Nonequilibrium Steady States and Entropy Production

A nonequilibrium steady state is stationary under the coupled dynamics even though it sustains nonzero fluxes. In algebraic statistical mechanics it is constructed from reservoirs prepared at different intensive parameters, not postulated as a Gibbs state. The central points are that the large-time limit needs a scattering or averaging hypothesis, and that nonnegative entropy production follows only after the current convention and the reference reservoir temperatures have been fixed.

Required background. C*-dynamical systems and the KMS condition supplies the reservoir equilibrium states; passivity, complete passivity, and ground states explains the work interpretation of KMS equilibrium; thermal nuclearity, return to equilibrium, and mixing separates existence of equilibrium from convergence toward it. Helpful background. Onsager reciprocity and entropy production gives the near-equilibrium comparison; the H-theorem gives the kinetic comparison; Landauer information engines supplies the information-thermodynamic interface.

Scattering construction of a steady state

Let the uncoupled algebra be a tensor product of reservoir and small-system algebras, with dynamics $\alpha_t^0$ and reference state

$$\omega_0=\omega_{\beta_1,1}\otimes\cdots\otimes\omega_{\beta_n,n}\otimes\omega_S.$$

Each reservoir state is KMS for its own dynamics. Switching on a bounded interaction $V$ produces a perturbed group $\alpha_t$. Suppose that the outgoing Møller morphism exists on a norm-dense subalgebra,

$$\gamma_+(A)=\lim_{t\to+\infty}\alpha_{-t}^0\!\left(\alpha_t(A)\right).$$

Because $\omega_0$ is invariant under $\alpha_t^0$ on the reservoir factors, the limiting state is

$$\omega_+(A)=\lim_{t\to\infty}\omega_0(\alpha_t(A)) =\omega_0(\gamma_+(A)).$$

Intertwining, $\gamma_+\circ\alpha_s=\alpha_s^0\circ\gamma_+$, implies $\omega_+\circ\alpha_s=\omega_+$ whenever the uncoupled reference state is invariant. This proves stationarity; it does not prove that $\omega_+$ is KMS for the coupled flow. With unequal $\beta_j$, it normally is not. Algebraic scattering constructions and their relation to natural nonequilibrium states are developed in Jakšić and Pillet 2002, §§2–5, pp. 790–818.

If the pointwise limit fails, one can instead take weak-* accumulation points of Cesàro means,

$$\overline\omega_T(A)=\frac1T\int_0^T\omega_0(\alpha_t(A))\,dt.$$

Every accumulation point is stationary, but different subsequences may give different states. Averaging therefore establishes existence of a steady state under compactness of the state space; it does not establish scattering, uniqueness, or loss of memory.

Currents and entropy production

Fix the sign before using the second law. If $\delta_j$ is the generator of the $j$th reservoir dynamics, define

$$\Phi_j=\delta_j(V)$$

as the energy current out of reservoir $j$ and into the junction. In a finite Hamiltonian approximation, $\Phi_j=-dH_j/dt$. Stationarity and absence of energy storage in the bounded junction give $\sum_j\omega_+(\Phi_j)=0$. The reservoir entropy-loss rate, with this convention, is

$$\sigma(\omega_+)=-\sum_j\beta_j\,\omega_+(\Phi_j).$$

Relative-entropy balance yields $\sigma(\omega_+)\geq0$ when the thermodynamic limit, differentiability, and steady-state limits needed in the balance law exist. The inequality is not a consequence of stationarity alone; it compares the evolved state with specified KMS reservoirs. Precise $C^*$-algebraic hypotheses and the relative-entropy argument are given in Jakšić and Pillet 2002, §§4–5, pp. 805–818.

For two reservoirs, let $J=\omega_+(\Phi_L)=-\omega_+(\Phi_R)$ be positive from left to right. Then

$$\sigma=(\beta_R-\beta_L)J.$$

Thus a hotter left reservoir, $\beta_L<\beta_R$, is compatible with the second law precisely when the average energy current is rightward.

Free-fermion transmission test

For two free fermionic leads at equal chemical potential, coupled through a finite scatterer, the Landauer energy current takes the form

$$J_E=\int_{-\infty}^{\infty}\frac{dE}{2\pi}\, E\,\mathcal T(E)\,[f_{\beta_L}(E)-f_{\beta_R}(E)], \qquad f_\beta(E)=\frac1{e^{\beta E}+1},$$

where $0\leq\mathcal T(E)\leq1$. If $\beta_L<\beta_R$, then $E[f_{\beta_L}(E)-f_{\beta_R}(E)]\geq0$ for every $E$, so $J_E\geq0$ and $(\beta_R-\beta_L)J_E\geq0$. This mode-by-mode check exposes both the sign convention and the role of transmission. At unequal chemical potentials, entropy production uses heat currents $J_E-\mu_jJ_N$ rather than energy currents alone.

This algebraic current convention is used at its first phenomenological application in Onsager reciprocity and entropy production.

Failure boundaries: what the construction does not prove

A finite collection of reservoirs has a discrete or recurrent dynamics in generic models; it does not supply an irreversible steady flux for infinite time. The thermodynamic limit must precede, or be controlled together with, the large-time limit. Likewise, a weak-* time-averaged NESS need not be mixing, unique, or reachable from every locally normal initial state. Finally, $\sigma=0$ does not by itself imply equilibrium: a perfectly reflecting junction has $\mathcal T=0$ and hence zero production while the two uncoupled reservoirs retain different temperatures.

There is a second nonconverse. Positivity of the entropy-production functional constrains the weighted sum of steady currents, not each current separately and not their fluctuations. In a multireservoir junction, circulating contributions may cancel in the conservation law. A fluctuation theorem, a Green–Kubo formula, or Onsager symmetry therefore needs its own time-reversal, regularity, and spectral assumptions. Nor may one exchange the thermodynamic, long-time, and weak-coupling limits without a uniform estimate. These distinctions are essential when a finite numerical simulation appears to settle into a plateau: the plateau can approximate a metastable pre-recurrence regime without defining the infinite-system state $\omega_+$.

Exercises

1. Stationarity from a time average. Show that every weak-* accumulation point of $\overline\omega_T$ is $\alpha$-invariant.

Solution

For fixed $s$,

$$\overline\omega_T(\alpha_s(A))-\overline\omega_T(A) =\frac1T\left(\int_T^{T+s}-\int_0^s\right)\omega_0(\alpha_t(A))\,dt.$$

Its absolute value is at most $2|s|\lVert A\rVert/T$. The difference therefore vanishes along every subnet on which $\overline\omega_T$ converges.

2. Two-reservoir sign check. Derive $\sigma=(\beta_R-\beta_L)J$ from the current convention above.

Solution

Stationarity gives $\omega_+(\Phi_R)=-\omega_+(\Phi_L)=-J$. Hence $\sigma=-\beta_LJ-\beta_R(-J)=(\beta_R-\beta_L)J$.

3. A zero-production nonequilibrium state. Give a case with $\beta_L\neq\beta_R$ but $\sigma=0$.

Solution

Take a decoupled or perfectly reflecting junction. Then $\Phi_L=\Phi_R=0$, equivalently $\mathcal T(E)=0$, so the entropy-production rate vanishes although the reservoir temperatures remain unequal. Extra irreducibility or transport assumptions are needed to infer equilibrium from zero production.

References

• Jakšić, V., and Pillet, C.-A. (2002). “Mathematical theory of non-equilibrium quantum statistical mechanics.” Journal of Statistical Physics 108, 787–829. DOI.
• Ruelle, D. (2000). “Natural nonequilibrium states in quantum statistical mechanics.” Journal of Statistical Physics 98, 57–75. DOI.