advanced29 min studyLesson 349 in Physics

Crooks relation and forward–reverse protocols

Compare correctly paired forward and reverse work distributions, recover free-energy differences from their ratios, and distinguish protocol reversal from simply replaying final data.

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What you’ll understand

  • Specify the preparation and control history of a reverse experiment.
  • Apply the work-distribution ratio with its required minus sign.
  • Use crossings and Gaussian examples without assuming universal work statistics.

The Jarzynski equality uses one driven ensemble. The Crooks relation compares that ensemble with a separately prepared reverse experiment and reveals where dissipative work enters their probability ratio. The comparison is powerful, but the reverse experiment is not obtained merely by reading a forward work list backward. It starts from equilibrium for the final Hamiltonian and reverses the prescribed controls. We will verify the ratio exactly in a two-state quench and then examine a compatible Gaussian example.

Reverse the protocol and reprepare the ensemble

Let the forward protocol take the control from A to B over a specified duration. Its reverse follows the reversed control schedule from B to A with the same duration. Each ensemble starts in canonical equilibrium for its own initial Hamiltonian at the same temperature. For underdamped systems, time reversal changes momenta as well; time-odd external parameters must be transformed consistently. The needed microscopic-reversibility condition relates probabilities of paired paths under these two experiments. It is stronger than observing a stationary histogram in either experiment.

Work is positive on the system in each direction. If a forward path has work W, its appropriately reversed partner has reverse work minus W. Consequently the denominator in the Crooks work ratio is PR(−W)P_R(-W), not PR(W)P_R(W). These quantities are probability densities for continuous work or probability masses for discrete work, used consistently. The free-energy change is always FB−FAF_B-F_A in the displayed forward convention, whereas the reverse experiment has the opposite equilibrium free-energy change.

PF(W)PR(−W)=eβ(W−ΔF),β=1kBT,ΔF=FB−FA\frac{P_F(W)}{P_R(-W)}=e^{\beta(W-\Delta F)},\qquad \beta=\frac1{k_BT},\qquad \Delta F=F_B-F_A
The standard relation assumes canonical preparation at each protocol's start, a common equilibrium bath, and microscopic reversibility with the correct time-reversal parity. It applies on common support and does not assume Gaussian work or slow driving.↔ Scroll wide equations horizontally.

The ratio exposes dissipative bias

A forward work value above the free-energy difference is more probable than its sign-reversed reverse counterpart by the stated exponential factor. At W equal to the free-energy difference the two probability curves coincide, provided they have overlapping support there. The crossing locates the equilibrium difference without equating the forward and reverse mean works. Integrating the Crooks ratio after rearrangement also gives the Jarzynski equality by normalization of the reverse distribution. This is a consistency check linking the two results, not a replacement for verifying their experimental assumptions.

For compatible Gaussian work distributions with equal variance, the ratio's logarithm is linear in work. Matching its slope and intercept imposes relations among the two means, variance, and free-energy difference. Gaussianity is useful in linear driven models or as an empirical approximation, but it is not required by the theorem. Strongly non-Gaussian distributions may satisfy the exact ratio. Conversely, visually smooth Gaussian-looking histograms can violate it if their preparation, work calibration, or assigned reverse protocol is inconsistent.

ln⁡PF(W)PR(−W)=βW−βΔF,PF(ΔF)=PR(−ΔF) when both densities exist there\ln\frac{P_F(W)}{P_R(-W)}=\beta W-\beta\Delta F,\qquad P_F(\Delta F)=P_R(-\Delta F)\ \text{when both densities exist there}
The log-ratio slope is beta and the intercept is −beta Delta F. A crossing is informative only where both distributions have support and sufficient statistical overlap; a discrete distribution need not contain a work value exactly at Delta F.↔ Scroll wide equations horizontally.

A reverse sample is not a forward final-state sample

A finite-time forward drive typically ends in a nonequilibrium distribution. Starting the reverse schedule immediately from that unrelaxed distribution generally does not meet the canonical initial condition of the standard Crooks work theorem. One must independently equilibrate at B or justify a more general relation with the required distribution boundary term. This difference is easy to miss because the control value B is correct in both preparations, while the state probabilities can be different. Control settings alone do not specify a statistical ensemble.

In measurements, forward and reverse distributions should be compared using matched work conventions and comparable density estimates. Unequal sample counts do not directly equal probability-density ratios without normalization. Sparse tails or disjoint practical samples make crossing estimates unreliable, even though the ideal ensemble relation remains well defined. Protocol changes in the interactive labs recompute their toy model at the retained clock value; they do not create a time-reversed thermodynamic protocol. Resetting and setting controls can compare fresh mathematical distributions, but no work history is thereby accumulated.

Pair the forward and reverse two-level quenches

Forward driving changes energies (0,0) to (0,ϵ)(0,\epsilon) with ϵ=kBTln⁡3\epsilon=k_BT\ln3. The reverse begins in canonical equilibrium for (0,ϵ)(0,\epsilon) and quenches back. Compute the work ratios at W = 0 and W = epsilon and use one ratio to recover Delta F.

  1. The forward work masses are PF(0)=PF(ϵ)=1/2P_F(0)=P_F(\epsilon)=1/2. At the reverse start, canonical weights are 3/43/4 and 1/41/4, so PR(0)=3/4P_R(0)=3/4 and PR(−ϵ)=1/4P_R(-\epsilon)=1/4.

  2. The ratios are PF(0)/PR(0)=2/3P_F(0)/P_R(0)=2/3 and PF(ϵ)/PR(−ϵ)=2P_F(\epsilon)/P_R(-\epsilon)=2. The second denominator uses negative reverse work, matching the paired path rather than an unrelated reverse outcome.

  3. At W = epsilon the theorem gives ΔF=ϵ−kBTln⁡2=kBTln⁡(3/2)\Delta F=\epsilon-k_BT\ln2=k_BT\ln(3/2). At W = 0 it gives e−βΔF=2/3e^{-\beta\Delta F}=2/3, independently agreeing. There is no discrete outcome exactly at Delta F, so a literal sampled crossing is unnecessary.

The result

Both exact forward–reverse ratios recover the same equilibrium increase kBT ln(3/2), while their unequal initial probabilities demonstrate why the reverse ensemble must be separately prepared.

Explore the model

Interactive lab

Interactive lab · probability

Compare paired outcomes using the correct reflection

Reset and set BOTH Success probability p = 0.25 and Trials n = 8. For a fresh run, Reset and set p = 0.75 and n = 8. Predict reflected exact histograms: the first probability at count k equals the second at count 8 − k. Inspect bars or Step to compare paired outcomes.

Compare paired outcomes using the correct reflectionThese 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 reflection illustrates careful pairing and normalization. It does not implement a reverse work protocol, equilibrium preparation, heat exchange, or the Crooks ratio.

Check your understanding

Which preparation is required for the standard reverse experiment in the Crooks work relation?

Try it yourself

Let e=kBTe=k_BT. A compatible illustrative pair has Gaussian forward work with mean 3e, reverse work with mean e, and common variance 4e24e^2. Derive the log ratio PF(W)/PR(−W)P_F(W)/P_R(-W), find Delta F, and locate its crossing.

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 reflected reverse density has mean minus e. Subtract the two Gaussian exponents; their normalization factors cancel.

Compare with the solution

Writing w=W/ew=W/e, the log ratio is [−(w−3)2+(w+1)2]/8=w−1[-(w-3)^2+(w+1)^2]/8=w-1. Matching the Crooks form gives βΔF=1\beta\Delta F=1, hence ΔF=e\Delta F=e. The distributions cross at w=1w=1. As an independent check, the Gaussian Jarzynski formula gives ΔF=⟨W⟩F−βVar⁡(W)/2=3e−2e=e\Delta F=\langle W\rangle_F-\beta\operatorname{Var}(W)/2=3e-2e=e. These parameter relations make this pair compatible; arbitrary independently chosen Gaussian means and variances need not satisfy the theorem.

Take it with you

  • The reverse protocol requires its own canonical initial ensemble and transformed controls.
  • Pair forward W with reverse minus W when evaluating the work-distribution ratio.
  • Distribution overlap and normalization matter when extracting a free-energy crossing from data.

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