Ergotropy and Passive States

Ergotropy is the maximum work that can be extracted from a quantum state by a cyclic unitary process. The Hamiltonian at the beginning and end is the same. The controller may act coherently on the system, but there is no heat bath, no measurement feedback, and no change in the system Hamiltonian after the cycle closes.

The concept is useful because it separates a sharp finite-system question from broader thermodynamic work statements. Given only a pair $(\rho,H)$ and full coherent control, which part of the state’s energy is ordered enough to be converted into work?

It is also easy to overinterpret. Ergotropy is not the same as nonequilibrium free energy, not the same as the random work distribution from a two-point measurement protocol, and not automatically available under thermal operations that lack an external phase reference.

Cyclic Unitary Work Extraction

Use a finite-dimensional system (or a finite-energy state with convergent sums). Its Hamiltonian at the beginning and end of the control cycle is

$$H = \sum_n E_n |E_n\rangle\langle E_n|,$$

with energy levels ordered as

$$E_1\le E_2\le\cdots .$$

A cyclic unitary protocol begins with state $\rho$, applies a unitary $U$, and ends with the same Hamiltonian $H$. The final state is

$$\rho' = U\rho U^\dagger .$$

Using the convention that positive work is extracted from the system, the extracted work is

$$W_{\mathrm{ext}}(U) = \operatorname{Tr}(\rho H) - \operatorname{Tr}(U\rho U^\dagger H).$$

The unitary does not change the eigenvalues of $\rho$. It can only rearrange the state’s eigenvectors relative to the energy eigenbasis. Therefore the minimization problem is:

$$\text{minimize } \operatorname{Tr}(U\rho U^\dagger H) \quad \text{over all unitaries } U .$$

The ergotropy of $(\rho,H)$ is the resulting maximum:

$$\mathcal W(\rho,H) = \operatorname{Tr}(\rho H) - \min_U \operatorname{Tr}(U\rho U^\dagger H).$$

Here $\mathcal W\ge0$. It vanishes exactly when $\rho$ is already passive with respect to $H$.

Passive Rearrangement

Write the spectral decomposition of the state as

$$\rho = \sum_k r_k |r_k\rangle\langle r_k|, \qquad r_1\ge r_2\ge\cdots .$$

For a nondegenerate Hamiltonian with ordered energies, the unitary that minimizes the final energy maps the largest eigenvalue of $\rho$ to the lowest energy eigenstate, the next largest eigenvalue to the next lowest energy eigenstate, and so on. The corresponding passive state is

$$\pi_\rho = \sum_k r_k |E_k\rangle\langle E_k|.$$

Thus

$$\mathcal W(\rho,H) = \operatorname{Tr}(\rho H) - \operatorname{Tr}(\pi_\rho H).$$

This is the quantum version of sorting. The spectrum of $\rho$ is fixed by unitary evolution, and the final energy is smallest when probability weight is placed as low in the energy ladder as the spectrum permits.

Degeneracies require a small convention. Inside a degenerate energy subspace, unitary rotations do not change the energy. Inside a degenerate eigenspace of $\rho$, the eigenvectors are not unique. The passive rearrangement is therefore unique only up to such rotations, but its energy and ergotropy are well defined.

Passive States

A state $\sigma$ is passive with respect to $H$ if no cyclic unitary can extract work:

$$\operatorname{Tr}(\sigma H) \le \operatorname{Tr}(U\sigma U^\dagger H) \qquad \text{for every unitary } U .$$

For a nondegenerate Hamiltonian, this is equivalent to two conditions:

• $\sigma$ is diagonal in the energy eigenbasis;
• the populations are nonincreasing with energy.

Explicitly, if

$$\sigma = \sum_n p_n |E_n\rangle\langle E_n|,$$

then passivity requires

$$E_i\le E_j \quad\Longrightarrow\quad p_i\ge p_j .$$

If a higher energy level is more populated than a lower one, a unitary swap lowers the mean energy and extracts work. Such a population inversion is active. For a nondegenerate Hamiltonian, any energy-basis coherence violates passivity, even when the diagonal populations alone are ordered.

Passivity is a statement about a specified Hamiltonian. The same density operator can be passive for one Hamiltonian and active for another.

Complete Passivity

Passivity of one copy is weaker than thermodynamic equilibrium. A state $\sigma$ is completely passive if every tensor power

$$\sigma^{\otimes N}$$

is passive with respect to the noninteracting Hamiltonian

$$H^{(N)} = \sum_{j=1}^N I\otimes\cdots\otimes H\otimes\cdots\otimes I .$$

Joint control of initially independent copies can unlock work even when each copy is passive. Pre-existing correlations are not required. In finite-dimensional systems under standard assumptions, Gibbs states

$$\tau_\beta = \frac{e^{-\beta H}} {Z}, \qquad \beta\ge0,$$

are completely passive. For a nontrivial finite-dimensional Hamiltonian, the completely passive states are these Gibbs states and states supported entirely on the ground-energy subspace. Ground degeneracy allows any state within that subspace.

For example, take energies $(0,\epsilon,3\epsilon)$ and populations $(0.5,0.3,0.2)$. One copy is passive. In two copies, however, $|0,2\rangle$ has energy $3\epsilon$ and probability $0.10$, while $|1,1\rangle$ has lower energy $2\epsilon$ and probability $0.09$. Swapping these two joint states extracts $0.01\epsilon$. The joint population ordering can fail even though each separate copy is ordered.

Relation to Free Energy

At fixed temperature $T$, the nonequilibrium free energy is

$$F_T(\rho) = \operatorname{Tr}(\rho H) - T S(\rho),$$

where

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

The difference

$$F_T(\rho)-F_T(\tau_\beta) = k_B T\,D(\rho\Vert\tau_\beta)$$

is the reversible average-work limit in ideal bath-assisted transformations with suitable control and entropy disposal. Exact single-copy thermal operations can impose additional constraints, as the thermal-operation example shows.

Ergotropy asks a different question. It allows a cyclic unitary on the system but no heat bath and no entropy disposal. Because unitary evolution preserves $S(\rho)$ and the eigenvalues of $\rho$, the only available work comes from arranging those eigenvalues more favorably relative to $H$.

A thermal state has zero ergotropy:

$$\mathcal W(\tau_\beta,H) = 0 \qquad (\beta\ge0).$$

This does not mean the state has zero thermodynamic free energy. It means that no work can be extracted from that state alone by a cyclic unitary.

Coherence and Work

Energy-basis coherence can contribute to ergotropy if the controller is allowed to implement arbitrary unitaries. For example, a pure state

$$|\psi\rangle = \sqrt{1-p}\,|g\rangle + e^{i\phi}\sqrt p\,|e\rangle$$

of a qubit with Hamiltonian $H=\epsilon |e\rangle\langle e|$ has mean energy $\epsilon p$. Since any pure state can be unitarily mapped to the ground state, its ergotropy is

$$\mathcal W(|\psi\rangle\langle\psi|,H) = \epsilon p .$$

The phase $\phi$ affects which unitary performs the extraction, even though the maximum extracted work depends only on the mean energy for this pure qubit example.

Operational restrictions change the statement. If allowed operations must commute with total energy, or if no phase reference is available, coherence between distinct energy eigenspaces is not freely convertible into mechanical work. In resource-theoretic thermodynamics, population imbalance and coherence are often separate resources. See Thermal Operations Preview for that viewpoint.

The useful rule is:

• ergotropy with arbitrary cyclic unitaries counts coherent control as part of the work-extraction apparatus;
• thermal operations without an external coherence reference do not treat energy-basis coherence as automatically extractable work.

Example: Diagonal Qubit

Let

$$H = \epsilon |e\rangle\langle e|, \qquad \epsilon>0,$$

with state

$$\rho = p_g |g\rangle\langle g| + p_e |e\rangle\langle e|, \qquad p_g+p_e=1 .$$

The mean energy is

$$E(\rho) = \epsilon p_e .$$

If $p_e\le p_g$, the state is passive: the larger population is already on the lower energy level. Its ergotropy is zero.

If $p_e>p_g$, the state is inverted. A unitary swap places population $p_e$ in $|g\rangle$ and $p_g$ in $|e\rangle$. The final energy is $\epsilon p_g$, so

$$\mathcal W(\rho,H) = \epsilon(p_e-p_g) = \epsilon(2p_e-1).$$

Combining the two cases,

$$\mathcal W(\rho,H) = \epsilon\max(0,2p_e-1).$$

This simple formula is a good check on sign conventions: population inversion is active, a positive-temperature thermal qubit is passive.

Example: Three-Level Sorting

Consider a three-level system with

$$E_1\le E_2\le E_3$$

and a diagonal state whose populations are

$$(p_1,p_2,p_3) = (0.2,0.5,0.3).$$

The state is not passive because $p_2>p_1$ while $E_2\ge E_1$. The passive rearrangement sorts the same probabilities in decreasing order over increasing energy:

$$(0.5,0.3,0.2).$$

The ergotropy is

$$\mathcal W = \sum_{n=1}^3 p_n E_n - \left( 0.5E_1+0.3E_2+0.2E_3 \right).$$

No entropy has been removed. The work comes only from undoing the population disorder relative to the energy ordering.

Example: Squeezed Oscillator State

A harmonic oscillator thermal state is passive:

$$\tau_\beta = \frac{e^{-\beta\hbar\omega a^\dagger a}} {\operatorname{Tr}(e^{-\beta\hbar\omega a^\dagger a})}.$$

For $\rho=S(r)\tau_\beta S^\dagger(r)$, the eigenvalues are still the thermal eigenvalues. Undoing the squeezing reaches the passive state $\tau_\beta$, so under arbitrary unitary control the entire excess energy is extractable:

$$\mathcal W=\hbar\omega(2\bar n+1)\sinh^2r.$$

The zero-point term cancels between the initial and final energies. A squeezed reservoir supplies this additional nonequilibrium resource; it is not equivalent to raising the temperature of a thermal reservoir.

The exact ergotropy depends on which Hamiltonian defines the energy and which unitaries the controller can implement. In open-system settings, be careful to distinguish a squeezed bath as an engineered nonequilibrium reservoir from a thermal bath at a higher temperature.

Relation to Work Distributions

Ergotropy is a state function of the pair $(\rho,H)$ under an allowed control class. It is a maximum extractable average work.

A work distribution is different. In a two-point measurement protocol, one samples initial and final energies and assigns

$$W = E_m^\tau-E_n^0 .$$

That random variable depends on projective measurements, driving, and a sign convention. It can obey fluctuation relations such as Jarzynski and Crooks identities when their assumptions hold. See Two-Point Measurement Scheme, Work Distributions, and Jarzynski and Crooks Relations.

The two languages meet in carefully designed protocols, but they should not be collapsed into one slogan. Ergotropy optimizes over unitaries for a given state; fluctuation relations compare probability distributions generated by specified protocols.

Exercises

A Mixed Coherent Qubit

Let $\rho$ in the $(g,e)$ basis have diagonal entries $1-p,p$ and off-diagonal entry $c$, with $|c|^2\le p(1-p)$. Find its ergotropy for $H=\epsilon|e\rangle\langle e|$.

Solution

The eigenvalues are $(1\pm R)/2$, where $R=\sqrt{(1-2p)^2+4|c|^2}$. The smaller eigenvalue goes on the excited level in the passive state. Therefore

$$\mathcal W=\epsilon\left[p-\frac{1-R}{2}\right].$$

For $c=0$ this reduces to $\epsilon\max(0,2p-1)$; for a pure state $R=1$, it gives $\epsilon p$. Coherence changes the eigenvalues at fixed populations and can make an otherwise passive population distribution useful for work extraction.

Pure Qubit Coherence

Let

$$|\psi\rangle = \sqrt{1-p}\,|g\rangle + e^{i\phi}\sqrt p\,|e\rangle, \qquad H=\epsilon |e\rangle\langle e|.$$

Assuming arbitrary cyclic unitaries are available, find the ergotropy.

Solution

The state is pure, so its eigenvalues are $1$ and $0$. A unitary can map $|\psi\rangle$ to the ground state $|g\rangle$. The passive state is therefore $|g\rangle\langle g|$, whose energy is zero.

The initial energy is

$$\langle\psi|H|\psi\rangle = \epsilon p.$$

Thus

$$\mathcal W = \epsilon p.$$

The phase $\phi$ affects the extracting unitary, but not the maximum value in this two-level pure-state example.

Gibbs Passivity

Show that a finite-dimensional Gibbs state

$$\tau_\beta = \frac{e^{-\beta H}}{Z}$$

with $\beta\ge0$ is passive.

Solution

Let $H|E_n\rangle=E_n|E_n\rangle$, with energies ordered increasingly. The Gibbs populations are

$$p_n = \frac{e^{-\beta E_n}}{Z}.$$

If $E_i\le E_j$ and $\beta\ge0$, then

$$e^{-\beta E_i} \ge e^{-\beta E_j}.$$

Therefore $p_i\ge p_j$. The Gibbs state is diagonal in the energy basis and its populations decrease with increasing energy, so it is passive.

Passive But Not Thermal

Consider a three-level system with equally spaced energies $0,\epsilon,2\epsilon$ and diagonal populations

$$(p_0,p_1,p_2) = (0.5,0.3,0.2).$$

Is the state passive? Is it necessarily a Gibbs state?

Solution

The populations decrease with energy, so the state is passive for one copy.

It is not necessarily Gibbs. For a Gibbs state with equally spaced levels, the ratios would obey

$$\frac{p_1}{p_0} = \frac{p_2}{p_1} = e^{-\beta\epsilon}.$$

Here

$$\frac{p_1}{p_0} = 0.6, \qquad \frac{p_2}{p_1} = \frac{2}{3}.$$

The ratios are unequal, so this specific state is passive but not Gibbs. This illustrates why passivity and thermal equilibrium are different statements.

References

• R. Alicki and M. Fannes, “Entanglement boost for extractable work from ensembles of quantum batteries,” Physical Review E 87, 042123 (2013).
• A. E. Allahverdyan, R. Balian, and Th. M. Nieuwenhuizen, “Maximal work extraction from finite quantum systems,” Europhysics Letters 67, 565-571 (2004).
• F. Binder, L. A. Correa, C. Gogolin, J. Anders, and G. Adesso, eds., Thermodynamics in the Quantum Regime: Fundamental Aspects and New Directions, Springer (2018).
• J. Goold, M. Huber, A. Riera, L. del Rio, and P. Skrzypczyk, “The role of quantum information in thermodynamics: a topical review,” Journal of Physics A: Mathematical and Theoretical 49, 143001 (2016).
• K. V. Hovhannisyan, M. Perarnau-Llobet, M. Huber, and A. Acín, “Entanglement generation is not necessary for optimal work extraction,” Physical Review Letters 111, 240401 (2013).
• A. Lenard, “Thermodynamical proof of the Gibbs formula for elementary quantum systems,” Journal of Statistical Physics 19, 575-586 (1978).
• W. Pusz and S. L. Woronowicz, “Passive states and KMS states for general quantum systems,” Communications in Mathematical Physics 58, 273-290 (1978).