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, and a reverse process satisfying microscopic reversibility. For the broader family of fluctuation relations, including open trajectories and feedback variants, see Fluctuation Theorems.

Setting

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 does not change the density operator. It only samples an initial energy value.

Forward Work Distribution

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

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

With projectors, replace the transition probability by the trace formula from 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 is exact under the assumptions above. It does not require slow or near-equilibrium driving.

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

Crooks’ relation compares a forward experiment with a reverse experiment. The reverse protocol begins in the Gibbs state of $H_\tau$ at the same inverse temperature:

$$\rho_\tau^{\mathrm{eq}} = \frac{e^{-\beta H_\tau}}{Z_\tau}.$$

It then runs the time-reversed driving protocol from $H_\tau$ back to $H_0$. In the simplest finite-dimensional presentation, microscopic reversibility gives paired transition probabilities

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

More generally, time reversal is represented by an antiunitary operator and the reverse Hamiltonian must also reverse time-odd external fields. See Time Reversal for the symmetry background.

The reverse branch probability paired with $m\to n$ is

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

Using microscopic reversibility,

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

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:

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

Integrating Crooks’ relation over $W$ recovers the Jarzynski equality.

Crossing Point

If the forward and reverse distributions have overlapping support, then the point where

$$P_F(W)=P_R(-W)$$

satisfies

$$W=\Delta F.$$

This crossing property is useful experimentally, but it is not guaranteed to be numerically stable if the overlap between forward and reverse distributions is poor. Strongly irreversible protocols can make the relevant overlap region rare in both experiments.

Two-Level Commuting Quench

Let

$$H_0 = \epsilon_0|e\rangle\langle e|, \qquad H_\tau = \epsilon_\tau|e\rangle\langle e|,$$

with the same eigenvectors before and after the quench. The initial Gibbs probabilities are

$$p_g^0 = \frac{1}{Z_0}, \qquad p_e^0 = \frac{e^{-\beta\epsilon_0}}{Z_0}, \qquad Z_0=1+e^{-\beta\epsilon_0}.$$

The work distribution is

$$P(W) = \frac{1}{Z_0}\delta(W) + \frac{e^{-\beta\epsilon_0}}{Z_0} \delta\!\left( W-\epsilon_\tau+\epsilon_0 \right).$$

The exponential average is

$$\begin{aligned} \left\langle e^{-\beta W}\right\rangle &= \frac{1}{Z_0} + \frac{e^{-\beta\epsilon_0}}{Z_0} e^{-\beta(\epsilon_\tau-\epsilon_0)} \\ &= \frac{1+e^{-\beta\epsilon_\tau}}{Z_0} = \frac{Z_\tau}{Z_0}. \end{aligned}$$

Thus

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

This example is algebraically simple because the energy eigenbasis does not change.

Two-Level Nonadiabatic Drive

Now let the same two-level system have transition probability $r$ between the lower and upper branches:

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

The four work values are $0$, $\epsilon_\tau$, $-\epsilon_0$, and $\epsilon_\tau-\epsilon_0$. The exponential average is

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

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.

What Can Go Wrong

The identities above are exact, but the assumptions are specific.

Common failure modes:

• The initial state is not Gibbs for $H_0$ at the stated temperature.
• The first energy measurement is omitted even though the state has energy coherence.
• The reverse protocol is not the correct time-reversed protocol.
• Magnetic fields or other time-odd controls are not reversed.
• The process is open, but bath energy changes are not included.
• The Hamiltonian used in $W$ differs from the Hamiltonian used in $\Delta F$.
• Finite sampling misses rare low-work events that dominate $\left\langle e^{-\beta W}\right\rangle$.

The equality failing in data can indicate experimental error, poor sampling, or a mismatch between the theorem’s assumptions and the implemented protocol. It does not by itself identify which issue occurred.

Exercises

Jarzynski from Unitarity

Starting from

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

derive the Jarzynski equality.

Solution

Compute

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

Substitute $p_F(m,n)$:

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

For each $m$,

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

so

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

Crooks Implies Jarzynski

Assume

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

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

Solution

Rearrange Crooks’ relation:

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

Integrate over $W$:

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

The reverse distribution is normalized, so

$$\int dW\,P_R(-W)=1.$$

Therefore

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

Nonadiabatic Two-Level Check

For the two-level nonadiabatic drive with transition probability $r$, verify that the Jarzynski equality does not depend on $r$.

Solution

Using the four branches,

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

The third term is $r$, and the fourth term is $(1-r)e^{-\beta\epsilon_\tau}$. Hence

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

The $r$ dependence cancels:

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

References

• C. Jarzynski, “Nonequilibrium equality for free energy differences,” Physical Review Letters 78, 2690, 1997.
• G. E. Crooks, “Entropy production fluctuation theorem and the nonequilibrium work relation for free energy differences,” Physical Review E 60, 2721, 1999.
• H. Tasaki, “Jarzynski relations for quantum systems and some applications,” arXiv:cond-mat/0009244, 2000.
• 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, 2007.
• M. Campisi, P. Hänggi, and P. Talkner, “Colloquium: Quantum fluctuation relations: Foundations and applications,” Reviews of Modern Physics 83, 771, 2011.