We have a family of potentials on , evolving with time, and their associated Gibbs measures , ie
The goal is to compute, at least numerically, the partition functions or equivalently the free energy . In many problems, the family of potentials connects an unknown potential , for which we would like to compute , and another potential for which is known.
An elegant way of estimating these free energies uses the Jarzynski identity, discovered in 1996. It is based on the Langevin dynamics associated with the evolving potential :
with a Brownian motion. Upon standard conditions on , this system – a standard Stochastic Differential Equation – has a solution. Note that the distribution of is not in general. However, there is a remarkable identity relating the distribution of with .
Jarzynski's identity states that
Here and after, the dot in or always represents a derivative with respect to time.
The consequence of this identity is that
giving rise to a natural estimator for the free energy difference : one samples many paths from the dynamics (2), then computes the path integral along each path, and then average the values of and take the log.
In general, exponential of paths integrals are not easy to study. The classical way is to use the Feynman-Kac representation of PDEs, but this proof seems quite involved; instead, there is a nice trick shown to me by Eric Vanden-Eijnden. The trick is as follows: if is the probability density of the system at , then
clearly, we want to compute ,
the quantity solves an explicit equation,
and from this equation we can see that .
The proof is then finished, since . From now on, we'll set , so that . Clearly, .
subject to the starting condition and . Then clearly,
We consider the system as a single SDE with
Let us note the density of , and write the associated Fokker-Planck equation; here and after, is a gradient and is the divergence, ie . Subscripts denote partial differentiation with respect to some variables.
Set , so that . Multiplying the last equation by , integrating, and swapping integrals and derivatives, we get
We have , hence
An integration by parts shows that , providing a sufficient decay of at infinity. We obtain the following equation:
It turns out that the density is the solution of this equation. Indeed, and , so that it is easy to check that (11) is satisfied. Moreover, , so the initial conditions are identical.
To conclude, we obtain