Jarzynski's equality and nonequilibrium methods

In this section, the relationship between work and free energy
will be explored in greater detail. We have already introduced
the inequality in eqn. (16), which
states that if an amount of work
is performed on a system,
taking from state to state , then
.
Here, equality holds only if the work is performed reversibly.
The work referred to here is thermodynamic quantity and,
as such, must be regarded as an ensemble average. In statistical
mechanics, we can also introduce the mechanical or microscopic
work
performed on one member of the ensemble to
drive it from state to state . Then,
is simply an ensemble average of
. However, we
need to be somewhat careful about how we define this ensemble
average because
the work is defined along a particular path or trajectory which
takes the system from state to state , and equilibrium
averages do not refer not to paths but to microstates. This distinction
is emphasized by the fact that the work could be carried out
irreversibly, such that the system is driven out of equilibrium.
Thus, the proper definition of the ensemble average follows along the
lines already discussed in the context of the free-energy perturbation
approach, namely, averaging over the canonical distribution for
the state . In this case, since we will be discussing actual
paths , we let the initial condition be the phase
space vector for the system in the (initial) state .
Recall that
is a unique function of the
initial conditions. Then

From such an inequality, it would seem that using the work
as a method for calculating the free energy is of limited
utility, since the work necessarily must be performed reversibly,
otherwise one obtains only upper bound on the free energy.
It turns out, however, that irreversible work can be used to
calculate free energy differences by virtue of a connection
between the two quantities first discovered in 1997 by
C. Jarzynski that as come to be known as the
*Jarzynski equality*. This equality
states that if, instead of averaging
over the
initial canonical distribution (that of state ), an
average of
is performed over the
same distribution, the result is
, i.e.

The Jarzynski equality be proved using different strategies.
Here, however, we will present a proof that is most relevant
for the finite-sized systems and techniques employed in
molecular dynamics calculations.
Consider a time-dependent
Hamiltonian of the form

where the phase space vector has been introduced. Integrating both sides over time from to a final time , we find

Eqn. (21) can be regarded as a microscopic version of the first law of thermodynamics, in which the first and second terms represent the heat absorbed by the system and the work done on the system over the trajectory, respectively. Note that the work is actually a function of the initial phase-space vector , which can be seen by writing this term explicitly as

where the fact that the work depends explicitly on in eqn. (22) is indicated by the subscript. In the present discussion, we will consider that each initial condition, selected from a canonical distribution in , evolves according to Hamilton's equations in isolation. In this case, the heat term , and we have the usual addition to Hamilton's equations .

With the above condition, we can write the microscopic work as

(23) |

The numerator in this expression becomes much more interesting if we perform a change of variables from to . Since the solution of Hamilton's equations for the time-dependent Hamiltonian uniquely map the initial condition onto , when , we have a new set of phase-space variables, and by Liouville's theorem, the phase-space volume element is preserved

(24) |

thus proving the equality. The implication of the Jarzynski equality is that the work can be carried out along a reversible or irreversible path, and the correct free energy will still be obtained.

Note that due to Jensen's inequality:

(25) |

(26) |

(27) |