# Introduction

About a year ago there was a reddit post on the Ising Model in Haskell. The discussion seems to have fizzled out but Ising models looked like a perfect fit for Haskell using repa. In the end it turns out that they are not a good fit for repa, at least not using the original formulation. It may turn out that we can do better with Swendson-Yang or Wolff. But that belongs to another blog post.

We can get some parallelism at a gross level using the Haskell parallel package via a one line change to the sequential code. However, this does not really show off Haskell’s strengths in this area. In any event, it makes a good example for “embarassingly simple” parallelism in Haskell, the vector package and random number generation using the random-fu package.

The Ising model was (by Stigler’s law) proposed by Lenz in 1920 as a model for ferromagnetism, that is, the magnetism exhibited by bar magnets. The phenomenon ferromagnetism is so named because it was first observed in iron (Latin ferrum and chemical symbol Fe). It is also exhibited, for example, by rare earths such as gadolinium. Ferromagnetic materials lose their magnetism at a critical temperature: the Curie temperature (named after Pierre not his more famous wife). This is an example of a phase transition (ice melting into water is a more familiar example).

The Ising model (at least in 2 dimensions) predicts this phase transition and can also be used to describe phase transitions in alloys.

Abstracting the Ising model from its physical origins, one can think of it rather like Conway’s Game of Life: there is a grid and each cell on the grid is updated depending on the state of its neighbours. The difference with the Game of Life is that the updates are not deterministic but are random with the randomness selecting which cell gets updated as well as whether it gets updated. Thus we cannot update all the cells in parallel as would happen if we used repa. The reader only interested in this abstraction can go straight to the implementation (after finishing this introduction).

The diagram below shows a 2 dimensional grid of cells. Each cell can either be in an (spin) up state or (spin) down state as indicated by the arrows and corresponding colours. The Ising model then applies a parameterized set of rules by which the grid is updated. For certain parameters the cells remain in a random configuration, that is the net spin (taking up = 1 and down = -1) remains near zero; for other parameters, the spins in the cells line up (not entirely as there is always some randomness). It is this lining up that gives rise to ferromagnetism.

On the other hand, the physics and the Monte Carlo method used to simulate the model are of considerable interest in their own right. Readers interested in the Monte Carlo method can skip the physics and go to Monte Carlo Estimation. Readers interested in the physics can start with the section on Magnetism.

Definitions, theorems etc. are in bold and terminated by $\blacksquare$.

# Magnetism

Following Ziman (Ziman 1964) and Reif (Reif 2009), we assume that each atom in the ferromagnetic material behaves like a small magnet. According to Hund’s rules, we would expect unpaired electrons in the $d$ and $f$ shells for example in the transition elements and rare earths and these would supply the magnetic moment. However, the magnetic interaction between atoms is far too small to account for ferromagnetism. For Iron, for example, the Curie temperature is 1043K and the magnetic interaction between atoms cannot account for this by some margin. There is a quantum mechanical effect called the exchange interaction which is a consequence of the Pauli exclusion principle. Two electrons with the same spin on neighbouring atoms cannot get too close to each other. If the electrons have anti-parallel spins then the exclusion principle does not apply and there is no restriction on how close they can get to each other. Thus the electrostatic interaction between electrons on neighbouring atoms depends on whether spins are parallel or anti-parallel. We can thus write the Hamiltonian in the form:

$\displaystyle E = -J\sum_{\langle i, j\rangle} \sigma_i \sigma_j - B\sum_k \sigma_k$

Where

• The notation $\langle i, j\rangle$ means we sum over all the nearest neighbours in the lattice;

• $B$ is the applied magnetic field (note we use E for the Hamiltonian), for all of this article we will assume this to be $0$;

• The range of each index is $1 \ldots M$ where $N = M \times M$ is the total number of atoms;

• And $J$ the coupling constant expressing the strength of the interaction between neighboring spins and depending on the balance between the Pauli exclusion principle and the electrostatic interaction energy of the electrons, this may be positive corresponding to parallel spins (ferromagnetism which is the case we consider in this article) or negative corresponding to anti parallel spins (antiferromagnetism or ferrimagnetism which we consider no further).

# Acknowledgments

This post uses the random-fu package for random number generation and has also benefitted from comments by the author of that package (James Cook).

All diagrams were drawn using the Haskell diagrams domain specific language; the inhabitants of #diagrams were extremely helpful in helping create these.

Internet sources too numerous to mention were used for the physics and Monte Carlo. Some are listed in the bibliography. Apologies if you recognize something which does not get its just acknowledgement. The advantage of blog posts is that this can easily be remedied by leaving a comment.

Pragmas and imports to which only the over-enthusiastic reader need pay attention.

> {-# OPTIONS_GHC -Wall                      #-}
> {-# OPTIONS_GHC -fno-warn-type-defaults    #-}
> {-# OPTIONS_GHC -fno-warn-unused-do-bind   #-}
> {-# OPTIONS_GHC -fno-warn-missing-methods  #-}
> {-# OPTIONS_GHC -fno-warn-orphans          #-}

> {-# LANGUAGE TypeFamilies                  #-}
> {-# LANGUAGE NoMonomorphismRestriction     #-}
> {-# LANGUAGE TypeOperators                 #-}
> {-# LANGUAGE ScopedTypeVariables           #-}

> module Ising (
>          energy
>        , magnetization
>        , McState(..)
>        , thinN
>        , gridSize
>        , nItt
>        , expDv
>        , tCrit
>        , singleUpdate
>        , initState
>        , multiUpdate
>        , temps
>        , initColdGrid
>        , exampleGrid
>        , notAperiodics
>        , main
>        ) where


We put all of our code for the diagrams in this blog post in a separate module to avoid clutter.

> import Diagrams

> import qualified Data.Vector.Unboxed as V
> import qualified Data.Vector.Unboxed.Mutable as M

> import Data.Random.Source.PureMT
> import Data.Random

> import Control.Parallel.Strategies

> import Graphics.Rendering.Chart.Backend.Cairo

> import Data.Array.Repa hiding ( map, (++), zipWith )
> import Data.Array.Repa.Algorithms.Matrix

> import PrettyPrint ()

> gridSize :: Int
> gridSize = 20
>
> exampleGrid :: Int -> V.Vector Int
> exampleGrid gridSize = V.fromList $f (gridSize * gridSize) > where > f m = > evalState (replicateM m (sample (uniform (0 :: Int) (1 :: Int)) >>= > \n -> return$ 2*n - 1))
>       (pureMT 1)

> couplingConstant :: Double
> couplingConstant = 1.0
>
> kB :: Double
> kB = 1.0
>
> mu :: Double
> mu = 1.0


# The Boltzmann Distribution

Statistical physics is an extremely large subject but there needs to be some justification for the use of the Boltzmann distribution. For an excellent introduction to the subject of Statistical Physics (in which the Boltzmann distribution plays a pivotal role) see David Tong’s lecture notes (Tong (2011)).

Suppose we have we 3 boxes (we use the more modern nomenclature rather than the somewhat antiquated word urn) and 7 balls and we randomly assign the balls to boxes. Then it is far more likely that we will get an assignment of 2,2,3 rather than 0,0,7. When the numbers of boxes and balls become large (which is the case in statistical physics where we consider e.g. $10^{23}$ numbers of atoms) then it becomes very, very, very likely that the balls (or atoms) will spread themselves out uniformly as in the example.

Now suppose we have $N$ balls and $M$ boxes and we associate an energy $k$ with the $k$-th box but restrict the total energy to be constant. Thus we have two constraints:

• The total number of balls $N = \sum_1^M n_k$ and

• The total energy $E = \sum_1^M k\, n_k$.

Let us assume the balls are allocated to boxes in the most likely way and let us denote the values in each box for this distribution as $\tilde{n}_k$.

Let us now move 2 balls from box $k$, one to box $k-1$ and one to box $k+1$. Note that both constraints are satisified by this move. We must therefore have

\displaystyle \begin{aligned} \frac{1}{(\tilde{n}_{k-1} + 1)!}\frac{1}{(\tilde{n}_{k} - 2)!}\frac{1}{(\tilde{n}_{k+1} + 1)!} &\le \frac{1}{\tilde{n}_{k-1}!}\frac{1}{\tilde{n}_{k}!}\frac{1}{\tilde{n}_{k+1}!} \\ \frac{\tilde{n}_k}{\tilde{n}_{k-1} + 1}\frac{\tilde{n}_k - 1}{\tilde{n}_{k+1} + 1} &\le 1 \end{aligned}

And let us start from the most likely distribution and move 1 ball from box $k-1$ and 1 ball from box $k+1$ into box $k$. Again both constraints are satisified by this move. Doing a similar calculation to the one above we get

\displaystyle \begin{aligned} \frac{1}{(\tilde{n}_{k-1} - 1)!}\frac{1}{(\tilde{n}_{k} + 2)!}\frac{1}{(\tilde{n}_{k+1} - 1)!} &\le \frac{1}{\tilde{n}_{k-1}!}\frac{1}{\tilde{n}_{k}!}\frac{1}{\tilde{n}_{k+1}!} \\ 1 &\le \frac{\tilde{n}_k + 1}{\tilde{n}_{k+1}}\frac{\tilde{n}_{k} + 2}{\tilde{n}_{k-1}} \end{aligned}

From this we deduce that as $n_k \rightarrow \infty$ then $n_k^2 \approx n_{k-1}n_{k+1}$ or that $n_{k+1} / n_k = B$ for some constant $B$.

Telecsoping we can write $n_k = n_1 B^{k-1} \propto B^k$. Thus the probability of a ball being in box $k$ is

$\displaystyle \mathbb{P}(k) = \frac{B^k}{\sum_{i=1}^M B^i}$

If we now set $B = e^{-\beta}$ and $Z = \sum_{i=1}^M B^i$ then we can re-write our distribution as

$\displaystyle \mathbb{P}(k) = \frac{e^{-\beta k}}{Z}$

It should therefore be plausible that the probability of finding a system with a given energy $E(\sigma)$ in a given state $\sigma$ is given by the Boltzmann distribution

$\displaystyle \mathbb{P}(\sigma) = \frac{\exp(-E(\sigma) / k_B T)}{Z(T)}$

where we have defined the temperature to be $T = 1 / k_B\beta$ with $k_B$ being Boltzmann’s constant and $Z(T)$ is another normalizing constant.

## Specific Heat

Using the Boltzmann distribution we can calculate the average energy of the system

\displaystyle \begin{aligned} \langle E \rangle &\triangleq \sum_\sigma E(\sigma) \mathbb{P}(\sigma) \\ &= \sum_\sigma E(\sigma) \frac{\exp(-E(\sigma) / k_B T)}{Z} \\ &= -\frac{1}{Z} \frac{\partial Z}{\partial \beta} \\ &= \frac{\partial}{\partial \beta} \log Z \end{aligned}

where it is understood that $Z$ depends on $T$.

If we let $C_V$ be the specific heat per particle (at constant volume) then

\displaystyle \begin{aligned} NC_V &\triangleq \frac{\partial \langle E\rangle}{\partial T} \\ &= \frac{\partial \langle E\rangle}{\partial \beta} \frac{\partial \beta}{\partial T} \\ &= -\frac{1}{k_B T^2} \frac{\partial \langle E\rangle}{\partial \beta} \\ &= -\frac{1}{k_B T^2} \frac{\partial^2 \log Z}{\partial \beta^2} \\ &= -\frac{1}{k_B T^2} \frac{\partial}{\partial \beta} \sum_\sigma E(\sigma) \frac{\exp(-E(\sigma) / k_B T)}{Z} \\ &= -\frac{1}{k_B T^2} \Bigg[\frac{\big(\sum_\sigma E(\sigma) \exp(-E(\sigma) / k_B T)\big)^2}{Z^2} + \frac{\sum_\sigma -E^2(\sigma) \exp(-E(\sigma) / k_B T)}{Z}\Bigg] \end{aligned}

We know that

$\displaystyle \Delta E^2 \triangleq \langle (E - \langle E \rangle)^2 \rangle = \langle E^2 \rangle - \langle E \rangle^2$

So we can write

$\displaystyle \Delta E^2 = k_BT^2C_V$

so if we can estimate the energy and the square of the energy then we can estimate the specific heat.

# Monte Carlo Estimation

Although Ising himself developed an analytic solution in 1 dimension and Onsager later developed an analytic solution in 2 dimensions, no-one has (yet) found an analytic solution for 3 dimensions.

One way to determine an approximate answer is to use a Monte Carlo method.

## Uniform Sampling

We could pick sample configurations at random according to the Boltzmann distribution

$\displaystyle \mathbb{P}(\sigma) = \frac{\exp(-E(\sigma) / k_B T)}{Z(T)}$

where the sum $T$ is the temperature, $k_B$ is Boltzmann’s constant, $E$ is the energy of a given state

$\displaystyle E(\sigma) = -J\sum_{i, j} \sigma_i \sigma_j - B \sum_k \sigma_k$

and $Z(T)$ is a normalizing constant (Z for the German word Zustandssumme, “sum over states”)

$\displaystyle Z(T) = \sum_\sigma \exp(-E(\sigma) / k_B T)$

We can evaluate the energy for one state easily enough as the Haskell below demonstrates. Note that we use so-called periodic boundary conditions which means our grid is actually a torus with no boundaries. In other words, we wrap the top of the grid on to the bottom and the left of the grid on to the right.

[As an aside, we represent each state by a Vector of Int. No doubt more efficient representations can be implemented. We also have to calculate offsets into the vector given a point’s grid co-ordinates.]

> energy :: (Fractional a, Integral b, M.Unbox b) => a -> V.Vector b -> a
> energy j v = -0.5 * j * (fromIntegral V.sum energyAux) > where > > energyAux = V.generate l f > > l = V.length v > > f m = c * d > where > i = m mod gridSize > j = (m mod (gridSize * gridSize)) div gridSize > > c = v V.! jc > jc = gridSize * i + j > > d = n + e + s + w > > n = v V.! jn > e = v V.! je > s = v V.! js > w = v V.! jw > > jn = gridSize * ((i + 1) mod gridSize) + j > js = gridSize * ((i - 1) mod gridSize) + j > je = gridSize * i + ((j + 1) mod gridSize) > jw = gridSize * i + ((j - 1) mod gridSize)  But what about the normalizing constant $Z$? Even for a modest grid size say $10 \times 10$, the number of states that needs to be summed over is extremely large $2^{10 \times 10}$. Instead of summing the entire state space, we could draw R random samples $(\sigma^{(i)})_{0 \le i < R}$ uniformly from the state space. We could then use $\displaystyle Z_R \triangleq \sum_0^{R-1} \exp (-\beta\sigma_i)$ to estimate e.g. the magnetization $\displaystyle \langle M \rangle = \sum_\sigma M(\sigma) \frac{\exp(-\beta E(\sigma))}{Z(T)} \mathrm{d} \sigma$ by $\displaystyle \widehat{\langle M \rangle} = \sum_{i=0}^{R-1} M(\sigma) \frac{exp(-\beta E(\sigma(i)))}{Z_R}$ However we know from statistical physics that systems with large numbers of particles will occupy a small portion of the state space with any significant probability. And according to (MacKay 2003, chap. 29), a high dimensional distribution is often concentrated on small region of the state space known as its typical set $T$ whose volume is given by $|T| \approx 2^H$ where $H$ is the (Shannon) entropy of the (Boltzmann) distribution which for ease of exposition we temporarily denote by $P$. $\displaystyle H = -\sum_\sigma P(\sigma)\log_2(P(\sigma))$ If almost all the probability mass is located in $T$ then the actual value of the (mean) magnetization will determined by the values that $M$ takes on that set. So uniform sampling will only give a good estimate if we make $R$ large enough that we hit $T$ at least a small number of times. The total size of the state space is $2^N$ and the $|T| \approx 2^H$, so there is a probability of $2^H / 2^N$ of hitting $T$. Thus we need roughly $2^{N - H}$ samples to hit $T$. At high temperatures, the Boltzmann distribution flattens out so roughly all of the states have an equal likelihood of being occupied. We can calculate the (Shannon) entropy for this. $\displaystyle H \approx \sum_\sigma \frac{1}{2^N}\log_2 2^N = N$ We can do a bit better than this. At high temperatures, $\beta J \ll 1$ and taking $B = 0$ (as we have been assuming all along) \displaystyle \begin{aligned} Z &= \sum_\sigma \exp\big(-\beta E(\sigma)\big) \\ &= \sum_\sigma \exp\bigg(\beta J \sum_{i, j} \sigma_i \sigma_j - B \sum_k \sigma_k\bigg) \\ &= \sum_\sigma \prod_{i, j} \exp\beta J \sigma_i \sigma_j \end{aligned} By definition \displaystyle \begin{aligned} \cosh x &= \frac{e^x + e^{-x}}{2} \\ \sinh x &= \frac{e^x - e^{-x}}{2} \end{aligned} Thus \displaystyle \begin{aligned} \exp \beta J \sigma_i \sigma_j &= \cosh\beta J + \sigma_i\sigma_j\sinh \beta J \\ &= \cosh\beta J (1 + \sigma_i\sigma_j\tanh \beta J) \end{aligned} We can therefore re-write the partition function as \displaystyle \begin{aligned} Z &= \sum_\sigma \prod_{i, j} \cosh\beta J (1 + \sigma_i\sigma_j\tanh \beta J) \\ &= (\cosh \beta J)^{4N/2} \sum_\sigma \prod_{i, j} (1 + \sigma_i\sigma_j\tanh \beta J) \\ \end{aligned} where the factor 2 is because we count the exchange interaction twice for each pair of atoms and the factor 4 is because each atom is assumed to only interact with 4 neighbours. Since at high temperatures, by assumption, we have $\beta J \ll 1$ then $\tanh \beta J \ll 1$ also. Thus $\displaystyle Z \approx (\cosh \beta J)^{2N}\sum_\sigma 1 = 2^N(\cosh \beta J)^{2N}$ Calculating the free energy \displaystyle \begin{aligned} F &= -k_B T \ln Z \\ &= -k_B T N \ln 2 -k_B T 2 N \ln (\cosh \beta J) \\ &\approx -k_B T N \ln 2 -k_B T N (\beta J)^2 \\ &= -k_B T N \ln 2 - N\frac{J^2}{k_B T} \\ \end{aligned} From this we can determine the (Boltzmann) entropy $\displaystyle S = k_B N (\ln 2 - (\beta J)^2) \approx k_B N \ln 2$ which agrees with our rather hand-wavy derivation of the (Shannon) entropy at high temperatures. At low temperatures the story is quite different. We can calculate the ground state energy where all the spins are in the same direction. $\displaystyle E_0 = -J\sum_{i,j}\sigma_i\sigma_j = -4JN / 2 = -2JN$ And we can assume that at low temperatures that flipped spins are isolated from other flipped spins. The energy for an atom in the ground state is -4J and thus the energy change if it flips is 8J. Thus the energy at low temperatures is $\displaystyle E_1 \approx E_0 + 8J\sum_i (\sigma_i + 1) / 2$ The partition function is \displaystyle \begin{aligned} Z &= 2\sum_\sigma \exp{-\beta \bigg(E_0 + 8J\sum_i (\sigma_i + 1) /2\bigg)} \\ &= 2\exp{-\beta E_0} \prod_i \sum_{\sigma_i = \pm 1} \exp{-8\beta J(\sigma_i + 1) / 2} \\ &= 2\exp{-\beta E_0} (1 + \exp{-8\beta J})^N \end{aligned} Again we can calculate the free energy and since we are doing this calculation to get a rough estimate of the entropy we can approximate further. \displaystyle \begin{aligned} F &= -k_B T \ln Z \\ &= -k_B (T \ln 2 - T E_0 + T N \ln (1 + \exp{-8\beta J})) \\ &\approx -k_B (T \ln 2 - T E_0 + T N \exp{-8\beta J}) \end{aligned} From this we can determine the (Boltzmann) entropy. Using the fact that $\beta = 1 / k_B T$ and thus that $\displaystyle \frac{\partial \beta}{\partial T} = - \frac{1}{k_B T^2} = -k_B\beta^2$ we have \displaystyle \begin{aligned} S &= - \frac{\partial F}{\partial T} \\ &= k_B\bigg(\ln 2 + N\frac{\partial}{\partial T} \exp{-8\beta J}\bigg) \\ &= k_B\bigg(\ln 2 + N\frac{\partial}{\partial T} \frac{1}{\beta}\exp{-8\beta J}\bigg) \\ &= k_B\bigg(\ln 2 + N\frac{\partial \beta}{\partial T} \frac{\partial}{\partial \beta}\frac{1}{\beta}\exp{-8\beta J}\bigg) \\ &= k_B\bigg(\ln 2 + N\frac{\partial \beta}{\partial T} \frac{\partial}{\partial \beta}\frac{1}{\beta}\exp{-8\beta J}\bigg) \\ &= k_B\bigg(\ln 2 + N\frac{\partial \beta}{\partial T} \bigg(-\frac{1}{\beta^2}\exp{-8\beta J} - \frac{8J}{\beta}\exp{-8\beta J}\bigg) \\ &= k_B\bigg(\ln 2 + N\exp{-8\beta J} + 8JN\beta\exp{-8\beta J}\bigg) \end{aligned} The critical temperature (as we shall obtain by simulation) given by Onsager’s exact result in 2d is $\displaystyle \frac{1}{\beta} = k_BT_{\text Crit} = \frac{2J}{\ln(\sqrt{2}+1)}$ Taking $J = 1$ > tCrit :: Double > tCrit = 2.0 * j / log (1.0 + sqrt 2.0) where j = 1  ghci> tCrit 2.269185314213022  Plugging this in we get ghci> exp (-8.0/tCrit) * (1.0 + 8.0 / tCrit) 0.1332181153896559  Thus the (Shannon) entropy is about 0.13N at the interesting temperature and is about N at high temperatures. So uniform sampling would require $\sim 2^{(N - N)}$ samples at high temperatures but $\sim 2^{(N - 0.13N)} \approx 2^{N /2}$ at temperatures of interest. Even for our modest $10 \times 10$ grid this is $2^{50} \approx 10^{17}$ samples! Fortunately, Metropolis and his team (Metropolis et al. 1953) discovered a way of constructing a Markov chain with a limiting distribution of the distribution required which does not require the evaluation of the partition function and which converges in a reasonable time (although theoretical results substantiating this latter point seem to be hard to come by). # Markov Chains Markov first studied the stochastic processes that came to be named after him in 1906. We follow (Norris 1998), (Beichl and Sullivan 2000), (Diaconis and Saloff-Coste 1995), (Chib and Greenberg 1995) (Levin, Peres, and Wilmer 2008) and (Gravner 2011). As usual we work on a probability measure space $(\Omega, {\mathbb F}, \mathbb{P})$ (that is $\mathbb{P}(\Omega) = 1$). Although this may not be much in evidence, it is there lurking behind the scenes. Let $S$ be a finite set. In the case of an Ising model with $N$ cells, this set will contain $2^N$ elements (all possible configurations). A Markov chain is a discrete time stochastic process $X_0, X_1, \ldots$ such that $\displaystyle \mathbb{P} (X_{n+1} = j \,|\, X_0 = i_0, X_1 = i_1, \dots X_n = i) = \mathbb{P} (X_{n+1} = j \,|\, X_n = i)$ $\blacksquare$ That is, where a Markov chain goes next only depends on where it is not on its history. A stochastic transition matrix is a matrix $P = \{ p_{ij} : i, j \in S \}$ such that $\displaystyle \sum_{j \in S} p_{ij} = 1 \, \forall i \in S$ $\blacksquare$ We can describe a Markov chain by its transition matrix $P$ and initial distribution $\pi_0(i) = \mathbb{P} (X_0 = i)$. In the case we say a stochastic process $(X_n)_{n \ge 0}$ is Markov $(\pi_0, P)$. We need to be able to discuss properties of Markov chains such as stationarity, irreducibility, recurrence and ergodicity. ## Stationarity A Markov chain has a stationary distribution $\pi(i)$ if $\displaystyle \sum_{i \in S} \pi(i) p_{ij} = \pi(j)$ $\blacksquare$ One question one might ask is whether a given Markov chain has such a distribution. For example, for the following chain, any distribution is a stationary distribution. That is $\pi P = \pi$ for any $\pi$. Any distribution is a stationary distribution for the unit transition matrix. $\displaystyle \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ The $n$-th transition matrix of a Markov chain is $P^n$. The corresponding matrix entries are $\displaystyle p^{(n)}_{ij} = (P^n)_{ij}$ Another key question is, if there is a unique stationary distribution, will the $n$-th transition probabilities converge to that distribution, that is, when does, $p^{(n)}_{ij} \rightarrow \pi(j)$ as $n \rightarrow \infty$. ## Irreducibility Write ${\mathbb P}_i(A) = {\mathbb P}(A \, | \, X_0 = i)$ We say that $i$ leads to $j$ and write $i \rightarrow j$ if $\displaystyle {\mathbb P}_i(X_n = j \, \text{eventually}) > 0$ Theorem For distinct states $i$ and $j$, the following are equivalent: • $i \rightarrow j$ • $p_{i_0i_1} p_{i_1i_2} \ldots p_{i_{n-1}i_n} > 0$ • $p_{ij}^{(n)} > 0$ for some $n \ge 0$ $\blacksquare$ This makes it clear that $\rightarrow$ is transitive and reflexive hence an equivalence relation. We can therefore partition the states into classes. If there is only one class then the chain is called irreducible. For example, $\displaystyle \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & \frac{1}{4} & \frac{3}{4} \\ 0 & 0 & 0 & 1 \end{bmatrix}$ has classes $\{1, 2\}$, $\{3\}$ and $\{4\}$ so is not irreducible. On the other hand $\displaystyle \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{1}{2} & \frac{1}{4} & \frac{1}{4} \\ 0 & \frac{1}{3} & \frac{2}{3} \end{bmatrix}$ is irreducible. ## Recurrence Let $(X_n)_{n \ge 0}$ be a Markov chain. A state $i$ is recurrent if $\displaystyle {\mathbb P} (X_n = i \, \text{infinitely often}) = 1$ The first passage time is defined as $\displaystyle T_j = \inf_{n \ge 1} \{X_n = j\}$ Note that the $\inf$ is taken over $n$ strictly greater than 1. Incidentally the first passage time is a stopping time but that any discussion of that would take this already long article even longer. The expectation of the $i$-th first passage time starting from $i$ is denoted $m_i = {\mathbb E}_i(T_i)$. Theorem Let $P$ be irreducible then the following are equivalent: • Every state is positive recurrent. • Some state is positive recurrent. • P has an invariant distribution $\pi$ and in this case $\pi(i) = 1 / m_i$. $\blacksquare$ A state $i$ is aperiodic if $p^{(n)}_{ii} > 0$ for all sufficiently large $n$. For example, the chain with this transition matrix is not periodic: $\displaystyle \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ as running the following program segment shows with the chain flip-flopping between the two states. > notAperiodic0 :: Array U DIM2 Double > notAperiodic0 = fromListUnboxed (Z :. (2 :: Int) :. (2 :: Int)) ([0,1,1,0] :: [Double]) > notAperiodics :: [Array U DIM2 Double] > notAperiodics = scanl mmultS notAperiodic0 (replicate 4 notAperiodic0)  ghci> import Ising ghci> import PrettyPrint ghci> import Text.PrettyPrint.HughesPJClass ghci> pPrint notAperiodics [[0.0, 1.0] [1.0, 0.0], [1.0, 0.0] [0.0, 1.0], [0.0, 1.0] [1.0, 0.0], [1.0, 0.0] [0.0, 1.0], [0.0, 1.0] [1.0, 0.0]]  Theorem Let $P$ be irreducible and aperiodic and suppose that $P$ has an invariant distribution $\pi$. Let $\pi_0$ be any distribution (on the state space???). Suppose that $(X_n)_{n \ge 0}$ is Markov $(\pi_0, P)$ then • $\mathbb{P}(X_n = j) \rightarrow \pi(j)$ as $n \rightarrow \infty$ for all $j$ $\blacksquare$ A proof of this theorem uses coupling developed by Doeblin; see (Gravner 2011) for more details. Corollary With the conditions of the preceding Theorem • $p_{ij}^{(n)} \rightarrow \pi(j)$ as $n \rightarrow \infty$ for all $i, j$ $\blacksquare$ If the state space is infinite, the existence of a stationary distribution is not guaranteed even if the Markov chain is irreducible, see (Srikant 2009) for more details. ## Detailed Balance A stochastic matrix $P$ and a distribution $\pi$ are said to be in detailed balance if $\displaystyle \pi(i) p_{ij} = \pi(j) p_{ji}$ $\blacksquare$ Theorem If a stochastic matrix $P$ and a distribution $\pi$ are in detailed balance then $\pi$ is a stationary distribution. $\blacksquare$ ## The Ergodic Theorem Define the number of visits to $i$ strictly before time $n$ as $\displaystyle V_i(n) \triangleq \sum_{k = 0}^{n - 1}{\mathcal I}_{\{X_k = i\}}$ $V_i(n) / n$ is the proportion of time before $n$ spent in state $i$. Theorem Let $(X_n)_{(n \ge 0)}$ be an irreducible Markov chain then $\displaystyle {\mathbb P} \Bigg(\frac{V_i(n)}{n} \rightarrow \frac{1}{m_i} \, {\text as} \, n \rightarrow \infty \Bigg) = 1$ If further the chain is positive recurrent then for any bounded function $f : {\mathbb I} \rightarrow {\mathbb R}$ then $\displaystyle {\mathbb P} \Bigg(\frac{1}{n} \sum_{k = 0}^{n - 1}f(X_i)\rightarrow \hat{f} \, {\text as} \, n \rightarrow \infty \Bigg) = 1$ If further still the chain is aperiodic then it has a unique stationary distribution, which is also the limiting distribution $\blacksquare$ A Markov chain satisfying all three conditions is called ergodic. ## The Metropolis Algorithm Thus if we can find a Markov chain with the required stationary distribution and we sample a function of this chain we will get an estimate for the average value of the function. What Metropolis and his colleagues did was to provide a method of producing such a chain. Algorithm Let $\pi$ be a probability distribution on the state space $\Omega$ with $\pi(i) > 0$ for all $i$ and let $(Q, \pi_0)$ be an ergodic Markov chain on $\Omega$ with transition probabilities $q(i,j) > 0$ (the latter condition is slightly stronger than it need be but we will not need fully general conditions). Create a new (ergodic) Markov chain with transition probabilities $\displaystyle p_{ij} = \begin{cases} q(i,j)\bigg[\frac{\pi(j) q(j,i)}{\pi(i) q(i,j)} \land 1 \bigg] & \text{if } j \ne i \\ 1 - \sum_{k : k \ne i} q(i,k) \bigg[\frac{\pi(j) q(j,i)}{\pi(i) q(i,j)} \land 1 \bigg] & \text{if } j = i \end{cases}$ where $\land$ takes the maximum of its arguments. Calculate the value of interest on the state space e.g. the total magnetization for each step produced by this new chain. Repeat a sufficiently large number of times and take the average. This gives the estimate of the value of interest. $\blacksquare$ Let us first note that the Markov chain produced by this algorithm almost trivially satisfies the detailed balance condition, for example, \displaystyle \begin{aligned} \pi(i) q(i,j)\bigg[\frac{\pi(j) q(j, i)}{\pi(i)q(i,j)} \land 1\bigg] &= \pi(i)q(i,j) \land \pi(j)q(j,i) \\ &= \pi(j)q(j,i)\bigg[\frac{\pi(i) q(i, j)}{\pi(j)q(j,i)} \land 1\bigg] \end{aligned} Secondly since we have specified that $(Q, \pi_0)$ is ergodic then clearly $(P, \pi_0)$ is also ergodic (all the transition probabilities are $> 0$). So we know the algorithm will converge to the unique distribution we specified to provide estimates of values of interest. Two techniques that seem to be widespread in practical applications are burn in and thinning. Although neither have strong theoretical justification (“a thousand lemmings can’t be wrong”), we follow the practices in our implementation. • “Burn in” means run the chain for a certain number of iterations before sampling to allow it to forget the initial distribution. • “Thinning” means sampling the chain every $n$ iterations rather than every iteration to prevent autocorrelation. ## Haskell Implementation Calculating the total magnetization is trivial; we just add up all the spins and multiply by the magnetic moment of the electron. > magnetization :: (Num a, Integral b, M.Unbox b) => a -> V.Vector b -> a > magnetization mu = (mu *) . fromIntegral . V.sum  We keep the state of the Monte Carlo simulation in a record. > data McState = McState { mcMagnetization :: !Double > , mcMAvg :: !Double > , mcEnergy :: !Double > , mcEAvg :: !Double > , mcEAvg2 :: !Double > , mcCount :: !Int > , mcNumSamples :: !Int > , mcGrid :: !(V.Vector Int) > } > deriving Show  As discussed above we sample every thinN iterations, a technique known as “thinning”. > thinN :: Int > thinN = 100  The total number of iterations per Monte Carlo run. > nItt :: Int > nItt = 1000000  There are only a very limited number of energy changes that can occur for each spin flip. Rather the recalculate the value of the Boltzmann distribution for every spin flip we can store these in a Vector. For example if spin is up and all its surrounding spins are up and it flips to down then energy change is $8J$ and the corresponding value of the Boltzmann distribution is $\exp -8J\beta$. Another example, the energy before is $-2J$ and the energy after is $2J$ so the energy change is $4J$. > expDv :: Double -> Double -> Double -> V.Vector Double > expDv kB j t = V.generate 9 f > where > f n | odd n = 0.0 > f n = exp (j * ((fromIntegral (8 - n)) - 4.0) * 2.0 / (kB * t))  The most important function is the single step update of the Markov chain. We take an Int representing the amount of thinning we wish to perform, the vector of pre-calculated changes of the Boltzmann distribution, the current state, a value representing the randomly chosen co-ordinates of the grid element that will be updated and a value sampled from the uniform distribution which will decide whether the spin at the co-ordinates will be updated. > singleUpdate :: Int -> V.Vector Double -> McState -> (Int, Int, Double) -> McState > singleUpdate thinN expDvT u (i, j, r) = > McState { mcMagnetization = magNew > , mcMAvg = mcMAvgNew > , mcEnergy = enNew > , mcEAvg = mcEAvgNew > , mcEAvg2 = mcEAvg2New > , mcCount = mcCount u + 1 > , mcNumSamples = mcNumSamplesNew > , mcGrid = gridNew > } > where > > (gridNew, magNew, enNew) = > if p > r > then ( V.modify (\v -> M.write v jc (-c)) v > , magOld - fromIntegral (2 * c) > , enOld + couplingConstant * fromIntegral (2 * c * d) > ) > else (v, magOld, enOld) > > magOld = mcMagnetization u > enOld = mcEnergy u > > (mcMAvgNew, mcEAvgNew, mcEAvg2New, mcNumSamplesNew) = > if (mcCount u) mod thinN == 0 > then ( mcMAvgOld + magNew > , mcEAvgOld + enNew > , mcEAvg2Old + enNew2 > , mcNumSamplesOld + 1 > ) > else (mcMAvgOld, mcEAvgOld, mcEAvg2Old, mcNumSamplesOld) > > enNew2 = enNew * enNew > > mcMAvgOld = mcMAvg u > mcEAvgOld = mcEAvg u > mcEAvg2Old = mcEAvg2 u > mcNumSamplesOld = mcNumSamples u > > v = mcGrid u > > p = expDvT V.! (4 + c * d) > > c = v V.! jc > jc = gridSize * i + j > > d = n + e + s + w > > n = v V.! jn > e = v V.! je > s = v V.! js > w = v V.! jw > > jn = gridSize * ((i + 1) mod gridSize) + j > js = gridSize * ((i - 1) mod gridSize) + j > je = gridSize * i + ((j + 1) mod gridSize) > jw = gridSize * i + ((j - 1) mod gridSize)  In order to drive our Markov chain we need a supply of random positions and samples from the uniform distribution. > randomUpdates :: Int -> V.Vector (Int, Int, Double) > randomUpdates m = > V.fromList
>   evalState (replicateM m x)
>   (pureMT 1)
>   where
>     x = do r <- sample (uniform (0 :: Int)    (gridSize - 1))
>            c <- sample (uniform (0 :: Int)    (gridSize - 1))
>            v <- sample (uniform (0 :: Double)            1.0)
>            return (r, c, v)


To get things going, we need an initial state. We start with a cold grid, that is, one with all spins pointing down.

> initColdGrid :: V.Vector Int
> initColdGrid = V.fromList $replicate (gridSize * gridSize) (-1) > > initState :: McState > initState = McState { mcMagnetization = magnetization mu initColdGrid > , mcMAvg = 0.0 > , mcEnergy = energy couplingConstant initColdGrid > , mcEAvg = 0.0 > , mcEAvg2 = 0.0 > , mcCount = 0 > , mcNumSamples = 0 > , mcGrid = initColdGrid > }  We will want to run the simulation over a range of temperatures in order to determine where the phase transition occurs. > temps :: [Double] > temps = getTemps 4.0 0.5 100 > where > getTemps :: Double -> Double -> Int -> [Double] > getTemps h l n = [ m * x + c | > w <- [1..n], > let x = fromIntegral w ] > where > m = (h - l) / (fromIntegral n - 1) > c = l - m  Now we can run a chain at a given temperature. > multiUpdate :: McState -> Double -> V.Vector (Int, Int, Double) -> McState > multiUpdate s t = V.foldl (singleUpdate thinN (expDv kB couplingConstant t)) s  For example running the model at a temperature of $3.0$ for 100, 1000 and 10,000 steps respectively shows disorder growing. On the other hand running the model at a temperature of $2.0$ shows a very limited disorder. For any given state we can extract the magnetization, the energy and the square of the energy. > magFn :: McState -> Double > magFn s = abs (mcMAvg s / (fromIntegral$ mcNumSamples s))
>
> magFnNorm :: McState -> Double
> magFnNorm s = magFn s / (fromIntegral (gridSize * gridSize))
>
> enFn :: McState -> Double
> enFn s  = mcEAvg s / (fromIntegral $mcNumSamples s) > > enFnNorm :: McState -> Double > enFnNorm s = enFn s / (fromIntegral (gridSize * gridSize))  > en2Fn :: McState -> Double > en2Fn s = mcEAvg2 s / (fromIntegral$ mcNumSamples s)


And we can also calculate the mean square error.

> meanSqErr :: McState -> Double
> meanSqErr s = e2 - e*e
>   where
>     e = enFn s
>     e2 = en2Fn s
>
> meanSqErrNorm :: McState -> Double
> meanSqErrNorm s = meanSqErr s / (fromIntegral (gridSize * gridSize))


Finally we can run our simulation in parallel using parMap rather than map (this is the one line change required to get parallelism).

> main :: IO ()
> main = do print "Start"
>
>           let rs = parMap rpar f temps
>                     where
>                       f t = multiUpdate initState t (randomUpdates nItt)
>
>           renderableToFile (FileOptions (500, 500) PNG)
>                            (errChart temps rs magFnNorm enFnNorm meanSqErrNorm)
>                           "diagrams/MagnetismAndEnergy.png"
>
>           print "Done"


## Performance and Parallelism

ghc -O2 Ising.lhs -threaded -o Ising -package-db=.cabal-sandbox/x86_64-osx-ghc-7.6.2-packages.conf.d/ -main-is Ising
time ./Ising +RTS -N1
"Start"
"Done"

real	0m14.879s
user	0m14.508s
sys	0m0.369s

time ./Ising +RTS -N2
"Start"
"Done"

real	0m8.269s
user	0m15.521s
sys	0m0.389s

time ./Ising +RTS -N4
"Start"
"Done"

real	0m5.444s
user	0m19.386s
sys	0m0.414s

## Bibliography and Resources

Beichl, Isabel, and Francis Sullivan. 2000. “The Metropolis Algorithm.” In Computing in Science and Engg., 2:65–69. 1. Piscataway, NJ, USA: IEEE Educational Activities Department. doi:10.1109/5992.814660.

Chib, Siddhartha, and Edward Greenberg. 1995. “Understanding the Metropolis-hastings Algorithm.” The American Statistician 49 (4) (November): 327–335. http://www.jstor.org/stable/2684568.

Diaconis, Persi, and Laurent Saloff-Coste. 1995. “What Do We Know About the Metropolis Algorithm?” In Proceedings of the Twenty-seventh Annual ACM Symposium on Theory of Computing, 112–129. STOC ’95. New York, NY, USA: ACM. doi:10.1145/225058.225095. http://doi.acm.org/10.1145/225058.225095.

Gravner, Janko. 2011. “MAT135A Probability.” https://www.math.ucdavis.edu/~gravner/MAT135A/resources/lecturenotes.pdf.

Levin, David A., Yuval Peres, and Elizabeth L. Wilmer. 2008. Markov Chains and Mixing Times. American Mathematical Society. http://pages.uoregon.edu/dlevin/MARKOV/markovmixing.pdf.

MacKay, David J. C. 2003. Information Theory, Inference, and Learning Algorithms. Cambridge University Press. http://www.cambridge.org/0521642981.

Metropolis, Nicholas, Arianna W. Rosenbluth, Marshall N. Rosenbluth, Augusta H. Teller, and Edward Teller. 1953. “Equation of State Calculations by Fast Computing Machines.” Journal of Chemical Physics 21: 1087–1092.

Norris, James R. 1998. Markov Chains. Cambridge Series in Statistical and Probabilistic Mathematics. Cambridge University Press.

Reif, F. 2009. Fundamentals of Statistical and Thermal Physics. McGraw-hill Series in Fundamentals of Physics. Waveland Press, Incorporated. http://books.google.co.uk/books?id=gYpBPgAACAAJ.

Srikant, R. 2009. “ECE534 Random Processes.” http://www.ifp.illinois.edu/~srikant/ECE534/Spring09/DTMC.pdf.

Tong, David. 2011. “Lectures on Statistical Physcs.” http://www.damtp.cam.ac.uk/user/tong/statphys.html.

Ziman, J.M. 1964. Principles of the Theory of Solids. London, England: Cambridge University Press.