**5. Entropy Production And Efficiency**

In 1865, Rudolf Clausius proposed a new state function, the thermodynamic entropy *S*, that turned out to be crucial to study the limits and the efficiency of different physical processes. The thermodynamic entropy is defined as the amount of heat *δQJ* that is transferred in a reversible thermodynamic process, *δS* = *δQ*rev/*T*. Here, we are assuming that each reservoir *r* is in its own equilibrium at a constant temperature *Tr* and chemical potential *μr* (we are not going to treat time-dependent *Tr* or *μr*). As the reservoirs are considered to be macroscopic, the aforementioned equilibrium state is not altered by the coupling to the local system. This assumption allows us to associate the reservoirs' heat flux with the variation in their thermodynamic entropy, i.e., *δQJr* = *TrδSr*. Therefore, from Equation (16) and the definition of the heat current given in Equation (20), we can write:

$$
\dot{M}\_r = T\_r \hat{\mathbf{S}}\_r + \dot{N}\_r \mu\_r. \tag{26}
$$

The information theory entropy, known as the Shannon or von Neumann entropy, times the Boltzmann constant equals the thermodynamic entropy for equilibrium states. There is a debate on whether this equality can be extended to nonequilibrium states; see, e.g., [84–86]. However, for the purpose of this article, only the change of the entropy of the reservoirs is needed, not that of the system. Besides, we will not need to evaluate the entropy from the density matrix. Therefore, the thermodynamic definition of entropy suffices in our case.

Now, let us consider the sum of the internal energy over the set of all reservoirs coupled to the local system,

$$
\sum\_{r} \dot{\mathcal{U}}\_{r} = \sum\_{r} T\_{r} \dot{\mathcal{S}}\_{r} + \sum\_{r} \dot{N}\_{r} \mu\_{r}. \tag{27}
$$

If we take *Tr* = *δTr* + *T*0 and *μr* = *μ*0 + *δμ<sup>r</sup>*, add and subtract the change of the internal energy of the local system *U* ˙ *s* and that of the couplings between the local system and the reservoirs ∑*r U* ˙ s,*r*, one can rearrange the above equation to the following:

$$T\_0 \sum\_{\mathbf{r}} \dot{S}\_{\mathbf{r}} = \left( \dot{L}\_{\mathbf{s}} + \sum\_{\mathbf{r}} \dot{L}\_{\mathbf{r}} + \sum\_{\mathbf{r}} \dot{L}\_{\mathbf{s},\mathbf{r}} \right) - \sum\_{\mathbf{r}} \dot{N}\_{\mathbf{r}} \delta \mu\_{\mathbf{r}} - \sum\_{\mathbf{r}} \dot{S}\_{\mathbf{r}} \delta T\_{\mathbf{r}} - \left( \dot{L}\_{\mathbf{s}} + \sum\_{\mathbf{r}} \dot{L}\_{\mathbf{s},\mathbf{r}} \right) - \mu\_0 \left( \sum\_{\mathbf{r}} \dot{N}\_{\mathbf{r}} \right) . \tag{28}$$

Note that the values of *μ*0 and *T*0 are completely arbitrary, and there is no need to identify them with the chemical potential and temperature of the central region, which can be ill-defined far from equilibrium.

Replacing *S* ˙ *r* by = *Jr*/*Tr* in the right-hand side of the above equation, using energy conservation (17) and particle number conservation (18), allows one to rewrite Equation (28) as:

$$\begin{array}{rcl} \mathrm{T\_0\dot{S}\_{\rm res}} & = & -F\_{\rm D}\Omega - \sum\_{r} I\_{r}\delta V\_{r} - \sum\_{r} I\_{r} \left( \frac{\delta T\_{r}}{T\_{r}} \right) - \left( \dot{\mathcal{U}}\_{\rm s} + \sum\_{r} \dot{\mathcal{U}}\_{\rm s,r} \right) + \mu\_{0} \dot{\mathcal{N}}\_{\rm s} \end{array} \tag{29}$$

where *S* ˙ res = ∑*r S* ˙ *r* is the variation of the entropy of the electrons of all reservoirs, and we used *eN* ˙ *r* = *Ir*. The CIF can be split into "equilibrium" and "nonequilibrium" terms, *F*(eq) and *<sup>F</sup>*(ne), respectively, where one can prove that *F*(eq) is always conservative [13,22,42,63]. We are interested in the steady-state situation of our local system. As discussed around Equation (19), the change of the internal energy of the electronic part of the local system and that of the coupling region must be zero after a cycle, as energy cannot be accumulated indefinitely within a finite region. The same argumen<sup>t</sup> is true for the number of particles accumulated in a cycle, which should be zero. At steady state, we therefore recognize the reversible component of the entropy variation as that given by:

$$\dot{S}\_{\rm res}^{(\rm rev)} = -\frac{1}{T\_0} \left( F\_{\Omega}^{(\rm eq)} \Omega + \dot{\mathcal{U}}\_{\rm s} + \sum\_r \dot{\mathcal{U}}\_{\rm s,r} - \mu\_0 \dot{\mathcal{N}}\_{\rm s} \right) \,. \tag{30}$$

Obviously, this quantity will not contribute to the total entropy production. Therefore, the rate of entropy production *S* ˙(irrev) res yields:

$$\mathcal{S}\_{\text{res}}^{\text{(irrev)}} = -\frac{1}{T\_0} \left( F\_{\Omega}^{\text{(ne)}} \Omega - I \cdot \delta V - f \cdot \delta \mathcal{T} \right), \tag{31}$$

where *δT* = (*<sup>δ</sup>T*1/*T*1, *δT*2/*T*2, ...)T, and the currents are defined through *I* = (*<sup>I</sup>*1, *I*2, ...) and *J* = (*J*1, *J*2, ...). Integrating Equation (31) over a cycle and taking into account the second law of thermodynamics, one finds:

$$0 \ge \mathcal{W}\_{\mathcal{F}} + \mathcal{Q}\_{I} \cdot \delta V + \mathcal{Q}\_{\mathcal{I}} \cdot \delta \mathcal{T} \,. \tag{32}$$

The above general formula, also valid far from equilibrium, can be used to set efficiency bounds for energy transfer processes between the electronic and mechanical degrees of freedom. For example, if we take *δT* = 0, *QI* · *δV* < 0, and W*F* > 0, then the system should operate as a nanomotor driven by electric currents, and the following relation holds:

$$1 \geq -\frac{\mathcal{W}\_{\text{F}}}{\mathbf{Q}\mathbf{i} \cdot \delta \mathbf{V}'} \tag{33}$$

while for *QI* · *δV* > 0 and W*F* < 0, the system operates as a charge pump, and Equation (32) implies:

$$1 \geq -\frac{\mathbb{Q}\_I \cdot \delta \mathcal{V}}{\mathcal{W}\_\mathcal{F}}.\tag{34}$$

Notice that, because of the steady-state condition, W*F* equals Wext, where Wext can be taken as the output or the input energy, depending on the considered type of process. Therefore, the above formulas describe the efficiency *η* of the device's process, defined as the ratio between the output and input energies per cycle. It is also interesting to note that the above equations reflect no more than energy conservation in this particular case. A different situation occurs for *δV* = 0 and *δT* = 0, where Equation (32) yields:

$$1 \ge -\frac{\mathcal{W}\_{\text{F}}}{\mathcal{Q}\_{\text{I}} \cdot \delta \mathcal{T}}, \quad \text{and} \quad 1 \ge -\frac{\mathcal{Q}\_{\text{J}} \cdot \delta \mathcal{T}}{\mathcal{W}\_{\text{F}}}.\tag{35}$$

The first equation thus corresponds to a quantum heat engine and the second one to a quantum heat pump, respectively. Now, because of the factor *δT* , the above formulas differ from what is expected from energy conservation solely. This is clear in a two-lead system, where *η* is limited by Carnot's efficiency of heat engines and refrigerators, respectively. To illustrate this, let us consider a hot and a cold reservoir and set the temperature of the cold reservoir as the reference. For the heat engine, this gives:

$$Q\_l \cdot \delta \mathcal{T} = Q\_{l\_{\text{hot}}} \left( 1 - \frac{T\_{\text{cold}}}{T\_{\text{hot}}} \right) < 0 \quad \Rightarrow \quad \left( 1 - \frac{T\_{\text{cold}}}{T\_{\text{hot}}} \right) \ge -\frac{\mathcal{W}\_{\text{F}}}{Q\_{l\_{\text{hot}}}} , \tag{36}$$

where the left-hand side of the second equation represents the Carnot limit for heat engines. Other energy transfer processes mixing voltage and temperature biases can also be analyzed in the context of Equation (32) to set the bounds of their associated efficiencies.
