p - (1-p) \ln (1-p)$ and βΔF = −ln2, this erasure protocol also achieves the generalized Landauer bound in (79).

We assume that any logical state in $\mathcal M$ is a pair of two logical states of sub-memories that we refer to as S and D; any input logical state $m \in \mathcal M$ is written as m = (s, d) where s and d are the logical states of memories S and D, respectively. Let $\mathcal M^{\rm S}$ ($\mathcal M^{\rm D}$) be the set of logical states of S (D) with $\mathcal M = \mathcal M^{\rm S} \times \mathcal M^{\rm D}$, where $s \in \mathcal M^{\rm S}$ and $d \in \mathcal M^{\rm D}$.

We also consider the physical states of the memories. Let $\mathcal Y^{\rm S}$ ($\mathcal Y^{\rm D}$) be the set of physical states of S (D) with $\mathcal Y = \mathcal Y^{\rm S} \times \mathcal Y^{\rm D}$. Let y : = (u, v) for any initial physical state $y \in \mathcal Y$ with $u \in \mathcal Y^{\rm S}$ and $v \in \mathcal Y^{\rm D}$. Let $\mathcal Y_m := \mathcal Y_s^{\rm S} \times \mathcal Y_d^{\rm D}$.

The conditional probability of (u, v) under the condition of (s, d) is given by P[u, v|s, d] that takes nonzero value only if $u \in \mathcal Y_s^{\rm S}$ and $v \in \mathcal Y_d^{\rm D}$. By applying the argument in section 5.1 to the present situation, we have

$$\begin{equation}P[u,v] = \sum _{(s,d) \in \mathcal M} P[u,v | s,d] P[s,d], \end{equation} \tag{ 81 }$$

whose Shannon entropy is given by equation (47).

The mutual information plays a crucial role in the presence of two memories; it characterizes the correlation between them [79, 80]. The mutual information between the physical states and that between the logical states are respectively given by

$$\begin{eqnarray}&&I (\mathcal Y^{\rm S} : \mathcal Y^{\rm D}) = S (\mathcal Y^{\rm S}) + S(\mathcal Y^{\rm D}) - S(\mathcal Y),\end{eqnarray} \tag{ 82 }$$

$$\begin{eqnarray}&&I (\mathcal M^{\rm S} : \mathcal M^{\rm D}) = H (\mathcal M^{\rm S}) + H(\mathcal M^{\rm D}) - H(\mathcal M). \end{eqnarray} \tag{ 83 }$$

The mutual information between the internal states is

$$\begin{equation}(\mathcal Y^{\rm S} : \mathcal Y^{\rm D} | s, d) = S (\mathcal Y^{\rm S} | s) + S(\mathcal Y^{\rm D} | d) - S(\mathcal Y | s,d), \end{equation} \tag{ 84 }$$

whose ensemble average over (s, d) is given by

$$\begin{eqnarray}I (\mathcal Y^{\rm S} : \mathcal Y^{\rm D} | \mathcal M) &:=& \sum _{(s,d) \in \mathcal M } I (\mathcal Y^{\rm S} : \mathcal Y^{\rm D} | s, d) P[s,d]\nonumber \\ &=& S (\mathcal Y^{\rm S} | \mathcal M^{\rm S}) + S(\mathcal Y^{\rm D} | \mathcal M^{\rm D}) -S(\mathcal Y | \mathcal M). \end{eqnarray} \tag{ 85 }$$

Therefore, we obtain

$$\begin{equation}I (\mathcal Y^{\rm S} : \mathcal Y^{\rm D}) = I (\mathcal M^{\rm S} : \mathcal M^{\rm D}) + I (\mathcal Y^{\rm S} : \mathcal Y^{\rm D} | \mathcal M), \label {two_decom1} \end{equation} \tag{ 86 }$$

which implies that the total correlation between the two memories can be decomposed into the correlation between their logical states and that between their internal physical states in the individual logical subspaces. We note that $I (\mathcal Y^{\rm S} : \mathcal Y^{\rm D} | \mathcal M) = 0$ if P[u, v|s, d] = P[u|s, d]P[v|s, d]. We also note that

$$\begin{eqnarray}&&S (\mathcal Y^{\rm S}) = H(\mathcal M^{\rm S}) + S(\mathcal Y^{\rm S} | \mathcal M^{\rm S}), \label {two_decom2}\end{eqnarray} \tag{ 87 }$$

$$\begin{eqnarray}&&S (\mathcal Y^{\rm D}) = H(\mathcal M^{\rm D}) + S(\mathcal Y^{\rm D} | \mathcal M^{\rm D}). \label {two_decom3} \end{eqnarray} \tag{ 88 }$$

The sum of equations (86), (87), and (88) leads to equation (47).

We also consider the probability distributions, the Shannon entropy, and the mutual information after the computation in the parallel manner to those before the computation; for example, we write $\mathcal M' := \mathcal M'_{\rm S} \times \mathcal M'_{\rm D}$, m^' : = (s^', d^') with $s' \in \mathcal M^{\prime {\rm S}}$ and $d' \in \mathcal M^{\prime {\rm D}}$. To show that a quantity is about the states after computation, we use the notation of a prime.

The change in the total Shannon entropy is then given by

$$\begin{equation} \Delta S = \Delta H^{\rm S} + \Delta H^{\rm D} - \Delta I + \Delta S_{\rm in}^{\rm S} + \Delta S_{\rm in}^{\rm D} - \Delta I_{\rm in}, \end{equation} \tag{ 89 }$$

where

$$\begin{eqnarray} &&\Delta S := S'(\mathcal Y') - S(\mathcal Y),\end{eqnarray} \tag{ 90 }$$

$$\begin{eqnarray} &&\Delta H^{\rm S} := H'(\mathcal M^{\prime {\rm S}}) - H(\mathcal M^{\rm S}), \quad\qquad \Delta H^{\rm D} := H'(\mathcal M^{\prime {\rm D}}) - H(\mathcal M^{\rm D}),\end{eqnarray} \tag{ 91 }$$

$$\begin{eqnarray} &&\Delta I := I' (\mathcal M^{\prime {\rm S}} : \mathcal M^{\prime {\rm D}}) - I (\mathcal M^{\rm S} : \mathcal M^{\rm D}),\end{eqnarray} \tag{ 92 }$$

$$\begin{eqnarray} &&\Delta S_{\rm in}^{\rm S} := S'(\mathcal Y^{\prime {\rm S}} | \mathcal M^{\prime {\rm S}} ) - S(\mathcal Y^{\rm S} | \mathcal M^{\rm S}), \quad\qquad \Delta S_{\rm in}^{\rm D} := S'(\mathcal Y^{\prime {\rm D}} | \mathcal M^{\prime {\rm D}}) - S(\mathcal Y^{\rm D} | \mathcal M^{\rm D}),\end{eqnarray} \tag{ 93 }$$

$$\begin{eqnarray} &&\Delta I_{\rm in} := I' (\mathcal Y^{\prime {\rm S}} : \mathcal Y^{\prime {\rm D}} | \mathcal M') - I (\mathcal Y^{\rm S} : \mathcal Y^{\rm D} | \mathcal M). \end{eqnarray} \tag{ 94 }$$

Here, ΔH^S and ΔH^D describe the changes in the logical entropies, ΔI describes the change in the logical correlation, $\Delta S_{\rm in}^{\rm S}$ and $\Delta S_{\rm in}^{\rm D}$ describe the change in the internal physical entropies in the individual logical subspaces, and ΔI_in describes the change in the internal correlation between the individual logical subspaces. Therefore, the total entropy production is given by

$$\begin{equation} \Delta S_{\rm tot} = \Delta H^{\rm S} + \Delta H^{\rm D} - \Delta I + \Delta S_{\rm in}^{\rm S} + \Delta S_{\rm in}^{\rm D} - \Delta I_{\rm in} - \beta Q. \label {two_total} \end{equation} \tag{ 95 }$$

We next assume that the energies of two memories are additive before the computation, E[y] = E^S[u] + E^D[v], where E^S[u] (E^D[v]) is the Hamiltonian of S (D). This assumption implies that the interaction Hamiltonian between the memories is negligible before the computation. The corresponding free energies are given by

$$\begin{equation}F_s^{\rm S} := - \beta ^{-1} \ln \int _{u \in \mathcal Y_s} {\rm e} ^{- \beta E^{\rm S}[u]}, \quad\qquad F_d^{\rm D} := - \beta ^{-1} \ln \int _{v \in \mathcal Y_d} {\rm e} ^{- \beta E^{\rm D}[v]}. \end{equation} \tag{ 96 }$$

In this case, the conditional distribution is given by

$$\begin{equation}P[u,v |s,d] = P[u|s]P[v|d], \end{equation} \tag{ 97 }$$

where, if $u \in \mathcal Y_s^{\rm S}$ and $v \in \mathcal Y_d^{\rm S}$,

$$\begin{equation}P[u|s] = {\rm e} ^{\beta (F_s^{\rm S} - E^{\rm S} [u])}, \quad\qquad P[v|d] = {\rm e} ^{\beta (F_d^{\rm D} - E^{\rm D} [v])}. \end{equation} \tag{ 98 }$$

We also consider the energies in the memories after the computation, where the energies are also assumed to be additive. The argument about energy and the free energy after the computation is parallel to that before the computation. To show that a quantity is about the states after the computation, we use the notation of a prime. We note that the energies are not necessarily additive during the computation, since the interaction Hamiltonian between the memories may become nonzero during the computation.

If the initial probability distribution is the conditional probability distribution, the total entropy production satisfies equation (76). Therefore, we obtain the lower bound of the work that is needed for the computation with two memories:

$$\begin{equation}W \geq - \Delta H + \Delta F, \end{equation} \tag{ 99 }$$

where

$$\begin{eqnarray} &&\Delta H = \Delta H^{\rm S} + \Delta H^{\rm D} - \Delta I, \end{eqnarray} \tag{ 100 }$$

$$\begin{eqnarray} &&\Delta F := \Delta F^{\rm S} + \Delta F^{\rm D},\end{eqnarray} \tag{ 101 }$$

$$\begin{eqnarray} &&\Delta F^{\rm S} := \sum _{s \in \mathcal M^{\rm S}} P[s] F_s^{\rm S}, \quad\qquad \Delta F^{\rm D} := \sum _{d \in \mathcal M^{\rm D}} P[d] F_d^{\rm D}. \end{eqnarray} \tag{ 102 }$$

In the special case of ΔH^S = ΔH^D = 0 and ΔF = 0, we have

$$\begin{equation} - W \leq - \Delta I, \end{equation} \tag{ 103 }$$

which is the work extraction by using the correlation, which can be demonstrated by Maxwell's demon [1].

In this section, we consider measurement and feedback as a special case of the argument in section 6.1 with the assumption of the initial and final conditional canonical distributions.

We first consider a measurement process, where memory D performs a measurement on memory S. In other words, the information in memory S is copied to memory D. The logical state of D is initially the standard state 0^D with unit probability, and the logical entropy of S is initially given by H^S. We assume that the initial logical entropy in D and the initial mutual information between the logical states are both zero. We also assume that the two memories are in the conditional canonical distributions before and after the measurement.

The two memories interact with each other, and make a correlation between the logical states. We assume that the logical state of S does not change in time during the measurement. When both of the two memories are one bit, a typical example of such measurement is given by CNOT (18), where the measurement is error-free (i.e., the copy of information is perfect).

We denote by I the correlation between the logical states after the measurement, and denote by H^D the logical entropy of D after the measurement. We note that the mutual information satisfies the following inequalities [79, 80]:

$$\begin{equation}0 \leq I \leq H^{\rm S}, \quad\qquad 0 \leq I \leq H^{\rm D}. \end{equation} \tag{ 104 }$$

Applying equation (76) to the measurement, the total entropy production is given by

$$\begin{equation} \Delta S_{\rm tot} = \beta (W_{\rm meas} - \Delta F_{\rm meas}^{\rm D}) + H^{\rm D} - I, \label {tot_meas} \end{equation} \tag{ 105 }$$

where W_meas is the work performed on the memories, and $\Delta F_{\rm meas}^{\rm D}$ is the change in the average free energy of D. Therefore, we obtain

$$\begin{equation}W_{\rm meas} \geq I - H^{\rm D} + \Delta F_{\rm meas}^{\rm D}, \label {meas_cost1} \end{equation} \tag{ 106 }$$

which gives the minimal work needed for the measurement. Several inequalities that are essentially equivalent to (106) have been obtained in [35, 65, 71].

If I = H^D holds and memory D is symmetric (i.e., $\Delta F_{\rm meas}^{\rm D} = 0$), inequality (106) reduces to

$$\begin{equation}W_{\rm meas} \geq 0, \label {bound_Bennett} \end{equation} \tag{ 107 }$$

which is the bound discussed by Bennett [10]. Since the energies of the memories do not change in time during the measurement if the memory is symmetric, inequality (107) is equivalent to

$$\begin{equation} - Q_{\rm meas} \geq 0, \label {bound_Bennett1} \end{equation} \tag{ 108 }$$

where Q_meas is the heat absorbed by the memories during the measurement. Inequalities (107) and (108) imply that the lower bounds of the work requirement and the heat emission during the measurement are both zero.

We consider a simple example where the logical states of S and D before and after the measurement are all one bit, which is a conventional setup of the measurement [10]. We assume that S is symmetric (i.e., $F^{\rm S}_0 = F^{\rm S}_1$) before and after the measurement; in this case, memory S can be modeled by two boxes (see also figure 6(a)). Before the measurement, the logical state (s, d) is (0,0) or (1,0) with equal probability 1/2. The measurement protocol is given by CNOT where S is the control bit and D is the target bit, and then the final logical state is (0,0) or (1,1) with equal probability 1/2. In this case, I = H^D = ln2 and $\Delta F_{\rm meas}^{\rm D} = 0$. Therefore, W_meas = 0 and − Q_meas = 0 are achieved in the quasi-static limit, where the measurement is thermodynamically reversible as well as logically reversible. We summarize the thermodynamic and logical reversibilities for the conventional setup of measurement in table 2, which is contrastive to table 1 for the information erasure.

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We next consider a feedback control process followed by the measurement, where memory D performs feedback control on memory S. This situation can be regarded as a typical setup of Maxwell's demon (see also section 6.3), where D is the demon and S is the engine to be controlled. We assume that the two memories are in the conditional canonical distributions before and after the feedback control. The logical state of S changes depending on the logical state of D after the measurement. We assume that memory D does not evolve in time during the feedback control. For simplicity, we assume that the final correlation between the logical states between the memories is zero after the feedback control, such that the change in the logical mutual information is given by − I. We also assume that the output logical state of S is the standard state 0^S with unit probability after feedback control such that the change in the logical entropy of S is given by − H^S. When both of the two memories are one bit, a typical example of such feedback control is also given by CNOT (19).

Applying equation (76) to the feedback control, the total entropy production is given by

$$\begin{equation} \Delta S_{\rm tot} = \beta (W_{\rm fb} - \Delta F_{\rm fb}^{\rm S}) - H^{\rm S} + I, \label {tot_feedback} \end{equation} \tag{ 109 }$$

where W_fb is the work performed on the memories, and $\Delta F_{\rm fb}^{\rm S}$ is the change in the average free energy of S. Therefore, we obtain

$$\begin{equation}W_{\rm fb} \geq - I + H^{\rm S} + \Delta F_{\rm fb}^{\rm S}, \label {feedback_cost1} \end{equation} \tag{ 110 }$$

which gives the minimal work needed for the feedback control. Several inequalities that are essentially equivalent to (110) have been obtained in [51, 54, 61, 65, 71]; in these previous works, $H^{\rm S} + \Delta F_{\rm fb}^{\rm S}$ is denoted by, for example, just $\Delta F_{\rm fb}^{\rm S}$.

We consider a simple situation where the logical states of S and D before and after feedback are all one bit. We assume that S is symmetric (i.e., $F^{\rm S}_0 = F^{\rm S}_1$) before and after the feedback (see also figure 6(b)). Before feedback, the logical state (s, d) is (0,0) or (1,1) with equal probability 1/2. The feedback protocol is given by CNOT that is logically reversible, where D is the control bit and S is the target bit, and then the final logical state is (0,0) or (0,1) with equal probability 1/2. In this case, I = H^S = ln2 and $\Delta F_{\rm fb}^{\rm S} = 0$ hold. Therefore, inequality (110) reduces to W_fb ≥ 0. In the quasi-static limit, W_fb = 0 is achieved, which implies that we can reduce the entropy of S (i.e., H^S = ln2) without performing any positive amount of work on S. We note that, in the conventional thermodynamics, we need a positive amount of work to isothermally decrease the entropy. In contrast, in the present situation, the mutual information plays the role of the resource to decrease the entropy of S.

We next consider another situation that there is a single logical state 0^S in S after the feedback, which corresponds to the whole phase space of S (see also figure 6(c)). In this case, I = H^S = ln2 and $\Delta F_{\rm fb}^{\rm S} = - \beta ^{-1} \ln 2$ hold. Therefore, inequality (110) reduces to W_fb ≥−β⁻¹ ln2. In the quasi-static limit, W_fb = −β⁻¹ ln2 is achieved, where we extract β⁻¹ ln2 of work by the feedback control; the mutual information is the resource of the work extraction. This situation is equivalent to the case of the conventional Szilard engine [3].

We consider Maxwell's demon as a special example of our general argument. A typical situation of Maxwell's demon consists of measurement and feedback control. In the setup of section 6.2, memory D plays the role of the demon and memory S is the system to be measured and controlled. By summing up inequalities (106) and (110), we obtain

$$\begin{equation}W_{\rm meas} + W_{\rm fb} \geq H^{\rm S} - H^{\rm D} + \Delta F_{\rm meas}^{\rm S} + \Delta F_{\rm fb}^{\rm D}, \label {total_work1} \end{equation} \tag{ 111 }$$

where the contribution of the mutual information vanishes in the right-hand side. If we perform the information erasure from D after the feedback control, the total work is given by

$$\begin{equation}W_{\rm meas} + W_{\rm fb} + W_{\rm eras} \geq H^{\rm S} + \Delta F_{\rm meas}^{\rm S}, \label {total_work2} \end{equation} \tag{ 112 }$$

where we have used the generalized Landauer principle (79). Here, $H^{\rm S} + \Delta F_{\rm meas}^{\rm S}$ in the right-hand side can be regarded as an effective free energy of S, which vanishes in, for example, the case of the Szilard engine discussed in section 6.2. If $H^{\rm S} + \Delta F_{\rm meas}^{\rm S} = 0$ holds, inequality (112) reduces to

$$\begin{equation}W_{\rm meas} + W_{\rm fb} + W_{\rm eras} \geq 0, \end{equation} \tag{ 113 }$$

which implies that we cannot extract any work from the entire process.

We stress that, before the information erasure, the total entropy productions (105) and (109) are always non-negative (i.e., ΔS_tot ≥ 0) for the individual processes of measurement and feedback. This confirms that measurement and feedback control are individually consistent with the second law of thermodynamics, without considering the information erasure.

As shown in section 6.2, both of measurement and feedback are logically reversible in typical situations, and can be thermodynamically reversible in the quasi-static limit. On the other hand, the information erasure is logically irreversible, while it can be thermodynamically reversible in the quasi-static limit. Therefore, the entire process of Maxwell's demon including the erasure is logically irreversible, but can be thermodynamically reversible in the quasi-static limit, where ΔS_tot = 0 holds for the individual processes of the measurement, feedback control, and information erasure.