advanced33 min studyLesson 40 in Physics

Nonequilibrium work and fluctuation relations

Derive the logic of exponential work averages, test Jarzynski's equality in an exact quench, and expose rare-event sampling limits.

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What you’ll understand

  • Distinguish work on a driven realization from equilibrium free-energy differences.
  • Verify Jarzynski's equality and the mean-work bound in a finite two-state quench.
  • State the preparation, reverse-protocol, and sampling conditions behind fluctuation relations.

A rapid drive need not keep a system near equilibrium, so its work can vary substantially from one repetition to another. Remarkably, a particular exponential average can still recover the equilibrium free-energy difference between the endpoint Hamiltonians. This does not mean that ordinary average work equals reversible work, nor that a few trajectories determine free energy reliably. We will examine an exactly soluble sudden quench where all work outcomes are visible, including negative-work realizations that remain compatible with the second law.

Define a protocol and its initial ensemble

Consider a prescribed time-dependent control parameter taking a Hamiltonian from an initial form to a final form. Work WW is defined as work done on the system, and the process starts from a canonical equilibrium ensemble at positive temperature T>0T>0 and inverse temperature β=1/(kBT)>0\beta=1/(k_BT)>0. The free-energy difference ΔF\Delta F compares the equilibrium ensembles of the endpoint Hamiltonians at that same temperature, even if the actual final distribution has not equilibrated. Mechanical driving, thermal exchange, and work measurement must be defined consistently.

Under the usual microreversible classical Hamiltonian or suitable heat-bath dynamics assumptions, Jarzynski's equality relates the exponential work average to that free-energy difference. A quantum formulation needs a specified work-measurement protocol, commonly two projective energy measurements, rather than treating work as an arbitrary Hermitian observable. Feedback, postselection, noncanonical initial conditions, and strong coupling can require modified relations. The equality is powerful because it allows far-from-equilibrium driving under its assumptions, not because it removes those assumptions.

⟨e−βW⟩F=e−βΔF,ΔF=Ffinal−Finitial,⟨W⟩F≥ΔF\left\langle e^{-\beta W}\right\rangle_F=e^{-\beta\Delta F},\qquad \Delta F=F_{\mathrm{final}}-F_{\mathrm{initial}},\qquad \langle W\rangle_F\geq\Delta F
The average is over repetitions of one prescribed forward protocol from its canonical initial preparation. The mean-work inequality follows from convexity through Jensen's inequality.↔ Scroll wide equations horizontally.

Reverse trajectories compare probabilities

Crooks's work relation compares a forward protocol with the appropriately time-reversed protocol, including reversal of odd variables and external fields when required. The reverse process starts in equilibrium for the forward final Hamiltonian. The ratio of forward probability for work WW to reverse probability for work −W-W is an exponential of the dissipated work. This is a statement about paired ensemble probabilities, not a claim that one observed path has reversed spontaneously with equal likelihood.

Integrating the probability-ratio relation gives Jarzynski's equality when the distributions and supports satisfy the required conditions. The forward and reverse work distributions cross at W=ΔFW=\Delta F where a well-defined common density or matching discrete masses exist. Rare negative dissipated-work outcomes are allowed; their relative probabilities are constrained. A small system can temporarily deliver more work than the free-energy drop without violating the average bound, because the distribution, rather than every realization separately, carries the thermodynamic statement.

PF(W)PR(−W)=eβ(W−ΔF),Wdiss=W−ΔF\frac{P_F(W)}{P_R(-W)}=e^{\beta(W-\Delta F)},\qquad W_{\mathrm{diss}}=W-\Delta F
PF and PR are corresponding work probability densities, or matching probability masses for a discrete model, for correctly paired forward and reverse protocols. Time-reversal conditions must be respected.↔ Scroll wide equations horizontally.

An exact identity can be a difficult estimator

The factor e−βWe^{-\beta W} gives unusually large weight to low-work realizations. Under strongly dissipative driving those realizations may be very rare, so a finite sample can underestimate the exponential mean and bias an inferred free energy. Increasing the number of typical high-work trajectories does not guarantee that the influential rare tail has been sampled. Forward–reverse methods, slower protocols, repeated uncertainty analysis, and overlap checks can be more informative than one apparently converged running average.

The endpoint free-energy difference is an equilibrium state function, while the work distribution depends on the driving protocol. Different protocols can therefore share the same exact exponential average and have very different mean dissipation and sampling difficulty. Finite systems are especially convenient for checking this distinction because every state can sometimes be enumerated. Macroscopic extensivity usually makes dissipation and rare-event costs grow with size, so an equality that remains formally valid can become impractical to estimate by naive repeated driving.

Instantaneously swap two energy labels

A two-state system starts in equilibrium with energies Ea=0E_a=0 and Eb=ϵE_b=\epsilon, where βϵ=ln⁡2\beta\epsilon=\ln2. Suddenly swap the two energies without changing the occupied label during the quench. Compute work outcomes, the endpoint free-energy change, and the exponential average.

  1. Initially pa=2/3p_a=2/3 and pb=1/3p_b=1/3. The sudden change gives work +ϵ+\epsilon for label a and −ϵ-\epsilon for label b. The partition sums before and after are both 1+1/2=3/21+1/2=3/2, so ΔF=0\Delta F=0.

  2. The exponential average is (2/3)e−βϵ+(1/3)eβϵ=(2/3)(1/2)+(1/3)(2)=1(2/3)e^{-\beta\epsilon}+(1/3)e^{\beta\epsilon}=(2/3)(1/2)+(1/3)(2)=1, exactly e−βΔFe^{-\beta\Delta F}.

  3. The ordinary mean work is (2/3)ϵ−(1/3)ϵ=ϵ/3=(kBTln⁡2)/3≈0.231049kBT>0(2/3)\epsilon-(1/3)\epsilon=\epsilon/3=(k_BT\ln2)/3\approx0.231049k_BT>0. One third of runs yield negative work, yet the average obeys the second-law bound.

The result

A nonzero mean dissipation and negative-work realizations coexist with an exact zero free-energy change and an exponential average equal to one.

Explore the model

Interactive lab

Interactive lab · probability

Count the work-extracting realizations

Set p near 0.33 as an approximation to the one-third probability of negative work in the sudden-swap example, and let n count independent repetitions. The binomial bars show how uncertain the count of those realizations remains in a finite sample.

Count the work-extracting realizationsThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe lab counts Bernoulli outcomes only; it does not calculate Jarzynski averages or generate physical work trajectories. The slider cannot set one third exactly. Changing p independently no longer describes the exact fixed quench unless preparation and gap are changed consistently.

Check your understanding

Why does observing W < ΔF in one realization not by itself violate the mean-work bound?

Try it yourself

For the sudden-swap example, imagine an unlucky three-run sample in which every run starts in the lower initial-energy label. What free-energy estimate follows from the sample exponential average? Compare it with the exact value and state the lesson.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Every sampled work value is +ε, so average its exponential first and then apply −kBT times the logarithm.

Compare with the solution

The sample exponential average is e−βϵ=1/2e^{-\beta\epsilon}=1/2, giving ΔF^=−kBTln⁡(1/2)=ϵ\widehat{\Delta F}=-k_BT\ln(1/2)=\epsilon. The exact value is zero. This sample occurs with probability (2/3)3=8/27(2/3)^3=8/27 and misses all negative-work runs, which carry large exponential weight. It demonstrates finite-sample error; it does not contradict the exact ensemble identity or justify replacing an uncertainty analysis with a single small sample.

Take it with you

  • Work depends on a protocol and realization, while endpoint free energy is an equilibrium state function.
  • Fluctuation relations constrain distributions under explicit preparation and reversibility assumptions.
  • Exponential work averages can be dominated by rare outcomes and be difficult to estimate reliably.

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