Computational Physics Lectures: Random walks, Brownian motion and the Metropolis algorithm

A Markov process allows in principle for a microscopic description of Brownian motion. As with the random walk studied in the previous section, we consider a particle which moves along the $x$-axis in the form of a series of jumps with step length $\Delta x = l$. Time and space are discretized and the subsequent moves are statistically independent, i.e., the new move depends only on the previous step and not on the results from earlier trials. We start at a position $x=jl=j\Delta x$ and move to a new position $x =i\Delta x$ during a step $\Delta t=\epsilon$, where $i\ge 0$ and $j\ge 0$ are integers. The original probability distribution function (PDF) of the particles is given by $w_i(t=0)$ where $i$ refers to a specific position on the grid in

For the Markov process we have a transition probability from a position $x=jl$ to a position $x=il$ given by $$\begin{equation*} W_{ij}(\epsilon)=W(il-jl,\epsilon)=\left\{\begin{array}{cc}\frac{1}{2} & |i-j| = 1\\ 0 & \mathrm{else} \end{array} \right. , \end{equation*}$$ where $W_{ij}$ is normally called the transition probability and we can represent it, see below, as a matrix. Here we have specialized to a case where the transition probability is known.

Our new PDF $w_i(t=\epsilon)$ is now related to the PDF at $t=0$ through the relation $$\begin{equation*} w_i(t=\epsilon) =\sum_{j} W(j\rightarrow i)w_j(t=0). \end{equation*}$$

This equation represents the discretized time-development of an original PDF with equal probability of jumping left or right.

Since both $W$ and $w$ represent probabilities, they have to be normalized, i.e., we require that at each time step we have $$\begin{equation*} \sum_i w_i(t) = 1, \end{equation*}$$ and $$\begin{equation*} \sum_j W(j\rightarrow i) = 1, \end{equation*}$$ which applies for all $j$-values. The further constraints are $0 \le W_{ij} \le 1$ and $0 \le w_{j} \le 1$. Note that the probability for remaining at the same place is in general not necessarily equal zero.

The time development of our initial PDF can now be represented through the action of the transition probability matrix applied $n$ times. At a time $t_n=n\epsilon$ our initial distribution has developed into $$\begin{equation*} w_i(t_n) = \sum_jW_{ij}(t_n)w_j(0), \end{equation*}$$ and defining $$\begin{equation*} W(il-jl,n\epsilon)=(W^n(\epsilon))_{ij} \end{equation*}$$ we obtain $$\begin{equation*} w_i(n\epsilon) = \sum_j(W^n(\epsilon))_{ij}w_j(0), \end{equation*}$$ or in matrix form $$\begin{equation} \label{eq:wfinal} \hat{w}(n\epsilon) = \hat{W}^n(\epsilon)\hat{w}(0). \end{equation}$$

The following simple example may help in understanding the meaning of the transition matrix $\hat{W}$ and the vector $\hat{w}$. Consider the $4\times 4$ matrix $\hat{W}$ $$\begin{equation*} \hat{W} = \left(\begin{array}{cccc} 1/4 & 1/9 & 3/8 & 1/3 \\ 2/4 & 2/9 & 0 & 1/3\\ 0 & 1/9 & 3/8 & 0\\ 1/4 & 5/9& 2/8 & 1/3 \end{array} \right), \end{equation*}$$ and we choose our initial state as $$\begin{equation*} \hat{w}(t=0)= \left(\begin{array}{c} 1\\ 0\\ 0 \\ 0 \end{array} \right). \end{equation*}$$

We note that both the vector and the matrix are properly normalized. Summing the vector elements gives one and summing over columns for the matrix results also in one. Furthermore, the largest eigenvalue is one. We act then on $\hat{w}$ with $\hat{W}$. The first iteration is $$\begin{equation*} \hat{w}(t=\epsilon) = \hat{W}\hat{w}(t=0), \end{equation*}$$

resulting in $$\begin{equation*} \hat{w}(t=\epsilon)= \left(\begin{array}{c} 1/4\\ 1/2 \\ 0 \\ 1/4 \end{array} \right). \end{equation*}$$

The next iteration results in $$\begin{equation*} \hat{w}(t=2\epsilon) = \hat{W}\hat{w}(t=\epsilon), \end{equation*}$$

resulting in $$\begin{equation*} \hat{w}(t=2\epsilon)= \left(\begin{array}{c} 0.201389\\ 0.319444 \\ 0.055556 \\ 0.423611 \end{array} \right). \end{equation*}$$ Note that the vector $\hat{w}$ is always normalized to $1$.

We find the steady state of the system by solving the set of equations $$\begin{equation*} w(t=\infty) = Ww(t=\infty), \end{equation*}$$ which is an eigenvalue problem with eigenvalue equal to one! This set of equations reads $$\begin{align} W_{11}w_1(t=\infty) +W_{12}w_2(t=\infty) +W_{13}w_3(t=\infty)+ W_{14}w_4(t=\infty)=&w_1(t=\infty) \nonumber \\ W_{21}w_1(t=\infty) + W_{22}w_2(t=\infty) + W_{23}w_3(t=\infty)+ W_{24}w_4(t=\infty)=&w_2(t=\infty) \nonumber \\ W_{31}w_1(t=\infty) + W_{32}w_2(t=\infty) + W_{33}w_3(t=\infty)+ W_{34}w_4(t=\infty)=&w_3(t=\infty) \nonumber \\ W_{41}w_1(t=\infty) + W_{42}w_2(t=\infty) + W_{43}w_3(t=\infty)+ W_{44}w_4(t=\infty)=&w_4(t=\infty) \nonumber \\ \label{_auto1} \end{align}$$ with the constraint that $$\begin{equation*} \sum_i w_i(t=\infty) = 1, \end{equation*}$$ yielding as solution $$\begin{equation*} \hat{w}(t=\infty)= \left(\begin{array}{c}0.244318 \\ 0.319602 \\ 0.056818 \\ 0.379261 \end{array} \right). \end{equation*}$$

The table here demonstrates the convergence as a function of the number of iterations or time steps. After twelve iterations we have reached the exact value with six leading digits.

Iteration	$w_1$	$w_2$	$w_3$	$w_4$
0	1.000000	0.000000	0.000000	0.000000
1	0.250000	0.500000	0.000000	0.250000
2	0.201389	0.319444	0.055556	0.423611
3	0.247878	0.312886	0.056327	0.382909
4	0.245494	0.321106	0.055888	0.377513
5	0.243847	0.319941	0.056636	0.379575
6	0.244274	0.319547	0.056788	0.379391
7	0.244333	0.319611	0.056801	0.379255
8	0.244314	0.319610	0.056813	0.379264
9	0.244317	0.319603	0.056817	0.379264
10	0.244318	0.319602	0.056818	0.379262
11	0.244318	0.319602	0.056818	0.379261
12	0.244318	0.319602	0.056818	0.379261
$\hat{w}(t=\infty)$	0.244318	0.319602	0.056818	0.379261

We have after $t$-steps $$\begin{equation*} \hat{w}(t) = \hat{W}^t\hat{w}(0), \end{equation*}$$ with $\hat{w}(0)$ the distribution at $t=0$ and $\hat{W}$ representing the transition probability matrix.

We can always expand $\hat{w}(0)$ in terms of the right eigenvectors $\hat{v}$ of $\hat{W}$ as $$\begin{equation*} \hat{w}(0) = \sum_i\alpha_i\hat{v}_i, \end{equation*}$$ resulting in $$\begin{equation*} \hat{w}(t) = \hat{W}^t\hat{w}(0)=\hat{W}^t\sum_i\alpha_i\hat{v}_i= \sum_i\lambda_i^t\alpha_i\hat{v}_i, \end{equation*}$$ with $\lambda_i$ the $i^{\mathrm{th}}$ eigenvalue corresponding to the eigenvector $\hat{v}_i$.

If we assume that $\lambda_0$ is the largest eigenvector we see that in the limit $t\rightarrow \infty$, $\hat{w}(t)$ becomes proportional to the corresponding eigenvector $\hat{v}_0$. This is our steady state or final distribution.

Let us recapitulate some of our results about Markov chains and random walks.

• The time development of our PDF $w(t)$, after

This equation represents the discretized time-development of an original PDF. We can rewrite this as a $$\begin{equation*} w_i(t=\epsilon) = W_{ij}w_j(t=0). \end{equation*}$$ with the transition matrix $W$ for a random walk given by $$\begin{equation*} W_{ij}(\epsilon)=W(il-jl,\epsilon)=\left\{\begin{array}{cc}\frac{1}{2} & |i-j| = 1\\ 0 & \mathrm{else} \end{array} \right. \end{equation*}$$

We call $W_{ij}$ for the transition probability and we represent it as a matrix.

• Both $W$ and $w$ represent probabilities and they have to be normalized, meaning that at each time step we have

The further constraints are $0 \le W_{ij} \le 1$ and $0 \le w_{j} \le 1$.

• We can thus write the action of $W$ as

Another way of phrasing this is $$\begin{equation} w(t=\infty) = Ww(t=\infty). \label{_auto2} \end{equation}$$

The question then is how can we model anything under such a severe lack of knowledge? The Metropolis algorithm comes to our rescue here. Since $W(j\rightarrow i)$ is unknown, we model it as the product of two probabilities, a probability for accepting the proposed move from the state $j$ to the state $j$, and a probability for making the transition to the state $i$ being in the state $j$. We label these probabilities $A(j\rightarrow i)$ and $T(j\rightarrow i)$, respectively. Our total transition probability is then $$\begin{equation*} W(j\rightarrow i)=T(j\rightarrow i)A(j\rightarrow i). \end{equation*}$$ The algorithm can then be expressed as

• We make a suggested move to the new state $i$ with some transition or moving probability $T_{j\rightarrow i}$.
• We accept this move to the new state with an acceptance probability $A_{j \rightarrow i}$. The new state $i$ is in turn used as our new starting point for the next move. We reject this proposed moved with a $1-A_{j\rightarrow i}$ and the original state $j$ is used again as a sample.

We wish to derive the required properties of the probabilities $T$ and $A$ such that $w_i^{(t\rightarrow \infty)} \rightarrow w_i$, starting from any distribution, will lead us to the correct distribution.

We can now derive the dynamical process towards equilibrium. To obtain this equation we note that after $t$ time steps the probability for being in a state $i$ is related to the probability of being in a state $j$ and performing a transition to the new state together with the probability of actually being in the state $i$ and making a move to any of the possible states $j$ from the previous time step.

We can express this as, assuming that $T$ and $A$ are time-independent, $$\begin{equation*} w_i(t+1) = \sum_j \left [ w_j(t)T_{j\rightarrow i} A_{j\rightarrow i} +w_i(t)T_{i\rightarrow j}\left ( 1- A_{i\rightarrow j} \right) \right ] \,. \end{equation*}$$

All probabilities are normalized, meaning that $\sum_j T_{i\rightarrow j} = 1$. Using the latter, we can rewrite the previous equation as $$\begin{equation*} w_i(t+1) = w_i(t) + \sum_j \left [ w_j(t)T_{j\rightarrow i} A_{j\rightarrow i} -w_i(t)T_{i\rightarrow j}A_{i\rightarrow j}\right ] \,, \end{equation*}$$ which can be rewritten as $$\begin{equation*} w_i(t+1)-w_i(t) = \sum_j \left [w_j(t)T_{j\rightarrow i} A_{j\rightarrow i} -w_i(t)T_{i\rightarrow j}A_{i\rightarrow j}\right ] . \end{equation*}$$

The last equation is very similar to the so-called Master equation, which relates the temporal dependence of a PDF $w_i(t)$ to various transition rates. The equation can be derived from the so-called Chapman-Einstein-Enskog-Kolmogorov equation. The equation is given as $$\begin{equation} \label{eq:masterequation} \frac{d w_i(t)}{dt} = \sum_j\left[ W(j\rightarrow i)w_j-W(i\rightarrow j)w_i\right], \end{equation}$$ which simply states that the rate at which the systems moves from a state $j$ to a final state $i$ (the first term on the right-hand side of the last equation) is balanced by the rate at which the system undergoes transitions from the state $i$ to a state $j$ (the second term). If we have reached the so-called steady state, then the temporal development is zero. This means that in equilibrium we have $$\begin{equation*} \frac{d w_i(t)}{dt} = 0. \end{equation*}$$

In the limit $t\rightarrow \infty$ we require that the two distributions $w_i(t+1)=w_i$ and $w_i(t)=w_i$ and we have $$\begin{equation*} \sum_j w_jT_{j\rightarrow i} A_{j\rightarrow i}= \sum_j w_iT_{i\rightarrow j}A_{i\rightarrow j}, \end{equation*}$$ which is the condition for balance when the most likely state (or steady state) has been reached. We see also that the right-hand side can be rewritten as $$\begin{equation*} \sum_j w_iT_{i\rightarrow j}A_{i\rightarrow j}= \sum_j w_iW_{i\rightarrow j}, \end{equation*}$$ and using the property that $\sum_j W_{i\rightarrow j}=1$, we can rewrite our equation as $$\begin{equation*} w_i= \sum_j w_jT_{j\rightarrow i} A_{j\rightarrow i}= \sum_j w_j W_{j\rightarrow i}, \end{equation*}$$ which is nothing but the standard equation for a Markov chain when the steady state has been reached.

However, the condition that the rates should equal each other is in general not sufficient to guarantee that we, after many simulations, generate the correct distribution. We may risk to end up with so-called cyclic solutions. To avoid this we therefore introduce an additional condition, namely that of detailed balance $$\begin{equation*} W(j\rightarrow i)w_j= W(i\rightarrow j)w_i. \end{equation*}$$ These equations were derived by Lars Onsager when studying irreversible processes. At equilibrium detailed balance gives thus $$\begin{equation*} \frac{W(j\rightarrow i)}{W(i\rightarrow j)}=\frac{w_i}{w_j}. \end{equation*}$$ Rewriting the last equation in terms of our transition probabilities $T$ and acceptance probobalities $A$ we obtain $$\begin{equation*} w_j(t)T_{j\rightarrow i}A_{j\rightarrow i}= w_i(t)T_{i\rightarrow j}A_{i\rightarrow j}. \end{equation*}$$

Since we normally have an expression for the probability distribution functions $w_i$, we can rewrite the last equation as $$\begin{equation*} \frac{T_{j\rightarrow i}A_{j\rightarrow i}}{T_{i\rightarrow j}A_{i\rightarrow j}}= \frac{w_i}{w_j}. \end{equation*}$$

In statistical physics this condition ensures that it is e.g., the Boltzmann distribution which is generated when equilibrium is reached.

We introduce now the Boltzmann distribution $$\begin{equation*} w_i= \frac{\exp{(-\beta(E_i))}}{Z}, \end{equation*}$$ which states that the probability of finding the system in a state $i$ with energy $E_i$ at an inverse temperature $\beta = 1/k_BT$ is $w_i\propto \exp{(-\beta(E_i))}$. The denominator $Z$ is a normalization constant which ensures that the sum of all probabilities is normalized to one. It is defined as the sum of probabilities over all microstates $j$ of the system $$\begin{equation*} Z=\sum_j \exp{(-\beta(E_i))}. \end{equation*}$$

From the partition function we can in principle generate all interesting quantities for a given system in equilibrium with its surroundings at a temperature $T$.

With the probability distribution given by the Boltzmann distribution we are now in a position where we can generate expectation values for a given variable $A$ through the definition $$\begin{equation*} \langle A \rangle = \sum_jA_jw_j= \frac{\sum_jA_j\exp{(-\beta(E_j)}}{Z}. \end{equation*}$$ In general, most systems have an infinity of microstates making thereby the computation of $Z$ practically impossible and a brute force Monte Carlo calculation over a given number of randomly selected microstates may therefore not yield those microstates which are important at equilibrium. To select the most important contributions we need to use the condition for detailed balance. Since this is just given by the ratios of probabilities, we never need to evaluate the partition function $Z$.

For the Boltzmann distribution, detailed balance results in $$\begin{equation*} \frac{w_i}{w_j}= \exp{(-\beta(E_i-E_j))}. \end{equation*}$$

Let us now specialize to a system whose energy is defined by the orientation of single spins. Consider the state $i$, with given energy $E_i$ represented by the following $N$ spins $$\begin{equation*} \begin{array}{cccccccccc} \uparrow&\uparrow&\uparrow&\dots&\uparrow&\downarrow&\uparrow&\dots&\uparrow&\downarrow\\ 1&2&3&\dots& k-1&k&k+1&\dots&N-1&N\end{array} \end{equation*}$$

We are interested in the transition with one single spinflip to a new state $j$ with energy $E_j$ $$\begin{equation*} \begin{array}{cccccccccc} \uparrow&\uparrow&\uparrow&\dots&\uparrow&\uparrow&\uparrow&\dots&\uparrow&\downarrow\\ 1&2&3&\dots& k-1&k&k+1&\dots&N-1&N\end{array} \end{equation*}$$ This change from one microstate $i$ (or spin configuration) to another microstate $j$ is the configuration space analogue to a random walk on a lattice. Instead of jumping from one place to another in space, we 'jump' from one microstate to another.

However, the selection of states has to generate a final distribution which is the Boltzmann distribution. This is again the same we saw for a random walker, for the discrete case we had always a binomial distribution, whereas for the continuous case we had a normal distribution. The way we sample configurations should result, when equilibrium is established, in the Boltzmann distribution. Else, our algorithm for selecting microstates is wrong.

As stated above, we do in general not know the closed-form expression of the transition rate and we are free to model it as $W(i\rightarrow j)=T(i\rightarrow j)A(i\rightarrow j)$. Our ratio between probabilities gives us $$\begin{equation*} \frac{A_{j\rightarrow i}}{A_{i\rightarrow j}}= \frac{w_iT_{i\rightarrow j}}{w_jT_{j\rightarrow i}}. \end{equation*}$$ The simplest form of the Metropolis algorithm (sometimes called for brute force Metropolis) assumes that the transition probability $T(i\rightarrow j)$ is symmetric, implying that $T(i\rightarrow j)=T(j\rightarrow i)$.

We obtain then (using the Boltzmann distribution) $$\begin{equation*} \frac{A(j\rightarrow i)}{A(i\rightarrow j)}= \exp{(-\beta(E_i-E_j))} . \end{equation*}$$ We are in this case interested in a new state $E_j$ whose energy is lower than $E_i$, viz., $\Delta E = E_j-E_i \le 0$. A simple test would then be to accept only those microstates which lower the energy. Suppose we have ten microstates with energy $E_0 \le E_1 \le E_2 \le E_3 \le \dots \le E_9$. Our desired energy is $E_0$.

At a given temperature $T$ we start our simulation by randomly choosing state $E_9$. Flipping spins we may then find a path from $E_9\rightarrow E_8 \rightarrow E_7 \dots \rightarrow E_1 \rightarrow E_0$. This would however lead to biased statistical averages since it would violate the ergodic hypothesis discussed in the previous section. This principle states that it should be possible for any Markov process to reach every possible state of the system from any starting point if the simulations is carried out for a long enough time.

Any state in a Boltzmann distribution has a probability different from zero and if such a state cannot be reached from a given starting point, then the system is not ergodic. This means that another possible path to $E_0$ could be $E_9\rightarrow E_7 \rightarrow E_8 \dots \rightarrow E_9 \rightarrow E_5 \rightarrow E_0$ and so forth. Even though such a path could have a negligible probability it is still a possibility, and if we simulate long enough it should be included in our computation of an expectation value.

Thus, we require that our algorithm should satisfy the principle of detailed balance and be ergodic. The problem with our ratio $$\begin{equation*} \frac{A(j\rightarrow i)}{A(i\rightarrow j)}= \exp{(-\beta(E_i-E_j))}, \end{equation*}$$ is that we do not know the acceptance probability. This equation only specifies the ratio of pairs of probabilities. Normally we want an algorithm which is as efficient as possible and maximizes the number of accepted moves. Moreover, we know that the acceptance probability has $0$ as its smallest value and $1$ as its largest. If we assume that the largest possible acceptance probability is $1$, we adjust thereafter the other acceptance probability to this constraint.

To understand this better, assume that we have two energies, $E_i$ and $E_j$, with $E_i < E_j$. This means that the largest acceptance value must be $A(j\rightarrow i)$ since we move to a state with lower energy. It follows from also from the fact that the probability $w_i$ is larger than $w_j$. The trick then is to fix this value to $A(j\rightarrow i)=1$. It means that the other acceptance probability has to be $$\begin{equation*} A(i\rightarrow j)= \exp{(-\beta(E_j-E_i))}. \end{equation*}$$

One possible way to encode this equation reads $$\begin{equation*} A(j\rightarrow i)=\left\{\begin{array}{cc} \exp{(-\beta(E_i-E_j))} & E_i-E_j > 0 \\ 1 & else \end{array} \right., \end{equation*}$$ implying that if we move to a state with a lower energy, we always accept this move with acceptance probability $A(j\rightarrow i)=1$. If the energy is higher, we need to check this acceptance probability with the ratio between the probabilities from our PDF. From a practical point of view, the above ratio is compared with a random number. If the ratio is smaller than a given random number we accept the move to a higher energy, else we stay in the same state.

Nothing hinders us obviously in choosing another acceptance ratio, like a weighting of the two energies via $$\begin{equation*} A(j\rightarrow i)=\exp{(-\frac{1}{2}\beta(E_i-E_j))}. \end{equation*}$$ However, it is easy to see that such an acceptance ratio would result in fewer accepted moves.

From experiment there are strong indications that the flux of particles $j(x,t)$, viz., the number of particles passing $x$ at a time $t$ is proportional to the gradient of $w(x,t)$. This proportionality is expressed mathematically through $$\begin{equation*} j(x,t) = -D\frac{\partial w(x,t)}{\partial x}, \end{equation*}$$ where $D$ is the so-called diffusion constant, with dimensionality length$^2$ per time.

If the number of particles is conserved, we have the continuity equation $$\begin{equation*} \frac{\partial j(x,t)}{\partial x} = -\frac{\partial w(x,t)}{\partial t}, \end{equation*}$$ which leads to $$\begin{equation} \label{eq:diffequation1} \frac{\partial w(x,t)}{\partial t} = D\frac{\partial^2w(x,t)}{\partial x^2}, \end{equation}$$ which is the diffusion equation in one dimension.

With the probability distribution function $w(x,t)dx$ we can evaluate expectation values such as the mean distance $$\langle x(t)\rangle = \int_{-\infty}^{\infty}xw(x,t)dx,$$ or $$\langle x^2(t)\rangle = \int_{-\infty}^{\infty}x^2w(x,t)dx,$$ which allows for the computation of the variance $\sigma^2=\langle x^2(t)\rangle-\langle x(t)\rangle^2$. Note well that these expectation values are time-dependent.

In a similar way we can also define expectation values of functions $f(x,t)$ as $$\begin{equation*} \langle f(x,t)\rangle = \int_{-\infty}^{\infty}f(x,t)w(x,t)dx. \end{equation*}$$ The normalization condition $$\begin{equation*} \int_{-\infty}^{\infty}w(x,t)dx=1 \end{equation*}$$ imposes significant constraints on $w(x,t)$.

We have $$\begin{equation*} w(x=\pm \infty,t)=0 \hspace{1cm} \frac{\partial^{n}w(x,t)}{\partial x^n}|_{x=\pm\infty} = 0, \end{equation*}$$ implying that when we study the time-derivative $\partial\langle x(t)\rangle/\partial t$, we obtain after integration by parts and using Eq. $\eqref{eq:diffequation1}$ $$\frac{\partial \langle x\rangle}{\partial t} = \int_{-\infty}^{\infty}x\frac{\partial w(x,t)}{\partial t}dx= D\int_{-\infty}^{\infty}x\frac{\partial^2w(x,t)}{\partial x^2}dx,$$ leading to $$\frac{\partial \langle x\rangle}{\partial t} = Dx\frac{\partial w(x,t)}{\partial x}|_{x=\pm\infty}- D\int_{-\infty}^{\infty}\frac{\partial w(x,t)}{\partial x}dx.$$

The result is $$\frac{\partial \langle x\rangle}{\partial t} = 0.$$ This means in turn that $\langle x\rangle$ is independent of time. If we choose the initial position $x(t=0)=0$, the average displacement $\langle x\rangle= 0$. If we link this discussion to a random walk in one dimension with equal probability of jumping to the left or right and with an initial position $x=0$, then our probability distribution remains centered around $\langle x\rangle= 0$ as function of time.

The variance is not necessarily 0. Consider first $$\frac{\partial \langle x^2\rangle}{\partial t} = Dx^2\frac{\partial w(x,t)}{\partial x}|_{x=\pm\infty}- 2D\int_{-\infty}^{\infty}x\frac{\partial w(x,t)}{\partial x}dx,$$ where we have performed an integration by parts as we did for $\frac{\partial \langle x\rangle}{\partial t}$.

Integration by parts results in $$\frac{\partial \langle x^2\rangle}{\partial t} = -Dxw(x,t)|_{x=\pm\infty}+ 2D\int_{-\infty}^{\infty}w(x,t)dx=2D,$$ leading to $$\langle x^2\rangle = 2Dt,$$ and the variance as $$\begin{equation} \label{eq:variancediffeq} \langle x^2\rangle-\langle x\rangle^2 = 2Dt. \end{equation}$$ The root mean square displacement after a time $t$ is then $$\begin{equation*} \sqrt{\langle x^2\rangle-\langle x\rangle^2} = \sqrt{2Dt}. \end{equation*}$$

This should be contrasted to the displacement of a free particle with initial velocity $v_0$. In that case the distance from the initial position after a time $t$ is $x(t) = vt$ whereas for a diffusion process the root mean square value is $\sqrt{\langle x^2\rangle-\langle x\rangle^2} \propto \sqrt{t}$. Since diffusion is strongly linked with random walks, we could say that a random walker escapes much more slowly from the starting point than would a free particle.

$$\begin{equation*} w(x,t)dx = \frac{1}{\sqrt{4\pi Dt}}\exp{(-\frac{x^2}{4Dt})}dx. \end{equation*}$$ At a time $t=2$s the new variance is $\sigma^2=4D$s, implying that the root mean square value is $\sqrt{\langle x^2\rangle-\langle x\rangle^2} = 2\sqrt{D}$. At a further time $t=8$ we have $\sqrt{\langle x^2\rangle-\langle x\rangle^2} = 4\sqrt{D}$. While time has elapsed by a factor of $4$, the root mean square has only changed by a factor of 2.

Consider $$\begin{equation*} \langle x(n)\rangle = \sum_{i}^{n} \Delta x_i = 0 \hspace{1cm} \Delta x_i=\pm l, \end{equation*}$$ since we have an equal probability of jumping either to the left or to right. The value of $\langle x(n)^2\rangle$ is $$\begin{equation*} \langle x(n)^2\rangle = \left(\sum_{i}^{n} \Delta x_i\right)\left(\sum_{j}^{n} \Delta x_j\right)=\sum_{i}^{n} \Delta x_i^2+ \sum_{i\ne j}^{n} \Delta x_i\Delta x_j=l^2n. \end{equation*}$$

For many enough steps the non-diagonal contribution is $$\begin{equation*} \sum_{i\ne j}^{N} \Delta x_i\Delta x_j=0, \end{equation*}$$ since $\Delta x_{i,j} = \pm l$. The variance is then $$\begin{equation} \langle x(n)^2\rangle - \langle x(n)\rangle^2 = l^2n. \label{eq:rwvariance} \end{equation}$$ It is also rather straightforward to compute the variance for $L\ne R$. The result is $$\begin{equation*} \langle x(n)^2\rangle - \langle x(n)\rangle^2 = 4LRl^2n. \end{equation*}$$

In Eq. $\eqref{eq:rwvariance}$ the variable $n$ represents the number of time steps. If we define $n=t/\Delta t$, we can then couple the variance result from a random walk in one dimension with the variance from the diffusion equation of Eq. $\eqref{eq:variancediffeq}$ by defining the diffusion constant as $$\begin{equation*} D = \frac{l^2}{\Delta t}. \end{equation*}$$

The main program reads the name of the output file from screen and sets up the arrays containing the walker's position after a given number of steps. The corresponding program for a two-dimensional random walk (not listed in the main text) is found under programs/chapter12/program2.cpp

/*
  1-dim random walk program. 
  A walker makes several trials steps with
  a given number of walks per trial
*/
#include <iostream>
#include <fstream>
#include <iomanip>
#include "lib.h"
using namespace  std;

// Function to read in data from screen, note call by reference
void initialise(int&, int&, double&) ;
// The Mc sampling for random walks
void  mc_sampling(int, int, double, int *, int *);
// prints to screen the results of the calculations 
void  output(int, int, int *, int *);

int main()
{
  int max_trials, number_walks; 
  double move_probability;
  // Read in data 
  initialise(max_trials, number_walks, move_probability) ;
  int *walk_cumulative = new int [number_walks+1];
  int *walk2_cumulative = new int [number_walks+1];
  for (int walks = 1; walks <= number_walks; walks++){   
    walk_cumulative[walks] = walk2_cumulative[walks] = 0;
  } // end initialization of vectors
  // Do the mc sampling  
  mc_sampling(max_trials, number_walks, move_probability, 
              walk_cumulative, walk2_cumulative);
  // Print out results 
  output(max_trials, number_walks, walk_cumulative, 
         walk2_cumulative);
  delete [] walk_cumulative; // free memory
  delete [] walk2_cumulative; 
  return 0; 
} // end main function
\end{lstlisting}
The  input and output functions are 
\begin{lstlisting} 
void initialise(int& max_trials, int& number_walks, double& move_probability) 
{
  cout << "Number of Monte Carlo trials ="; 
  cin >> max_trials;
  cout << "Number of attempted walks=";
  cin >> number_walks;
  cout << "Move probability=";
  cin >> move_probability;
}  // end of function initialise   

The algorithm tests the probability of moving to the left or to the right by generating a random number.

void mc_sampling(int max_trials, int number_walks, 
                 double move_probability, int *walk_cumulative, 
                 int *walk2_cumulative)
{
  long idum;
  idum=-1;  // initialise random number generator
  for (int trial=1; trial <= max_trials; trial++){
    int position = 0;
    for (int walks = 1; walks <= number_walks; walks++){   
      if (ran0(&idum) <= move_probability) {
	position += 1;
      } 
      else {
	position -= 1;
      }
      walk_cumulative[walks] += position;
      walk2_cumulative[walks] += position*position;
    }  // end of loop over walks
  } // end of loop over trials
}   // end mc_sampling function