# Cluster correlations for a perfectly random state in the Alloy Theoretic Automated Toolkit

In their paper, Van de Walle et al. describe how their code mcsqs treats cluster correlations in the algorithm to generate special quasirandom structures (SQS). In the third section, titled 'Algorithm', they describe how cluster correlations in a perfectly random state are determined. Somehow I can't seem to understand the equation. Can someone help me understand it?

If a responder wishes to use illustrations, a perfect test case would be the correlation for nearest neighbour (NN) pairs.

Let's understand it with an example of a binary $$AB$$ compound.

A correlation function for a particular type of cluster (pair, triangle, tetrahedron, etc.), $$\rho$$, is defined as

$$$$\rho_\alpha\: =\: _\alpha,\tag{1}$$$$

Where $$\alpha$$ represents the type of cluster and $$<...>$$ represents an average over all clusters equivalent to $$\alpha$$ by symmetry.

$$$$cluster function = \prod_i{\gamma_{\alpha_i,M_i}}_{(\sigma_i)} \tag{2}$$$$

The cluster function is a product of a quantity which in the sense of binary alloys is called a spin variable (inspired by the Ising model), whereas, in general, for multicomponent systems seems to have no name. The paper cited in the question denotes it as $$\gamma_{\alpha_i,M_i}$$, where $$i$$ represents a lattice point, $$\alpha_i$$ takes two values, 0 for when $$i$$ is not in a certain cluster and 1 when it is, and $$M_i$$ is simply the number of type of atoms, 2 for binary, 3 for ternary, and so on. Note: $$\alpha$$ and $$\alpha_i$$ aren't the same.

Convention: $$\gamma$$ equals 1 for any site $$i$$ that isn't part of the particular cluster $$\alpha$$ ($$\alpha_i\: =\: 0$$).

$$\sigma_i$$ in equation 2 is an occupation variable which holds the information about which atom occupies lattice site $$i$$. It takes values from $$0$$ to $$M_i - 1$$. For example, in a ternary alloy, $$\sigma_i\: =\: 0, 1,$$ or $$2$$.

$$$$\sigma_i\: =\: \begin{cases} -1, & \text{atom A on i}\\ +1, & \text{atom B on i} \end{cases} \tag{3}$$$$

Let's calculate the pair correlation for $$AB$$:

$$\alpha\: =\: 2$$ (pair)

$$M_i\: =\: 2$$ (binary)

In a perfectly random phase, each lattice point $$i$$ is equally likely to be occupied by $$A$$ and $$B$$ since it is an equiatomic alloy.

$$\alpha_i$$ will be $$0$$ for all lattice points that aren't in the particular pair we are calculating the cluster function for and $$1$$ for those that are part of the pair. So the four possible values of $${\gamma_{\alpha_i,M_i}}_{(\sigma_i)}$$ for a binary are:

$${\gamma_{0,2}}_{(0)}\: =\: 1$$

$${\gamma_{0,2}}_{(1)}\: =\: 1$$

$${\gamma_{1,2}}_{(0)}\: =\: -1$$

$${\gamma_{1,2}}_{(1)}\: =\: +1$$

In a random phase of $$AB$$, every lattice site $$i$$ on average has a $$0.5$$ occupation of $$A$$ and $$0.5$$ of $$B$$. To calculate the correlation function, $$\rho_\alpha$$, we need to take an average of cluster functions for all equivalent clusters $$\alpha$$. Every pair (even higher clusters) in a perfectly random phase is equivalent, so all cluster functions are the same and equal to the average, hence equal to the correlation function.

The value of $${\gamma_{1,2}}_{(\sigma_i = 0,1)}\: =\: 0.5\cdot (-1) + 0.5\cdot (+1) = 0$$ for each site $$i$$.

So the cluster function for a pair which contains two lattice points $$=\: 0\cdot 0\: = 0$$.

Finally, the correlation function, $$\rho_2$$, is equal to $$0$$.