Open boundary condition and Glauber Dynamics

Question

Warning: by background is in math, not physics. I've just recently started working with things that are close to theoretical physics. So please note that

1. I'm still very confused by the jargon.
2. Maybe I'm asking something trivial, in which case please just give me some reference and I'll be happy to read them.

I am reading some notes about Glauber Dynamics for discrete spin models (mainly stuff from Martinelli, Olivieri and Schonmann), and I'm surprised by how little space is given to models with "open boundary conditions" (which as far as I understood is the current terminology for no boundary conditions).

Googling the web didn't helped me understand if this is a choice of the authors, or there is some more stringent motivation beyond.

Let's suppose that we are working with a regular lattice (i.e. $\mathbb{Z}^d$), discrete spin space ($\{\pm 1\}$ is enough), and finite range interactions ($H(\sigma) = \sum_{A} J_A(\sigma_A)$, where the sum is taken over finite sets $A$ of diameter smaller than the interaction length).

1. Are there any problems with defining the Gibbs measure for finite volumes $V \subset \mathbb Z^d$ without boundary conditions, i.e. $$ \mu_V(\sigma) = Z_V^{-1} \exp \left( \sum_{A \subset V} J_A(\sigma_A) \right) $$ where by "problem" I think of things like "is not a measure on the right space of configurations", or "does not verify some important compatibility rule" (DLR?), or "is not unique", etc.
2. Are known conditions for which Glauber Dynamics on this finite region $V$ will converge quickly to the above-defined distribution will be uniformly gapped?

Answer 1

There are no problems--- I don't know a proof of the convergence of the Glauber dynamics to the Boltzmann distribution which doesn't work independent of boundary conditions. You just need to check two things:

1. The Glauber dynamics preserves the Boltzmann distribution (this is easy to check by starting with the Boltzmann distribution, and showing it is obeys detailed balance, i.e. that $\rho_i K_{i\rightarrow j} = \rho_j K_{j\rightarrow i}$, in words: the rate of the i->j transition in equilibrium is balanced by the rate of the j->i transition. This shows that Glauber dynamics has a stationary distribution which is the Boltzmann distribution.
2. Show ergodicity: this says that starting from any two configurations at time zero, there is a path with total nonzero Glauber probability which ends at the same state, i.e. the two paths collide with nonzero probability after finite time.

Under these conditions, it is easy to prove by a coupling argument that the stationary distribution is unique. The proof goes as follows: you consider the coupled walk, where the two walks are independent until they happen to be the same, and then stay the same forever. When there is a finite probability for combining, this will guarantee eventual convergence of the two walks, and therefore the final distribution is independent of the starting distribution, and must be the Boltzmann distribution (this is covered in one of my answers here in more detail, but I can't remember which).

The estimates from this argument are usually not optimal. The mixing time of the Glauber dynamics is usually reasonably fast away from a critical point, where there is critical slowing down in large systems. To get around this, you should use block updates.