Passivity, Work, and Information in QFT

Passivity is the precise statement that a state cannot power a cyclic process: after the externally controlled Hamiltonian has returned to its initial form, the system’s mean energy cannot be lower. The word “cyclic” is essential. A decrease of energy in one field region, detector, or mode is not by itself extracted work, because the switching source and the rest of the apparatus may have supplied more energy than that decrease. This page develops the finite-system formula, its algebraic QFT version, and the extra bookkeeping needed for localized field operations.

Required background. Thermal density operators and the KMS condition provide the equilibrium states; self-adjointness and unitary evolution provide the Hamiltonian domains; the first law of entanglement supplies the relative-entropy comparison used below; and stress tensors and charge algebras explain how field energy is defined.

Helpful background. Energy-constrained channel distances illustrate why an operation on an infinite-dimensional field needs an energy domain, not only an operator norm.

For the chapter-wide dictionary, see Enter this chapter; for the hypotheses attached to each claim, see the claim-validity summary; and for controls that invalidate tempting conclusions, see failure controls.

Cyclic work and passive states

Let a regulated system have a self-adjoint reference Hamiltonian $H_0$, bounded below, and an initial density operator $\rho_0$ with finite mean energy. During $0\le t\le T$, an external controller applies $V(t)$, so

$$H(t)=H_0+V(t), \qquad V(0)=V(T)=0,$$

and $\dot\rho_t=-i[H(t),\rho_t]$. We use the convention that positive $W_{\rm on}$ is work supplied by the controller. Differentiating the instantaneous mean energy gives

$$\frac{d}{dt}\operatorname{tr}(\rho_tH(t)) =\operatorname{tr}(\rho_t\dot H(t)),$$

because the commutator term has zero trace. Integration and cyclicity of the Hamiltonian therefore give

$$W_{\rm on} =\int_0^Tdt\,\operatorname{tr}(\rho_t\dot H(t)) =\operatorname{tr}(\rho_TH_0)-\operatorname{tr}(\rho_0H_0), \qquad W_{\rm ext}=-W_{\rm on}.$$

A state is passive for $H_0$ and a declared class of cyclic controls if $W_{\rm on}\ge0$ for every admissible control. In finite dimension, if all unitaries are reachable, this is equivalent to

$$\operatorname{tr}(U\rho_0U^\dagger H_0) \ge \operatorname{tr}(\rho_0H_0) \quad\text{for every unitary }U.$$

Ordering the energy eigenvalues as $E_0\le E_1\le\cdots$, passivity requires $[\rho_0,H_0]=0$ and populations that do not increase with energy: $p_0\ge p_1\ge\cdots$. Coherences within a degenerate eigenspace are harmless because they do not change energy. A population inversion lets one swap a more populated high-energy vector with a less populated low-energy vector and lower the mean energy. This finite-system characterization and its relation to cyclic processes are established by Lenard 1978, pp. 576–580 and Pusz and Woronowicz 1978, Definition 1.1 and Theorem 1.2, pp. 275–277.

The largest decrease on a unitary orbit is often called ergotropy,

$$\mathcal W(\rho,H_0) =\operatorname{tr}(\rho H_0) -\min_U\operatorname{tr}(U\rho U^\dagger H_0).$$

It is an energy difference under an idealized unitary operation, not automatically the net useful work of a physical engine. The distinction between this kinematic optimum and an implemented work source is explicit in Allahverdyan, Balian, and Nieuwenhuizen 2004, Eqs. (5)–(8), pp. 566–568.

Why a Gibbs state is passive

For $\rho_\beta=e^{-\beta H_0}/Z$ with $\beta>0$, unitary invariance of entropy gives a short information-theoretic derivation. Set $\rho_U=U\rho_\beta U^\dagger$. Since $\log\rho_\beta=-\beta H_0-\log Z$,

$$\begin{aligned} D(\rho_U\Vert\rho_\beta) &=\operatorname{tr}\!\left[\rho_U(\log\rho_U-\log\rho_\beta)\right]\\ &=-S(\rho_\beta)+\beta\operatorname{tr}(\rho_UH_0)+\log Z\\ &=\beta\!\left[ \operatorname{tr}(\rho_UH_0)-\operatorname{tr}(\rho_\beta H_0) \right]. \end{aligned}$$

Positivity of relative entropy yields $W_{\rm on}=\beta^{-1}D(\rho_U\Vert\rho_\beta)\ge0$. Equality holds precisely when $\rho_U=\rho_\beta$ on its support. This proof assumes a trace-class Gibbs density operator. A thermal state of an infinite field volume is generally represented instead as a KMS state on an algebra, so the finite-dimensional trace formula is then a regulator or local approximation, not the definition.

In a $C^*$-dynamical system $(\mathcal A,\alpha_t)$, let $\delta$ be the generator, $\delta(A)=\lim_{t\to0}(\alpha_t(A)-A)/t$, on its dense domain. For a differentiable cyclic unitary $U$ connected to the identity, the algebraic work supplied is

$$W_{\rm on}(\omega,U)=-i\,\omega\!\left(U^*\delta(U)\right).$$

In a Hamiltonian representation with $\delta(A)=i[H_0,A]$, this reduces to $\omega(U^*H_0U-H_0)$. Algebraic passivity demands $W_{\rm on}(\omega,U)\ge0$ for the admissible differentiable unitaries. This formulation does not require $H_0$ or a thermal density matrix to belong to the observable algebra.

Reproducible two-mode Gaussian benchmark

Take two regulated bosonic field modes, set $\hbar=1$, and subtract the state-independent zero-point energy:

$$H_0=\omega_1a_1^\dagger a_1+\omega_2a_2^\dagger a_2, \qquad \rho_\beta=\rho_{\beta,1}\otimes\rho_{\beta,2}, \qquad \bar n_j=\frac{1}{e^{\beta\omega_j}-1}.$$

Apply the cyclic Gaussian unitary $U=\bigotimes_jD_j(\alpha_j)S_j(r_j)$, with real squeezing for simplicity. Using

$$S_j^\dagger a_jS_j=a_j\cosh r_j-a_j^\dagger\sinh r_j, \qquad D_j^\dagger a_jD_j=a_j+\alpha_j,$$

and the vanishing first and anomalous moments of a thermal state gives

$$\Delta E =\sum_{j=1}^2\omega_j \left[|\alpha_j|^2+(2\bar n_j+1)\sinh^2r_j\right] \ge0.$$

This is a useful line-by-line test of signs and normalizations: displacement and squeezing both cost energy when the input is thermal. Choose the dimensionless values

$$\beta=1, \quad (\omega_1,\omega_2)=(1,2), \quad (\alpha_1,\alpha_2)=(0.30,-0.20), \quad (r_1,r_2)=(0.20,0.10).$$

Then $\bar n_1=0.5819767$ and $\bar n_2=0.1565176$. The two contributions are $0.1777184$ and $0.1063484$, so

$$E_i=0.8950120, \qquad E_f=1.1790788, \qquad W_{\rm on}=0.2840668, \qquad W_{\rm ext}=-0.2840668.$$

Recompute these numbers by evaluating the displayed formula directly; retaining ten digits in the Bose occupations reproduces the quoted result to better than $10^{-7}$. The calculation tests one experimentally natural class of controls. The all-unitary conclusion comes from Gibbs passivity, not from extrapolating this Gaussian example. Restricted Gaussian passivity can be strictly weaker than ordinary passivity, as characterized by Brown, Friis, and Huber 2016, §§2–4.

Local operations and the full energy account

A typical field control has the form

$$V(t)=\chi(t)\int d^{d-1}\mathbf x\, f(\mathbf x)\mathcal O(t,\mathbf x),$$

with smooth switching $\chi$, spatial smearing $f$, and a renormalized field observable $\mathcal O$. Compact spacetime support localizes the interaction algebraically, but it does not make the global Hamiltonian bounded or guarantee that every vector has finite post-operation energy. A defensible operation class must preserve the relevant energy form domain, or else define work through a controlled approximation whose limit exists.

The controller’s contribution is not optional. If the field loses energy while a detector gains energy and the source performs switching work, conservation applies to the closed field–detector–controller model, not to the field alone. A local negative $\Delta E_F$ can therefore coexist with $W_{\rm on}>0$. Passivity also depends on the operation class: global passivity implies passivity under a subset of local controls, while failure to extract work locally does not prove global passivity.

Two adversarial controls are especially useful.

Shift the energy zero consistently. Replacing $H_0$ by $H_0+c\mathbf1$ leaves $\operatorname{tr}[(\rho_T-\rho_0)H_0]$ unchanged because both states have unit trace. The benchmark remains $0.2840668$ even for $c=10^6$. A changed answer signals that initial and final energies used different references, or that the alleged cycle did not return to the same Hamiltonian.

Remove the switching account. If one reports only $-\Delta E_F$ while omitting $\int dt\,\langle\partial_tV(t)\rangle$, one has computed a subsystem energy transfer, not net extracted work. Restore the detector, source, and interaction terms and check the first-law balance before invoking passivity.

Common pitfalls

Calling a state passive after testing one pulse. Passivity quantifies over a declared class of cyclic operations. A null result for Gaussian, local, or perturbative controls establishes only passivity relative to that restricted class.

Using arbitrary unitaries in QFT without a domain. A bounded unitary can carry a finite-energy vector outside the form domain of an unbounded Hamiltonian. The algebraic differentiability condition or an explicit energy cutoff is part of the claim.

Equating local energy decrease with useful work. Work belongs to a protocol with a controller and reference Hamiltonian. Stress-energy redistribution alone does not define a work repository.

Exercises

1. A three-level system has energies $(0,\epsilon,2\epsilon)$ and populations $(0.4,0.2,0.4)$. Find a cyclic permutation that extracts work.

Solution

Swap the populations of levels one and two. Initially $E_i=0.2\epsilon+0.8\epsilon=\epsilon$. Afterwards the populations are $(0.4,0.4,0.2)$ and $E_f=0.4\epsilon+0.4\epsilon=0.8\epsilon$. Thus $W_{\rm ext}=E_i-E_f=0.2\epsilon>0$, so the original state was not passive.

2. Show directly that adding $c\mathbf1$ to $H_0$ cannot change the ergotropy.

Solution

For every unitary $U$,

$$\operatorname{tr}(\rho(H_0+c\mathbf1)) -\operatorname{tr}(U\rho U^\dagger(H_0+c\mathbf1)) =\operatorname{tr}(\rho H_0)-\operatorname{tr}(U\rho U^\dagger H_0),$$

because both density operators have trace one. Taking the maximum over $U$ preserves the equality.

3. A thermal mode with $\beta\omega=\log 3$ is squeezed by $r=\operatorname{arsinh}(1/2)$ and not displaced. Find the supplied work in units of $\omega$.

Solution

$\bar n=(3-1)^{-1}=1/2$ and $\sinh^2r=1/4$. Hence

$$\frac{W_{\rm on}}{\omega}=(2\bar n+1)\sinh^2r =2\times\frac14=\frac12.$$

The extracted-work convention gives $W_{\rm ext}/\omega=-1/2$.

4. Why does passivity under all global cyclic unitaries imply passivity under localized cyclic unitaries, while the converse need not hold?

Solution

Localized admissible unitaries form a subset of the global admissible class, so a nonnegative work inequality on the larger set holds on the subset. The converse is a quantifier error: correlations or population inversions may be accessible only to a joint, nonlocal operation, even though every operation confined to one region fails to lower the energy.

References

• Allahverdyan, A. E., R. Balian, and Th. M. Nieuwenhuizen. “Maximal Work Extraction from Finite Quantum Systems.” Europhysics Letters 67, no. 4 (2004): 565–571. DOI.
• Brown, Eric G., Nicolai Friis, and Marcus Huber. “Passivity and Practical Work Extraction Using Gaussian Operations.” New Journal of Physics 18 (2016): 113028. DOI.
• Lenard, Andrew. “Thermodynamical Proof of the Gibbs Formula for Elementary Quantum Systems.” Journal of Statistical Physics 19, no. 6 (1978): 575–586. DOI.
• Pusz, W., and S. L. Woronowicz. “Passive States and KMS States for General Quantum Systems.” Communications in Mathematical Physics 58, no. 3 (1978): 273–290. DOI.