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  Effect of boundary conditions on partition functions

+ 7 like - 0 dislike

While computing partition functions in statistical mechanics models (say) on a 2d lattice one usually makes use of "circular boundary conditions" which thus gives the lattice topology of a torus. It makes the expressions, and calculations simpler; and one usually assumes that in the infinite volume limit boundary effects will be negligible. However I have never come through any general proof of this physical argument. What if we choose more complicated boundary identifications so that lattice has topology of some higher genus surface ? How will it affect the final answer for partition function or other correlation functions ?

This post imported from StackExchange Physics at 2015-03-30 13:26 (UTC), posted by SE-user user10001
asked Jul 10, 2012 in Theoretical Physics by user10001 (635 points) [ no revision ]

2 Answers

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If the interaction is short range, it wouldn't make a difference when the size of the system goes to infinity, i.e. the thermodynamic limit. The boundary conditions that you refer to are global, "topological" effects, while (short range) interactions are local, "metric" effects.

I find it hard to believe that there's a general proof of this for an arbitrary Hamiltonian and an arbitrary geometry, but it is quite simple to see this for specific systems.

For example, consider a 1D spin chain model (Ising, XY, Heisenberg, or whatnot) with $N$ sites and nearest neighbours interaction. For a given state of the system, the energy depends on the boundary conditions only through the interaction between two spins. Since there are $N-1$ other spins, it is very easy to convince yourself that the contribution from the boundary is negligible when $N\to\infty$.

By the way, when you calculate the partition function of an ideal gas, you usually assume it is confined in a box, and then state that the result is true for an arbitrary geometry. This is basically the same reasoning.

This post imported from StackExchange Physics at 2015-03-30 13:26 (UTC), posted by SE-user yohBS
answered Jul 10, 2012 by yohBS (10 points) [ no revision ]
+ 1 like - 0 dislike

You won't find any good proof in the literature, I suspect, because it'd be a nuisance to write up. You could prove it for any model which admits a cluster expansion, though.

The general case is quite general: the universality class of a statistical model can depend on the boundary conditions which appear in thermodynamic limit, and this dependence can be as complicated as the limits of boundary conditions allow. For example, the Ising Model's magnetization can be either +1 or -1 below the critical temperature; the mixture you get depends on how you dial down the magnetic field in the IR limit. You can break translational invariance by choosing wacky boundary conditions. You can land in different topological classes. You can make non-abelian gauge theory deconfine at high temperature.

This post imported from StackExchange Physics at 2015-03-30 13:26 (UTC), posted by SE-user user1504
answered Jul 10, 2012 by user1504 (1,110 points) [ no revision ]

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