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 a full-rank stationary state $\pi$ (or restrict to a common invariant support):

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

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. Gibbs stationarity suffices for this positivity argument; quantum detailed balance is a stronger condition, useful when deriving the generator microscopically.

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.

For a driven Hamiltonian with instantaneous stationary state $\pi_t=e^{-\beta H(t)}/Z_t$, Spohn’s expression still applies at each time if $\mathcal L_t(\pi_t)=0$. However, the reference state now moves, so

$$\dot\Sigma=-k_B\frac{d}{dt}D(\rho_t\Vert\pi_t) -k_B\operatorname{Tr}[\rho_t\,\partial_t\ln\pi_t].$$

The extra term must be retained. The relative entropy to a moving equilibrium state need not decrease even when the instantaneous entropy production is nonnegative.

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

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

For a monitoring scheme and physical reverse process that reproduce the thermal entropy balance, its average gives

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

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.

An arbitrary mathematical unraveling need not reproduce the same thermodynamic entropy production. For energy-resolved thermal jump unravelings, local detailed balance of jump rates supplies the connection. 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 unital energy-dephasing generator. Such a generator fixes every Gibbs state of the same Hamiltonian, so an infinite-temperature assumption is unnecessary.

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.

Correlations and Bath Disturbance

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

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

the bath Hamiltonian is fixed, and the total evolution is unitary. Define the energy lost by the bath as $Q=-\Delta E_B$. Then 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 exact even for a finite bath and strong interaction. It gives $\Sigma(t)\ge0$ relative to the initially factorized state, but its time derivative can be negative during a return of correlations or energy. At strong coupling, $Q=-\Delta E_B$ need not equal $\Delta E_S-W$ because interaction energy can change. Also, a finite bath obeys $\Delta S_B=\Delta E_B/T-k_BD(\rho_B(t)\Vert\rho_B^\beta)$, not the ideal-reservoir equality without the relative-entropy correction.

The identity 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.

Exercises

Thermal Qubit Relaxation

A diagonal qubit with gap $\epsilon$ has excitation and relaxation flows $a=\gamma_\uparrow p_g$ and $b=\gamma_\downarrow p_e$, with $\gamma_\uparrow/\gamma_\downarrow=e^{-\beta\epsilon}$. Show that $\dot\Sigma=k_B(a-b)\ln(a/b)\ge0$.

Solution

Since $\dot p_e=a-b$, $\dot S=k_B(a-b)\ln(p_g/p_e)$ and $\dot Q=\epsilon(a-b)$. Thus

$$\dot\Sigma=k_B(a-b)[\ln(p_g/p_e)-\beta\epsilon] =k_B(a-b)\ln(a/b).$$

Both factors have the same sign because the logarithm increases monotonically. The result vanishes at detailed balance, $a=b$; endpoint cases are understood by limits.

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.

References

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• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002).
• H.-P. Breuer, “Quantum jumps and entropy production,” Physical Review A 68, 032105 (2003).
• M. Esposito and C. Van den Broeck, “Three detailed fluctuation theorems,” Physical Review Letters 104, 090601 (2010).
• M. Esposito, K. Lindenberg, and C. Van den Broeck, “Entropy production as correlation between system and reservoir,” New Journal of Physics 12, 013013 (2010). arXiv:0908.1125.
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• U. Seifert, “Stochastic thermodynamics, fluctuation theorems and molecular machines,” Reports on Progress in Physics 75, 126001 (2012).
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• H. Spohn, “Entropy production for quantum dynamical semigroups,” Journal of Mathematical Physics 19, 1227-1230 (1978).