Passivity, Work, and Information in QFT

Passivity asks whether an allowed cyclic operation can lower a state’s energy. It is a statement about a state, a dynamics, and an operation class together. In QFT the Hamiltonian is unbounded as an operator, local operations need not preserve its domain, and a detector’s switching energy is not automatically included in the field-energy change.

Required background. The KMS condition supplies equilibrium dynamics, self-adjoint evolution supplies the Hamiltonian domain, modular first-law response separates linear response from a finite work inequality, and the stress tensor and charges supply the energy observable.

Helpful background. Energy-constrained channel distances explain why an energy domain must accompany an infinite-dimensional operation.

Cyclic work extraction

Let $H$ generate the reference time evolution and choose its additive constant once. A cyclic perturbation returns the external control parameters to their initial values and implements an admissible unitary $U$. When both energy expectations exist,

$$W_{\rm ext}(U,\rho) =\operatorname{tr}(\rho H)-\operatorname{tr}(U\rho U^\dagger H).$$

The state is passive for the allowed class $\mathcal U$ when $W_{\rm ext}\le0$ for every $U\in\mathcal U$. This definition does not say that $\rho$ minimizes energy absolutely. It forbids a cyclic rearrangement that moves population toward lower energies. A ground state is passive, and a Gibbs state is passive; some nonthermal states are passive for one copy but fail under collective operations on several copies.

In an algebraic QFT formulation, one can define work from a time-dependent perturbation and the derivation generating the dynamics. The classic characterization by Pusz and Woronowicz 1978, Theorems 1.1 and 1.4 makes the operation class and equilibrium structure explicit. Local passivity is weaker than global passivity: forbidding energy extraction by operations in one bounded region need not forbid extraction by a nonlocal cyclic unitary.

A finite-mode check

Take a regulated mode with $H=\omega a^\dagger a$ and a diagonal state $\rho=\sum_n p_n|n\rangle\langle n|$. It is passive exactly when the populations do not increase with energy: $p_n\ge p_{n+1}$. A unitary that swaps $|n\rangle$ and $|m\rangle$, $m>n$, changes the energy by

$$\Delta E=(E_m-E_n)(p_n-p_m).$$

If $p_m>p_n$, the swap lowers the energy and extracts work. A thermal sequence $p_n\propto e^{-\beta\omega n}$ passes every such test. For a finite collection of modes, apply a cyclic Gaussian unitary and verify the same inequality from the covariance matrix, while retaining the energy cutoff used to make every trace finite.

Passivity occupies only the cyclic-dynamics branch. It supplies a resource baseline for localized protocols but does not itself include sampling bounds, entropy variations, or apparatus costs. The diagram is schematic.

Field energy is not net useful work

For a localized detector coupling, a decrease $\Delta E_F<0$ in the field does not by itself imply positive useful work. A complete account includes the detector, switching source, controller, stored classical record, and any reference system. With total Hamiltonian $H_{\rm tot}(t)$,

$$W_{\rm drive}=\int dt\,\operatorname{tr}\!\left[\rho(t)\,\partial_t H_{\rm tot}(t)\right]$$

must be reconciled with subsystem energy changes. Changing the additive zero of a time-dependent subsystem Hamiltonian can also change a naive “work” number. Fix the reference and report all terms.

A passivity claim fails operationally if the admissible domain or energy reference changes, or if switching and control work are omitted. Those checks are independent of whether the system-energy decrease is calculated correctly. The map is schematic.

Common pitfalls

Treating every energy decrease as work. A subsystem may lose energy while the external controller supplies more. Define the cyclic protocol and the work repository before assigning a sign.

Ignoring domains. An abstract unitary can move a finite-energy vector outside the form domain of $H$. Restrict the operation class so both energy expectations are meaningful.

Exercises

For a three-level Hamiltonian with energies $0,\epsilon,2\epsilon$ and probabilities $(0.4,0.2,0.4)$, find an energy-lowering cyclic permutation.

Solution

Swap levels one and two. The initial mean energy is $\epsilon$; the final probabilities are $(0.4,0.4,0.2)$ and the mean is $0.8\epsilon$, so $0.2\epsilon$ is extracted.

References

• Pusz, Władysław, and Stanisław L. Woronowicz. “Passive States and KMS States for General Quantum Systems.” Communications in Mathematical Physics 58 (1978): 273–290. DOI.