Passivity, Complete Passivity, and Ground States

Passivity expresses the second law as a statement about cyclic operations on a specified C*-dynamical system: no such operation can lower the mean energy of the state and deliver positive net work. One-copy passivity is weaker than equilibrium. Requiring the same property for every finite tensor power—complete passivity—selects precisely KMS states and ground states under the Pusz–Woronowicz hypotheses.

Required background. States, GNS representations, and folia supplies states and tensor-product representations; C*-dynamical systems and the KMS condition supplies the equilibrium boundary condition. Helpful background. Passivity, work, and information in QFT interprets localized work protocols, while complete passivity, KMS structure, and resource conversion develops the resource-theoretic consequences.

Cyclic work in a C*-dynamical system

Let $(\mathcal A,\alpha)$ have densely defined generator

$$\delta(A)=\lim_{t\to0}\frac{\alpha_t(A)-A}{t}.$$

For a unitary $U$ in the connected component of the identity and in the domain of $\delta$, define the work supplied to the system by

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

This sign agrees with $\alpha_t(A)=e^{itH}Ae^{-itH}$. In a type-I system,

$$-iU^*\delta(U)=U^*HU-H,$$

so $W_\omega(U)$ is the final minus initial energy for the cyclic unitary operation. A state is passive when $W_\omega(U)\geq0$ for every admissible cyclic $U$. Equivalently, one may use differentiable, time-dependent perturbations that vanish before and after a finite interval; the resulting cocycle endpoint gives the same class of tests.

Passivity immediately forces stationarity. If $\omega$ changed under $\alpha_t$, sufficiently small cyclic perturbations with opposite signs would produce a negative first-order work in one direction. The exact variational formulation and invariance theorem are Pusz and Woronowicz 1978, Definition 1.1 and Theorem 1.1, pp. 275–277.

Complete passivity and the equilibrium theorem

For $n$ independent copies, use

$$\mathcal A^{(n)}=\mathcal A^{\otimes n}, \qquad \alpha_t^{(n)}=\alpha_t^{\otimes n}, \qquad \omega^{(n)}=\omega^{\otimes n}.$$

The state $\omega$ is completely passive when every $\omega^{(n)}$ is passive for every finite $n$. The theorem is

A state of a C*-dynamical system is completely passive if and only if it is a ground state or a $\beta$-KMS state for some $\beta\geq0$, with the tracial case included at $\beta=0$.

The forward and reverse directions require different ideas. KMS and ground states satisfy the work inequality by their analytic or positive-energy structure Pusz and Woronowicz 1978, Theorem 1.2, pp. 277–279. For the converse, passivity constrains the joint energy–modular spectrum in the GNS representation. Tensor powers add those spectral sets. Complete passivity makes the union additive, forcing either a supporting line with finite slope $-\beta$—the KMS case—or a positive-energy half-line—the ground-state case. The spectral argument and conclusion are Pusz and Woronowicz 1978, Theorems 3.1 and 1.4 with Lemma 4.1, pp. 282–287.

One-copy passivity plus a phase assumption can also suffice: weak clustering under a suitable amenable symmetry makes the relevant spectral set semi-additive. That is a separate theorem, not permission to omit tensor powers in an arbitrary mixed state. Likewise, a convex mixture of passive states can be passive without being KMS at one temperature.

Free-scalar displacement test

Take the massive free-scalar Weyl system in its $\beta$-KMS state. For a real Cauchy datum $f$ in the domain of the one-particle energy $h$, the Weyl unitary $W(f)$ implements a coherent displacement. With the normalization

$$W(f)=e^{i\phi(f)},$$

its action on the second-quantized Hamiltonian has the form

$$W(f)^*d\Gamma(h)W(f)-d\Gamma(h) =\text{a field term}+\frac12\langle f,hf\rangle.$$

The centered quasifree KMS state annihilates the field term. Therefore

$$W_{\omega_\beta}(W(f))=\frac12\langle f,hf\rangle\geq0.$$

Compactly supported Cauchy data make this a localized operation; differentiability requires the stated $h$-domain. Products of independent thermal copies remain KMS for the product dynamics, so the same reasoning is compatible with complete passivity. This verifies one family of operations, while the general all-unitary conclusion comes from the theorem—not from extrapolating the coherent calculation. The physical interpretation and phase qualifications belong with infinite-volume KMS states, passivity, and phase multiplicity.

Adversarial test: passive but activatable

Consider a three-level Hamiltonian with energies $(0,1,2)$ and a diagonal state with populations

$$(p_0,p_1,p_2)=\left(\frac12,\frac13,\frac16\right).$$

The populations decrease with energy, so no one-copy unitary can lower the mean energy: the state is passive. It is not Gibbs, because

$$\frac{p_1}{p_0}=\frac23 \neq \frac{p_2}{p_1}=\frac12.$$

Five copies reveal the missing conclusion. Compare the product basis configurations with occupation counts

$$A:(n_0,n_1,n_2)=(0,5,0), \qquad B:(3,0,2).$$

They have $E_A=5>E_B=4$, but

$$p_A=\left(\frac13\right)^5 >\left(\frac12\right)^3\left(\frac16\right)^2=p_B.$$

A unitary swapping these two vectors lowers the average energy. Thus the state is not completely passive. The surviving claim is one-copy passivity; the missing hypothesis is the all-tensor-powers requirement.

Common pitfalls

Testing only Gaussian or local unitaries. Positivity on a convenient family is evidence for that family. Passivity quantifies over the full admissible connected unitary class of the stated algebra.

Equating passivity with a unique temperature. A passive state can fail to be KMS, and mixtures can obscure a phase temperature. Complete passivity or an appropriate clustering hypothesis supplies the extra rigidity.

Exercises

1. For a finite-dimensional Hamiltonian with nondegenerate increasing energies, show that a diagonal state is passive exactly when its populations are nonincreasing with energy.

Solution

If $E_i<E_j$ but $p_i<p_j$, swapping the two eigenvectors changes the energy by $(p_j-p_i)(E_i-E_j)<0$, so the state is not passive. Conversely, the rearrangement inequality says that pairing the largest population with the lowest energy minimizes $\sum_i p_iE_i$ among unitary rotations; coherences cannot do better because the diagonal of $U\rho U^*$ is doubly stochastic relative to the eigenvalues of $\rho$.

2. Why is a ground state completely passive even though it need not satisfy a finite-temperature KMS condition?

Solution

The product of ground states is a ground state for the sum dynamics, whose implementing generator is a sum of nonnegative generators. No cyclic unitary can lower its energy expectation below the ground value. Its analytic structure is a half-plane positive-energy condition, corresponding heuristically to $\beta=\infty$, not a finite KMS strip.

References

• Pusz, Wiesław, and Stanisław L. Woronowicz. “Passive States and KMS States for General Quantum Systems.” Communications in Mathematical Physics 58 (1978): 273–290. DOI.
• Haag, Rudolf, Daniel Kastler, and Ewa B. Trych-Pohlmeyer. “Stability and Equilibrium States.” Communications in Mathematical Physics 38 (1974): 173–193. DOI.