advanced28 min studyLesson 348 in Physics

Jarzynski equality in driven models

Recover an equilibrium free-energy difference from exponential work averaging, verify a discrete quench exactly, and distinguish the ensemble theorem from a finite-sample estimator.

On this page

What you’ll understand

  • State the preparation and work definition required for the Jarzynski equality.
  • Compute an exact exponential average in a two-state quench.
  • Explain why a finite work dataset can miss rare but strongly weighted trajectories.

Driving a system quickly usually requires more average work than the equilibrium free-energy change. The Jarzynski equality nevertheless recovers that equilibrium difference from a nonlinear average over nonequilibrium work. It does not require every trajectory to be reversible, or even require the final distribution to be equilibrium. It does require the correct initial ensemble, dynamics, and definition of work. A two-state quench makes every probability and exponential weight visible, avoiding the misleading impression that the equality is merely an approximate near-equilibrium formula.

Prepare equilibrium, then repeat one protocol

Use work W positive on the system and β=1/(kBT)\beta=1/(k_BT). The standard isothermal relation assumes initial canonical equilibrium at control value A, a fixed externally prescribed protocol ending at B, and thermal dynamics consistent with microscopic reversibility or an equivalent equilibrium-preserving framework. For the usual weak-coupling description, ΔF\Delta F refers to the system's equilibrium Helmholtz free energies at the same bath temperature. Strong coupling, feedback, or noncanonical preparation needs a separately justified extension rather than direct reuse of this formula.

Each realization begins with a new draw from the A equilibrium distribution and follows the same protocol timing. Inclusive work is the energy supplied by changing the Hamiltonian's control parameter along that trajectory. The average is over the resulting full ensemble of work values. The system may finish far from equilibrium; the final free energy in the equality still refers to the equilibrium state associated with the final Hamiltonian. Adding a control-dependent constant to all energy levels shifts both work and free-energy difference, leaving the equality consistent.

⟨e−βW⟩=e−βΔF,ΔF=−kBTln⁡ZBZA,W=∫∂λU(xt,λt) λ˙t dt\left\langle e^{-\beta W}\right\rangle=e^{-\beta\Delta F},\qquad \Delta F=-k_BT\ln\frac{Z_B}{Z_A},\qquad W=\int \partial_\lambda U(x_t,\lambda_t)\,\dot\lambda_t\,dt
The displayed work formula is for a classical potential-driven model; analogous Hamiltonian formulas include all relevant degrees of freedom. Beta has inverse-energy units, so the exponential arguments are dimensionless.↔ Scroll wide equations horizontally.

A sudden quench is exactly computable

During an instantaneous quench the microscopic state does not change, but its assigned energy can. The trajectory work equals the final energy minus the initial energy evaluated at that same state. Multiplying its exponential by the initial canonical probability cancels the initial energy factor. Summing over states then produces the final partition function divided by the initial one. This algebra proves the equality for the quench without any relaxation assumption afterward. It also shows why sampling initial states uniformly is valid only when their initial energies make the canonical distribution uniform.

The exponential average is not the exponential of average work. Convexity gives ⟨W⟩≥ΔF\langle W\rangle\ge\Delta F, with equality in an appropriate reversible limit, while individual work values can lie below the free-energy difference. Low-work trajectories receive disproportionately large exponential weights and are essential to the equality. The inequality does not mean every realization dissipates a positive amount of work. Stochastic violations of that trajectory-level claim are part of the mechanism by which the exact ensemble identity holds.

⟨e−βW⟩≥e−β⟨W⟩⟹⟨W⟩≥ΔF,ΔF^N=−kBTln⁡[1N∑r=1Ne−βWr]\left\langle e^{-\beta W}\right\rangle\ge e^{-\beta\langle W\rangle}\quad\Longrightarrow\quad\langle W\rangle\ge\Delta F,\qquad \widehat{\Delta F}_N=-k_BT\ln\left[\frac1N\sum_{r=1}^N e^{-\beta W_r}\right]
The inequality is Jensen's inequality. The estimator uses N work realizations; it is random at finite N and need not equal the exact equilibrium free-energy difference in a particular dataset.↔ Scroll wide equations horizontally.

Exact identities can have difficult estimators

For independent correctly sampled trajectories, the finite exponential sample mean is an unbiased estimator of the ensemble exponential average. Taking its logarithm introduces finite-sample bias in the free-energy estimator; under suitable integrability its expected estimate is no smaller than the true free-energy difference. A particular dataset can deviate either way. Rare low-work trajectories may dominate the exact average but be absent from a practical sample. Faster driving can make this overlap problem much worse even though the mathematical equality remains exact.

Useful analysis therefore reports preparation, work calibration, protocol repeatability, statistical dependence, and evidence for tail coverage. A narrow-looking work histogram alone cannot prove that an unobserved tail contributes negligibly to an exponential average. Reverse-process data can help, motivating the Crooks relation next. Instrument parameter changes that recompute a toy distribution are also not a physical driving experiment: a work protocol requires a time-ordered energy change experienced along each realized trajectory, with its history included in the work calculation.

Raise one of two initially degenerate levels

Two states initially have energies (0,0). An instantaneous quench changes them to (0,ϵ)(0,\epsilon) with ϵ=kBTln⁡3\epsilon=k_BT\ln3. Find the work distribution, exponential average, free-energy change, and average work.

  1. Initial canonical probabilities are one-half for each state. The state cannot jump during the instantaneous quench, so W equals zero with probability one-half and epsilon with probability one-half.

  2. The exponential average is 12(1+e−βϵ)=12(1+1/3)=2/3\tfrac12(1+e^{-\beta\epsilon})=\tfrac12(1+1/3)=2/3. Independently, ZA=2Z_A=2 and ZB=1+1/3=4/3Z_B=1+1/3=4/3, so their ratio is also two-thirds.

  3. Therefore ΔF=kBTln⁡(3/2)\Delta F=k_BT\ln(3/2). Average work is (kBT/2)ln⁡3(k_BT/2)\ln3, approximately 0.5493kBT0.5493k_BT, exceeding the free-energy change of approximately 0.4055kBT0.4055k_BT despite one branch having zero work.

The result

The exact exponential work average gives the equilibrium free-energy increase kBT ln(3/2), while the larger mean work reflects dissipative driving from the initial ensemble.

Explore the model

Interactive lab

Interactive lab · probability

Exact outcome probabilities are different from a finite dataset

Reset and set BOTH Success probability p = 0.1 and Trials n = 10; predict mean count 1 and probability (0.9)^10 of zero successes. For a fresh run, Reset and set p = 0.1 and n = 30; predict mean count 3 and a smaller zero-success probability (0.9)^30. Inspect exact bars or Step.

Exact outcome probabilities are different from a finite datasetThese 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 binomial instrument computes a distribution without resampling. It illustrates missing a rare outcome, but does not generate work trajectories, perform exponential reweighting, or estimate a free-energy difference.

Check your understanding

Which statement about the standard Jarzynski equality is correct?

Try it yourself

A two-state system starts with energies (0,ϵ)(0,\epsilon), where ϵ=kBTln⁡2\epsilon=k_BT\ln2. A sudden quench swaps them to (ϵ,0)(\epsilon,0). Find both work outcomes and probabilities, the exponential average, and mean work.

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

Give me a hint

The initial states are not equally probable. The partition function is unchanged by swapping the two levels.

Compare with the solution

The initial canonical probabilities are 2/32/3 for the zero-energy state and 1/31/3 for the raised state. The work outcomes are +ϵ+\epsilon and −ϵ-\epsilon with those probabilities. Thus ⟨e−βW⟩=(2/3)(1/2)+(1/3)(2)=1\langle e^{-\beta W}\rangle=(2/3)(1/2)+(1/3)(2)=1, giving ΔF=0\Delta F=0, consistent with unchanged partition function. Mean work is (2/3−1/3)ϵ=ϵ/3>0(2/3-1/3)\epsilon=\epsilon/3>0. The negative-work branch is less probable but receives the larger exponential weight.

Take it with you

  • Exponential work averaging recovers an equilibrium difference from a specified nonequilibrium ensemble.
  • Canonical preparation and the inclusive work definition are essential assumptions.
  • Rare low-work trajectories can make a finite-sample estimate difficult even when the theorem is exact.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.