I am having a very basic problem understanding the idea of detailed balance, particularly in the context of the Ising model. Most references I have found contain the following phrase:

"In equilibrium, each elementary process must be equilibrated by its reverse process".

What does it mean for one process to equilibrate another?

In particular, I am trying to understand the section here: http://en.wikipedia.org/wiki/Ising_model#Algorithm_Specification in which they state that the given form of selection probabilities is required by detailed balance, but I suspect that that will follow once I understand what detailed balance actually means.



1 Answer 1


There are really two questions here.

One question is, "What does detailed balance mean for equilibrium systems?" and the second question is, "What does detailed balance mean in the context of selecting a move for a metropolis monte carlo algorithm like the one I linked to?" I will start with the first one.

Imagine we have an ensemble of particles that can be in one of three internal states, $A$, $B$, and $C$, (say a spin that can have $z$ component $-1$, $0$, or $1$) and suppose that this system in thermal equilibrium. Then as the system evolves, the internal state of each particle may change with time. However, since the system is equilibrated, the fraction of particles in each state must be constant. Thus if one particle, say, goes from state $A$ to state $B$, then there must be one particle going from state $B$ to state $A$. Thus detailed balance says that for any two states $i$ and $j$, the transition rate for $i \to j$ must equal the rate for $j \to i$. This is the story of detailed balance for systems in thermodynamic equilibrium.

The above paragraph, however, is not the explanation for what you were reading about in the Ising model article. The part of the article you were reading about had to do with simulating ising systems. One method of simulation is a metropolis monte carlo. If you want to simulate a system with many states, like an Ising system, what you can do is start the system out in a random state $A$, then iterate as follows. You pick another state $B$ so that you have the candidate move $A \to B$. You then accept this move with a probability which gets small if $B$ has a much higher energy than $A$.

If this is done right, then the probability of being at a given state in a given iteration is just the boltzmann probability, so this method gives you botlzmann statistics. Now there are two things that have to be done right. First, you must pick the candidate moves fairly, and second you must pick the acceptance probability of the candidate move correctly.

Picking the acceptance probability is easy, if the energy of the new state is lower, automatically accept the move. If the energy is higher, accept the move with probability $\exp(-\Delta E / T)$, which gets small when $\Delta E$ is big.

Picking the candidate moves fairly is when detailed balance comes in. Suppose you always decided the candidate move would always be to the same state. Clearly this will destroy boltzmann statistics because your system will always be in the given state regardless of its energy or any other energies. Clearly there must be a better way to pick moves. The correct way to choose candidate moves is for them to satisfy the detailed balance condition: $p(A \to B) = p(B \to A)$, where $p(A \to B)$ is the probability picking the candidate move to $B$ if you are in state $A$. This is what was meant by detailed balance in the article you linked to. Note it does not refer to the actual moves which are accepted by the algorithm, just the moves that are proposed.

  • $\begingroup$ So in this context detailed balance is then satisfied simply (once boltzmann statistics are also satisfied) by choosing a random spin to flip, since then there is equal probability of any particular spin flip at each step, hence p(a->B)=p(B->A) = 1/N, and independence of the spin chosen at each single spin-flip step takes care of the rest. Thank you for the thorough answer. $\endgroup$
    – KBriggs
    Commented Oct 7, 2013 at 16:43

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