Entropy Production

Entropy production measures irreversibility. In open quantum systems it is the part of the entropy balance that cannot be assigned merely to heat exchanged with ideal reservoirs. It is the quantity that becomes nonnegative in a thermodynamically consistent Markovian model and that appears in fluctuation theorems as the log ratio of forward and reverse probabilities.

The most common weak-coupling convention is:

$$\dot\Sigma = \frac{dS}{dt} - \sum_\alpha \frac{\dot Q_\alpha}{T_\alpha}, \qquad \dot\Sigma\ge0.$$

Here $\dot Q_\alpha$ is heat current into the system from bath $\alpha$. With this sign convention, the entropy change of bath $\alpha$ is $-\dot Q_\alpha/T_\alpha$. The formula is powerful only when the currents and temperatures are defined by a physically consistent system-bath model.

Entropy Balance

The von Neumann entropy of a density operator is

$$S(\rho) = -k_B\operatorname{Tr}(\rho\ln\rho).$$

For trace-preserving differentiable dynamics,

$$\frac{dS}{dt} = -k_B \operatorname{Tr} \left[ \dot\rho\ln\rho \right],$$

with the usual support qualifications when $\rho$ has zero eigenvalues.

A closed system evolving unitarily has

$$\dot\rho = - \frac{i}{\hbar}[H,\rho].$$

Its eigenvalues are constant, so $S(\rho)$ is constant. Entropy production for a reduced system appears because one traces out degrees of freedom, coarse-grains records, couples to reservoirs, or uses an irreversible effective equation.

For a Markovian open system,

$$\dot\rho = - \frac{i}{\hbar}[H,\rho] + \sum_\alpha \mathcal L_\alpha(\rho),$$

the unitary commutator does not change $S(\rho)$. The entropy balance is governed by the dissipators.

Heat Currents

When $\mathcal L_\alpha$ represents a thermal bath at temperature $T_\alpha$, the standard heat current into the system is

$$\dot Q_\alpha = \operatorname{Tr} \left[ H\,\mathcal L_\alpha(\rho) \right].$$

This definition assumes that $H$ is the Hamiltonian used by the bath rates and that interaction energy can be neglected or consistently absorbed into a renormalized Hamiltonian. It is safest for weak-coupling, secular, global thermal master equations. See Thermal Master Equations and Detailed Balance.

With heat currents into the system, the entropy flux from the reservoirs into the system is

$$\dot\Phi = \sum_\alpha \frac{\dot Q_\alpha}{T_\alpha}.$$

The entropy balance is often written

$$\frac{dS}{dt} = \dot\Sigma+\dot\Phi,$$

or equivalently

$$\dot\Sigma = \frac{dS}{dt} - \dot\Phi.$$

The notation varies across books. The sign convention should always be stated before comparing formulas.

Spohn Inequality

Spohn’s inequality is the standard positivity statement for quantum dynamical semigroups. Let $\mathcal L$ be a Gorini–Kossakowski–Sudarshan–Lindblad generator with stationary state $\pi$:

$$\mathcal L(\pi)=0.$$

Then

$$- \operatorname{Tr} \left[ \mathcal L(\rho) \left( \ln\rho-\ln\pi \right) \right] \ge0$$

under the usual finite-dimensional support assumptions. Multiplying by $k_B$ gives a nonnegative entropy-production rate associated with the relaxation generated by $\mathcal L$:

$$\dot\Sigma = - k_B \operatorname{Tr} \left[ \mathcal L(\rho) \left( \ln\rho-\ln\pi \right) \right] \ge0.$$

For a thermal generator with

$$\pi=\rho_\beta = \frac{e^{-\beta H}}{Z},$$

this becomes the familiar Clausius form. Since

$$\ln\rho_\beta = -\beta H-\ln Z,$$

and $\operatorname{Tr}[\mathcal L(\rho)]=0$, one finds

$$\dot\Sigma = - k_B \operatorname{Tr} \left[ \mathcal L(\rho)\ln\rho \right] - \frac{1}{T} \operatorname{Tr} \left[ H\mathcal L(\rho) \right].$$

The first term is the dissipative contribution to $dS/dt$, and the second is $-\dot Q/T$. Thus

$$\dot\Sigma = \frac{dS}{dt} - \frac{\dot Q}{T} \ge0$$

for a single thermal bath.

Multiple Baths

If each bath contribution $\mathcal L_\alpha$ has its own Gibbs stationary state for the same Hamiltonian $H$,

$$\rho_{\beta_\alpha} = \frac{e^{-\beta_\alpha H}}{Z_\alpha},$$

then the total entropy-production rate is

$$\dot\Sigma = - k_B \sum_\alpha \operatorname{Tr} \left[ \mathcal L_\alpha(\rho) \left( \ln\rho-\ln\rho_{\beta_\alpha} \right) \right] \ge0.$$

Using the same heat-current convention, this reduces to

$$\dot\Sigma = \frac{dS}{dt} - \sum_\alpha \frac{\dot Q_\alpha}{T_\alpha}.$$

At a nonequilibrium steady state, $dS/dt=0$, so

$$\dot\Sigma_{\mathrm{ss}} = - \sum_\alpha \frac{\dot Q_\alpha}{T_\alpha}.$$

For two baths with heat current $J$ entering the system from the hot bath and leaving into the cold bath,

$$\dot Q_h=J, \qquad \dot Q_c=-J.$$

Then

$$\dot\Sigma_{\mathrm{ss}} = J \left( \frac{1}{T_c} - \frac{1}{T_h} \right),$$

which is nonnegative for $T_h>T_c$ and $J\ge0$. This is the Clausius statement that heat flows spontaneously from hot to cold.

Relative-Entropy Form

The quantum relative entropy is

$$D(\rho\Vert\sigma) = \operatorname{Tr} \left[ \rho(\ln\rho-\ln\sigma) \right],$$

when the support of $\rho$ lies inside the support of $\sigma$. For a time-independent thermal semigroup with stationary state $\pi$, Spohn’s inequality says that

$$\frac{d}{dt} D(\rho(t)\Vert\pi) \le0.$$

The entropy-production rate is

$$\dot\Sigma = - k_B \frac{d}{dt} D(\rho(t)\Vert\pi) \ge0.$$

Thus relaxation to equilibrium is monotonic in relative entropy, not necessarily in every intuitive distance or every observable. This is one reason relative entropy appears so often in rigorous thermodynamic statements.

For a thermal state $\rho_\beta$, the relative entropy identity

$$F(\rho)-F(\rho_\beta) = k_B T\,D(\rho\Vert\rho_\beta)$$

shows the same structure in equilibrium: dissipative relaxation lowers nonequilibrium free energy.

Stochastic Entropy Production

The ensemble rate $\dot\Sigma$ is not the only useful object. In quantum-jump or continuous-measurement descriptions, one can assign a trajectory entropy production

$$\sigma[\gamma] = \ln \frac{P_F[\gamma]}{P_R[\tilde\gamma]}.$$

Its average gives the ensemble entropy production in the appropriate limit:

$$\langle\sigma\rangle = \frac{\Sigma}{k_B}.$$

This is the bridge to Fluctuation Theorems. Individual trajectories may have $\sigma\lt0$, but the average is nonnegative when the forward and reverse ensembles satisfy the theorem’s assumptions.

For jump unravelings, local detailed balance of jump rates is essential. Without a physical identification of the jumps with reservoir energy exchange, a trajectory entropy-production formula may be only formal.

Coherence and Dephasing

Quantum entropy production is not only about population relaxation. Pure dephasing can produce entropy without exchanging energy if the Hamiltonian is fixed and the dissipator destroys coherence in the energy basis. Then

$$\dot Q=0, \qquad \dot\Sigma=\frac{dS}{dt}$$

for a consistent infinite-temperature or nondemolition dephasing model.

This does not mean coherence is “just entropy.” Coherence can be a resource for work extraction under suitable controls, can be converted into correlations, and can be hidden by the measurement scheme used to define work. The entropy-production statement depends on which operations, reservoirs, and records are included.

Global Unitary Accounting

For a system coupled to a single initially thermal bath, suppose the initial state is factorized,

$$\rho_{SB}(0) = \rho_S(0)\otimes\rho_B^\beta,$$

and the total evolution is unitary. If heat into the system is $Q=-\Delta E_B$, the entropy production over a finite process is

$$\frac{\Sigma}{k_B} = \frac{\Delta S_S}{k_B} - \beta Q.$$

Under the same assumptions, this can be written as

$$\frac{\Sigma}{k_B} = I_{S:B}(t) + D(\rho_B(t)\Vert\rho_B^\beta),$$

where

$$I_{S:B} = \frac{ S(\rho_S)+S(\rho_B)-S(\rho_{SB}) }{k_B}$$

is the mutual information. The two terms on the right are nonnegative: irreversibility comes from system-bath correlations and from the bath being displaced away from its initial thermal state.

This identity is useful because it shows what a Markovian reservoir approximation discards. In the ideal reservoir limit the bath disturbance is unobserved and correlations are continuously coarse-grained, leaving an effective entropy-production rate for the reduced system.

Strong-Coupling and Non-Markovian Caveats

The formula

$$\dot\Sigma = \frac{dS}{dt} - \sum_\alpha \frac{\dot Q_\alpha}{T_\alpha}$$

is not automatically valid for every reduced equation. Problems arise when:

• the interaction energy is not negligible;
• the bath is finite or far from equilibrium;
• the dissipator is local but the system Hamiltonian has strong internal coupling;
• the generator is not derived from thermal correlations;
• memory effects make heat temporarily reside in structured environmental modes;
• initial system-bath correlations make the reduced map preparation-dependent.

At strong coupling one may need an enlarged system boundary, a reaction-coordinate mapping, an inclusive energy balance, or a Hamiltonian of mean force. For memory effects, see CP Divisibility and Reaction-Coordinate Mapping.

Common Mistakes

• Identifying $dS/dt$ with entropy production even when heat flows are present.
• Forgetting that $\dot Q_\alpha$ must have a sign convention.
• Applying Spohn’s inequality to a dissipator without checking its stationary state.
• Calling a positive-rate Lindblad equation thermodynamic without detailed balance.
• Using local dissipators for an interacting system and assuming the second law follows automatically.
• Treating negative trajectory entropy production as a violation of the second law.
• Ignoring system-bath correlations in strong-coupling or non-Markovian regimes.
• Mixing entropy in units of $k_B$ with dimensionless entropy production without saying which is used.

Exercises

Unitary Entropy Conservation

Show that a closed quantum system evolving under $\dot\rho=-i[H,\rho]/\hbar$ has constant von Neumann entropy.

Solution

Unitary evolution has the form

$$\rho(t) = U(t)\rho(0)U^\dagger(t).$$

Unitary conjugation preserves the eigenvalues of $\rho$. Since

$$S(\rho) = -k_B\sum_n p_n\ln p_n$$

depends only on those eigenvalues, $S(\rho(t))$ is constant.

Spohn to Clausius

For a single thermal generator with stationary state $\rho_\beta=e^{-\beta H}/Z$, show that Spohn’s expression gives $\dot\Sigma=dS/dt-\dot Q/T$.

Solution

Start from

$$\dot\Sigma = - k_B \operatorname{Tr} \left[ \mathcal L(\rho) \left( \ln\rho-\ln\rho_\beta \right) \right].$$

Use

$$\ln\rho_\beta=-\beta H-\ln Z.$$

Because $\mathcal L$ is trace preserving,

$$\operatorname{Tr}[\mathcal L(\rho)]=0,$$

so the $\ln Z$ term drops out. Then

$$\dot\Sigma = - k_B \operatorname{Tr}[\mathcal L(\rho)\ln\rho] - k_B\beta \operatorname{Tr}[H\mathcal L(\rho)].$$

The first term is $dS/dt$ from the dissipator, and

$$k_B\beta = \frac{1}{T}.$$

With $\dot Q=\operatorname{Tr}[H\mathcal L(\rho)]$,

$$\dot\Sigma = \frac{dS}{dt} - \frac{\dot Q}{T}.$$

Two-Bath Steady State

A system in a steady state absorbs heat current $J$ from a hot bath at $T_h$ and dumps the same heat current into a cold bath at $T_c$. Compute $\dot\Sigma$.

Solution

At steady state $dS/dt=0$. With heat into the system as positive,

$$\dot Q_h=J, \qquad \dot Q_c=-J.$$

Therefore

$$\dot\Sigma = - \frac{J}{T_h} - \frac{-J}{T_c} = J \left( \frac{1}{T_c} - \frac{1}{T_h} \right).$$

For $T_h>T_c$ and $J\ge0$, this is nonnegative.

Global Correlation Identity

For an initially factorized system and thermal bath, explain why entropy production can be written as mutual information plus bath relative entropy.

Solution

The total entropy is conserved by unitary evolution. Starting from a product state, this gives

$$I_{S:B}(t) = \frac{\Delta S_S+\Delta S_B}{k_B}.$$

For the bath,

$$D(\rho_B(t)\Vert\rho_B^\beta) = - \frac{\Delta S_B}{k_B} + \beta\Delta E_B.$$

With heat into the system $Q=-\Delta E_B$,

$$D(\rho_B(t)\Vert\rho_B^\beta) = - \frac{\Delta S_B}{k_B} - \beta Q.$$

Adding the two equations gives

$$I_{S:B}(t) + D(\rho_B(t)\Vert\rho_B^\beta) = \frac{\Delta S_S}{k_B} - \beta Q = \frac{\Sigma}{k_B}.$$

Both terms on the left are nonnegative, so the finite-process entropy production is nonnegative.

Cross-Links

• Entropy in Quantum Statistical Mechanics
• Quantum Thermodynamics
• Fluctuation Theorems
• Thermal Master Equations
• Detailed Balance
• Lindblad–GKSL Equation
• Quantum Dynamical Semigroups
• System–Bath Hamiltonians
• Fluctuation–Dissipation Relation
• Reaction-Coordinate Mapping
• Approximation Checklist

References

• H. Spohn, “Entropy production for quantum dynamical semigroups,” Journal of Mathematical Physics 19, 1227-1230 (1978).
• H. Spohn and J. L. Lebowitz, “Irreversible thermodynamics for quantum systems weakly coupled to thermal reservoirs,” Advances in Chemical Physics 38, 109-142 (1978).
• R. Alicki, “The quantum open system as a model of the heat engine,” Journal of Physics A: Mathematical and General 12, L103-L107 (1979).
• H.-P. Breuer, “Quantum jumps and entropy production,” Physical Review A 68, 032105 (2003).
• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002).
• M. Esposito, U. Harbola, and S. Mukamel, “Nonequilibrium fluctuations, fluctuation theorems, and counting statistics in quantum systems,” Reviews of Modern Physics 81, 1665-1702 (2009).
• M. Esposito and C. Van den Broeck, “Three detailed fluctuation theorems,” Physical Review Letters 104, 090601 (2010).
• R. Kosloff, “Quantum thermodynamics: A dynamical viewpoint,” Entropy 15, 2100-2128 (2013).
• U. Seifert, “Stochastic thermodynamics, fluctuation theorems and molecular machines,” Reports on Progress in Physics 75, 126001 (2012).