Jarzynski Equality and Crooks Relation

The Jarzynski equality and Crooks relation are nonequilibrium work identities. They compare a driven process to equilibrium free-energy differences even when the driving is fast and irreversible. In quantum mechanics, their standard closed-system form uses the two-point measurement protocol for work.

This page gives the focused finite-dimensional derivation. It assumes an initially thermal state, unitary driven evolution, ideal projective energy measurements, with a reverse process satisfying microscopic reversibility additionally required for the Crooks relation. For the broader family of fluctuation relations, including open trajectories and feedback variants, see Fluctuation Theorems.

Thermal Initial State and Driven Protocol

Let the forward protocol drive a closed quantum system from $H_0$ to $H_\tau$ with unitary $U_F$. Write

$$H_0 = \sum_n E_n^0\Pi_n^0, \qquad H_\tau = \sum_m E_m^\tau\Pi_m^\tau.$$

The initial state is Gibbs at inverse temperature $\beta$:

$$\rho_0 = \frac{e^{-\beta H_0}}{Z_0}, \qquad Z_0=\operatorname{Tr}(e^{-\beta H_0}).$$

The final equilibrium partition function for the same temperature is

$$Z_\tau = \operatorname{Tr}(e^{-\beta H_\tau}),$$

and the equilibrium free-energy difference is

$$\Delta F = F_\tau-F_0 = - \frac{1}{\beta} \ln \frac{Z_\tau}{Z_0}.$$

The two-point measurement work value for outcomes $n$ then $m$ is

$$W_{mn} = E_m^\tau-E_n^0.$$

Because $\rho_0$ commutes with $H_0$, the first energy measurement preserves the unread density operator. The conditional state for a recorded outcome is the corresponding normalized energy block.

Forward Work Distribution

For nondegenerate notation, the forward probability of the branch $n\to m$ is

$$p_F(m,n) = \frac{e^{-\beta E_n^0}}{Z_0} \left| \langle m^\tau|U_F|n^0\rangle \right|^2.$$

For possibly degenerate energy levels, the joint probability is instead

$$p_F(m,n)=\frac{e^{-\beta E_n^0}}{Z_0} \operatorname{Tr}(\Pi_m^\tau U_F\Pi_n^0U_F^\dagger).$$

The trace includes every state in the initial energy block; it is not itself a normalized conditional probability. This is the Gibbs-state specialization of the Two-Point Measurement Scheme. The work distribution is

$$P_F(W) = \sum_{m,n} p_F(m,n) \delta\!\left( W-E_m^\tau+E_n^0 \right).$$

The average work is generally larger than the equilibrium free-energy difference, but the exact statement is stronger than an average inequality.

Jarzynski Equality

Compute the exponential work average:

$$\left\langle e^{-\beta W}\right\rangle_F = \sum_{m,n} p_F(m,n) e^{-\beta(E_m^\tau-E_n^0)}.$$

Substituting the branch probability gives

$$\left\langle e^{-\beta W}\right\rangle_F = \frac{1}{Z_0} \sum_{m,n} e^{-\beta E_m^\tau} \left| \langle m^\tau|U_F|n^0\rangle \right|^2.$$

For each final state $m$, unitarity gives

$$\sum_n \left| \langle m^\tau|U_F|n^0\rangle \right|^2 = 1.$$

Therefore

$$\left\langle e^{-\beta W}\right\rangle_F = \frac{1}{Z_0} \sum_m e^{-\beta E_m^\tau} = \frac{Z_\tau}{Z_0} = e^{-\beta\Delta F}.$$

This is the Jarzynski equality:

$$\left\langle e^{-\beta W}\right\rangle_F = e^{-\beta\Delta F}.$$

It requires neither slow driving nor a reverse experiment. Degeneracy does not change the proof: replace the squared matrix element by $\operatorname{Tr}(\Pi_m^\tau U_F\Pi_n^0U_F^\dagger)$. Summing over $n$ gives $\operatorname{Tr}\Pi_m^\tau$, the final degeneracy, so the remaining sum is exactly $Z_\tau$.

Average Work Bound

Jensen’s inequality gives

$$e^{-\beta\langle W\rangle_F} \le \left\langle e^{-\beta W}\right\rangle_F.$$

Using the Jarzynski equality,

$$e^{-\beta\langle W\rangle_F} \le e^{-\beta\Delta F}.$$

For positive temperature this implies

$$\langle W\rangle_F \ge \Delta F.$$

Define dissipated work by

$$W_{\mathrm{diss}} = W-\Delta F.$$

Then

$$\left\langle e^{-\beta W_{\mathrm{diss}}} \right\rangle_F = 1, \qquad \langle W_{\mathrm{diss}}\rangle_F \ge 0.$$

The equality constrains the whole distribution. The inequality is only its average consequence.

Reverse Protocol

Let $\Theta$ be the antiunitary time-reversal operator. The physical reverse drive is

$$H_R(t)=\Theta H_F(\tau-t)\Theta^\dagger, \qquad U_R=\Theta U_F^\dagger\Theta^\dagger.$$

It starts in $\Theta e^{-\beta H_\tau}\Theta^\dagger/Z_\tau$ and pairs $|n^0\rangle\to|m^\tau\rangle$ with $\Theta|m^\tau\rangle\to\Theta|n^0\rangle$. Time-odd controls, including magnetic fields, must be reversed. The spectra and partition functions are unchanged by $\Theta$. For rank-one energy outcomes, microscopic reversibility gives

$$p_F(m|n)=p_R(n|m).$$

For degenerate energy outcomes, define $d_n^0=\operatorname{Tr}\Pi_n^0$ and $d_m^\tau=\operatorname{Tr}\Pi_m^\tau$. A Gibbs state conditioned on outcome $n$ is $\Pi_n^0/d_n^0$, so the corresponding block probabilities obey

$$d_n^0 p_F(m|n)=d_m^\tau p_R(n|m) =\operatorname{Tr}(\Pi_m^\tau U_F\Pi_n^0U_F^\dagger).$$

The reverse projectors are $\Theta\Pi_m^\tau\Theta^\dagger$ and $\Theta\Pi_n^0\Theta^\dagger$. Antiunitarity and $U_R=\Theta U_F^\dagger\Theta^\dagger$ equate the unnormalized transition traces; dividing by different block dimensions need not give equal conditional probabilities. Campisi, Hänggi, and Talkner (2011), Secs. IV.B and IV.E, make the distinction between energy-block probabilities and their nondegenerate specialization. See Time Reversal for the antiunitary transformation.

Including the Gibbs multiplicity, the reverse joint probability is

$$p_R(n,m) = \frac{d_m^\tau e^{-\beta E_m^\tau}}{Z_\tau} p_R(n|m).$$

The forward initial probability is $d_n^0e^{-\beta E_n^0}/Z_0$. The multiplicities cancel using the block identity, giving, on each nonzero paired branch,

$$\frac{p_F(m,n)} {p_R(n,m)} = \frac{ e^{-\beta E_n^0}/Z_0 }{ e^{-\beta E_m^\tau}/Z_\tau } = e^{\beta(E_m^\tau-E_n^0-\Delta F)}.$$

Since $W_{mn}=E_m^\tau-E_n^0$,

$$\frac{p_F(m,n)} {p_R(n,m)} = e^{\beta(W_{mn}-\Delta F)}.$$

Summing paired branches with the same work value gives the Crooks work relation:

$$\frac{P_F(W)} {P_R(-W)} = e^{\beta(W-\Delta F)}.$$

Multiplying by $e^{-\beta W}$ and integrating recovers the Jarzynski equality: the right side becomes $e^{-\beta\Delta F}\int dW\,P_R(-W)=e^{-\beta\Delta F}$.

Crossing Point

For continuous distributions, a crossing with $P_F(W)=P_R(-W)>0$ must occur at $W=\Delta F$. This can be useful experimentally, but poor overlap makes the crossing difficult to estimate.

For a finite system, work distributions are often sums of delta peaks. Crooks’ relation then compares the masses of matching peaks; one should not divide delta functions. There need not be any allowed work value equal to $\Delta F$, so a literal crossing is not guaranteed. For example, a commuting qubit gap change has only the two work values given in the TPM quench example, while $\Delta F$ generally lies between them.

Two-Level Nonadiabatic Drive

Let a two-level system have initial gap $\epsilon_0>0$, final gap $\epsilon_\tau>0$, and unitary transition probability $r$ between the lower and upper branches:

$$P_{g\to e} = P_{e\to g} = r, \qquad P_{g\to g} = P_{e\to e} = 1-r.$$

Write $Z_0=1+e^{-\beta\epsilon_0}$. The four work values are $0$, $\epsilon_\tau$, $-\epsilon_0$, and $\epsilon_\tau-\epsilon_0$. The exponential average is

$$\left\langle e^{-\beta W}\right\rangle = \frac{1}{Z_0} \big[ (1-r) + r e^{-\beta\epsilon_\tau} + e^{-\beta\epsilon_0}r e^{\beta\epsilon_0} + e^{-\beta\epsilon_0}(1-r) e^{-\beta(\epsilon_\tau-\epsilon_0)} \big].$$

Collecting terms,

$$\left\langle e^{-\beta W}\right\rangle = \frac{1}{Z_0} \left[ 1+e^{-\beta\epsilon_\tau} \right] = \frac{Z_\tau}{Z_0}.$$

The distribution depends on the nonadiabatic transition probability $r$, and the mean work generally changes with $r$. The Jarzynski equality still holds because the exponential average uses unitarity and the initial Gibbs weights.

Interpreting a Failed Equality

If a measured exponential average disagrees with the predicted partition-function ratio, check the physical assumptions and statistical convergence separately. An initially nonthermal state, untracked reservoir exchange, or a different measurement protocol changes the theoretical prediction. Missing rare low-work events changes the estimate even when the theorem applies. Work Distributions gives a qubit example of this sampling problem.

Exercises

Why the Initial Gibbs State Matters

A qubit with gap $\epsilon>0$ starts in $|e\rangle$ and undergoes a cyclic unitary swap of ground and excited states. Does its TPM work obey $\langle e^{-\beta W}\rangle=1$ at finite positive $\beta$?

Solution

Every run has $W=-\epsilon$, so the exponential average is $e^{\beta\epsilon}>1$. The Hamiltonian is cyclic and evolution is unitary, but the initial state is not the Gibbs state at the stated temperature. This is not a counterexample to the equality; it shows why the initial thermal weights are essential.

Degeneracy and Reverse Probabilities

Take $H_0=H_\tau=\epsilon|2\rangle\langle2|$, with $\epsilon>0$ and the two-dimensional ground space spanned by $|0\rangle,|1\rangle$. Let $U_F$ swap $|0\rangle$ and $|2\rangle$ while leaving $|1\rangle$ fixed. Use spinless time reversal $\Theta=K$ in this basis, and prepare a Gibbs state at inverse temperature $\beta$ in both protocols.

Find the conditional probabilities $p_F(e|g)$ and $p_R(g|e)$. Then calculate the work-peak probabilities and verify Crooks and Jarzynski. Why would setting the two conditional probabilities equal give the wrong answer?

Solution

The unitary is real and symmetric, so $U_R=U_F$. Conditioning on ground energy gives $(|0\rangle\langle0|+|1\rangle\langle1|)/2$: only half of this state reaches the excited level. Conversely, the excited state always reaches the ground space. Thus

$$p_F(e|g)=\frac12,\qquad p_R(g|e)=1, \qquad d_g p_F(e|g)=d_e p_R(g|e)=1.$$

Writing $Z=2+e^{-\beta\epsilon}$, the initial energy probabilities are $p_g=2/Z$ and $p_e=e^{-\beta\epsilon}/Z$. The work distribution in either protocol has three peaks:

$$P(W)=\frac1Z\delta(W) +\frac1Z\delta(W-\epsilon) +\frac{e^{-\beta\epsilon}}Z\delta(W+\epsilon).$$

Since $\Delta F=0$, the positive- to negative-work peak ratio is $e^{\beta\epsilon}$, while the zero-work ratio is one, as required by Crooks. The exponential average is

$$\langle e^{-\beta W}\rangle =\frac{1+e^{-\beta\epsilon}+1}{2+e^{-\beta\epsilon}}=1.$$

Equating the block conditional probabilities would discard the normalization of the two-dimensional initial ground space. Microscopic reversibility pairs individual transitions; coarse-graining them into unequal-dimensional energy blocks preserves the dimension-weighted identity instead.

Mean Dissipated Work in a Cyclic Swap

Apply a cyclic ground–excited swap to an initially thermal qubit with gap $\epsilon>0$. Find the mean work and verify that it is nonnegative for $\beta>0$.

Solution

With $p_e=1/(1+e^{\beta\epsilon})$, the work values are $+\epsilon$ with probability $1-p_e$ and $-\epsilon$ with probability $p_e$. Therefore

$$\langle W\rangle=\epsilon(1-2p_e) =\epsilon\tanh(\beta\epsilon/2)>0.$$

Since $\Delta F=0$, this is also the mean dissipated work. The negative-work branch is required for the exact exponential average despite the positive mean.

References

• M. Campisi, P. Hänggi, and P. Talkner, “Colloquium: Quantum fluctuation relations: Foundations and applications,” Reviews of Modern Physics 83, 771–791 (2011). arXiv:1012.2268.
• G. E. Crooks, “Entropy production fluctuation theorem and the nonequilibrium work relation for free energy differences,” Physical Review E 60, 2721–2726 (1999).
• C. Jarzynski, “Nonequilibrium equality for free energy differences,” Physical Review Letters 78, 2690–2693 (1997).
• J. Kurchan, “A quantum fluctuation theorem,” arXiv:cond-mat/0007360 (2000).
• P. Talkner, E. Lutz, and P. Hänggi, “Fluctuation theorems: Work is not an observable,” Physical Review E 75, 050102(R) (2007).
• H. Tasaki, “Jarzynski relations for quantum systems and some applications,” arXiv:cond-mat/0009244 (2000).